Solutions Manual for Open Channel Hydraulics 1st Edition by A. Osman Akan

CHAPTER 1. Basic Principles 1.1.

Classify each of the following flows as steady or unsteady from the viewpoint of the observer: Flow (a) Flow of river around bridge piers

Observer (1) Standing on bridge (2) In boat, drifting (1) Standing on bank (2) Moving with surge

(b) Movement of flood surge downstream Solution.

(a) Over relatively short time intervals, the observer standing on the bridge will see a steady flow even though the flow may be unsteady over longer time intervals; however, the observer drifting in the boat will see an unsteady flow as the boat passes under the bridge because the velocity increases around the bridge piers even if the approach flow is steady. (b) An observer standing on the river bank will see an unsteady flow as the surge passes but a steady flow while riding on the surge if the flow is uniform in the direction of movement.

1.2.

At the crest of an ogee spillway as shown in Figure 1.1c, would you expect the pressure on the face of the spillway to be greater than, less than, or equal to the hydrostatic value? Explain your answer.

Solution.

The convex curvature near the crest of the spillway results in a centripetal acceleration toward the center of curvature and a corresponding pressure gradient with decreasing pressure toward the center of curvature. The decreasing pressure reduces the equivalent hydrostatic pressure for a parallel flow so that the pressure is less than hydrostatic on the face of the spillway. The decrease in pressure can be so severe that vapor pressure is reached and cavitation occurs with pitting and erosion of the concrete spillway surface (see Chapter 6).

1.3.

The river flow at an upstream gauging station is measured to be 1500 m3/s, and at another gauging station 3 km downstream, the discharge is measured to be 750 m3/s at the same instant of time. If the river channel is uniform with a width of 300 m, estimate the rate of change in the water surface elevation in meters per hour. Is it rising or falling?

Solution.

Use Equation 1.6, the continuity equation:

750 − 1500 ∂A ∂y ∂Q =B =− ≅− = 0.25 m 2 /s ∂t ∂t ∂x 3000 ∂y 0.25 0.25 = = = 8.33 × 10 −4 m/s or 3.0 m/hr (rising) 300 ∂t B

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Open Channel Hydraulics

1.4.

CHAPTER 1

A paved parking lot section has a uniform slope over a length of 100 m (in the flow direction) from the point of a drainage area divide to the inlet grate, which extends across the lot width of 30 m. Rainfall is occurring at a uniform intensity of 10 cm/hr. If the detention storage on the paved section is increasing at the rate of 60 m3/hr, what is the runoff rate into the inlet grate?

Solution.

Utilize the continuity equation for a finite control volume given by Equation 1.3 for an incompressible fluid so that the fluid density ρ cancels on both sides of the equation. Then we have

d∀ = − ∑ Qout + ∑ Qin dt 60 = − Q runoff +

10 × 100 × 30 100 cm/m

Q runoff = 300 − 60 = 240 m 3 /hr 1.5.

A rectangular channel 6 m wide with a depth of flow of 3 m has a mean velocity of 1.5 m/s. The channel undergoes a smooth, gradual contraction to a width of 4.5 m. (a) Calculate the depth and velocity in the contracted section. (b) Calculate the net fluid force on the walls and floor of the contraction in the flow direction.

In each case, identify any assumptions that you make.

Solution.

1 6m

F

2 4.5 m

(a) Apply the energy equation from the approach section 1 to the contracted section 2 with negligible head losses and assuming a horizontal channel bottom:

V12 q22 = y2 + y1 + 2g 2 gy 22 where q 2 = V 2 y 2 = (6/4.5)q 1 = (6/4.5)(1.5)(3.0) = 6.0 m2/s. Substituting and solving, we have

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CHAPTER 1

3.0 +

1.5 2 6.0 2 = y2 + 19.62 19.62 × y 22

y2 +

1.835 = 3.115 y 22

from which y 2 = 2.90 m by trial and error and V 2 = q 2 /y 2 = 6.0/2.90 = 2.07 m/s. Note that there are two solutions, but this is the subcritical solution and the correct one as discussed in more detail in Chapter 2. (b) Apply the momentum equation in the flow direction in which F = the resultant force of the walls and floor on the flow. Assume a hydrostatic pressure distribution at sections 1 and 2.

Because the transition is horizontal, there is no component of the gravity force in the flow direction. The momentum equation becomes

− F + b1γ

y12 y2 − b2γ 2 = ρQ (V2 − V1 ) 2 2

− F + 6.0 × 9810 ×

3.0 2 2.90 2 − 4.5 × 9810 × = 1000 × (1.5 × 3.0 × 6.0) × (2.07 − 1.5) 2 2

from which F = 63.8 kN. 1.6.

A bridge has cylindrical piers 1 m in diameter and spaced 15 m apart. Downstream of the bridge where the flow disturbance from the piers is no longer present, the flow depth is 2.9 m and the mean velocity is 2.5 m/s. (a) Calculate the depth of flow upstream of the bridge assuming that the pier coefficient of drag is 1.2. (b) Determine the head loss caused by the piers.

Solution.

In part (a), apply the momentum equation with the control volume boundaries halfway between the piers; then apply the energy equation in part (b).

s = 15 m Fp2

D

Fp1 1

2

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CHAPTER 1

(a) The momentum equation, neglecting boundary friction, is

Fp1 − Fp 2 − D = ρQ (V2 − V1 )

sγ

y12 y2 V2 − sγ 2 − C D ρA f 1 = ρQ (V2 − V1 ) 2 2 2

in which D = drag force on the pier; F p = hydrostatic force; A f = frontal area of the pier at section 1 on a plane perpendicular to the flow direction = ay 1 ; a = pier diameter = 1.0 m; s = pier spacing = 15.0 m; C D = drag coefficient =1.2; and Q = A 2 V 2 = (15)(2.9)(2.5) = 108.8 m3/s . Using continuity and substituting, we have

y 2 2.9 2 1 1 (108.8) 2 = 1000 × 108.8 2 ( − ) 15 × 9810 × ( 1 − ) − 1.2 × 1000 × (1.0 y1 ) × 2 15 × 2.9 15 y1 2 2 2(15 y1 ) which reduces to

y12 +

10.30 − 12.11 = 0 y1

and the solution is y 1 = 2.932 m, or a backwater of 0.032 m, and V 1 = Q/A 1 = 108.8/[(15)(2.932)] = 2.474 m/s. (b) The head loss, h L , is obtained from the energy equation assuming a negligible change in channel bed elevation from point 1 to 2:

h L = y1 + 1.7.

V12 V2 2.474 2 2.52 − y 2 − 2 = 2.932 + − 2.90 − = 0.0254 m 2g 2g 19.62 19.62

A symmetric compound channel in overbank flow has a main channel with a bottom width of 30 m, side slopes of 1:1, and a flow depth of 3 m. The floodplains on either side of the main channel are both 300 m wide and flowing at a depth of 0.5 m. The mean velocity in the main channel is 1.5 m/s, while the floodplain flow has a mean velocity of 0.3 m/s. Assuming that the velocity variation within the main channel and the floodplain subsections is much smaller than the change in mean velocities between subsections, find the value of the kinetic energy correction coefficient α.

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CHAPTER 1

Solution.

0.5 m

3m

300 m

1 1

300 m

30 m

Flow area of the floodplains, A f :

A f = 2 × (300 × 0.5) = 300 m 2

Flow area of the main channel, A m :

Am =

1 (30 + 35) × 2.5 + 35 × 0.5 = 98.75 m 2 2

Then for the entire channel, we have a total discharge, Q = 300(0.3 m/s) + 98.75(1.5 m/s) = 238 m3/s, and mean velocity V = Q/A = 238/398.75 = 0.597 m/s. The kinetic energy flux correction coefficient, α, then is given by

∑ vi3 ai 2 × (0.3) 3 ×150 + (1.5) 3 × 98.75 = 4.02 α= 3 = (0.597) 3 × (398.75) V A 1.8.

The power law velocity distribution for fully-rough turbulent flow in an open channel is given by

 z  1/ 6 u = a   u*  ks  in which u = point velocity at a distance z from the bed; u * = shear velocity = (τ 0 /ρ)1/2; τ 0 = bed shear stress; ρ = fluid density; k s = equivalent sand grain roughness height; and a = constant. (a) (b)

Find the ratio of the maximum velocity, which occurs at the free surface where z = the depth, y 0 , to the mean velocity for a very wide channel.

Calculate the values of the kinetic energy correction coefficient α and the momentum flux correction coefficient β for a very wide channel.

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CHAPTER 1

Solution. (a) Substituting for u = u max at z = y 0 , we have

u max = au* (

y0 1/ 6 ) ks

The mean velocity, V, is obtained by integration of the point velocity distribution over the depth for a very wide channel:

[ ]

7 / 6 y0

y au* z 1 y0 z 6 0 V= au* ( )1 / 6 dz = = au* ( 0 )1 / 6 1 / 6 ∫ y0 0 ks ks 7/6 7 y0 k s Then by comparison, it is obvious that u max /V = 7/6. (b) From the definition of α for a very wide channel, we have

α=

∫

y0 0

z 1/ 2 ) dz 0 ks (2 / 3) y 03 / 2 = = = 1.059 (6 / 7) 3 y 03 / 2 3 3 3 y0 1/ 2 a u * (6 / 7 ) ( ) y 0 ks y0

a 3 u* ∫ ( 3

u 3 dz

V 3 y0

Similarly, β is given by

β=

1.9.

∫

y0 0

z 1/ 3 ) dz 0 ks (3 / 4) y 04 / 3 = = = 1.021 y (6 / 7) 2 y 04 / 3 a 2 u*2 (6 / 7) 2 ( 0 )1 / 3 y 0 ks y0

u 2 dz

V 2 y0

a 2 u*2 ∫ (

An alternative expression for the velocity distribution in fully-rough turbulent flow is given by the logarithmic distribution

u 1  z  = ln   u* κ  z0 

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CHAPTER 1

in which κ = the von Karman constant = 0.40; z 0 = k s / 30; and the other variables are the same as defined in Exercise 1.8. Show that α and β for this distribution in a very wide channel are given by

α = 1 + 3ε 2 − 2ε 3 β = 1+ ε 2 in which ε = (u max / V) – 1; u max = maximum velocity; and V = mean velocity.

Solution.

First, find the mean velocity, V, by integrating over the depth with a change of variable, η = z/z 0 , so that dz = z 0 dη and the upper limit, η 0 = y 0 /z 0. Then we have

V=

y0 uz u* y0 u z ln( ) dz = * ∫ z 0 lnη dη = * 0 [η lnη − η ]η00 ∫ z0 κy0 0 κy0 0 κy0

so, the mean velocity is

V=

u*

κ

(lnη 0 − 1)

and the maximum velocity is given by

u max =

u*

κ

lnη 0

so we conclude that

ε=

u max 1 −1 = V lnη 0 − 1

Now we can calculate α from its definition, using the same change of variable to η, to produce

u*3 z 0 η0 α = 3 3 ∫ (lnη ) 3 dη κ V y0 0 From a math handbook, we can look up the integral, which is given by

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Open Channel Hydraulics

CHAPTER 1

∫ (lnη ) dη = η[(lnη ) − 3(lnη ) + 6 lnη − 6] 3

3

2

Substituting the integral, with the limits applied, and the expression for mean velocity into the equation for α, we have

α=

x 3 − 3x 2 + 6 x − 6 ( x − 1) 3 + 3x − 5 = ( x − 1) 3 ( x − 1) 3

in which x = ln η 0 = 1/ε + 1. Substituting for x in terms of ε, we have finally

α = 1 + 3ε 2 − 2ε 3 Similarly, we have for β

β=

η0 u*2 z 0 (lnη ) 2 dη 2 2 ∫ 0 κ V y0

and the indefinite integral is given in this case by

∫ (lnη ) dη = η[(lnη ) − 2 lnη + 2] 2

2

so that upon substitution into the equation for β we have

x 2 − 2 x + 2 ( x − 1) 2 + 1 β= = 1+ ε 2 = 2 2 ( x − 1) ( x − 1) in which x = ln η 0 = 1/ε +1 as before.

1.10.

In a hydraulic jump in a rectangular channel of width b, the depth after the jump y 2 is known to depend on the following variables:

y 2 = f [ y1 , q , g ] in which y 1 = depth before the jump; q = discharge per unit width = Q/b; and g = gravitational acceleration. Complete the dimensional analysis of the problem.

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CHAPTER 1

Solution.

We have n = 4 variables but only two fundamental dimensions, L and T, so that m = 2 and n–m = 2 Π groups. Choose y 1 and g as repeating variables. Then by inspection, the first Π group is given by

Π1 =

y2 y1

The second Π group is found from

[Π 2 ] = L0 T 0 = [ y1 ]c [ g ]d q = Lc (LT −2 ) d (L2 T −1 ) in which the square brackets denote "the dimensions of" the enclosed variables. Equating exponents on length, L, and time, T, we have

L: 0 = c+d +2 T : 0 = −2 d − 1 from which d = –1/2 and c = –3/2 so that

Π2 =

q = F1 g y13 / 2 1/ 2

which is the Froude number for a rectangular channel. Finally we can write from the dimensional analysis that

y2 = φ (F1 ) y1 1.11

The backwater ∆y caused by bridge piers in a bridge opening is thought to depend on the pier diameter and spacing, d and s, respectively; downstream depth, y 0 ; downstream velocity, V; fluid density, ρ; fluid viscosity, µ; and gravitational acceleration, g. Complete the dimensional analysis of the problem.

Solution. First, write the functional relationship as

∆y = f ( d , s, y 0 ,V , ρ , µ , g ) We have 8 variables and 3 fundamental dimensions so there must be 5 Π groups. Choose ρ, V, and y 0 as repeating variables. Then by inspection, the first 3 Π groups are Π 1 = ∆y/y 0 ; Π 2 = d/y 0 ; and Π 3 = s/y 0 . The next Π group is found from

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Open Channel Hydraulics

CHAPTER 1

[Π 4 ] = M 0 L0 T 0 = [ ρ ]c [V ]d [ y0 ]e µ = (ML−3 ) c (LT −1 ) d (L) e (ML−1T −1 ) in which the square brackets denote "the dimensions of" the enclosed variables. Equating exponents, we have

M : 0 = c +1 L : 0 = −3c + d + e − 1 T : 0 = −d − 1 from which c = –1; d = –1; and e = –1. Then Π 4 = ρVy 0 /µ, which is the Reynolds number of the flow. The final Π group is obtained from

[Π 5 ] = M 0 L0 T 0 = [ ρ ]c [V ]d [ y0 ]e g = (ML−3 ) c (LT −1 ) d (L) e (LT −2 ) Equating the exponents, we have

M: 0=c L : 0 = −3c + d + e + 1 T : 0 = −d − 2 so that c = 0; d = –2; and e = 1. Then Π 5 can be written as V/(gy 0 )1/2, which is the Froude number.

The final functional relationship, with some rearrangement of the Π groups, is

y d ρVd ∆y V = φ[ 0 , , , ] d d s µ ( gy 0 )1 / 2 1.12

The longitudinal velocity u near the fixed bed of an open channel depends on the distance from the bed, z; the kinematic viscosity, ν; and the shear velocity u * = (τ 0 /ρ)0.5 in which τ 0 is the wall shear stress. Develop the dimensional analysis for the point velocity, u.

Solution.

The functional relationship is

u = f ( z , ν , u* )

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Open Channel Hydraulics

CHAPTER 1

There are 4 variables and only 2 fundamental dimensions (L and T), so we should expect 2 Π groups. Choose z and u * as repeating variables. Then we have

Π 1 = z c u*d u which, by inspection, gives Π 1 = u/u * . The second Π group comes from

Π 2 = z c u*dν L0 T 0 = Lc (LT −1 ) d L2 T −1 L: 0 = c+d +2 T : 0 = −d − 1 from which d = –1 and c = –1 so that Π 2 = u * z/ν. The final relationship is

uz u = φ( * ) u* ν 1.l3

In very slow motion of a fluid around a sphere, the drag force on the sphere, D, depends on the sphere diameter, d; the velocity of the approach flow, V; and the fluid viscosity, µ. Complete the dimensional analysis. How many dimensionless groups are there and what are the implications for the corresponding values of the group(s)? Why was the fluid density not included in the list of variables?

Solution.

The functional relationship is

D = f (d , V , µ ) There are 4 variables and 3 fundamental dimensions, so we should expect only one Π group.

Choose d, V, and µ as repeating variables, which do not themselves form a Π group. Then we have

Π 1 = d aV b µ c D M 0 L0 T 0 = La (LT −1 ) b (ML−1T −1 ) c MLT −2 M : 0 = c +1 L : 0 = a + b − c +1 T : 0 = −b − c − 2

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Open Channel Hydraulics

CHAPTER 1

from which c = –1; b = –1; a = –1 and Π 1 = D/µVd. In this case, according to the Buckingham Π theorem, we can set a function of the single Π group to zero, which can only be true if the Π group itself is a constant. This constant was determined analytically by Stokes to be 3π, but in the general case it could be determined experimentally. The fluid density was not included in the list of variables because the acceleration terms were neglected as a result of the condition of very slow, or creeping, motion.

1.14

The discharge over a sharp-crested weir, Q, is a function of the head on the weir crest, H; the crest length, L; the height of the crest, P; density, ρ; viscosity, µ; surface tension, σ ; and gravitational acceleration, g. Carry out the dimensional analysis using ρ, g, and H as repeating variables. If it is known that Q is directly proportional to crest length, L, how would you alter the dependent Π group?

Solution.

Write the functional relationship as

Q = f ( H , L, P, ρ , µ ,σ , g ) in which there are 8 variables and 3 fundamental dimensions resulting in 5 Π groups. Use ρ, g, and H as repeating variables as suggested. Then each Π group is determined as follows: By inspection, Π 1 = H/L; and Π 2 = H/P. Then we have

Π3 = ρ a g bH cµ M 0 L0 T 0 = ( ML−3 ) a ( LT −2 ) b Lc ML−1T −1 M : 0 = a +1 L : 0 = −3a + b + c − 1 T : 0 = −2 b − 1 from which a = –1; b = –1/2; and c = –3/2 so that Π 3 = ρH(gH)1/2/µ.

Π 4 = ρ a g b H cσ M 0 L0 T 0 = ( ML−3 ) a ( LT −2 ) b Lc MT −2 M : 0 = a +1 L : 0 = −3a + b + c T : 0 = −2 b − 2 from which a = –1; b = –1; c = –2 so that Π 4 = ρgH 2 /σ.

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Π 5 = ρ a g b H cQ M 0 L0 T 0 = ( ML−3 ) a ( LT −2 ) b Lc (L3T −1 ) M: 0=a L : 0 = −3a + b + c + 3 T : 0 = −2 b − 1 from which a =0; b = –1/2; and c = –5/2 so that Π 5 = Q/(g1/2H 5/2 ). The final functional relationship can be written as

Q H H ρH ( gH )1 / 2 ρgH 2 ( , , , ) = φ L P µ σ g 1/ 2 H 5 / 2 If it is known that Q varies directly with the crest length, L, then the dependent Π group can be multiplied by (H/L) with the result

Q H H ρH ( gH )1 / 2 ρgH 2 ( , , , ) = φ L P µ σ g 1 / 2 LH 3 / 2 The function φ becomes a discharge coefficient that depends on the Π groups listed, and the form of its functional dependence can be determined by experiment. See Chapter 6.

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CHAPTER 1

14

CHAPTER 2. Specific Energy 2.1

10 ft/s

Water is flowing at a depth of 10 ft with a velocity of 10 ft/s in a channel of rectangular section.

Find the depth and change in water surface elevation caused by a smooth upward step in the channel bottom of 1 ft. What is the maximum allowable step size so that choking is prevented?

(Use a head loss coefficient = 0.)

y2

10 ft

∆z = 1 ft

Solution.

q = V1 y1 = (10)(10) = 100 ft 2 /s 1/ 3

1/ 3

 q2   100 2   = 6.77 ft ⇒ subcritical approach flow y c =   =   g   32.2  3 E c = y c = (1.5)(6.77) = 10.16 ft 2 V2 10 2 E1 = y1 + 1 = 10 + = 11.55 ft 64.4 2g Now because E 1 – ∆z > E c , there is no choking. So write the energy equation from 1 to 2 and find a subcritical solution for y 2 :

E1 = y 2 +

q2 + ∆z 2 gy 22

11.55 = y 2 + y2 +

100 2 + 1.0 64.4 y 22

155.3 = 10.55 y 22

which can be solved by trial and error or an equation solver to give y 2 = 8.29 ft. The water surface elevation drops by 10 – (8.29 +1) = 0.71 ft.

For the limiting choking case, set the specific energy at section 2 equal to E c :

E1 = 11.55 = E c + ∆z c = 10.16 + ∆z c so ∆z c = 11.55 – 10.16 = 1.39 ft.

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Open Channel Hydraulics

2.2

CHAPTER 2

The upstream conditions are the same as in Exercise 2.1, but there is a smooth contraction in width from 10 ft to 9 ft and a horizontal bottom. Find the depth of flow and change in water surface elevation in the contracted section. What is the greatest allowable contraction in width so that choking is prevented? (Head loss coefficient = 0.)

10 ft

9 ft 2

1

Solution.

From Exercise 2.1, q 1 = 100 cfs/ft; y c1 = 6.77 ft; and E 1 = 11.55 ft. Then from continuity, we have q 2 = (10/9) q 1 = (10/9)(100) = 111.1 cfs/ft and y c2 = (111.12/32.2)1/3 = 7.26 ft. Writing the energy equation from 1 to 2:

E1 = y 2 +

q 22 2 gy 22

11.55 = y 2 +

(111.1) 2 64.4 y

2 2

= y2 +

191.7 y 22

from which the subcritical solution is y 2 = 9.36 ft with a water surface elevation drop of (10 – 9.36) = 0.64 ft.

For the limiting choking case, E 1 = E c2 = 1.5 y c2 , so that 1/ 3

 q2  11.55 = 1.5 2   g  q2 = [( 2 / 3)(11.55)]3 / 2 (32.2)1 / 2 = 121.2 cfs/ft But b 2 = Q/q 2 = 1000/121.2 = 8.25 ft.

2.3

Determine the downstream depth in the transition and the change in water surface elevation if the channel bottom rises 0.15 m and the upstream conditions are a velocity of 4.5 m/s and a depth of 0.6 m.

Solution.

Assuming a rectangular channel of constant width with negligible head loss, as in Exercise 2.1, check the approach conditions:

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CHAPTER 2

q = ( 4.5)(0.6) = 2.7 m 2 /s 1/ 3

1/ 3

 q2   2.7 2   = 0.906 m y c =   =   g   9.81  V12 4.52 E1 = y1 + = 0.6 + = 1.63 m 2g 19.62 The approach flow is supercritical, and the minimum specific energy, E c = 1.5y c = (1.5)(0.906) = 1.36 m. Because E 1 – ∆z = 1.63 – 0.15 = 1.48 m > E c , choking is not expected. Solve the energy equation for y 2 in the supercritical flow regime:

E1 = 1.63 = y 2 + y2 +

q2 2.7 2 z y + ∆ = + + 0.15 2 2 gy 22 19.62 y 22

0.3716 = 1.48 y 22

and the solution is y 2 = 0.683 m. The water surface elevation rises by 0.683 + 0.15 – 0.60 = 0.233 m. The limiting choking case is given by ∆z = E 1 – E c = 1.63 – 1.36 = 0.27 m.

As in Figure 2.9 for a width contraction with a supercritical approach flow, there is a second mode of choking in this example with a hydraulic jump upstream of the transition and critical depth in the transition. As discussed in Chapter 3, the sequent depth for an upstream hydraulic jump can be calculated to be 1.30 m corresponding to a specific energy after the jump of E 1 = 1.52 m; however, in this event, E 1 – ∆z = 1.37 m, which is still greater than E c , so no choking occurs by this mode either.

2.4

Determine the downstream depth in a subcritical transition if Q=262 cfs and the channel bottom rises 3.279 ft in going from an upstream circular channel to a downstream rectangular channel.

The upstream circular channel has a diameter of 9.18 ft and a depth of flow of 7.34 ft. The downstream rectangular channel has a width of 6.56 ft. Neglect the head loss.

7.34 ft 3.279 ft d

b

d = 9.18 ft

b = 6.56 ft

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Solution.

For the upstream circular channel, calculate the flow area and top width for the given depth of 7.34 ft:

y d

7.34 ) = 4.4264 rad 9.18 d2 9.18 2 A = (θ − sin θ ) = [4.4264 − sin(4.4264)] = 56.73 ft 2 8 8 B = d sin(θ / 2) = 9.18 sin(4.4264 / 2) = 7.35 ft

θ = 2 cos −1 (1 − 2 ) = 2 cos −1 (1 − 2

Then the approach specific energy and Froude number are given by

Q2 262 2 = 7.34 + = 7.67 ft E1 = y1 + 64.4 × 56.732 2 gA2 262 × 7.351 / 2 QB 1 / 2 = 0.29 F1 = 1 / 2 3 / 2 = 32.21 / 2 × 56.733 / 2 g A

The approach flow is subcritical. Check for choking with critical depth occurring in the downstream rectangular section: 1/ 3

1/ 3

 q2   ( 262 / 6.56) 2   y c 2 =   =  32.2  g    E c 2 = 1.5 y c = 1.5 × 3.67 = 5.51 ft

= 3.67 ft

Note that from the energy equation, E 2 = E 1 – ∆z = 7.67 –3.279 = 4.39 ft < E c2 . Thus, choking will occur with y 2 = 3.67 ft, and y 1 has to be recalculated from

E1 = E c 2 + ∆z = 5.51 + 3.279 = 8.79 ft y1 +

1066 Q2 = y1 + 2 = 8.79 2 2 gA1 A1

which can be solved by assuming a value of y 1 ; calculating for the circular cross-section, θ = 2 cos-1(1 – 2y 1 /d), and the flow area, A 1 = (θ - sin θ)d 2/8; substituting to obtain E 1 ; and repeating until E 1 = 8.79 ft. The result is y 1 = 8.53 ft; θ = 5.2058 rad; A 1 = 64.12 ft2; and E 1 = 8.789 ft. Choking causes the upstream depth to increase by (8.53 – 7.34) = 1.19 ft.

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Open Channel Hydraulics

2.5

CHAPTER 2

Determine the upstream depth of flow in a subcritical transition from an upstream rectangular flume that is 49 ft wide to a downstream trapezoidal channel with a width of 75 ft and side slopes of 2:1. The transition bottom drops 1 ft from the upstream flume to the downstream trapezoidal channel. The flow rate is 12,600 cfs, and the depth in the downstream trapezoidal channel is 22 ft.

Use a head loss coefficient of 0.5.

y1

22 ft

1.0 ft 1 49 ft

75 ft

2

Solution.

First, calculate the downstream conditions:

A2 = y 2 (b2 + my 2 ) = ( 22.0)(75 + 2 × 22.0) = 2618 ft 2 B2 = b2 + 2my 2 = 75 + 2 × 2 × 22.0 = 163 ft V2 =

Q 12,600 = = 4.81 ft/s A2 2618

F2 =

V2 4.81 = = 0.21 1/ 2 [ gA2 / B2 ] [32.2 × 2618 / 163]1 / 2

E2 = y2 +

Q2 12,600 2 = + = 22.36 ft 22 .

0 64.4 × 2618 2 2 gA22

The downstream flow is subcritical, and we are looking for a subcritical depth upstream. Writing the energy equation including the head loss, and assuming that V 1 > V 2 , we have

E1 + ∆z = E 2 + hL Q2 Q2 Q2 y1 + + 1.0 = 22.36 + 0.5 − 2 gA12 2 gA12 2 gA22 y1 + (1.0 − 0.5) y1 +

12,600 2 4.812 = − 21 .

36 0 .

5 64.4 64.4 × ( 49 y1 ) 2

513.4 = 21.18 y12

Solving, the result is y 1 = 19.88 ft where y c1 = (q2/g)1/3 = [(12,600/49)2/32.2]1/3 = 12.7 ft. The corresponding head loss is 1.12 ft, where E 1 = 22.48 ft and V 1 = 12.94 ft/s.

19

Open Channel Hydraulics

2.6

CHAPTER 2

In a horizontal rectangular flume, suppose that a smooth "bump" with a height of 0.33 ft has been placed on the channel bottom. The discharge per unit width in the flume is 0.4 cfs/ft. Determine the depth at the obstruction for a tailwater depth of 1.0 ft and negligible head losses. Sketch the results on a specific energy diagram.

y2

y1

y3 = 1.0 ft

∆z = 0.33 ft

Solution.

Calculate critical conditions and downstream specific energy: 1/ 3

1/ 3

 0.4 2   q2   = 0.171 ft y c =   =   32.2   g  E c = 1.5 y c = 1.5 × 0.171 = 0.256 ft E3 = y3 +

0.4 2 q2 = + = 1.0025 ft 1 .

0 64.4 × 1.0 2 2 gy 32

Because E 3 – ∆z = 1.0025 – 0.33 = 0.67 ft > E c , there is no choking, and the depth at point 2 is subcritical. Writing the energy equation from point 2 to point 3, we have

y2 +

q2 = E 3 − ∆z = 1.0025 − 0.33 = 0.6725 2 gy 22

0.4 2 0.002484 y2 + = y2 + = 0.6725 2 y 22 64.4 × y 2 Solving for y 2 , we obtain y 2 = 0.667 ft. As a consequence, the dip in the water surface over the bump is barely perceptible because we are so high up on the upper limb of the specific energy curve. Furthermore, because there is no energy loss, the depth at point 1 is identical to that at point 3. If the tailwater (y 3 ) is dropped to 0.6 ft, we get y 2 = 0.23 ft and a dip in the water surface elevation of 0.04 ft over the bump. Continuing to drop the tailwater elevation results in choking with passage through critical depth at the bump to supercritical flow followed by a hydraulic jump downstream of the bump. The specific energy diagram follows.

20

Open Channel Hydraulics

CHAPTER 2

y, ft 1.0

1

y=E 3

2

0.667

y = (2/3)E

0.33 ft

0.672

0.256

2.7

1.002

E, ft

A rectangular channel 3.6 m wide contracts to a 1.8-m wide rectangular channel and then expands back to the 3.6-m width. The contraction is gradual enough that head losses can be neglected, but the expansion loss coefficient is 0.5. The discharge through the transition is 10 m3/s. If the downstream depth at the re-expanded section is 2.4 m, calculate the depths at the approach section and the contracted section. Show the positions of the depth and specific energy for all three sections on a specific energy diagram.

3

1 2 3.6 m

3.6 m

1.8 m

Solution.

From continuity, q 3 = q 1 = 10/3.6 = 2.778 m2/s and q 2 = 10/1.8 = 5.556 m2/s. Check critical conditions and calculate downstream specific energy:

 q2  y c 3 =  3   g 

1/ 3

1/ 3

 2.778 2   =  9 .

81  

1/ 3

= 0.923 m 1/ 3

 q22   5.556 2   = 1.465 m y c 2 =   =   9.81   g  E c 2 = 1.5 y c 2 = 1.5 × 1.465 = 2.198 m q32 2.778 2 = + = 2.468 m 2 .

4 E3 = y3 + 19.62 × 2.4 2 2 gy 32 Now because E 3 > E c2 , choking cannot occur, and we are seeking a subcritical depth at section 2.

The energy equation from 2 to 3, including the head loss, is

21

Open Channel Hydraulics

CHAPTER 2

q22 y2 + = E 3 + hL 2 gy 22 y2 +

 q22 q32  5.556 2  = + − 2 .

468 0 .

5  2 gy 2 2 gy 2  19.62 × y 22 2 3  

y 2 + 0.5 × y2 +

1.573 2.778 2 = − × 2 .

468 0 .

5 y 22 19.62 × 2.4 2

0.786 = 2.434 y 22

from which y 2 = 2.28 m and E 2 = y 2 + (q 2 )2/(2gy 2 2) = 2.28 + 5.5562/[(19.62)(2.28)2] = 2.583 m.

The head loss is (E 2 – E 3 ) = 0.12 m. The energy equation from point 1 to point 2, neglecting head loss, is

y1 +

q12 = E 2 = 2.583 2 gy12

y1 +

0.3933 2.778 2 = y1 + = 2.583 2 19.62 × y1 y12

The solution is y 1 = 2.52 m, which is 0.12 m higher than the downstream depth of 2.4 m. The increase in water surface elevation is approximately equal to the head loss because of the small difference in the velocity heads. See Figure 2.11 in the text for the specific energy diagram.

2.8

Determine the discharge in a circular culvert on a steep slope if the diameter is 1.0 m and the upstream head is 1.3 m with an unsubmerged entrance. Also calculate the critical depth. Neglect entrance losses. Repeat for a box culvert that is 1.0-m square.

Solution.

The entrance depth for a steep culvert is critical depth. Write the energy equation from a point just upstream of the entrance to the culvert entrance, and set the Froude number squared equal to one for critical conditions:

Q2 1.3 = y c + 2 gAc2 Q 2 Bc = 1.0 gAc3

22

Open Channel Hydraulics

CHAPTER 2

Solve by trial by assuming a value of y c ; calculating θ c , A c , and B c from the circular channel geometry; solving for Q from the second equation; and substituting back into the first equation to determine if the upstream head is equal to 1.3 m. The equations needed for the circular channel geometry are

θ c = 2 cos −1 (1.0 − 2

yc ) d

d2 Ac = (θ − sin θ ) 8 Bc = d sin(θ / 2) in which d = diameter = 1.0 m. Organizing the computations in a table or a spreadsheet, we have

θc

yc, m 0.8 0.9 0.833

4.4286 4.9962 4.5993

A c , m2 0.674 0.744 0.699

Bc, m 0.800 0.600 0.746

Q, m3/s 1.94 2.60 2.12

H, m 1.22 1.52 1.30

The final iteration gives a discharge of 2.12 m3/s with critical depth equal to 0.833 m. The solution can be obtained approximately from Figure 2.14. In this case, E c /d = 1.3, and from Figure 2.14, we read Z = Q/(g1/2 d 5/2) = 0.7; therefore, Q = (0.7)(9.81)1/2 = 2.2 m3/s, which is acceptable considering the graphical error.

For a 1.0-m square box culvert, the channel shape is rectangular. Then for this case, we have

2 2 H = (1.30) = 0.867 m 3 3 3 1/ 2 q = ( gy c ) = (9.81 × 0.867 3 )1 / 2 = 2.53 m 2 /s yc =

and Q = bq = 2.53 m3/s.

2.9

An open channel has a semicircular bottom and vertical, parallel walls. If the diameter, d, is 3 ft, calculate the critical depth and the minimum specific energy for two discharges, 10 cfs and 30 cfs.

3.0 ft

Solution.

First, calculate the critical discharge for the water level just filling the semicircular portion of the cross section. For this case, y c = 1.5 ft; A c = πd2/8 = π(3)2/8 = 3.53 ft2; and B c = 3.0 ft. Then from the Froude number criterion, we have

23

Open Channel Hydraulics

CHAPTER 2

g 1 / 2 Ac3 / 2 32.21 / 2 × 3.533 / 2 Qc = = = 21.7 cfs Bc1 / 2 3.01 / 2 Hence, the lower discharge is for a circular geometry, and the higher discharge has a critical depth above the semicircular portion of the cross-section. For the discharge of 10 cfs, set the Froude number squared for a circular section to unity to obtain

Ac3 Q 2 10 2 = = = 3.106 Bc g 32.2 [(θ − sin θ )d 2 / 8]3 = 3.106 d sin(θ / 2) (θ − sin θ ) 3 = 6.544 sin(θ / 2) from which we find θ = 2.462 rad and y c = (d/2)[1 – cos (θ /2)] = 1.00 ft. The corresponding values of area and top width are A c = 2.063 ft2 and B c = 2.828 ft. Then E c = y c + D c /2 = 1.00 + 2.063/[(2)(2.828)] = 1.365 ft.

For Q = 30 cfs, we add the value of flow area for the semicircular section (3.53 ft2) to the area for a rectangular section stacked on top. The top width remains constant at 3.0 ft. Setting the Froude number squared to unity results in

Ac3 Q 2 30 2 = = = 27.95 Bc g 32.2 [3.53 + 3 × ( y c − 1.5)]3 = 27.95 3.0 and solving we obtain y c = 1.78 ft for which A c = 4.37 ft2 and B c = 3.0 ft. Then E c = 1.78 + 4.37/[(2)(3.0)] = 2.508 ft.

2.10

Derive an exact solution for critical depth in a parabolic channel and place it in dimensionless form. Repeat for a triangular channel.

Solution.

Referring to Table 2-1 for the parabolic channel, the bank-full depth and width are designated by y 1 and B 1 , respectively. Setting the Froude number squared to a value of unity, and substituting the expressions for flow area and top width from the table into the equation, the result for critical depth in a parabolic channel is obtained:

24

Open Channel Hydraulics

CHAPTER 2

Ac3 Q 2 = Bc g [( 2 / 3)( B1 / y11 / 2 ) y c3 / 2 ]3 Q 2 = g ( B1 / y11 / 2 ) y 1c / 2 4

 yc   3  Q 2   =   2 3  y1   2  gB1 y1 3

  yc Q = 1.355  1 / 2 3/ 2  y1  g B1 y1 

1/ 2

(parabolic)

For the triangular channel, refer again to Table 2-1, and repeat the procedure above using the triangular channel geometric expressions in which m = side-slope ratio:

Ac3 Q 2 = Bc g (myc2 ) 3 Q 2 = g 2myc  Q  yc = 1.149  1 / 2   g m

2/5

(triangular)

While y c could be nondimensionalized by the bank-full top width, for example, the only geometric factor needed is the side-slope ratio, so the expression is left in this form.

2.11

A parabolic-shaped irrigation canal has a top width of 10 m at a bank-full depth of 2 m. Calculate the critical discharge, Q c (i.e., the discharge for which the depth of uniform flow is equal to critical depth) for a uniform flow depth of 1.0 m. If Q < Q c for the uniform flow depth of 1.0 m, will the uniform flow be supercritical or subcritical?

Solution.

Calculate the flow area and top width of flow for y = 1.0 m using the geometric expressions from Table 2-1.

B=

B1 1 / 2 10 y = 1 / 2 (1.0)1 / 2 = 7.07 m 1/ 2 2 y1

A = ( 2 / 3) By = ( 2 / 3)(7.07)(1.0) = 4.71 m 2

25

Open Channel Hydraulics

CHAPTER 2

Then from the definition of the Froude number, the critical discharge is defined for the Froude number equal to unity and y = y c to give

Qc =

g 1 / 2 A3 / 2 (9.81)1 / 2 × ( 4.71) 3 / 2 = = 12.04 m 3 /s B1/ 2 7.071 / 2

If the uniform flow depth is equal to 1.0 m for Q < Q c , then the Froude number is less than one, and the flow is subcritical.

2.12

A USGS study of natural channel shapes in the western United States reports an average ratio of maximum depth to hydraulic depth in the main channel (with no overflow) of y/D = 1.55 for 761 measurements. (a)

Calculate the ratio of maximum depth to hydraulic depth for a (1) triangular channel; (2) parabolic channel; (3) rectangular channel. What do you conclude?

(b)

Calculate the discharge for a bank-full Froude number of F 1 = 1.0 if y/D = 1.55 and B 1 = 100 ft for y 1 = 10 ft. What is the significance of this discharge?

Solution. (a)

For each geometric shape, substitute the expressions for flow area and top width into the definition of hydraulic depth: Triangular:

D=

A my 2 y = = B 2my 2

⇒

y = 2.0 D

Parabolic:

D=

A ( 2 / 3) By y = = ( 2 / 3) y ⇒ = 1.5 B B D

Rectangular:

D=

A by y = = y ⇒ = 1.0 B b D

The implication is that natural channels from this data set are more nearly parabolic in shape.

26

Open Channel Hydraulics

CHAPTER 2

(b) For y/D = 1.55 at bank-full flow, we can show that

y1 = 1.55 A1 / B1 A1 =

B1 y1 (100)(10) = = 645 ft 2 1.55 1.55

Set the Froude number equal to unity and solve for Q c1 = critical bank-full discharge:

Q c1 =

g 1 / 2 A13 / 2 (32.2)1 / 2 (645) 3 / 2 = = 9295 cfs B11 / 2 1001 / 2

If the actual Q > Q c1 , which is referred to as the upper limiting discharge Q U in the text, then there is only one critical depth that occurs in overbank flow. In this case, any flows in the main channel alone would be supercritical.

2.13

A natural channel cross-section has a bank-full cross-sectional area of 45 m2 and a top width of 37.5 m. The maximum value of F c /F 1 has been calculated to be 1.236. Find the discharge range, if any, within which multiple critical depths could be expected.

Solution.

From the data given, A 1 = 45 m2 and B 1 = 37.5 m. First calculate the upper limiting discharge, Q U , as

g 1 / 2 A11 / 2 (9.81)1 / 2 ( 45) 3 / 2 QU = = = 154 m 3 /s 1/ 2 1/ 2 B1 37.5 Then the lower limiting discharge is calculated by

QU Fc max = = 1.236 F1 QL QL =

154 = 125 m 3 / s 1.236

So for Q in the range of 125 to 154 m3/s, there are two values of critical depth, one in main channel flow alone, and one in overbank flow for this cross section.

27

Open Channel Hydraulics

Design a broad-crested weir for a laboratory flume with a width of 15 in. The discharge range is 0.1 to 1.0 cfs. The maximum approach flow depth is 18 in. Determine the height of the weir and the weir length in the flow direction. Plot the expected head-discharge relationship.

Solution.

With reference to Figure 2.25, we require H/(H + P) ≤ 0.35 and 0.08 ≤ H/l ≤ 0.33 for broadcrested behavior. The head-discharge relationship is solved for H max when Q = Q max = 1.0 cfs to produce

  Q max H max =  1/ 2   C d ( 2 / 3)( 2 g / 3) L 

2/3

  1.0 =  1/ 2  (0.848)( 2 / 3)( 2 × 32.2 / 3) (1.25) 

2/3

= 0.454 ft

in which the coefficient of discharge has been taken to be 0.848, and the crest length is 1.25 ft.

Similarly, the minimum head is determined to be 0.098 ft for Q = 0.1 cfs. For broad-crested behavior, set H max /l = 0.33, and calculate l = 0.454/0.33 = 1.38 ft, which is the length of the weir in the flow direction. For the minimum head this gives H min /l =0.071, which is only slightly less than the allowable value. In addition, set H/(H + P) = 0.35 for H = H max and solve for the weir height, P = (H max /0.35) – H max = (0.454/0.35) – 0.454 = 0.843 ft. So the weir should have a length in the flow direction, l = 1.4 ft,and a height, P = 0.84 ft.

To plot the head-discharge relationship, use Equation 2.47 with C d = 0.848; L = 1.25 ft; and C v from Equation 2.48, which has to be solved for a calculated value of C d A*/A 1 for each head. The result is shown in the following figure:

0.5

0.4

0.3

H , ft

2.14

CHAPTER 2

0.2

0.1

0.0 0.0

0.2

0.4

0.6 Q , cfs

28

0.8

1.0

Open Channel Hydraulics

Plot and compare the head-discharge relationships for a rectangular sharp-crested weir having a crest length of 1.0 ft in a 5-ft wide channel with that for a 90o V-notch, sharp-crested weir if both weir crests are 1 ft above the channel bottom. Consider a head range of 0–0.5 ft.

Solution.

For the rectangular, sharp-crested weir, P = 1.0 ft; L = 1.0 ft; and b = 5 ft. Then L/b = 0.2 and from Table 2-3, C de = 0.589 – 0.0018 H/P. In addition, k L = 0.0082 ft (0.0025 m) from Figure 2.23 and k h = 0.003 ft. The head-discharge relationship is given by

2 ( 2 g )1 / 2 C de Le H e3 / 2 3 Q = 5.35 × (0.589 − 0.0018H / P ) × 1.0082 × ( H + 0.003) 3 / 2 Q=

The triangular weir has θ = 90o and P = 1.0 ft. From Figure 2.24, C de = 0.58 and k h = 0.0033 ft (1 mm). The head-discharge relationship is given by

8 (2 g )1 / 2 tan(θ / 2)( H + k h ) 5 / 2 15 8 Q = 0.58 × × 64.41 / 2 × tan(45°) × ( H + 0.0033) 5 / 2 15 Q = 2.48( H + 0.0033) 5 / 2 Q = C de

The head-discharge relationships are compared in the following figure.

0.5

0.4

0.3

H , ft

2.15

CHAPTER 2

0.2

0.1

0.0 0.0

0.2

0.4

0.6

0.8

Q , cfs Rect.

Triang.

29

1.0

1.2

Open Channel Hydraulics

2.16

CHAPTER 2

Derive the head-discharge relationship for a triangular, broad-crested weir and a corresponding relationship for C v analogous to Equation 2.48.

Solution.

The shape of the weir cross section is triangular with a notch angle of θ as shown in Figure 2.24.

(The head H is measured from the crest to the free surface upstream of the weir plate.) As a broad-crested weir, however, the crest width is long enough in the flow direction that the theoretical value of critical depth occurs on the crest. The energy equation is written from the approach section to the critical section at the crest to yield

H+

D V12 = H e = yc + c 2g 2 Ac y c2 tan(θ / 2) 5 = yc + = yc 2 Bc 2[2 y c tan(θ / 2)] 4

H e = yc +

in which H e = total energy head; V 1 = approach velocity; D c = hydraulic depth; A c = flow area at the crest; and B c = flow top width at the crest. In Exercise 2.10, it was shown that critical depth for the triangular cross section is given by

yc = 2

1/ 5

  Q  g 1 / 2 tan(θ / 2)   

2/5

in which the side-slope ratio, m = tan(θ/2). This expression for critical depth is substituted back into the equation, H e = (5/4)y c ; and the approach velocity coefficient, C v = (H e /H)5/2 , and discharge coefficient, C d , are incorporated to obtain the solution for Q:

Q = 0.405 Cv C d g 1 / 2 tan(θ / 2) H 5 / 2 Write the definition of the total energy head, H e ; substitute the solution for Q into the equation; and solve to give

He = H +

Q2 2 gA12

He = H +

[0.405 Cv C d g 1 / 2 tan(θ / 2) H 5 / 2 ]2 2 gA12

H   e  H 

5/ 2

2   C d A*  2   Cv  = Cv = 1 + 0.0820    A1 

(

)

1/ 2

C d A* C 2/5 −1 = v 0.286 Cv A1

30

5/ 2

Open Channel Hydraulics

CHAPTER 2

in which A* = flow area in the control section for a depth equal to H, and A 1 = flow area in the approach cross section.

2.17

Derive the head-discharge relationship for a truncated, triangular, sharp-crested weir with notch angle θ and vertical walls that begin at a height of h 1 above the triangular crest. Assume that H > h1.

L η

H

θ

h1

Solution.

To obtain the theoretical discharge , Q t , integrate the velocity distribution over the uncontracted flow area:

Qt = ∫

H − h1 0

vL dη + ∫

H H − h1

v [2( H − η ) tan(θ / 2)]dη

Neglecting the approach velocity head and assuming atmospheric pressure across the uncontracted nappe at the weir section, the velocity, v = (2gη)1/2. Substituting into the previous equation, we have

Qt = ∫

H − h1 0

2 g Lη 1 / 2 dη + ∫

H H − h1

2 2 g tan(θ / 2)[( Hη 1 / 2 − η 3 / 2 ]dη

Integrating, the result is

Qt = 2 g L

η 3/ 2

H − h1

3/ 2 0

H  Hη 3 / 2 H  η 5/ 2 + 2 2 g tan(θ / 2)  −   3 / 2 H −h1 5 / 2 H −h1 

Evaluating the limits and collecting terms, and introducing the coefficient of discharge, C d , to compensate for the earlier assumptions, the final result for the actual discharge, Q, is

Q=

2.18

2 2 2 4  2 g C d L( H − h1 ) 3 / 2 + 2 2 g tan(θ / 2) C d  H 5 / 2 − H ( H − h1 ) 3 / 2 + ( H − h1 ) 5 / 2  3 3 5 15 

Modify the computer program Y0YC in Appendix B to calculate the critical depth in a circular channel.

Solution.

The primary change to be made to Y0YC is the reformulation of the geometric elements of area, top width, and hydraulic radius for a circular section as given in Table 2-1. However, the

31

Open Channel Hydraulics

CHAPTER 2

relationship between θ and y/d requires the cos–1 function, which is not available in Visual Basic, but the tan–1 function is available. Hence, the following trigonometric identity is utilized:

cos −1 ( x ) =

 x   − tan −1  2  2  1− x 

π

in which x = (1 – 2y/d). This is shown in the following program listing in the function procedure F. In addition, upper limits must be placed on the root search for both critical and normal depth in the circular cross section. In the case of critical depth, the top width approaches zero as y/d approaches 1.0, so Q approaches infinity. In addition, the tan–1 argument goes to infinity as y/d approaches 1.0, so the upper limit Y2 is set equal to 0.9999d. For the normal depth function, it is clear from Figure 4.9 that there is a maximum in AR 2/3/d 8/3 above which there are two possible solutions, so the upper limit Y2 is set equal to 0.938d at which this maximum occurs. Finally, the initial test for the existence of a root in Sub BISECTION requires an indication of when the normal depth exceeds the maximum, and this is indicated by setting it equal to a very large number such 1.0E6. The program listing for the Sub procedure follows. The form module is similar to that shown for the program Y0YC in Appendix B except that the diameter d is entered for the circular section instead of b and m for the trapezoidal section.

Option Explicit Dim Q As Single, S As Single, d As Single Dim n As Single, Y0 As Single, YC As Single Sub Y0YC(Q, S, d, n, Y0, YC) Dim Y1 As Single, Y2 As Single, ER As Single Dim NF As Integer Y1 = 0.0001 Y2 = 0.9999 * d ER = 0.0001 NF = 1 Call BISECTION(Y1, Y2, NF, ER, Q, S, d, n, YC) Y1 = 0.0001 Y2 = 0.938 * d NF = 2 Call BISECTION(Y1, Y2, NF, ER, Q, S, d, n, Y0) End Sub Sub BISECTION(Y1, Y2, NFUNC, ER, Q, S, d, n, Y3) Dim FY1 As Single, FY2 As Single, FY3 As Single, FZ As Single Dim I As Integer FY1 = F(Y1, NFUNC, Q, S, d, n) FY2 = F(Y2, NFUNC, Q, S, d, n) If FY1 * FY2 > 0 Then Y3 = 1000000# Exit Sub End If For I = 1 To 50 Y3 = (Y1 + Y2) / 2 FY3 = F(Y3, NFUNC, Q, S, d, n) FZ = FY1 * FY3 If FZ = 0 Then Exit Sub

32

Open Channel Hydraulics

CHAPTER 2

If FZ < 0 Then Y2 = Y3 Else Y1 = Y3 If Abs((Y2 - Y1) / Y3) < ER Then Exit Sub Next I End Sub Function F(Y, NFUNC, Q, S, d, n) As Single Dim A As Single, P As Single, R As Single, T As Single Dim PI As Double, THETA As Double, X As Double X = 1 - 2 * Y / d PI = 3.141592654 THETA = 2 * (PI / 2 - Atn(X / (1 - X ^ 2) ^ 0.5)) A = (THETA - Sin(THETA)) * d ^ 2 / 8 P = THETA * d / 2 R = A / P T = d * Sin(THETA / 2) If NFUNC = 1 Then F = Q - Sqr(32.2) * A ^ 1.5 / T ^ 0.5 Else F = Q - (1.486 / n) * A * R ^ (2 / 3) * S ^ (1 / 2) End If End Function

2.19

Write a computer program that computes the depth in a width contraction and the upstream depth given a subcritical tailwater depth as in Figure 2.11. Assume that the channel is rectangular at all three sections and make provision for a head-loss coefficient that is nonzero; include a check for possible choking.

Solution.

Referring to Fig. 2.11, the downstream depth, y 3 ; channel widths, b 3 and b 2 ; discharge, Q; and the expansion and contraction loss coefficients, K Lexp and K Lcont , respectively, must be entered in the form module. It is assumed that b 1 = b 3 . In the program listing that follows, the Sub procedure YTRANS is called from the form module. First in YTRANS, the possibility of choking is checked from the following inequality:

E c 2 < E 3 + h L ( 2 c − 3)

in which E 3 = the known specific energy at the downstream section; E c2 = minimum specific energy at the contracted section 2; and h L(2c-3) = head loss from section 2 to 3 assuming that critical depth occurs in the contracted section. If this inequality is true, then choking does not occur and computations for the depth at section 2 can proceed by calling the BISECTION procedure. If it is true, on the other hand, then choking occurs and the depth at section 2 is set equal to the critical depth for the contracted section. For the nonchoking case, the equation to be solved is the energy equation written from section 2 to 3 as

33

Open Channel Hydraulics

y2 +

CHAPTER 2

Q2 Q2 Q2 E K = + − 3 L 2 gA22 2 gA22 2 gA32

in which y 2 is unknown while E 3 and A 3 are known from the specified downstream depth, y 3 .

The solution to this equation is obtained from Sub BISECTION used with the function subprocedure F, which is defined by the right hand side of the equation above minus the left hand side as shown in the program listing. The correspondence between the variables in the equation and the variable names in the program listing is given by: Y = y 2 ; Eds = E 3 ; Yds = y 3 ; and HL = the head loss term. Once this equation has been solved for y 2 , control returns to the calling procedure, YTRANS, which calls BISECTION again with section 2 as the downstream section and section 1 as the upstream section where the depth is unknown. Then a solution is obtained for y 1 . The program listing is given below.

Option Explicit Dim Q As Single, KLEXP As Single, KLCONT As Single Dim B3 As Single, B2 As Single, G As Single, YC3 As Single Dim Y3 As Single, Y2 As Single, Y1 As Single, YC2 As Single Private Sub cmdCalculate_Click() Q = Val(txtQ.Text) KLEXP = Val(txtHlexp.Text) KLCONT = Val(txtHlcont.Text) B3 = Val(txtB3.Text) B2 = Val(txtB2.Text) Y3 = Val(txtY3.Text) G = 32.2 YC3 = ((Q / B3) ^ 2 / G) ^ (1 / 3) YC2 = ((Q / B2) ^ 2 / G) ^ (1 / 3) txtYC3.Text = Format(YC3, "###.000") txtYC2.Text = Format(YC2, "###.000") Call YTRANS(Q, KLEXP, KLCONT, B3, B2, Y3, Y2, Y1) If Y2 = 0 Then txtY2.Text = "Enter y3 > YC3" txtY1.Text = "0" Else txtY2.Text = Format(Y2, "###.000") txtY1.Text = Format(Y1, "###.000") End If End Sub Private Sub cmdExit_Click() End End Sub Option Explicit Dim Q As Single, KLEXP As Single, KLCONT As Single Dim BE As Single, BC As Single Dim Y3 As Single, Y2 As Single, Y1 As Single Sub YTRANS(Q, KLEXP, KLCONT, BE, BC, Y3, Y2, Y1) Dim YL As Single, YR As Single, ER As Single Dim YC3 As Single, YC2 As Single, EC2 As Single Dim G As Single, HLC As Single, E3 As Single

34

Open Channel Hydraulics

CHAPTER 2

G = 32.2 ER = 0.0001 YC3 = ((Q / BE) ^ 2 / G) ^ (1 / 3) If Y3 <= YC3 Then Y2 = 0: Exit Sub YC2 = ((Q / BC) ^ 2 / G) ^ (1 / 3) EC2 = 1.5 * YC2 E3 = Y3 + Q ^ 2 / (2 * G * BE ^ 2 * Y3 ^ 2) HLC = KLEXP * Abs(Q ^ 2 / (2 * G * (BE * Y3) ^ 2) _ - Q ^ 2 / (2 * G * (BC * YC2) ^ 2)) If EC2 < (E3 + HLC) Then YL = YC2 YR = 50 Call BISECTION(YL, YR, ER, Q, KLEXP, BE, BC, Y3, Y2) Else Y2 = YC2 End If YL = YC3 YR = 50 Call BISECTION(YL, YR, ER, Q, KLCONT, BC, BE, Y2, Y1) End Sub Sub BISECTION(YL, YR, ER, Q, KL, Bds, Bus, Yds, Y) Dim FYL As Single, FYR As Single, FY As Single, FZ As Single Dim I As Integer FYL = F(YL, Q, KL, Bds, Bus, Yds) FYR = F(YR, Q, KL, Bds, Bus, Yds) If FYL * FYR > 0 Then Y = 1000000# Exit Sub End If For I = 1 To 50 Y = (YL + YR) / 2 FY = F(Y, Q, KL, Bds, Bus, Yds) FZ = FYL * FY If FZ = 0 Then Exit Sub If FZ < 0 Then YR = Y Else YL = Y If Abs((YR - YL) / Y) < ER Then Exit Sub Next I End Sub Function F(Y, Q, KL, Bds, Bus, Yds) As Single Dim Eds As Single, CQ As Single, HL As Single, G As Single G = 32.2 CQ = Q ^ 2 / (2 * G) Eds = Yds + CQ / (Bds * Yds) ^ 2 HL = KL * Abs(CQ / (Bus * Y) ^ 2 - CQ / (Bds * Yds) ^ 2) F = Eds + HL - (Y + CQ / (Bus * Y) ^ 2) End Function

35

Open Channel Hydraulics

2.20

CHAPTER 2

A laboratory experiment has been conducted in a horizontal flume in which a sharp-crested weir plate has been installed to determine the head-discharge relationship for a rectangular, sharpcrested weir. With reference to Figure 2.23, P = 0.506 ft, L = 0.25 ft, and b = 1.25 ft. The discharge was measured by a bend meter for which the calibration is given by Q = 0.075 ∆h0.523, in which Q = discharge in cubic feet per second; ∆h = manometer deflection in inches of water; and the uncertainty in the calibration is +0.003 cfs. The head on the crest of the weir was measured by a point gauge and is given in the data table that follows. An upstream view of the weir nappe can be seen in Figure 2.28. ∆h, in.

13.2 11.5 11.2 8.3 8.0 6.2 6.1 4.3 4.2 2.4 2.0

(a)

(b)

H, ft 0.498 0.476 0.474 0.425 0.421 0.384 0.386 0.333 0.334 0.272 0.257

Plot the head on the vertical scale and the discharge on the horizontal scale of log-log axes and obtain a least-squares regression fit forcing the inverse slope to be the theoretical value of 3/2. What are the single best-fit value of C d and the standard error in C d ? Compare the standard error of the "Q estimate" with the uncertainty in the bendmeter calibration.

Calculate the discharge using the Kindsvater-Carter relationship and using the single best-fit value of C d . Compare both sets of results with the measured discharges by calculating the percent differences and also plotting the measured vs. calculated discharges.

Solution. (a) The spreadsheet calculations are given next in which the Q values are calculated from the values of ∆h using the calibration equation for the bend meter. Linear regression analysis between the Q values and H 3/2 is performed to force the exponent on the head to be equal to its theoretical value of 3/2. The best fit value of C = Q/H 3/2 = 0.819 ± 0.002. The standard error of estimate of the Q values is ± 0.0019 cfs, which is slightly smaller than the estimate of uncertainty in the bend meter calibration. The coefficient of determination, R2 = 0.999. The coefficient C = (2/3)(2g)1/2C d L from which we can solve for C d to obtain C d = 0.612 ± 0.003. (The standard error also has to be converted from its value on C to the value on C d .)

36

Open Channel Hydraulics

∆h, in.

13.2 11.5 11.2 8.3 8.0 6.2 6.1 4.3 4.2 2.4 2.0

CHAPTER 2

H,ft 0.498 0.476 0.474 0.425 0.421 0.384 0.386 0.333 0.334 0.272 0.257

Q, cfs 0.289 0.269 0.265 0.227 0.223 0.195 0.193 0.161 0.159 0.119 0.108

H^3/2 0.351 0.328 0.326 0.277 0.273 0.238 0.240 0.192 0.193 0.142 0.130

Q, Best Fit 0.288 0.269 0.267 0.227 0.224 0.195 0.196 0.157 0.158 0.116 0.107

Q, K.-C. % Dif.: K.-C. 0.288 -0.52% 0.269 -0.03% 0.267 0.73% 0.227 0.17% 0.224 0.69% 0.195 0.35% 0.197 1.99% 0.158 -1.67% 0.159 -0.01% 0.117 -1.19% 0.108 -0.07%

SS 0.03661108 3.71335E-05 0.036648214

MS 0.03661108 3.7133E-06

F 9859.31359

Coefficients Standard Error 0 #N/A 0.8189643 0.002279472

t Stat #N/A 359.278001

P-value Lower 95% Upper 95% #N/A #N/A #N/A 6.865E-22 0.81388531 0.82404328

SUMMARY OUTPUT Regression Statistics Multiple R 0.99949325 R Square 0.99898676 Adjusted R Sq. 0.89898676 Standard Error 0.00192701 Observations 11 ANOVA df Regression Residual Total

Intercept X Variable 1

1 10 11

37

Significance F 5.4068E-15

Open Channel Hydraulics

CHAPTER 2

The plot of H vs. Q on log-log scales is shown in the following figure.

H , ft

1.00

Rectangular, sharp-crested weir L/b = 0.2

C = 0.819

0.10 0.10

1.00 Q , cfs Data

Best Fit

(b) The values of Q predicted by the Kindsvater-Carter relationship are determined from a coefficient of discharge obtained from Table 2-3 for L/b = 0.2 and given by

C de = 0.589 − 0.0018

H P

The correction on H is k H = 0.003 ft, and the crest length correction from Figure 2.23c is k L = 0.0082 ft. The Kindsvater-Carter values of Q along with the Q values obtained from a single best-fit C d value of 0.612 are shown in the spreadsheet given previously. The best-fit relationship and the Kindsvater-Carter formula give virtually identical results. The percent difference between the Q values predicted by the Kindsvater-Carter formula and the measured Q values are also shown in the spreadsheet. The maximum difference is 2.0 percent. The same comparison is shown graphically in the following figure.

38

Open Channel Hydraulics

CHAPTER 2

0.40

Rectangular, sharp-crested weir predictions by Kindsvater-Carter formula

Predicted Q , cfs

0.30

0.20

0.10

0.00 0.00

0.10

0.20

0.30

Measured Q , cfs

39

0.40

Open Channel Hydraulics

CHAPTER 2

40

CHAPTER 3. Momentum 3.1

A hydraulic jump is to be formed in a trapezoidal channel with a base width of 20 ft and side slopes of 2:1. The upstream depth is 1.25 ft and Q=1000 cfs. Find the downstream depth and the head loss in the jump. Solve by Figure 3.2 and verify by manual calculations. Compare the results for the sequent depth ratio and relative head loss with those in a rectangular channel of the same bottom width and approach Froude number.

Solution.

First, calculate the critical depth to limit the root search for the sequent depth by setting the Froude number squared equal to unity:

Ac3 [ y c ( 20 + 2 y c )]3 Q 2 1000 2 = = = = 31,056 20 + 4 y c 32.2 Bc g from which y c = 3.74 ft. Then from Table 3-1, the momentum function for a trapezoidal channel is set equal upstream and downstream of the hydraulic jump. Its value is calculated for the upstream sequent depth, and the resulting equation is solved for the downstream sequent depth, y 2 .

M1 = M 2 by12 my13 by 22 my 23 Q2 Q2 + + = + + 2 3 2 3 gy1 (b + my1 ) gy 2 (b + my 2 ) 20 × 1.252 2 × 1.253 1000 2 + + = 1121 2 3 32.2 × 1.25 × ( 20 + 2 × 1.25) 1121 = 10 y 22 +

2 y 23 31,056 + 3 y 2 ( 20 + 2 y 2 )

The result is y 2 = 8.12 ft. To use Figure 3.2, calculate Z trap as

Qm 3 / 2 1000 × 2 3 / 2 Z trap = 1 / 2 5 / 2 = = 0.28 g b 32.21 / 2 × 20 5 / 2 and my 1 /b = 2×1.25/20 = 0.125. Then from Figure 3.2, y 2 /y 1 = 6.5 and y 2 = 8.1 ft in agreement with the manual calculation. The upstream Froude number is

F1 =

QB11 / 2 1000 × (20 + 2 × 2 × 1.25)1 / 2 = = 5.91 g 1 / 2 A13 / 2 32.21 / 2 × [1.25 × (20 + 2 × 1.25)]3 / 2

and in a similar fashion, the downstream Froude number is 0.25. The energy head loss is calculated as the difference between the upstream and downstream energy values and is given by

41

Open Channel Hydraulics

∆E = y1 +

CHAPTER 3

Q2 Q2 y − − 2 2 gA12 2 gA22

∆E = 1.25 +

1000 2 1000 2 − − 8 .

12 64.4 × [1.25 × (20 + 2 × 1.25)]2 64.4 × [8.12 × (20 + 2 × 8.12)]2

∆E = 1.25 + 19.63 − 8.12 − 0.179 = 12.6 ft For a rectangular channel having the same upstream Froude number of 5.91 and a width of 20 ft, the upstream depth is 1.305 ft for Q = 1000 cfs. The sequent depth and energy loss are calculated from

y1 1.305 [ −1 + 1 + 8F12 ] = [ −1 + 1 + 8 × 5.912 ] = 10.27 ft 2 2 1000 2 1000 2 ∆E = 1.305 + − 10 .

27 − = 13.5 ft 64.4 × [1.305 × 20]2 64.4 × [10.27 × 20]2 y2 =

So for the trapezoidal channel, y 2 /y 1 = 6.50 and ∆E/E 1 = 12.6/20.9 = 0.60; while for the rectangular channel, the results are y 2 /y 1 = 7.87 and ∆E/E 1 = 13.5/24.1 = 0.56.

3.2

Determine the sequent depth for a hydraulic jump in a 3-ft diameter storm sewer with a flow depth of 0.6 ft at a discharge of 5 cfs. Solve by Figure 3.3 and verify by manual calculations.

Solution.

Calculate the critical depth by setting the Froude number squared equal to one, and rearranging to produce

Ac3 [(θ − sin θ )d 2 / 8]3 Q 2 52 = = = = 0.7764 Bc d sin(θ / 2) g 32.2 (θ − sin θ ) 3 = 1.636 sin(θ / 2) which can be solved to give θ = 2.0166 rad; and y c = (d/2)[1–cos(θ/2)] = 0.700 ft, where d = 3.0 ft. The given depth of 0.6 ft is supercritical, and we are seeking the sequent depth by equating the values of the momentum function upstream and downstream of the jump. Utilizing the definition from Table 3-1, the momentum function upstream of the jump is determined for θ 1 = 2 cos-1(1 – 2y 1 /d) = 2 cos-1(1 – 2×0.6/3.0) = 1.8546 rad. The result is

d3 Q2 M 1 = [3 sin(θ / 2) − sin (θ / 2) − 3(θ / 2) cos(θ / 2)] + 2 = 1.01767 24 gd (θ − sin θ ) / 8 3

42

Open Channel Hydraulics

CHAPTER 3

where θ = 1.8546 rad; d =3.0 ft; and Q = 5.0 cfs. Equating M 2 to 1.01767, we have

M 2 = [3 sin(θ / 2) − sin 3 (θ / 2) − 3(θ / 2) cos(θ / 2)]

33 52 + = 1.01767 24 32.2 × 32 × (θ − sin θ ) / 8

which can be solved to obtain the downstream value of θ = 2.1875 rad and y 2 = 0.811 ft. Note that the root search was limited by θ > θ c = 2.0166 rad. Utilizing Figure 3.3, Z circ is calculated from

Z circ =

Q 5.0 = = 0.0565 5/ 2 1/ 2 g d 32.2 × 3.05 / 2 1/ 2

and y 2 /y 1 is estimated to be 1.3 so that y 2 = 0.8 ft, which is within the graphical error.

3.3

Derive the relationship between the sequent depth ratio and approach Froude number for a triangular channel and verify with Figure 3.5. Repeat the derivation for a parabolic channel.

Solution.

Develop an expression for the Froude number for a triangular channel using the geometric elements from Table 3-1:

F2 =

Q 2 B Q 2 ( 2my ) 2Q 2 = = gA3 g ( my 2 ) 3 gm 2 y 5

in which m = side-slope ratio. Then set M 1 = M 2 for the triangular channel; divide by my 1 3/3; substitute for the Froude number squared; and rearrange:

43

Open Channel Hydraulics

CHAPTER 3

my13 my 23 Q2 Q2 + = + gmy12 gmy 22 3 3 3

 y2  3Q 2 3Q 2  y1      = + 1+ gm 2 y15  y1  gm 2 y15  y 2 

2

3

3F12 3 2  y2  1 + F1 =   + 2 2  y1  2( y 2 / y1 ) 3  2  y 2    − 1 3  y1   F12 =  1 1− ( y 2 / y1 ) 2

As an example, take F 1 = 8.0 and solve to obtain y 2 /y 1 = 4.52, which agrees with Figure 3.5.

Repeating the same procedure for the parabolic channel, the Froude number is given by

F2 =

2

Q 2 B0

y y0

27 Q 2 Q B = = 3 8 B02 4 gA3  2 B0 3 / 2  g y y  g y0 3 y  0  

in which B 0 = channel top width at a bank-full depth of y 0 . Now set M 1 = M 2 and substitute the geometric expressions for a parabolic channel from Table 3-1. Then divide by (4/15)B 0 y 1 5/2/y 0 1/2; substitute for the Froude number squared; and rearrange to obtain the final result.

4 B0 5 / 2 3 y1 + 2 15 y 0 45 1+ 8

4 B0 5 / 2 3 Q2 Q2 = y2 + B B 2 15 y 0 g 0 y 23 / 2 g 0 y13 / 2 y0 y0

y  Q2 =  2  2 B y g 0 y14  1  y0

5/ 2

45 + 8

Q 2  y1    B02 4  y 2  g y1 y0

44

3/ 2

Open Channel Hydraulics

CHAPTER 3

y  5 1 + F12 =  2  3  y1   y2     y1 

3 F = 5 1− 2 1

5/ 2

F12 5 + 3 ( y 2 / y1 ) 3 / 2

5/ 2

−1

1 ( y 2 / y1 ) 3 / 2

Again taking F 1 = 8.0, the solution for the parabolic channel is y 2 /y 1 = 6.33 in agreement with Figure 3.5.

3.4

A flume with triangular cross section contains water flowing at a depth of 0.15 m and at a discharge of 0.30 m3/s. The side slopes of the flume are 2:1. Determine the sequent depth for a hydraulic jump.

Solution.

Calculate the Froude number for the given conditions using the expression developed in the previous exercise for a triangular channel:

Q 2 B Q 2 (2my ) 2Q 2 2 × 0.30 2 F = = = = = 60.4 gA3 g (my 2 ) 3 gm 2 y 5 9.81× 2 2 × 0.155 2

from which F = 7.8, so the given depth is supercritical. The critical depth is obtained from setting the Froude number squared equal to one:

 2Q 2   y c =  2   gm 

1/ 5

 2 × 0.30 2   =  2   9.81 × 2 

1/ 5

= 0.341 m

Now we are looking for a sequent depth greater than 0.341 m that can be obtained either by equating the momentum function upstream and downstream of the jump, or by using the expression developed in the previous exercise. Taking the latter approach, we have

45

Open Channel Hydraulics

CHAPTER 3

3  2  y 2    − 1 3  y1   = 60.4 F12 =  1 1− ( y 2 / y1 ) 2

from which y 2 /y 1 = 4.43 and y 2 = 0.66 m.

3.5

A parabolic channel has a bank-full depth of 2.0 m and a bank-full width of 10.0 m. If the downstream sequent depth of a hydraulic jump in the channel is 1.5 m for a flow rate of 8.5 m3/s, what is the upstream sequent depth?

Solution.

In order to limit the root search for the sequent depth, calculate the critical depth for the parabolic channel by equating the Froude number squared to a value of one.

F2 =

Q 2 Bc = 1.0 gAc3

Q 2 B0

yc y0

 2 B0 3 / 2  g y  3 y c  0  

= 3

27 8

Q2 = 1.0 B02 4 g yc y0

    2 27 8.5   yc = ×  8 10 2  9.81 ×   2.0  

1/ 4

= 0.84 m

Hence, the given depth of 1.5 m is indeed subcritical, and the sequent depth is supercritical and less than 0.84 m. Calculate the momentum function for the downstream depth:

4 B0 5 / 2 3 M2 = y2 + 15 y 0 2 M2 =

Q2 B g 0 y 23 / 2 y0

4 10 3 8.52 × × 1.55 / 2 + × = 6.0466 m 3 10 15 2 2 9.81 × × 1.53 / 2 2

46

Open Channel Hydraulics

CHAPTER 3

Now set M 1 = M 2 , and solve for the sequent depth:

M1 =

Q2 = 6.0466 B0 3 / 2 g y1 y0

4 B0 5 / 2 3 y1 + 15 y 0 2

4 10 5 / 2 3 8.52 × y1 + × = 6.0466 15 2 9.81 × 10 y 3 / 2 2 1 2 1.562 1.8856 y15 / 2 + 3 / 2 = 6.0466 y1 Solving, we get y 1 = 0.415 m.

3.6

A hydraulic jump occurs on a sloping rectangular channel that has an angle of inclination, θ. The sequent depths are d 1 and d 2 measured perpendicular to the channel bottom. Assume that the jump has a length, L j , and a linear profile. Derive the solution for the sequent depth ratio, and show that it is identical to the solution for a horizontal slope if the upstream Froude number, F 1 , is replaced by the dimensionless number, G 1 , given by

G1 =

F1 L j sin θ cos θ − d 2 − d1

Solution.

Lj

d1

d2 F2

F1

θ W

Apply the momentum equation in the direction of flow down the slope; divide by γb/2; write the difference of two squares on the left as the product of the sum and difference while writing the right hand side over a common denominator; then divide by (d 1 – d 2 ) and multiply by d 2 /d 1 2 to produce

47

Open Channel Hydraulics

CHAPTER 3

F1 − F2 + W sin θ = ρA1V1 (V2 − V1 )

γ d 12 cosθ

γ d 22 cosθ

d1 + d 2 bL j sin θ = ρ by1V1 (V2 − V1 ) 2 2 2 2q 2  1 1  −  ( d 12 − d 22 ) cosθ + ( d 1 + d 2 ) L j sin θ = g  d 2 d1  b−

b +γ

( d 1 − d 2 )( d 1 + d 2 ) cosθ + ( d 1 + d 2 ) L j sin θ =

2q 2  d 1 − d 2    g  d 1d 2 

L j sin θ   2q 2 = ( d 1 + d 2 ) cosθ + d 1 − d 2  gd 1d 2  2

 d2   d2  2F12   +   − =0  d 1   d 1   cosθ − L j sin θ   d 2 − d 1   in which F 1 is defined here as V 1 /(gd 1 )0.5. The last equation can be solved by the quadratic formula. If the third term on the left hand side is defined as 2G 1 2 as indicated in the problem statement, the solution is

(

d2 1 = − 1 + 1 + 8G 12 d1 2 3.7

)

The discharge of water over a spillway 40 ft wide is 10,000 cfs into a stilling basin of the same width. The lake level behind the spillway has an elevation of 200 ft, and the river water surface elevation downstream of the stilling basin is 100 ft. Assuming a 10 percent energy loss in the flow down the spillway, find the invert elevation of the floor of the stilling basin so that the hydraulic jump forms in the basin. Select the appropriate U.S. Bureau of Reclamation (USBR) stilling basin and sketch it showing all dimensions.

Solution.

y1

y2

200 ft

TW = 100 ft Z

DATUM

48

Open Channel Hydraulics

CHAPTER 3

Let Z = the unknown elevation of the floor of the stilling basin, and write the energy equation from the upstream reservoir surface to section 1 assuming an energy loss that is 10 percent of the difference between the reservoir water surface elevation and the water surface elevation just upstream of the hydraulic jump:

200 = Z + y1 +

q2 + hL 2 gy12

Z = 200 − y1 −

(10,000 / 40) 2 − 0.10 × ( 200 − Z − y1 ) 64.4 × y12

Z = 200 − y1 −

1078 y12

(1)

Then from section 1 to section 2, apply the momentum equation to the hydraulic jump to yield

)

(

y2 1 q 2  1  2 = − 1 + 1 + 8F1 = −1+ 1+ 8 3 y1 2 2  gy1  y2 =

y1  15,528   −1+ 1+  2  y13 

(2)

Finally, the tailwater must match the elevation of the sequent depth which is expressed by

Z + y 2 = 100

(3)

Equations 1, 2, and 3 are solved simultaneously. If solved by trial, a value of y 1 can be assumed, and Equation 1 is solved for Z. Then Equation 2 is solved for y 2 , and the value of Z from Equation 1 is checked with the value obtained from Equation 3. The results are y 1 = 2.85 ft; y 2 = 35.5 ft; and Z = 64.5 ft.

The stilling basin design is based on the Froude number at section 1, which is given by

F1 =

q gy13

=

(10,000 / 40) 32.2 × 2.853

= 9.16

and the velocity, V 1 = 10,000/(40 × 2.85) = 87.7 ft/s. These criteria require a Type II stilling basin as shown in Figure 3.13 with the various dimensions depending on the values of y 1 and y 2 .

49

Open Channel Hydraulics

3.8

CHAPTER 3

A spillway chute and the hydraulic jump stilling basin at the end of the chute are rectangular in shape with a width of 80 ft. The floor of the stilling basin is at an elevation 787.6 ft above the datum. The incoming flow has a depth of 2.60 ft at a design discharge of 9500 cfs. Within the basin are 15 baffle blocks that are each 2.5 ft high and 2.75 ft wide. (a) Assuming an effective coefficient of drag of 0.5 for the baffle blocks, based on the upstream velocity and combined frontal area of the blocks, calculate the sequent depth and compare with the sequent depth without baffle blocks. (b) What is the energy loss in the basin with and without the blocks? (c) If the tailwater elevation for Q=9500 cfs is 797.6, will the stilling basin perform as designed? Explain your answer.

Solution. (a) The incoming velocity, V 1 = 9500/(80 × 2.60) = 45.7 ft/sec, which gives an approach Froude number, F 1 = 45.7/(32.2 × 2.60)1/2 = 5.0. Write the momentum equation from point 1 just upstream of the jump to point 2 just downstream of the jump, including the drag force of the baffle blocks on the flow. The result is

γy12

ρA f V12

γy 22

b− b − CD = ρQ (V2 − V1 ) 2 2 2 9500 1.94 × (15 × 2.75 × 2.5) × 45.7 2 62.4 × 80 × ( 2.60 2 − y 22 ) − 0.5 × = 1.94 × 9500 × ( − 45.7) 80 y 2 2 2 y 22 +

876.8 = 302.4 y2

which can be solved to give y 2 = 15.7 ft and V 2 = 9500/(80 × 15.7) = 7.56 ft/s. Without the baffle blocks, the sequent depth is given by

y2 =

y1 2.60 (−1 + 1 + 8F12 ) = × (−1 + 1 + 8 × 5.0 2 ) = 17.1 ft 2 2

(b) The energy loss with the baffle blocks in place is calculated from the energy equation as

∆E = y1 +

V12 V2 45.7 2 7.56 2 − y 2 − 2 = 2.60 + − 15.7 − = 18.4 ft 2g 2g 64.4 64.4

Similarly, the energy loss without baffle blocks is 17.2 ft. So the blocks reduce the sequent depth and increase the energy loss. (c) A tailwater depth of 797.6 corresponds to a downstream basin depth of 10 ft, which is less than the sequent depth so that the hydraulic jump will sweep out of the stilling basin with or without the baffle blocks. The floor of the stilling basin should be lowered if this is the governing tailwater elevation.

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Open Channel Hydraulics

3.9

CHAPTER 3

In a short, horizontal, rectangular flume in the laboratory, the depth just downstream of a sluice gate at the upstream end of the flume is 1.0 cm and the depth just upstream of the sluice gate is 60 cm. The width of the flume is 38 cm. If the tailgate height is 15 cm, over which there is a free overfall, will a hydraulic jump occur or will it be submerged?

Solution.

First, write the energy equation from point 1 just upstream of the sluice gate to point 2 just downstream of the sluice gate. Neglecting any energy loss, the result is

q2 q2 = y2 + y1 + 2 gy12 2 gy 22 q=

y1 y 2 y −y 2 1

2 2

0.60 × 0.01

2 g ( y1 − y 2 ) =

0.6 − 0.01 2

2

19.62 × (.60 − 0.01) = 0.03403 m 2 /s

Applying continuity, V 1 = q/y 1 = 0.03403/0.60 = 0.0567 m/s, and similarly, V 2 = 3.40 m/s. The Froude number just downstream of the sluice gate is 3.40/(9.81 × 0.01)1/2 = 10.86, and the total discharge, Q = 0.03403 × 0.38 = 0.0129 m3 /s. Next, assume that the tailgate behaves as a sharpcrested weir so that the tailwater depth y 3 is given by

  Q  y3 = 0.15 +   (2 / 3) 2 g C L  d  

2/3

  0.0129  = 0.15 +    (2 / 3) × 19.62 × 0.636 × 0.38 

2/3

= 0.219 m

in which C d = 0.636 from Table 2-3 for L/b = 1.0 and after some iteration. The corrections on H and L are small and have been neglected. Finally, calculate the sequent depth, y 3 ′, for a hydraulic jump in the flume with an upstream depth of 1.0 cm:

0.01 ′ y × (−1 + 1 + 8 ×10.86 2 ) = 0.149 m y3 = 2 (−1 + 1 + 8F22 ) = 2 2 Now because the tailwater depth, y3, is greater than the sequent depth, the jump will be drowned out just downstream of the sluice gate.

3.10

A steady flow is occurring in a rectangular channel, and it is controlled by a sluice gate. The upstream depth is 1.0 m, and the upstream velocity is 3.0 m/sec. If the gate is abruptly slammed shut, determine the depth and speed of the resulting surge.

Solution.

Vs

y1

V1

y2

V2 = 0

51

Open Channel Hydraulics

CHAPTER 3

Write the continuity and momentum equations for the unsteady flow problem made stationary by superimposing a velocity of Vs to the right:

(V1 + Vs ) y1 = Vs y 2  (V1 + Vs ) 2 1 y 2  y 2  + 1 = gy1 2 y1  y1  in which the boundary condition V2 = 0 has been imposed owing to the closed gate. Solve the continuity equation in terms of Vs to obtain

Vs =

V1 y1 3.0 = y 2 − y1 y 2 − 1.0

(1)

Substitute the values of V1 and y1 into the momentum equation and rearrange to give

Vs = 2.2147 y 2 ( y 2 + 1) − 3.0

(2)

Equating (1) and (2) and solving, we get y2 = 2.12 m and Vs = 2.69 m/s.

3.11

The depths upstream and downstream of a sluice gate in a rectangular channel are 8 ft and 2 ft, respectively, for steady flow. (a) What is the value of the flow rate per unit width q? (b) If q in part (a) is reduced by 50 percent by an abrupt partial closure of the gate, what will be the height and speed of the surge upstream of the gate?

Solution. (a) Write the energy equation from point 1 immediately upstream of the gate to point 3 just downstream of the gate to produce

y1 + q=

q2 q2 = y + 3 2 gy12 2 gy32 y1 y3 y −y 2 1

2 3

2 g ( y1 − y3 ) =

8× 2 8 −2 2

2

64.4 × (8 − 2) = 40.60 ft 2 /s

Then from continuity, V1 =q/y1 = 40.6/8.0 = 5.08 ft/s. Write the continuity and momentum equations for the surge made stationary as in Exercise 3.10, but apply the boundary condition, V2y2 = 40.60/2 = 20.3 ft2/s at point 2 at the upstream face of the gate:

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Open Channel Hydraulics

CHAPTER 3

(V1 + Vs ) y1 = (V2 + Vs ) y 2  (V1 + Vs ) 2 1 y 2  y 2  + 1 = gy1 2 y1  y1  Substituting and rearranging the continuity equation, we have

Vs =

V2 y 2 − V1 y1 20.3 − 5.08 × 8.0 20.3 = = y1 − y 2 y 2 − 8.0 8.0 − y 2

Similarly, the momentum equation becomes

Vs = 4.0125 y 2 (

y2 + 1) − 5.08 8

Equating the right hand sides of the continuity and momentum equations, which are both equal to Vs, the solution is y2 = 9.53 ft; Vs = 13.3 ft/s; and V2 = 2.13 ft/s.

3.12

Write both the momentum and energy equations for the subcritical case of flow through bridge piers of diameter a and spacing s. If the head loss in the energy equation is written as KL V12/2g, in which KL is the head loss coefficient and V1 is the approach velocity, show that KL = CDa/s for the special case that (y1 – y4) is very small (Figure 3.16).

The momentum equation written from the approach section 1 to the exit section 4 with a control volume of width s as shown in Figure 3.16a is given by

 1 y 42 ρ ay1V12 y12 1 γ s −γ s − CD = ρQ(V4 − V1 ) = ρ q 2 s −  2 2 2  y 4 y1  Divide through by γ s/2, and algebraically manipulate the result to obtain 2 2 q 2 ( y1 − y 4 ) C a y V ( y1 − y 4 )( y1 + y 4 ) −  D  1 1 = g y1 y 4  s  g

Divide through by (y1 + y4); multiply numerator and denominator of the term on the right hand side by (y1 + y4); and assume y1 ≅ y4 to yield

53

Open Channel Hydraulics

CHAPTER 3

2 q2  C D a  V1 = y1 − y 4 −  ( y12 − y 42 )  2 2  s  2 g 2 gy1 y 4

Applying the energy equation from point 1 to point 4 assuming that the head loss is written in terms of the approach velocity head, and rearranging, we have

y1 − y 4 − K L

V12 q 2  1 1  q2  2 − 2  = ( y12 − y 42 ) = 2 2 2 g 2 g  y 4 y1  2 gy1 y 4

Upon comparison of the two equations, it is clear that KL = CD a/s under this set of rather restrictive assumptions.

3.13

For a river flow between bridge piers 3 m in diameter with a spacing of 20 m, determine the backwater using the momentum method if the downstream depth is 4.0 m and the downstream velocity is 1.9 m/s. Assume a coefficient of drag of 2.0 for the bridge piers.

Solution.

Apply Equation 3.20, in which CD a/s = 2.0×3.0/20 = 0.30, and the downstream Froude number, F4 = V4/(gy4)1/2 = 1.9/(9.81 × 4.0)1/2 = 0.303, with the result given by

λ ( λ + 1)( λ + 2) C D a / s + 2λ λ ( λ + 1)( λ + 2) 0.3032 = 0.0918 = 0.30 + 2λ F42 =

Solving for λ, we obtain λ = h1*/y4 = 0.0148 in agreement with Figure 3.17, and thus, h1* = 0.0148 × 4.0 = 0.0592 m.

3.14

A straight-walled contraction connects two rectangular channels 12 ft and 6 ft wide. The discharge through the contraction is 200 cfs, and the depth of the approach flow is 0.7 ft.

Calculate the downstream depth, Froude number, and the length of the contraction that will minimize standing waves. Will choking be a problem?

Solution.

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Open Channel Hydraulics

CHAPTER 3

From continuity, the approach velocity, V1 = 200/(0.7 × 12) = 23.81 ft/s, and the approach Froude number, F1 = V1/(gy1)1/2 = 23.81/(32.2 × 0.7)1/2 = 5.02. The contraction ratio, r = 0.5. Equation 3.25 is solved for β with an assumed value of θ. Find the zero of the function defined by

 1  1 tan β 1  tan β 1  f (β ) =  + 1 F1  2 tan( β 1 − θ )  tan( β 1 − θ ) 

1/ 2

− sin β 1

Either write a computer program to find the zero of the function, or use a spreadsheet as shown below.

j

Fj

θ, deg.

β, deg.

1 2

5.020 4.014

5.12 5.12

15.82 18.82

f(β) 0.00009 -0.00003

yj+1/yj

Fj+1

r

1.500 1.398

4.014 3.307

0.500

Start with an assumed value of θ estimated from Figure 3.20. For j = 1, iterate on β until f (β) is close to zero. Then y2/y1 is calculated from Equation 3.24, and F2 from Equation 3.26. In the second row where j = 2, the calculations are repeated with F2 as the initial value of the Froude number. The value of r is calculated from Equation 3.28 with y3/y1 = y3/y2 × y2/y1. If it is not equal to the given value, the entire calculation is repeated with a new value of θ. The final result is θ = 5.12°, and F3 = 3.31. The length of the transition is calculated from Equation 3.27as

L=

b1 − b3 12 − 6 = = 33.5 ft 2 tan θ 2 tan(5.12°)

From Figure 3.20, the solution lies just to the right of curve A indicating that choking may or may not occur.

3.15

For Exercise 3.14, calculate the wave-front angles β1 and β2. What variables do they depend on?

Produce a generalized plot for β in terms of the dimensionless variables on which it depends.

Solution.

The values of β1 and β2 are shown in the previous exercise. From Equation 3.25, it is observed that β depends on the contraction angle, θ, and the approach Froude number. A spreadsheet is set up for a series of values of β and a specified value of θ. Then Equation 3.25 is solved directly for the corresponding Froude number values. This procedure is repeated for several additional fixed values of θ. The plot is shown next.

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Open Channel Hydraulics

CHAPTER 3

SUPERCRITICAL CONTRACTION

50 45 40

Beta, degrees

35 30

θ , degrees = 25

25 20 20 15 15 10 10 5 5 0 0

5

10

15

20

25

30

Froude number

3.16

A supercritical transition expands from a width of 1.0 m to 3.0 m, and the approach flow depth and velocity are 0.64 m and 10 m/s for maximum design conditions, respectively. Calculate the downstream depth, and design and plot the transition geometry. What would be the length of the expansion if it were designed for a Froude number that is 40 percent of the design value?

Solution.

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Open Channel Hydraulics

CHAPTER 3

The expansion ratio, re, is 3.0, and the approach Froude number is given by

F1 =

V1 gy1

=

10 = 3.99 9.81 × 0.64

Assuming negligible energy loss and minimization of standing waves by the design, the downstream depth is obtained from the energy equation:

V12 q22 10 2 y1 + = 0.64 + = 5.74 = y 2 + 19.62 2g 2 gy 22 in which q2 = Q/b2 = (10 × 0.64 × 1.0)/3.0 = 2.13 m2/s. Solving, we have y2 = 0.204 m.

The lengths of the first portion of the transition, Lp, and the total length of the transition, Lt, are calculated from Equations 3.31 and 3.32 as

L p = 0.7re × b1F1 = 0.7 × 3.0 × 1.0 × 3.99 = 8.38 m Lt = [1 + 3.25( re − 1)] × b1F1 = [1 + 3.25 × 2] × 1.0 × 3.99 = 29.92 m The variable half-width, z, for the first portion of the transition as a function of distance along the transition, x, comes from Equation 3.30:

b 1  x   z = 1   2  4  b1F1  

3/ 2

 + 1 

1.0  1  x  z=    2  4  1.0 × 3.99 

3/ 2

 1 + 1 = 0.01568 x 3 / 2 + 2 

At the end of the first portion of the curved wall, x = Lp = 8.38 m, and zp = 0.88 m. The reverse curvature downstream of point P in Figure 3.21 is given by Equation 3.33:

 x − Lp  z − zP = sin 90°  b2 / 2 − z p Lt − L p   x − 8.38   z = 0.88 + 0.62 sin 90° 21.54  

57

Open Channel Hydraulics

CHAPTER 3

The plot of one wall of the transition is given next.

SUPERCRITICAL EXPANSION

1.6 1.4 1.2

z, m

1.0 0.8 0.6 0.4 0.2 Centerline

0.0 0

5

10

15

20

25

30

x, m

If the transition is designed for a Froude number that is 40 percent of the maximum value, the length is reduced by 40 percent according to Equation 3.32.

3.17

Write a computer program that finds the sequent depth for a hydraulic jump in a trapezoidal channel. First compute critical depth and determine if the given depth is subcritical or supercritical to limit the root search. Then use the bisection method to find the sequent depth Solution.

The procedure YSEQ follows this discussion. It computes the sequent depth for a hydraulic jump in a trapezoidal channel regardless of whether the given sequent depth is supercritical or subcritical. First, the data are input through the form code (not shown but very similar to that for Y0YC in Appendix B of the textbook) and assigned to single precision variable names. Then the momentum function, MI, is calculated for the given sequent depth,YSI. The critical depth is determined next by the call to the BISECTION subprocedure with NF = 1 to indicate the function F (equivalent to f(y) = F2 – 1) to be used when the function procedure is invoked. Once the critical depth is returned to YSEQ, the limits on the root search are determined for finding the unknown sequent depth. If the given sequent depth is less than the critical depth, the flow is supercritical and the root search is between critical depth and some large value to find the subcritical sequent depth. Otherwise, the root search is executed between some very small depth and critical depth to find the unknown supercritical sequent depth. The BISECTION procedure is

58

Open Channel Hydraulics

CHAPTER 3

called again with NF = 2 to designate the function F = M1 – M2 to find the unknown sequent depth. The computed sequent depth, YSO, is sent back to the form code module for screen output.

Option Explicit Dim Q As Single, b As Single, m As Single Dim YC As Single, YSI As Single, YSO As Single Sub YSEQ(Q, b, m, YC, YSI, YSO) Dim Y1 As Single, Y2 As Single, ER As Single Dim MI As Single, AI As Single, G As Single, NF As Integer G = 32.2 AI = YSI * (b + m * YSI) MI = b * YSI ^ 2 / 2 + m * YSI ^ 3 / 3 + Q ^ 2 / (G * AI) Y1 = 0.0001 Y2 = 100 ER = 0.0001 NF = 1 Call BISECTION(Y1, Y2, NF, ER, Q, b, m, MI, YC) If YSI < YC Then Y1 = YC Y2 = 100# Else Y1 = 0.0001 Y2 = YC End If NF = 2 Call BISECTION(Y1, Y2, NF, ER, Q, b, m, MI, YSO) End Sub Sub BISECTION(Y1, Y2, NFUNC, ER, Q, b, m, MI, Y3) Dim FY1 As Single, FY2 As Single, FY3 As Single, FZ As Single Dim I As Integer FY1 = F(Y1, NFUNC, Q, b, m, MI) FY2 = F(Y2, NFUNC, Q, b, m, MI) If FY1 * FY2 > 0 Then Exit Sub For I = 1 To 50 Y3 = (Y1 + Y2) / 2 FY3 = F(Y3, NFUNC, Q, b, m, MI) FZ = FY1 * FY3 If FZ = 0 Then Exit Sub If FZ < 0 Then Y2 = Y3 Else Y1 = Y3 If Abs((Y2 - Y1) / Y3) < ER Then Exit Sub Next I End Sub Function F(Y, NFUNC, Q, b, m, MI) As Single Dim A As Single, T As Single, G As Single A = Y * (b + m * Y)

59

Open Channel Hydraulics

3.18

CHAPTER 3

T = b + 2 * m * Y G = 32.2 If NFUNC = 1 Then F = Q - Sqr(G) * A ^ 1.5 / T ^ 0.5 Else F = MI - (b * Y ^ 2 / 2 + m * Y ^ 3 / 3 + Q ^ 2 / (G * A)) End If End Function The data given below for a hydraulic jump have been measured in a laboratory flume by two different lab teams. The flume has a width of 38.1 cm. The upstream sluice gate was set to produce a supercritical flow for a given measured discharge, and the tailgate was adjusted until the hydraulic jump was positioned at the desired location in the flume. The depths measured by a point gauge upstream and downstream of the jump, y1 and y2, respectively, are given in the following table as is the discharge measured by a calibrated bend meter with an uncertainty of + 0.0001 m3/s. The uncertainty in the upstream depths is + 0.02 cm, while the downstream depths have a larger uncertainty of +0.30 cm due to surface waves. Photographs of the flume and hydraulic jumps at selected Froude numbers are shown in Figure 3.22.

Team A

Team B

y1, cm

y2, cm

Q, m3/s

y1, cm

y2, cm

Q, m3/s

1.41 1.62 2.22 3.10 1.30 1.38 1.59 2.10

15.8 14.2 12.3 10.2 13.3 13.0 11.9 9.9

0.0165 0.0165 0.0165 0.0165 0.0131 0.0131 0.0131 0.0131

1.47 1.76 2.20 2.63 1.21 1.45 1.70 1.87

15.4 14.5 12.8 11.0 13.0 11.7 10.9 10.0

0.0166 0.0166 0.0166 0.0166 0.0125 0.0125 0.0125 0.0125

(a) Calculate and plot the sequent depth ratios as a function of the Froude numbers of the experimental data and compare with the theoretical relationship for a hydraulic jump in a rectangular channel. (b) Calculate and plot the dimensionless energy loss EL /E1 as function of Froude number for the experimental data and compare it with the theoretical relationship. (c) Estimate the experimental uncertainty in y2/y1 and the Froude number and plot error bars on your graphs. Does the estimated uncertainty account for the differences between measured and theoretical values? Are the results for Team A and Team B consistent? (d) Based on the photos in Figure 3.22, describe the appearance of the jump as a function of Froude number and indicate the relative energy loss EL /E1 for each photo.

Solution.

The spreadsheet calculations are shown in the following table. The experimental value of the Froude number is obtained from

60

Open Channel Hydraulics

F1 =

CHAPTER 3

Q /b gy13

where Q is the measured discharge in m3/s; y is the measured depth converted to m; and b is the flume width = 0.381 m. The energy loss, EL, is calculated as (E1 – E2) where E1 and E2 are the calculated values of specific energy upstream and downstream of the jump based on the measured depths and discharge. The theoretical values of y2/y1 are calculated from Equation 3.8, and the corresponding theoretical values of EL/E1 come from Equation 3.9.

Team

y1, cm

A

1.41 1.62 2.22 3.10 1.30 1.38 1.59 2.10 1.47 1.76 2.20 2.63 1.21 1.45 1.70 1.87

B

y2, cm Q, m3/s

y2/y1

F1

y2/y1 theo.

EL /E1

EL /E1

15.80 14.20 12.30 10.20 13.30 13.00 11.90 9.90 15.40 14.50 12.80 11.00 13.00 11.70 10.90 10.00

11.21 8.77 5.54 3.29 10.23 9.42 7.48 4.71 10.48 8.24 5.82 4.18 10.74 8.07 6.41 5.35

8.26 6.71 4.18 2.53 7.41 6.77 5.48 3.61 7.81 5.96 4.26 3.26 7.87 6.00 4.73 4.10

11.19 9.00 5.43 3.12 9.99 9.09 7.26 4.63 10.55 7.94 5.55 4.14 10.64 8.00 6.20 5.31

0.673 0.614 0.402 0.148 0.631 0.596 0.515 0.333 0.658 0.547 0.397 0.290 0.656 0.561 0.451 0.399

0.673 0.605 0.412 0.181 0.639 0.609 0.528 0.343 0.656 0.561 0.421 0.296 0.659 0.564 0.466 0.402

0.0165 0.0165 0.0165 0.0165 0.0131 0.0131 0.0131 0.0131 0.0166 0.0166 0.0166 0.0166 0.0125 0.0125 0.0125 0.0125

theo.

The plots comparing the experimental data and the theoretical relationships for y2/y1 and EL/E1 are shown in the next two figures. Error bars are shown in the figure for y2/y1 and the Froude number calculated from the theorem of error propagation as

2

2

SY = Y

 S y1   S y 2    +    y1   y 2 

SF = F

 S Q  9  S y1    +    Q  4  y1 

2

2

in which S = standard error; Y = y2/y1 and F = Froude number. Using the estimated errors in Q, y1, and y2 stated in the problem, the average values of SY and SF are 0.2 and 0.1, respectively. The root-mean-square deviation between the theoretical and measured values of y2/y1 is 0.2, which

61

Open Channel Hydraulics

CHAPTER 3

means that this deviation is explained by the experimental uncertainty, SY. Comparing the estimated uncertainties in y2/y1 between the two teams shows no significant differences.

The theoretical values of relative energy loss for the hydraulic jumps shown in Figure 3.22 are EL/E1 = 0.08, 0.44, and 0.63 in Figure 3.22 (a), 3.22 (b), and 3.22 (c), respectively.

12

10

y2/y1

8

6

4

2

0 0

2

4

6

8

Froude no.

Team A

Team B

62

Theoretical

10

Open Channel Hydraulics

CHAPTER 3

1.0

0.8

EL/E1

0.6

0.4

0.2

0.0 0

2

4

6

8

Froude no.

Team A

Team B

63

Theoretical

10

CHAPTER 4. Uniform Flow 4.1

Determine the normal depth and critical depth in a trapezoidal channel with a bottom width of 40 ft, side slopes of 3:1, and a bed slope of 0.002 ft/ft. The Manning's n value is 0.025 and the discharge is 3,000 cfs. Is the slope steep or mild? Repeat for n=0.012.

Did the critical depth change? Why or why not?

Solution.

Rearrange Manning's equation and solve for the normal depth:

nQ 0.025 × 3000 = = 1125.5 1/ 2 1.49 S 1.49 × 0.0021 / 2 [ y 0 (b + my 0 )]5 / 3 [ y ( 40 + 3 y 0 )]5 / 3 = 0 = 1125.5 2/3 [b + 2 y 0 1 + m 2 ]2 / 3 [40 + 2 10 y 0 ] AR 2 / 3 =

with the result, y 0 = 6.59 ft. The critical depth is obtained from equating the Froude number squared to one:

F2 =

Q 2 Bc = 1.0 gAc3

Ac3 / 2 [ y c (b + my c )]3 / 2 [ y c ( 40 + 3 y c )]3 / 2 Q 3000 = = = = = 528.7 1/ 2 1/ 2 1/ 2 Bc [b + 2my c ] [40 + 6 y c ] g 32.2 from which y c = 4.91 ft. Therefore, this is a mild slope because y 0 > y c . Alternatively, calculate the critical slope from

Sc =

n 2Q 2 = 1.49 2 Ac2 Rc4 / 3

0.0252 × 3000 2 = 0.00595 [4.91 × ( 40 + 3 × 4.91)]10 / 3 2 1.49 × [40 + 2 10 × 4.91]4 / 3

The slope is mild because S 0 < S c . If the Manning's n value is changed to 0.012, Manning's equation becomes

[ y 0 ( 40 + 3 y 0 )]5 / 3 0.012 × 3000 = = 540.3 1/ 2 [40 + 2 10 y 0 ]2 / 3 1.49 × 0.002 from which y 0 = 4.43 ft. In this case, the slope is steep because y 0 < y c . The critical depth depends only on the channel geometry and discharge, and does not change with a change in Manning's n.

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Open Channel Hydraulics

4.2

CHAPTER 4

Compute normal and critical depths in a concrete culvert (n = 0.015) with a diameter of 36 in. and a bed slope of 0.002 ft/ft if the design discharge is 15 cfs. Is the slope steep or mild? Repeat for S = 0.02 ft/ft.

Solution.

Rearrange Manning's equation and solve for normal depth:

nQ .015 × 15 = = 3.3766 1/ 2 1.49 S 1.49 × 0.0021 / 2 [(θ − sin θ )d 2 / 8]5 / 3 (θ − sin θ ) 5 / 3 = = 3.3766 0 .

92867 θ 2/3 [θ d / 2]2 / 3 AR 2 / 3 =

from which θ = 3.3255 rad and y 0 /d = (1/2)[1 – cos(3.3255/2)] = 0.546; thus, y 0 = 1.64 ft.

Next solve for the critical depth by setting the Froude number equal to unity to obtain

Ac3 / 2 [(θ − sin θ )d 2 / 8]3 / 2 Q = = = 1/ 2 1/ 2 Bc [d sin(θ / 2)] g

15 = 2.643 32.2

[θ − sin θ ]3 / 2 = 3.837 [sin(θ / 2)]1 / 2 which gives θ = 2.7861; y c /d = (1/2)[1 – cos(2.7861/2)] = 0.412; and y c = 1.24 ft. Thus, the slope is mild because y 0 > y c . If the slope is 0.02, there is no change in critical depth, but Manning's equation becomes

(θ − sin θ ) 5 / 3

θ 2/3

= 1.1498

from which θ = 2.2734 rad; y 0 /d = (1/2)[1 – cos(2.2734/2)] = 0.2897; and y 0 = 0.869 ft.

This slope is steep because y 0 < y c .

4.3.

For a discharge of 12.0 m3/s, determine the normal and critical depths in a parabolic channel that has a bank-full width of 10 m and a bank-full depth of 2.0 m. The channel has a slope of 0.005 and Manning=s n = 0.05.

Solution.

The expression for critical depth in a parabolic channel is derived in Exercise 2.10 and given by

  yc Q = 1.355  1 / 2 3/ 2  y1  g B1 y1 

1/ 2

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Open Channel Hydraulics

CHAPTER 4

in which y 1 = bank-full depth and B 1 = bank-full top width. Substituting, we have

yc 12   = 1.355  1/ 2 3/ 2  y1  9.81 × 10 × 2 

1/ 2

= 0.4987

and y c = 0.4987 × 2.0 = 0.997 m. The solution for normal depth requires Manning's equation and the geometric expressions for a parabolic channel given in Table 2-1. The wetted perimeter is given by

P=

B 1 [ 1 + x 2 + ln( x + 1 + x 2 )] 2 x

in which B = B 1 (y/y 1 )1/2 = 10 × (y/2.0)1/2 = 7.071y1/2; and x = 4y/B = 0.566y1/2. The crosssectional area, A = (2/3)By = 4.714 y3/2. Substituting into Manning's equation, we obtain

A5 / 3 nQ 0.05 × 12.0 = 1/ 2 = = 8.485 2/3 P S 0.0051 / 2 13.25 y 5 / 2   1 1/ 2 ln(0.566 y 1 / 2 + 1 + 0.320 y )] 3.536 y [ 1 + 0.320 y + 1/ 2 0.566 y  

2/3

= 8.485

which can be solved for y = y 0 to produce y 0 = 1.52 m, the normal depth. The slope is mild.

4.4

For the horseshoe conduit shape defined in Figure 4.7, derive the relationships for A/A f , R/R f , and Q/Q f , where A f , R f , and Q f represent the full flow values of area, hydraulic radius and discharge, respectively. On the same graph, plot the relationships together with those for a circular conduit. Plot y/d on the vertical axis. Note that for the horseshoe conduit, A f = 0.8293 d 2 and R f = 0.2538 d.

Solution. We will derive first the geometric relationships for area, A; wetted perimeter, P; and top width, B; for three separate regions of the horseshoe conduit. y/d ≤ 0.088562:

R

θ d

d d–y d

d

0.088562 d

B S

T y

65

Open Channel Hydraulics

CHAPTER 4

From the Pythagorean theorem, the top width B is given by

B = 2 d 2 − (d − y ) 2 B y  y = 2 2 −  d d d 

2

From the geometry note that

d−y d y θ = 2 cos −1 (1 − ) d cos(θ / 2) =

so the wetted perimeter becomes

P =θ d P y = 2 cos −1 (1 − ) d d Finally, the flow area is the difference between the area of the sector of the circle with radius d, and the triangle RST:

A=

1 2 B  d θ −  ( d − y ) 2 2 

Substituting for B and θ from the foregoing formulas, we obtain

A y  y y  y  = cos −1 1 −  − 1 −  2 −   2 d  d   d  d d  0.088562 ≤ y/d ≤ 0.5d

c

O

θ

M N

β

y

d

0.088562 d

d/2

A0

66

2

Open Channel Hydraulics

CHAPTER 4

To determine the top width, we need the length c, which is obtained from the Pythagorean theorem as

c = d 2 − [( d / 2) − y ]2 c = d 0.75 +

y  y −  d d 

2

Then the top width, B, at the free surface is 2(c – d/2), from which we can show that 2

B y  y = 2 0.75 + −   − 1 d d d  Calculation of the wetted perimeter requires expressions for the angles β and θ shown in the figure. Note that d sin θ = (d/2) – y, from which we have

 

θ = sin −1  0.5 −

y  d

Similarly, d sin β = (d/2) – 0.088562 d, so that β = 0.424031 rad. The wetted perimeter becomes

P = P0 + 2d ( β − θ ) in which P 0 = wetted perimeter from the previous expression with y/d = 0.088562. The result is

P y   = 2 0.848062 − sin −1 (0.5 − ) d d   The flow area is the sum of A 0 , which is the area for y/d = 0.088562, the area of the segment of the circular arcs on either side of the cross section, and the area of the remaining trapezoid:

 d 2 (β − θ )  ( B + B) − A1  + 0 A = A0 + 2 ( y − 0.088562d ) 2 2   in which B 0 is the top width for y/d = 0.088562, and A 1 is the area of the triangle OMN, shown in the figure, which has a base f and a height h given by

67

Open Channel Hydraulics

CHAPTER 4

 β −θ  h = d cos   2   β −θ  f = 2 d 2 − d 2 cos 2    2  By substituting for β and θ from the expressions developed previously, and employing a trigonometric identity, the result for the area, A 1 = (1/2)fh, is

A1 1  y  = sin 0.424031 − sin −1 (0.5 − ) 2 d d  2  Substituting into the expression for the total area, A, and collecting terms, we have 2

A y  y y y  = ξ − sin ξ +  − 0.088562  0.75 + −   − 0.088562 + 0.056874 2 d d d  d d  in which ξ = β – θ = 0.424031 – sin-1(0.5 – y/d).

0.5 ≤ y/d ≤ 1

A3

α y

d/2 0.088562 d d/2

A2

The top width comes from the Pythagorean theorem:

B = 2

d2  d −y−  4 2 

B =2 d

y  y −  d d 

2

2

The wetted perimeter depends on the angle α, which is expressed by

68

Open Channel Hydraulics

CHAPTER 4

d α d cos = y − 2 2 2  y  α = 2 cos −1  2 − 1  d  Then the wetted perimeter is

P = P2 + π

d d −α 2 2

in which P 2 is the wetted perimeter of the lower portion of the cross section (y/d ≤ 0.5).

Substituting for P 2 and α, the result is

P π  y   y  = 1.69612 + − cos −1  2 − 1 = 1.69612 + sin −1  2 − 1 d 2  d   d  The total area is the sum of the area of the lower portion of the cross section, A 2 ; and the area of the upper semi-circle minus the area of the segment, A 3 :

A = A2 +

π d2 8

  1  d 2 B −    α − ( y − 0.5d ) 2   2  2 

Substituting for A 2 , α, and B, the result is

A 1  y   y 1 y  y = 0.436624 + sin −1  2 − 1 +  −  −  2 d 4  d  d 2 d d 

2

These expressions for area and wetted perimeter, and hence hydraulic radius (=A/P), for the three different regions of the horseshoe cross section can be placed in a spreadsheet and divided by the full flow values. They are plotted in the following figure with relative depth, y/d, on the vertical axis. The value of Q/Q f = (A/A f )(R/R f )2/3 from Manning's equation. Shown for comparison in the figure are the corresponding expressions for a circular cross section.

69

Open Channel Hydraulics

CHAPTER 4

HORSESHOE and CIRCULAR CONDUITS Horseshoe: A f = 0.8293 d 2; R f = 0.25386 d Circular: A f = 0.7854 d 2; R f = 0.25 d

1.0

A /A f

0.9 Q /Q f

0.8 R /R f

0.7

y /d

0.6 0.5 0.4 0.3 0.2

Horseshoe Circular

0.1 0.0 0.0

0.2

0.4

0.6

0.8

R /R f , A /A f , Q /Q f

70

1.0

1.2

1.4

Open Channel Hydraulics

4.5

CHAPTER 4

A diversion tunnel has a horseshoe shape with a diameter of 11.8 m. The slope of the tunnel is 0.022 and it is lined with gunite (n = 0.023). Find the normal depth for a discharge of 950 m3/s. Is the slope steep or mild?

Solution.

Calculate the full-flow discharge from Manning's equation to yield

1.0 A f R 2f / 3 S 1 / 2 n 1.0 Qf = × (0.8293 × 11.82 ) × (0.25386 × 11.8) 2 / 3 × 0.0221 / 2 = 1547 m 3 /s 0.023 Qf =

Then Q/Q f = 950/1547 = 0.614, and from the figure developed in the previous exercise, y 0 /d = 0.55 so that y 0 = 6.5 m. Alternatively, solve Manning's equation in the form

nQ ( A / d 2 )5 / 3 0.023 × 950 = 1/ 2 8 / 3 = = 0.2041 2/3 S d (P / d ) (.022)1 / 2 (11.8) 8 / 3 2  y 1 y  y  1 −1  y   0.436624 + sin  2 − 1 +  −  −   4  d   d 2  d  d    2/3 y   −1  1.69612 + sin  2 d − 1   

5/ 3

= 0.2041

in which the expressions for A/d 2 and P/d from the previous exercise for y/d ≥ 0.5 have been substituted. The solution is y/d = 0.551 from which y 0 = 6.50 m.

The critical depth is determined from equating the Froude number to unity and substituting the appropriate geometric expressions from the previous exercise (it is assumed initially that y/d ≥ 0.5):

Q ( A / d 2 )3/ 2 950 = 5/ 2 = = 0.6341 1/ 2 5/ 2 (B / d ) d g 11.8 × 9.81 2  y 1 y  y  1 −1  y   0.436624 + sin  2 − 1 +  −  −   4  d   d 2  d  d    1/ 2  y  y 2  2 −    d  d  

71

3/ 2

= 0.6341

Open Channel Hydraulics

CHAPTER 4

The solution is y/d = 0.773, confirming the choice of geometric expressions, so that y c = 9.12 m. The slope is steep.

4.6

A trapezoidal channel has vegetated banks with Manning's n = 0.040 and a stable bottom with Manning's n = 0.025. The channel bottom width is 10 ft and the side slopes are 4:1.

Find the composite value of Manning's n using the four methods given in this chapter if the flow depth is 3.0 ft.

Solution.

The cross section is divided into three subsections with imaginary vertical lines at the intersection of the channel bottom and the two side slopes. Assuming that the vertical dividing lines do not contribute to the wetted perimeter, the values of the wetted perimeter and hydraulic radius of the side-slope subsections, indicated by a subscript of ss, are given by

Pss = y 1 + m 2 = 3.0 1 + 4 2 = 12.37 ft my 2 / 2 4.0 × 3.0 2 / 2 = 1.455 ft = 12.37 12.37

Rss =

For the main channel subsection, indicated by a subscript of mc, we have

Pmc = b = 10 ft Rmc =

by 10 × 3.0 = = 3.0 ft 10 10

The total wetted perimeter, P = 2 × 12.37 + 10 = 34.74 ft; and the hydraulic radius of the whole cross section, R = A/P = (4 × 32 + 30)/34.74 = 1.90 ft. Substituting into Horton's formula (Equation 4.27), the result is

 ∑ P n3/ 2  nc =  i i   P 

2/3

 2 × 12.37 × 0.04 3 / 2 + 10 × 0.0253 / 2  =  2 × 12.37 + 10  

2/3

= 0.036

Equation 4.28, which is the Einstein and Banks formula, produces

 ∑ Pi ni2  nc =    P 

1/ 2

 2 × 12.37 × 0.04 2 + 10 × 0.025 2  =  2 × 12.37 + 10  

1/ 2

= 0.036

in agreement with Horton's formula. Lotter's formula (Equation 4.29) yields

nc =

PR 5 / 3 34.74 × 1.90 5 / 3 = = 0.028 Pi Ri5 / 3 2 × 12.37 × 1.4555 / 3 10 × 3.0 5 / 3 + ∑ n 0 .

04 0.025 i

72

Open Channel Hydraulics

CHAPTER 4

Finally, Equation 4.30, which was suggested by Krishnamurthy and Christensen, results in

∑ P y ln n = 2 × 12.37 × 1.5 × ln(0.040) + 10 × 3 ln n = 2 × 12.37 × 1.5 + 10 × 3 ∑Py i

3/ 2 i

c

i

3/ 2

3/ 2

i

3/ 2 i

3/ 2

× ln(0.025)

3/ 2

= −3.470

in which the depth in the side-slope subsections has been taken to be the mean depth there. Taking the antilog of the previous equation, the result is n c = 0.031 in fairly close agreement with Lotter's formula. The first two formulas suggest dominance of the sideslope roughness in the composite roughness coefficient, while the last two formulas indicate a greater influence of the channel bottom roughness. Such composite n formulas must be used with care.

4.7

Find the normal depth in a 12-in. diameter PVC storm sewer flowing at a discharge of 1.2 cfs if it has a slope of 0.001. Treat the pipe as smooth, and use the Chezy equation.

Verify your solution with the graphical solution given in the text. What is the equivalent value of Manning's n for your solution? Would the value of n be the same for other pipe diameters?

Solution.

From the Chezy equation, we have

AR1 / 2 =

Q f 8 gS

 d2  θ θ − ( sin )   8  A3 / 2  = P1 / 2 [θ d / 2]1 / 2

φ (θ ) =

(θ − sin θ ) 3 / 2

θ 1/ 2

3/ 2

=

1.2 8 × 32.2 × 0.001

f = 2.364 f

= 37.82 f

The solution proceeds by assuming a friction factor; solving for θ, A, and R; calculating the Reynolds number, Re = (Q/A)(4R)/ν; and obtaining a new friction factor from the smooth channel relationship

1 = 2.0 log(Re f ) − 0.8 f which is implicit in f. It is imperative that R be understood as the hydraulic radius of the partly full flow, not d/4. The calculations can be organized in a spreadsheet, as shown next, in which iterations are carried out until φ (θ ) = 37.82 f 0.5 for a given friction factor.

An outer iteration on the friction factor is continued until there is no change in f. The Reynolds number calculation assumes a water temperature of 60°F (ν = 1.21 × 10–5 ft2/s).

73

Open Channel Hydraulics

f 0.015

0.0155

CHAPTER 4

θ, rad

φ (θ )

3.1416 3.5 3.75 3.747 3.8 3.780

3.14 4.04 4.64 4.632 4.75 4.708

A, ft2

R, ft

Re

f

0.5395

0.288

2.12E5

0.0155

0.5470

0.289

2.10E5

0.0155

The solution converges for a friction factor of 0.0155. The normal depth is calculated from θ = 3.780 rad to give y 0 /d = (1/2)[1 – cos(3.780/2)] = 0.657 and y 0 = 0.657 ft. To check the graphical solution of Figure 4.10, calculate Re* and Q* as

1.0 × 32.2 × 1.0 × 0.001 = 1.5 × 10 4 −5 ν 1.21 × 10 Q 1.2 Q* = 2 = = 6.7 2 d gdS 1.0 32.2 × 1.0 × 0.001 Re * =

d gdS

=

Then from Figure 4.10, y 0 /d is approximately 0.65 with graphical interpolation in close agreement with the calculated value. The equivalent value of Manning's n is calculated from Manning's equation as

n=

1.49 AR 2 / 3 S 1 / 2 1.49 × 0.547 × 0.289 2 / 3 × 0.0011 / 2 = = 0.0094 Q 1.2

In general, the value of Manning's n is expected to change with pipe diameter because of the change in Reynolds number and friction factor for the smooth pipe.

4.8

A circular PVC plastic (smooth) storm sewer has a diameter of 18 in. At the design flow, it is intended to have a relative depth of 0.8. At what minimum slope can it be laid such that the velocity is at least 2 ft/s at design flow and deposited solids will be scoured out by the design flow? How would your answer for the minimum slope change for a concrete sewer?

Solution.

Solve the Chezy equation for the minimum slope, but first find the hydraulic radius and friction factor:

74

Open Channel Hydraulics

CHAPTER 4

θ = 2 cos −1 (1 − 2 y / d ) = 2 cos −1 (1 − 2 × 0.8) = 4.4286 rad A (θ − sin θ )d 2 / 8 [4.4286 − sin(4.4286)] × 1.52 / 8 R= = = = 0.456 ft θ d /2 P 4.4286 × 1.5 / 2 V × 4 R 2.0 × 4 × 0.456 Re = = = 3.02 × 105 −5 ν 1.21 × 10 1 = 2 log(Re f ) − 0.8 = 2 log(3.02 × 105 × f ) − 0.8 f

Solving the last equation for friction factor in a smooth conduit, we obtain f = 0.0144.

Then, solving the Chezy equation for the slope, we have

S=

f V 2 0.0144 × (2.0) 2 = 0.00049 = 8 gR 8 × 32.2 × 0.456

For a concrete sewer of the same diameter, the minimum slope can be determined from Manning's equation with n = 0.015:

S=

n2 V 2 0.015 2 × 2.0 2 = = 0.00116 1.49 2 R 4 / 3 1.49 2 × 0.456 4 / 3

These conditions satisfy Equation 4.22 for applicability of Manning’s equation.

4.9

Design a concrete sewer that has a maximum design discharge of 1.0 m3/s and a minimum discharge of 0.2 m3/s if its slope is 0.0018. Check the velocity for selfcleansing.

Solution.

Substitute into Equation 4.35 to obtain a first estimate of diameter for the maximum design discharge of 1.0 m3/s occurring at y/d = 0.8:

 nQ  d = 1.56 1/ 2   Kn S 

3/8

 0.015 × 1.0  = 1.56 1/ 2 1.0 × 0.0018 

3/8

= 1.056 m

Round up to a commercial diameter of 1.067 m (42 in.). Next calculate the normal depth for the selected commercial size from Manning's equation:

A5 / 3 [(θ − sin θ )d 2 / 8]5 / 3 nQ 0.015 × 1.0 = = = = 0.354 2/3 1/ 2 2/3 P [θ d / 2] Kn S 0.00181 / 2 (θ − sin θ ) 5 / 3

θ 2/3

= 6.003

75

Open Channel Hydraulics

CHAPTER 4

from which θ = 4.341 rad and y 0 = (1.067/2)[1 – cos(4.341/2)] = 0.835 m. The corresponding cross-sectional area, A = [4.341 – sin(4.341)]× 1.0672/8 = 0.7504 m2; and V = Q/A = 1.0/0.7504 = 1.33 m/s. From Figure 4.11, V c * = 0.82 for y 0 /d = 0.835/1.067 = 0.78. Using the definition of V c * from Equation 4.36, and an assumed critical shear stress of 2.0 Pa, we have

Vc = 0.82

K n u*c d 1 / 6 2.0 / 1000 × 1.0671 / 6 = 0.82 = 0.79 m/s n g 0.015 × 9.81

so that the design velocity at maximum discharge is greater than the critical velocity.

Repeating these calculations for the minimum discharge of 0.2 m3/s, we obtain θ = 2.301 rad, which gives y 0 = 0.316 m and A = 0.2214 m2. The velocity V = Q/A = 0.2/0.2214 = 0.90 m/s. For the minimum discharge, y 0 /d = 0.316/1.067 = 0.30 and V c * = 0.74 from Figure 4.11. Then the critical velocity, V c = 0.71 m/s, and once again, V > V c .

4.10

Determine the design depth of flow in a trapezoidal roadside drainage ditch with a design discharge of 3.75 m3/s if the ditch is lined with grass having a retardance of class C. The slope of the ditch is 0.004 and it has a bottom width of 2.0 m with side slopes of 3:1. Is the channel stable?

Solution.

The permissible shear stress, τ p , for retardance class C is 48 Pa. Estimate a hydraulic radius of 1.0 m; then Figure 4.14 indicates a Manning's n value of approximately 0.05 for a slope of 0.004. Solve Manning's equation for normal depth:

[ y 0 ( 2.0 + 3 y 0 )]5 / 3 nQ 0.05 × 3.75 A5 / 3 = = 1/ 2 = = 2.965 2/3 2 2 / 3 P S 0.0041 / 2 [2.0 + 2 y 0 1 + 3 ] to obtain y 0 = 0.918 m and R = 0.56 m. For this value of hydraulic radius, Figure 4.14 or Equation 4.43 gives an n value of 0.061. Solving Manning's equation again, we obtain y 0 = 1.01 m and R = 0.606 m. In the final iteration, n = 0.059; y 0 = 0.992 m and R = 0.597 m. To check for stability, calculate the maximum bottom shear stress as

τ 0 = γ y 0 S 0 = 9810 × 0.992 × 0.004 = 38.9 Pa which shows that the channel is stable because τ 0 < τ p .

4.11

A very wide rectangular channel is to be lined with a tall stand of Bermuda grass to prevent erosion. If the channel slope is 0.01 ft/ft, determine the maximum allowable flow rate per unit width and velocity for channel stability.

Solution.

The tall Bermuda grass has a retardance of Class B and a permissible shear stress of 100 Pa. The maximum allowable depth for stability is

76

Open Channel Hydraulics

y0 =

CHAPTER 4

τp 100 = = 1.02 m γ S 9810 × 0.01

Equation 4.43 provides the value of Manning's n for Class B retardance as

n=

R1 / 6 1.021 / 6 = = 0.0564 30.7 + 16.4 log( R1.4 S 0.4 ) 30.7 + 16.4 log(1.021.4 × 0.010.4 )

Then from Manning's equation for a very wide channel, we have the flow rate per unit width, q, given by

q=

1.0 5 / 3 1 / 2 1.0 y0 S0 = × 1.02 5 / 3 × 0.011 / 2 = 1.83 m 2 /s n 0.0564

Finally, the maximum allowable velocity, V = q/y 0 = 1.83/1.02 = 1.79 m/s.

4.12

Derive a relationship between the trapezoidal channel side slope and the angle of repose of the channel riprap lining such that failure of the rock riprap occurs simultaneously on the bed and banks. Allow the angle of repose to vary between 30° and 42°. What is the minimum value of the side slope, m:1, so that failure always would occur on the bed first?

Solution.

The criterion of balanced design for the channel bottom and side slope requires that

(τ 0 ) max

τ 0c

=

(τ 0w ) max

τ 0wc

which can be rearranged as

(τ 0w ) max τ 0wc = = Kr (τ 0 ) max τ 0 c From Equations 4.39 and 4.40, the left hand side of the preceding equation has a value of 0.8. Substituting the definition of K r from Equation 4.41, we have 1/ 2

 sin 2 θ  1 − sin 2 φ  = 0.8   −1 θ = sin (0.6 sin φ )

77

Open Channel Hydraulics

CHAPTER 4

This equation is plotted in Figure 4.13b along with the side slope ratio, m = 1/tan θ. Note that if m > 3.2, the side slope will be flatter than the slope that causes side slope failure for all values of φ in Figure 4.13b, so failure will occur first on the channel bed.

4.13

Design a riprap-lined trapezoidal channel that has a capacity of 1000 cfs and a slope of 0.0005 ft/ft. Crushed rock is to be used and the channel bottom width is not to exceed 15 ft. Determine the riprap size, the side slopes, and the design depth of flow Solution.

Choose a riprap size of 10 mm (0.033 ft). The angle of repose of crushed rock is 39° from Figure 4.13a, and the suggested side slope ratio is approximately 2.5 (θ = 21.8°) from Figure 4.13b. The critical bed shear stress from Equation 4.38 is τ 0c = 4 × 0.033 = 0.132 lbs/ft2. The critical wall shear stress follows from Equation 4.41:

τ

w 0c

 sin 2 ( 21.8°)  = K rτ 0 c = 1 − sin 2 (39°)  

1/ 2

× 0.132 = 0.107 lbs/ft 2

Equation 4.37 gives a value of Manning's n = 0.04 × 0.0331/6 = 0.0226. Solve Manning's equation for normal depth:

nQ A5 / 3 [ y 0 (15 + 2.5 y 0 )]5 / 3 0.0226 × 1000 = = = = 678 2/3 1/ 2 2/3 P 1.49 S 1.49 × 0.00051 / 2 [15 + 2 7.25 y 0 ] The result is y 0 = 7.40 ft and R = 4.52 ft. The maximum shear stresses come from Equations 4.39 and 4.40:

(τ 0 ) max = 1.5γRS = 1.5 × 62.4 × 4.52 × 0.0005 = 0.211 lbs/ft 2 (τ 0w ) max = 1.2γRS = 1.2 × 62.4 × 4.52 × 0.0005 = 0.169 lbs/ft 2 These values of maximum shear stress are considerably larger than the critical values, so a larger rock size is needed. After several iterations, which are best done with a spreadsheet or computer program, try a size of 17 mm (0.0558 ft). Then φ = 40° and use the same value of θ = 21.8°. The critical shear stresses are τ 0c = 0.223 lbs/ft2 and τ 0c w = 0.182 lbs/ft2; and Manning's n = 0.0247. Solving Manning's equation, we get y 0 = 7.72 ft and R = 4.68 ft. The maximum shear stresses are 0.219 lbs/ft2 on the bed and 0.178 lbs/ft2 on the wall, which are both just slightly less than the critical values, so this is an acceptable design. The final result is d 50 = 17 mm; m = 2.5, and y 0 = 7.72 ft. If Equation 4.42 is used for obtaining Manning's n, the required rock size increases to 18 mm and the depth increases to 8.2 ft.

78

Open Channel Hydraulics

4.14

CHAPTER 4

A rectangular channel has a width of 10 ft and a Manning's n value of 0.020. Determine the channel slope such that uniform flow will always have a Froude number less than or equal to 0.5 regardless of the discharge.

Solution.

First calculate the limit slope from Equation 4.47:

SL =

2.67 g n 2 2.67 × 32.2 0.020 2 = × = 0.00719 K n2 b1 / 3 1.49 2 101 / 3

Next solve Equation 4.50 for the channel bed slope: 2 S 0 = S L Fmax = 0.00719 × 0.5 2 = 0.0018

4.15

A rectangular channel in a laboratory flume has a width of 1.25 ft and a Manning's n of 0.017. In order to erode a sediment sample, the shear stress needs to be 0.15 lbs/ft2. A supercritical uniform flow is desired with the Froude number less than or equal to 1.5 to avoid roll waves. (a) Calculate the maximum and minimum slopes to satisfy the Froude number criterion. (b) Choose a slope in the range determined from part (a) and calculate the depth and discharge to achieve the desired shear stress.

Solution. (a) The limit slope from Equation 4.47 is

SL =

2.67 g n 2 2.67 × 32.2 0.017 2 = × = 0.0104 1.49 2 1.251 / 3 K n2 b1 / 3

For a maximum Froude number of 1.5, the maximum bed slope from Equation 4.50 is 2 S 0 = S L Fmax = 0.0104 × 1.52 = 0.0234

while for a Froude number of 1.0, the minimum bed slope is the limit slope, S 0 = 0.0104.

Therefore, the bed slope should fall in the range of 0.0104 to 0.0234. (b) Choose a bed slope of 0.02 and solve for the hydraulic radius for the given shear stress from the uniform flow equation:

R=

τ0 0.15 = = 0.120 ft γS 62.4 × 0.02

The corresponding depth for the rectangular cross section, calculated from the definition of the hydraulic radius as R = by/(b + 2y), is y = 0.148 ft. The required discharge is calculated from Manning's equation as

79

Open Channel Hydraulics

Q=

CHAPTER 4

1.49 1.49  1.25 × 0.148  AR 2 / 3 S 1 / 2 = × (1.25 × 0.148) ×  n 0.017 1.25 + 2 × 0.148 

2/3

× 0.021 / 2

with the result given by Q = 0.557 cfs.

4.16

A mountain stream has boulders with a median size (d 50 ) of 0.50 ft. The stream can be considered approximately rectangular in shape and very wide, with a slope of 0.01. You may assume that k s = 2 d 50 . (a) For a discharge of 7.0 cfs/ft, calculate the normal depth and critical depth and classify the slope as steep or mild. (b) Discuss how a Manning's n that is variable with depth affects the critical slope and the slope classification.

Solution.

As an initial value of Manning's n, use the minimum value in the fully-rough turbulent range from Keulegan's equation as shown in Figure 4.4:

n = 0.032k s1 / 6 = 0.032 × ( 2 × 0.50)1 / 6 = 0.032 The critical depth for a rectangular channel is y c = (q2/g)1/3 = (72/32.2)1/3 = 1.15 ft. The normal depth from Manning's equation for a very wide channel (R ≅ y) is

 nq  y0 =  1/ 2   1.49 S 

0.6

 0.032 × 7.0  = 1/ 2   1.49 × 0.01 

0.6

= 1.28 ft

The value of R/k s = 1.28/(2 × 0.5) = 1.28, which is not in the range in which Manning’s n is constant with depth. Therefore, Equation 4.21 must be iterated with Manning's equation to find the normal depth. Assume n = 0.040, then Manning's equation gives y 0 = 1.46 ft. Then calculate n from Equation 4.21:

Kn  R    1/ 2   ks  1 / 6 (8 g )

n = ks

1/ 6

 12 R   2.0 log  ks 

1.49  1.46   1/ 2  (8 × 32.2)  2 × 0.5  = ( 2 × 0.5)1 / 6  12 × 1.46  2.0 log   2 × 0.5 

1/ 6

= 0.040

The final value is y 0 = 1.46 ft and n = 0.040; the slope is mild. (b) For a slope near the critical value, small values of R/k s cause an increase in Manning's n and normal depth, and increase the likelihood of the slope being mild rather than steep.

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Open Channel Hydraulics

4.17

CHAPTER 4

Find the best hydraulic section for a trapezoidal channel. Express the wetted perimeter of a trapezoidal channel in terms of area, A, and depth, y, then differentiate P with respect to y, setting the result to zero to show that R = y/2. Also differentiate P with respect to the side-slope ratio, m, and set the result to zero. What is the best value of m and what do you conclude is the best trapezoidal shape?

Solution.

A = y (b + my ) A b = − my y P = b + 2 y 1 + m2 =

A + y (2 1 + m 2 − m) y

dP A = − 2 + 2 1 + m2 − m = 0 dy y 2A y A or R = ⇒ = 2 y 1 + m 2 − my and P = 2 y y Now rewrite P as 2A/y and differentiate with respect to m:

P = 4 y 1 + m 2 − 2my dP 1 = 4 y × (1 + m 2 ) −1 / 2 × 2m − 2 y = 0 2 dm 2 −1 / 2 =2 4m(1 + m ) 16m 2 = 4(1 + m 2 ) m=

3 3

The side-slope angle corresponding to this value of m is θ = tan–1(1/m) = 60°.

Furthermore, substituting into the expression for bottom width, b, we have

2

 3 3 A 3  −2 b = − my = 2 y 1 + m 2 − 2my = 2 y 1 +  y = 2 y  y 3 3  3  In comparison, the wetted perimeter of the side slope is y(1 + m2)1/2 = (2/31/2)y, which is identical in length to the channel bottom. Therefore, the best trapezoidal section is a halfhexagon, which has a side slope angle of 60°, and a bottom and sides of equal length.

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Open Channel Hydraulics

4.18

CHAPTER 4

A compound channel has symmetric floodplains, each of which is 100 m wide with Manning's n = 0.06, and a main channel, which is trapezoidal with a bottom width of 10 m, side slopes of 1.5:1, a bank-full depth of 2.5 m, and a Manning’s n of 0.03. If the channel slope is 0.001 and the total depth is 3.7 m, compute the uniform flow discharge using the divided channel method, first with a vertical interface both with and without wetted perimeter included for the main channel, then with a diagonal interface with wetted perimeter excluded.

Solution. n = 0.06

n = 0.03

n = 0.06

1.5

3.7 m 1

2.5 m

100 m 10 m

Given a Manning's n = 0.03 for the main channel and n = 0.06 for the floodplains, calculate area and wetted perimeter for the floodplains and main channel assuming a vertical interface with wetted perimeter included (VI) as shown in Figure 4.12:

Amc = 2.5(10 + 1.5 × 2.5) + (10 + 2 × 1.5 × 2.5)(1.2) = 55.4 m 2 Pmc = 10 + 2( 2.5) 1 + 1.52 + 2(1.2) = 21.4 m A fp = 100 × 1.2 = 120 m 2 Pfp = 100 + 1.2 = 101.2 m Then from Manning's equation, the discharge is given by

Q = 2×

1 1 2 / 3 1/ 2 A fp R 2fp/ 3 S 1 / 2 + Amc Rmc S n fp nmc

2  120  Q= ×120 ×   0.06  101.2 

2/3

× 0.001

1/ 2

1  55.4  + × 55.4 ×   0.03  21.4 

2/3

× 0.0011 / 2

Q = 141.7 + 110.1 = 252 m 3 /s Excluding the wetted perimeter of the vertical interface (VE), the wetted perimeter of the main channel, P mc = 19.0 m, and Q = 141.7 + 119.2 = 261 m3/s.

Using the method of diagonal subdivision with wetted perimeter excluded as shown in Figure 4.12, we have

Amc = 2.5(10 + 1.5 × 2.5) + 0.5 × (10 + 2 × 1.5 × 2.5)(1.2) = 44.9 m 2 A fp = 100 × 1.2 + 0.25 × (10 + 2 × 1.5 × 2.5)(1.2) = 125.25 m 2

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The wetted perimeters in the DE method are the same as for the VE method; that is, P mc = 19.0 m and P fp = 101.2 m. Substituting into Manning's equation, the total discharge, Q = 152.2 + 84.0 = 236 m3/s.

4.19

The power-law velocity distribution in a very wide open channel in uniform flow is given by

 z = a   u*  ks  u

m

in which u is the point velocity; u * is the shear velocity; a is a constant; z is the distance above the channel bed; k s is the equivalent sand-grain roughness; and m is the given fractional exponent that is constant. (a) Find the mean velocity, V, in terms of u max and m, where u max is the maximum velocity at z = y 0 ; y 0 = depth of flow; and m = exponent in power law. (b) Write the expression for V from part (a) in the form of a uniform flow formula and deduce the value of the exponent m that is compatible with Manning's equation.

Solution. (a) Integrate the point velocity distribution to find the mean velocity:

m

 z  au*  y0  1 y0   V = ∫ au*   dz = y0 0 (m + 1) y0  k s   ks  au*  y0    V= (m + 1)  k s 

m +1

m

The maximum velocity occurs at z = y 0 , and is given by

y  u max = au*  0   ks 

m

Comparing the expressions for V and u max , we have

V=

u max m +1

83

ks

Open Channel Hydraulics

CHAPTER 4

(b) Rewrite the expression for mean velocity, substituting u * = (gy 0 S)1/2, to obtain

V=

a g y 0m +1 / 2 S 1 / 2 m ( m + 1)k s

If m = 1/6, note that this equation for V agrees with Manning's and Strickler's equations in the exponents on y 0 , k s , and S. In other words, Manning's equation is compatible with a 1/6 power law velocity distribution.

4.20

Velocity data have been measured at the centerline of a tilting flume having a bed of crushed rock with d 50 = 0.060 ft. Consider the Run 12 data that follows, for which Q = 2.40 cfs and S = 0.00281. The elevations, z t , are given with respect to the bottom of the flume on which the rocks have been laid one layer thick. The total water depth above the flume bottom is 0.437 ft. The average bed elevation of the rocks is 0.055 ft above the flume bottom. Taking this elevation as the origin of the logarithmic velocity distribution, determine the shear stress from the velocity distribution and compare it with the value obtained from the uniform flow formula. Discuss the results. z t , ft 0.430 0.368 0.327 0.285 0.265 0.244 0.224 0.203 0.182 0.170 0.157 0.145 0.133 0.120 0.108 0.096 0.087 0.079

Velocity, ft/s 2.19 2.10 2.03 1.95 1.87 1.83 1.80 1.69 1.64 1.58 1.56 1.49 1.35 1.30 1.15 1.05 0.97 0.88

Solution.

The reference bed elevation of 0.055 ft is subtracted from the values of z t , and the point velocities are plotted on the horizontal scale with the corrected elevations of z on the vertical scale, which is a log scale, in the figure given next.

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1.000

1

z, ft

N = 1.142

0.100

0.010 0.00

0.50

1.00

1.50

2.00

2.50

u, ft/s

A linear least-squares regression analysis of u as a function of log z gives the slope of N = 1.142 as shown in the figure. By definition, N is given by

N=

2.3 du du du = = 2.3z d(log z ) d(ln z ) dz

From Equation 4.10, however, we can show that

du u * = dz κ z Substituting, it follows that

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CHAPTER 4

u N = 2.3 *

κ

or u* =

Nκ 1.142 × 0.4 = = 0.199 ft/s 2.3 2.3

From its definition, u * = (τ 0 /ρ)1/2, from which τ 0 = ρ u * 2 = 1.94 × 0.1992 = 0.0768 lbs/ft2.

The actual depth of flow was measured to be 0.437 ft relative to the reference elevation, so the uniform flow formula gives

τ 0 = γy 0 S = 62.4 × 0.437 × 0.00281 = 0.0766 lbs/ft 2 This assumes that the aspect ratio is large enough that no secondary currents influence a central region of the flow, which requires b/y 0 > 5. The flume is 2.5 ft wide, so the aspect ratio is 2.5/0.437 = 5.7.

4.21

Write a computer program, in the language of your choice, that computes the normal depth and critical depth in a circular channel using the bisection method. The input data should include the conduit diameter, Manning’s roughness coefficient, slope, and discharge. Illustrate your program with an example and verify the results for normal and critical depth.

Solution.

For convenience, the solution to Exercise 2.18 is a computer program that calculates normal depth as well as critical depth for a circular cross section, even though critical depth alone was asked for in that exercise. As an additional exercise, the student can be asked to expand the program to include the horseshoe cross section using the equations for geometry developed in Exercise 4.4.

4.22

Write a computer program to design a trapezoidal channel with a vegetative or rock riprap lining if the slope and design discharge are given. Allow the user to adjust the vegetal retardance class, or the rock size and angle of repose, and the channel bottom width and side slope interactively.

Solution.

The program can be written starting from the program Y0YC given in Appendix B of the textbook. The input consists of the discharge, Q; slope, S; channel bottom width, b; and side-slope ratio, m. In addition, as shown in the form module in the program given next, a grass-lined or rock riprap lined channel is chosen using the option button controls named "optGR" and "optRR". If the lining is grass, the user is asked to enter a vegetal retardance class, which is stored in the variable, strVR. Choosing the rock riprap option requires entry of the median size of the riprap in ft, D50; and the angle of repose, Phi. In contrast to the program Y0YC, Manning's n is not specified but is computed as part of the solution because it depends on the depth as given by Equation 4.42.

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The Sub procedure, STABLE, is called to carry out the stability computations. First, if grass lining has been chosen, the values of the permissible shear stress, τ p , and the constant, a 0 , are entered from Table 4-2 in the textbook using a Select Case decision structure. For a rock riprap lining, the value of the critical shear stress is computed from Equation 4.38, and the tractive force ratio, KR, is calculated from Equation 4.41. The critical shear stresses for the bed and banks are stored in the array variable TAU as TAU(1) and TAU(3), respectively.

Next, the Sub procedure BISECTION is called to find the critical depth, and then it is called a second time to find normal depth. To calculate Manning's n, the function F utilizes Equation 4.43 for the grass-lined channel, and Equation 4.42 for the rock riprap channel, as a part of the functional evaluation for depth. The final solution includes the value of Manning's n as well as the normal depth, Y0. Once normal depth has been determined, control returns to Sub STABLE where the maximum shear stresses are calculated according to Equations 4.39 and 4.40 for a rock riprap channel, and from τ max = γy 0 S for the grass-lined channel as given on p. 136 of the textbook. The maximum shear stresses are stored in TAU(2) and TAU(4) for the bed and banks, respectively.

Finally, control is returned to the form module where the normal depth, critical depth, and Manning's n are printed in text boxes, while the critical and maximum shear stresses are given for comparison in the picture box, picResults. The Sub Form_Load at the end of the form code initializes the option buttons to False when the program is run.

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Open Channel Hydraulics

CHAPTER 4

Option Explicit Dim Q As Single, S As Single, b As Single, n As Single Dim m as Single, Y0 As Single, YC As Single Dim D50 As Single, Phi As Single Dim J As Integer Dim strVR As String, strLIN As String Dim TAU(4) Private Sub cmdCalculate_Click() Q = Val(txtQ.Text) S = Val(txtS.Text) b = Val(txtB.Text) m = Val(txtSS.Text) If optGR Then strLIN = "Grass" strVR = txtVegret.Text ElseIf optRR Then strLIN = "Rock" D50 = Val(txtSize.Text) Phi = Val(txtAngleRepose.Text) End If Call STABLE(Q, S, b, m, strLIN, strVR, D50, Phi, n, Y0, _ YC, TAU()) txtY0.Text = Format(Y0, "#0.000") txtYC.Text = Format(YC, "#0.000") txtn.Text = Format(n, "0.0000") For J = 1 To 4 TAU(J) = Format(TAU(J), "0.000") Next J picResults.Cls picResults.Print strLIN; " Channel" picResults.Print picResults.Print Spc(14); "SHEAR STRESS IN LBS/FT^2" picResults.Print picResults.Print "BED:", "TAUc ="; TAU(1), "TAUmax=";TAU(2) If strLIN = "Rock" Then picResults.Print "BANKS:", "TAUc ="; TAU(3), "TAUmax="; _ TAU(4) End If End Sub Private Sub cmdExit_Click() End End Sub Private Sub Form_Load() optGR.Value = False optRR.Value = False End Sub

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CHAPTER 4

Sub STABLE(Q, S, b, m, strLIN, strVR, D50, Phi, n, Y0, _ YC, TAU()) Dim Y1 As Single, Y2 As Single, ER As Single Dim Theta As Single Dim KR As Single, gamma As Single, a0 As Single Dim NF As Integer, Pi As Single, R As Single gamma = 62.4 Pi = 3.1415926 If strLIN = "Grass" Then strVR = UCase(strVR) Select Case strVR Case "A" TAU(1) = 3.7: a0 = 24.7 Case "B" TAU(1) = 2.1: a0 = 30.7 Case "C" TAU(1) = 1#: a0 = 36.4 Case "D" TAU(1) = 0.6: a0 = 40# Case "E" TAU(1) = 0.35: a0 = 42.7 Case Else TAU(1) = 10000: a0 = 0 End Select ElseIf strLIN = "Rock" Then TAU(1) = 4 * D50 Theta = Atn(1 / m) Phi = Phi * Pi / 180# KR = Sqr(1 - (Sin(Theta)) ^ 2 / (Sin(Phi)) ^ 2) TAU(3) = KR * TAU(1) End If Y1 = 0.0001 Y2 = 100 ER = 0.0001 NF = 1 Call BISECTION(Y1, Y2, NF, ER, Q, S, b, m, strLIN, a0, _ D50, n, YC) Y1 = 0.0001 Y2 = 100 NF = 2 Call BISECTION(Y1, Y2, NF, ER, Q, S, b, m, strLIN, a0, _ D50, n, Y0) R = Y0 * (b + m * Y0) / (b + 2 * Y0 * Sqr(1 + m ^ 2)) If strLIN = "Grass" Then TAU(2) = gamma * Y0 * S ElseIf strLIN = "Rock" Then TAU(2) = 1.5 * gamma * R * S TAU(4) = 0.8 * TAU(2) End If End Sub

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Open Channel Hydraulics

CHAPTER 4

Sub BISECTION(Y1, Y2, NFUNC, ER, Q, S, b, m, strLIN, a0, _ D50, n, Y3) Dim FY1 As Single, FY2 As Single, FY3 As Single, FZ As _ Single Dim I As Integer FY1 = F(Y1, NFUNC, Q, S, b, m, strLIN, a0, D50, n) FY2 = F(Y2, NFUNC, Q, S, b, m, strLIN, a0, D50, n) If FY1 * FY2 > 0 Then Exit Sub For I = 1 To 50 Y3 = (Y1 + Y2) / 2 FY3 = F(Y3, NFUNC, Q, S, b, m, strLIN, a0, D50, n) FZ = FY1 * FY3 If FZ = 0 Then Exit Sub If FZ < 0 Then Y2 = Y3 Else Y1 = Y3 If Abs((Y2 - Y1) / Y3) < ER Then Exit Sub Next I End Sub Function F(Y, NFUNC, Q, S, b, m, strLIN, a0, D50, n) _ As Single Dim A As Single, P As Single, R As Single, T As Single Dim Kn As Single, G As Single, Rm As Single, C As Single A = Y * (b + m * Y) P = b + 2 * Y * Sqr(1 + m ^ 2) R = A / P T = b + 2 * m * Y Kn = 1.486 G = 32.2 If NFUNC = 1 Then F = Q - Sqr(G) * A ^ 1.5 / T ^ 0.5 ElseIf NFUNC = 2 Then If strLIN = "Grass" Then Rm = R / 3.281 n = Rm ^ (1 / 6) / (a0 + 16.4 * 0.4343 * _ Log(Rm ^ 1.4 * S ^ 0.4)) ElseIf strLIN = "Rock" Then C = D50 ^ (1 / 6) * (Kn / Sqr(8 * G)) n = C * (R / D50) ^ (1 / 6) / (0.794 + _ 1.85 * 0.4343 * Log(R / D50)) End If F = Q - (Kn / n) * A * R ^ (2 / 3) * S ^ (1 / 2) End If End Function

90

Open Channel Hydraulics

4.23

CHAPTER 4

Using the computer program Ycomp in Appendix B, find the normal and critical depths for the following compound channel section given in a data file. The discharge is 5000 cfs, and the slope is 0.009. Plot the cross-section. Is this a subcritical or supercritical flow?

"DUCKCR",19,4 -480,796 -440,788 -420,786 -305,784 -175,782 -95,780 -50,778 -30,776 -25,774 2,772 17,772 20,774 28,780 50,780 670,780 990,782 1070,784 1120,786 1260,810 -95 , .1 , 28 , .04 , 670 , .08 , 1260 , .05

Solution.

The data file given is used with Ycomp in Appendix B. The results are Lower critical depth = 7.699 ft Upper critical depth = 8.657 ft Lower limiting discharge = 4406 cfs Upper limiting discharge = 5448 cfs Normal depth = 8.337 ft The cross section is plotted as shown next. Bank-full depth is elevation 780 – 772 = 8 ft.

The given discharge of 5000 cfs falls in the interval in which multiple critical depths exist.

The normal depth lies between the upper critical depth and bank-full depth. The normal depth in the right floodplain is 8.337 – 8.0 = 0.337 ft, which gives a relative depth of 0.337/8.337 = 0.040. Theoretically, this is a supercritical overbank flow, but the floodplain depth is so small that maintenance of one-dimensional flow is unlikely in this case.

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Open Channel Hydraulics

CHAPTER 4

Duck Creek 815 810 805

Elevation, ft

800

n = 0.10

0.04

0.08

0.05

795 790 785 780 775 770 -600

-400

-200

0

200

400

600

Horizontal Station, ft

92

800

1,000

1,200

1,400

CHAPTER 5. Gradually Varied Flow 5.1

Prove from the equation of gradually varied flow that S2 and S3 profiles asymptotically approach normal depth in the downstream direction.

Solution.

The gradually varied flow equation is given by

dy S 0 − S e = dx 1 − F 2 Now for an S2 profile, y 0 < y < y c (see Figure 5.2), so it must be true that S e < S 0 and F2 > 1 according to Equations 5.6 and 5.7. This implies that dy/dx < 0 from the equation of gradually-varied flow, and that the depth is decreasing in the downstream direction. As x becomes very large, the normal depth is approached such that dy/dx approaches zero because S e → S 0 while the Froude number approaches a finite value. This is an asymptotic approach to normal depth. Similarly, we have for an S3 profile that y < y 0 < y c , which implies that S e > S 0 and F2 > 1. Then it follows that dy/dx > 0 with an increasing depth in the downstream direction. As normal depth is approached from below for very large values of x, dy/dx goes to zero because S e → S 0 , so that once again we have an asymptotic approach to normal depth.

5.2

A reservoir discharges into a long trapezoidal channel that has a bottom width of 20 ft, side slopes of 3:1, a Manning's n of 0.025, and a bed slope of 0.001. The reservoir water surface is 10 ft above the invert of the channel entrance. Determine the channel discharge.

Solution.

Assume that the channel is steep. Then we have from the energy equation

H = 10 = y c +

Dc y ( 20 + 3 y c ) = yc + c 2 2( 20 + 2 × 3 y c )

from which y c = 7.54 ft. Now setting the Froude number equal to one, and solving for the discharge, Q, the result is

Q=

g Ac3 / 2 32.2 [7.54( 20 + 3 × 7.54)]3 / 2 = = 4047 cfs Bc1 / 2 [20 + 2 × 3 × 7.54]1 / 2

Check if the slope is steep by calculating the critical slope as

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Open Channel Hydraulics

CHAPTER 5

n 2Q 2 Sc = = 1.49 2 Ac2 Rc4 / 3

0.0252 × 4047 2 [7.54( 20 + 3 × 7.54)]10 / 3

1.49 × 2

= 0.0056

[20 + 2 × 7.54 × 1 + 32 ]4 / 3

Because the actual slope of 0.001 is smaller than the critical slope, the slope is not steep.

Try again with the energy equation assuming that the channel is very long so that the entrance depth is normal depth. In addition, Manning's equation must be satisfied, so we must solve simultaneously the equations given by

H = 10 = y 0 + Q=

Q2 2 gA02

1.49 1.49 [ y 0 ( 20 + 3 y 0 )]5 / 3 A0 R02 / 3 S 1 / 2 = (0.001)1 / 2 n 0.025 [20 + 2 y 0 1 + 32 ]2 / 3

Q = 1.885

[ y 0 ( 20 + 3 y 0 )]5 / 3 [20 + 6.325 y 0 ]2 / 3

in which Manning's equation has been simplified to the extent possible. The solution can be obtained by substituting Q from Manning's equation into the energy equation and using a nonlinear algebraic equation solver to solve for normal depth. Alternatively, assume y 0 and solve for Q from Manning's equation. Then substitute back into the energy equation and solve for the head. Iterate on the normal depth until the energy equation gives the known head of 10 ft. The result is y 0 = 9.44 ft and Q = 2751 cfs.

5.3

A reservoir discharges into a long, steep channel followed by a long channel with a mild slope. Sketch and label the possible flow profiles as the tailwater rises. Explain how you could determine if the hydraulic jump occurs on the steep or mild slope.

Solution.

S1 NDL S2

M1 hydraulic jump

Steep

hydraulic jump

M3

M2

NDL CDL

Mild

Starting at the steep slope, the S2 profile approaches normal depth because the channel is very long. Using the normal depth on the steep slope, calculate the sequent depth for a hydraulic jump and compare with the normal depth on the mild slope. If the normal depth on the mild slope is greater than the sequent depth, then the jump occurs on the steep slope because the S1 profile has an increasing depth in the downstream direction. If the normal depth on the mild slope is less than the sequent depth, then the S2 will

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Open Channel Hydraulics

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continue to the mild slope followed by an M3 profile with the hydraulic jump on the mild slope. A similar argument can be made starting with the normal depth on the mild slope and working upstream.

5.4

Compute the water surface profile of Table 5-1 in the text using the method of numerical integration with the trapezoidal rule. Use the same step sizes as in the table and determine the distance required to reach a depth of 1.74 m. Discuss the results.

Solution.

The computations are given in the following spreadsheet:

Q, cms= n= S0= b,m = m:1 (1) y m 1.03 1.04 1.06 1.08 1.10 1.12 1.14 1.16 1.18 1.20 1.22 1.24 1.26 1.28 1.30 1.32 1.34 1.36 1.38 1.40 1.42 1.44 1.46 1.48 1.50 1.52 1.54 1.56 1.58 1.60

30 0.025 0.001 8 2 (2) A sq. m 10.36 10.48 10.73 10.97 11.22 11.47 11.72 11.97 12.22 12.48 12.74 13.00 13.26 13.52 13.78 14.04 14.31 14.58 14.85 15.12 15.39 15.67 15.94 16.22 16.50 16.78 17.06 17.35 17.63 17.92

y0, m yc, m

(3) R m 0.822 0.829 0.842 0.855 0.868 0.882 0.895 0.908 0.921 0.934 0.947 0.959 0.972 0.985 0.998 1.010 1.023 1.035 1.048 1.060 1.073 1.085 1.097 1.110 1.122 1.134 1.146 1.158 1.170 1.182

1.7538 1.0298

(4) D m 0.855 0.862 0.876 0.891 0.905 0.919 0.933 0.947 0.961 0.975 0.989 1.003 1.017 1.030 1.044 1.058 1.071 1.085 1.098 1.112 1.125 1.139 1.152 1.165 1.179 1.192 1.205 1.218 1.231 1.244

(5) V m/s 2.90 2.86 2.80 2.73 2.67 2.62 2.56 2.51 2.45 2.40 2.36 2.31 2.26 2.22 2.18 2.14 2.10 2.06 2.02 1.98 1.95 1.91 1.88 1.85 1.82 1.79 1.76 1.73 1.70 1.67

(6) Se

(7) F

(8) g(y)

6.80E-03 6.58E-03 6.15E-03 5.75E-03 5.39E-03 5.06E-03 4.75E-03 4.47E-03 4.20E-03 3.96E-03 3.73E-03 3.52E-03 3.32E-03 3.14E-03 2.97E-03 2.81E-03 2.67E-03 2.53E-03 2.40E-03 2.28E-03 2.16E-03 2.06E-03 1.96E-03 1.86E-03 1.77E-03 1.69E-03 1.61E-03 1.54E-03 1.47E-03 1.40E-03

0.9997 0.9840 0.9538 0.9249 0.8974 0.8712 0.8461 0.8222 0.7992 0.7773 0.7562 0.7361 0.7167 0.6981 0.6803 0.6632 0.6467 0.6308 0.6155 0.6008 0.5866 0.5729 0.5597 0.5470 0.5347 0.5228 0.5114 0.5003 0.4895 0.4791

-9.102E-02 -5.679E+00 -1.754E+01 -3.039E+01 -4.430E+01 -5.938E+01 -7.574E+01 -9.351E+01 -1.128E+02 -1.338E+02 -1.568E+02 -1.818E+02 -2.092E+02 -2.393E+02 -2.724E+02 -3.089E+02 -3.494E+02 -3.944E+02 -4.446E+02 -5.009E+02 -5.644E+02 -6.365E+02 -7.189E+02 -8.138E+02 -9.244E+02 -1.054E+03 -1.209E+03 -1.397E+03 -1.629E+03 -1.922E+03

95

(9) (10) Del x Sum m Del x, m 0 -0.029 -0.029 -0.232 -0.261 -0.479 -0.740 -0.747 -1.49 -1.037 -2.52 -1.351 -3.88 -1.692 -5.57 -2.063 -7.63 -2.467 -10.1 -2.906 -13.0 -3.386 -16.4 -3.910 -20.3 -4.485 -24.8 -5.117 -29.9 -5.814 -35.7 -6.583 -42.3 -7.438 -49.7 -8.389 -58.1 -9.454 -67.6 -10.65 -78.2 -12.01 -90.2 -13.55 -104 -15.33 -119 -17.38 -137 -19.79 -156 -22.64 -179 -26.06 -205 -30.26 -235 -35.50 -271

Open Channel Hydraulics

1.62 18.21 1.194 1.64 18.50 1.206 1.66 18.79 1.218 1.68 19.08 1.230 1.70 19.38 1.242 1.72 19.68 1.254 1.74 19.98 1.266 A=y*(b+m*y) R=A/(b+2*y*sqrt(1+m^2)) D=A/(b+2*m*y) V=Q/A

CHAPTER 5

1.258 1.65 1.34E-03 0.4691 -2.303E+03 -42.24 -313 1.271 1.62 1.28E-03 0.4593 -2.819E+03 -51.22 -364 1.284 1.60 1.22E-03 0.4499 -3.557E+03 -63.76 -428 1.297 1.57 1.17E-03 0.4408 -4.696E+03 -82.53 -511 1.309 1.55 1.12E-03 0.4319 -6.685E+03 -113.8 -624 1.322 1.52 1.07E-03 0.4233 -1.103E+04 -177.1 -801 1.335 1.50 1.03E-03 0.4150 -2.793E+04 -389.6 -1191 Se =n^2*V^2/R^(4/3) Del x = (y2 - y1)*(g(y1)+g(y2))/2 F = V/sqrt(9.81*D) g(y) = (1 - F^2)/(S0 - Se)

In contrast to Table 5-1 (direct step method), the hydraulic depth, D = A/B, is needed to compute the Froude number, F = V/(gD)1/2, which appears in column 7. The function g(y), which is the integrand as defined in Equation 5.16, is calculated in column 8. The trapezoidal rule is applied in column 9 to compute ∆x = ∆y[g(y 1 ) + g(y 2 )]/2, and the cumulative sum of the distances is shown in column 10. The depth of 1.74 m is reached at x = 1191 m in comparison to x = 1100 m in Table 5-1. Alternatively, by linear interpolation, the depth from the spreadsheet given above at x = 1100 m is 1.735 m, which is 0.3 percent less than the value of 1.74 m in Table 5-1. As the normal depth is approached asymptotically, very small depth changes result in very large distance changes, and so a better comparison is made between the depths at a specified distance.

5.5

A rectangular channel 6.1 m wide with n = 0.014 is laid on a slope of 0.001 and terminates in a free overfall. Upstream 300 m from the overfall is a sluice gate that produces a depth of 0.47 m immediately downstream. For a discharge of 17.0 m3/s, with a spreadsheet, compute the water surface profiles and the location of the hydraulic jump using the direct step method. Verify with the program WSP, or with a program that you write.

Solution.

Calculate normal depth from Manning's equation:

A5 / 3 nQ 0.014 × 17 = 1/ 2 = = 7.526 2/3 P S 0.0011 / 2 (by 0 ) 5 / 3 (6.1 y 0 ) 5 / 3 = = 7.526 (b + 2 y 0 ) 2 / 3 (6.1 + 2 y 0 ) 2 / 3 which gives y 0 = 1.308 m. Critical depth for a rectangular channel is obtained directly from Equation 2.6 as follows

 q2  y c =    g 

1/ 3

 (17 / 6.1) 2  =    9.81 

1/ 3

= 0.925 m

96

Open Channel Hydraulics

CHAPTER 5

The slope is mild which means that there will be an M2 profile ending in critical depth at the free overfall. The depth downstream of the sluice gate is supercritical resulting in an M3 profile. If a hydraulic jump occurs, it will be at a location where the momentum function values for the M2 and M3 profiles are equal. The spreadsheets containing the direct step computations for the M2 and M3 profiles are given next.

M2 Profile Q, cms= 17.0 y0, m 1.308 n= 0.014 yc, m 0.925 S0= 0.001 b,m = 6.10 m:1 0 (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) y A R V E Se S e bar Del E Del x Sum Mom. Fn. m sq. m m m/s m m Del x, m cu m 0.926 5.65 0.710 3.010 1.3877 2.80E-03 300 7.831 0.93 5.67 0.713 2.997 1.3877 2.76E-03 2.78E-03 3.73E-05 -0.021 299.98 7.831 0.94 5.73 0.719 2.965 1.3880 2.68E-03 2.72E-03 3.14E-04 -0.182 299.80 7.833 0.95 5.80 0.724 2.934 1.3886 2.59E-03 2.63E-03 6.18E-04 -0.378 299.42 7.836 0.96 5.86 0.730 2.903 1.3895 2.51E-03 2.55E-03 9.10E-04 -0.586 298.83 7.842 0.97 5.92 0.736 2.873 1.391 2.43E-03 2.47E-03 1.19E-03 -0.807 298.03 7.849 0.98 5.98 0.742 2.844 1.392 2.36E-03 2.40E-03 1.46E-03 -1.043 296.98 7.857 0.99 6.04 0.747 2.815 1.394 2.29E-03 2.33E-03 1.72E-03 -1.294 295.69 7.868 1.00 6.10 0.753 2.787 1.396 2.22E-03 2.26E-03 1.96E-03 -1.563 294.13 7.879 1.01 6.16 0.759 2.759 1.398 2.16E-03 2.19E-03 2.20E-03 -1.850 292.28 7.893 1.02 6.22 0.764 2.732 1.400 2.09E-03 2.12E-03 2.43E-03 -2.159 290.12 7.908 1.03 6.28 0.770 2.706 1.403 2.03E-03 2.06E-03 2.65E-03 -2.490 287.63 7.925 1.04 6.34 0.776 2.680 1.406 1.98E-03 2.00E-03 2.86E-03 -2.847 284.78 7.943 1.05 6.41 0.781 2.654 1.409 1.92E-03 1.95E-03 3.06E-03 -3.232 281.55 7.962 1.06 6.47 0.787 2.629 1.412 1.87E-03 1.89E-03 3.26E-03 -3.649 277.90 7.983 1.07 6.53 0.792 2.605 1.416 1.81E-03 1.84E-03 3.45E-03 -4.102 273.80 8.005 1.08 6.59 0.798 2.580 1.419 1.76E-03 1.79E-03 3.63E-03 -4.595 269.20 8.029 1.09 6.65 0.803 2.557 1.423 1.72E-03 1.74E-03 3.80E-03 -5.133 264.07 8.054 1.10 6.71 0.808 2.534 1.427 1.67E-03 1.69E-03 3.97E-03 -5.724 258.35 8.081 1.11 6.77 0.814 2.511 1.431 1.63E-03 1.65E-03 4.13E-03 -6.373 251.97 8.109 1.12 6.83 0.819 2.488 1.436 1.58E-03 1.60E-03 4.29E-03 -7.092 244.88 8.138 1.13 6.89 0.825 2.466 1.440 1.54E-03 1.56E-03 4.44E-03 -7.891 236.99 8.168 1.14 6.95 0.830 2.445 1.445 1.50E-03 1.52E-03 4.59E-03 -8.783 228.21 8.200 1.15 7.02 0.835 2.423 1.449 1.46E-03 1.48E-03 4.73E-03 -9.787 218.42 8.233 1.16 7.08 0.840 2.402 1.454 1.43E-03 1.45E-03 4.86E-03 -10.92 207.50 8.267 1.17 7.14 0.846 2.382 1.459 1.39E-03 1.41E-03 4.99E-03 -12.22 195.3 8.303 1.18 7.20 0.851 2.362 1.464 1.36E-03 1.37E-03 5.12E-03 -13.71 181.6 8.340 1.19 7.26 0.856 2.342 1.470 1.32E-03 1.34E-03 5.24E-03 -15.45 166.1 8.377 1.20 7.32 0.861 2.322 1.475 1.29E-03 1.31E-03 5.36E-03 -17.49 148.6 8.417 1.21 7.38 0.866 2.303 1.480 1.26E-03 1.27E-03 5.47E-03 -19.94 128.7 8.457 1.22 7.44 0.871 2.284 1.486 1.23E-03 1.24E-03 5.59E-03 -22.90 105.8 8.498 1.23 7.50 0.877 2.266 1.492 1.20E-03 1.21E-03 5.69E-03 -26.59 79.2 8.541 1.24 7.56 0.882 2.247 1.497 1.17E-03 1.19E-03 5.80E-03 -31.27 47.9 8.584 1.25 7.63 0.887 2.230 1.503 1.14E-03 1.16E-03 5.90E-03 -37.44 10.5 8.629 1.26 7.69 0.892 2.212 1.509 1.12E-03 1.13E-03 5.99E-03 -45.92 -35.4 8.675

97

Open Channel Hydraulics

CHAPTER 5

M3 Profile (1) (2) y A m sq. m 0.470 2.87 0.48 2.93 0.49 2.99 0.50 3.05 0.51 3.11 0.52 3.17 0.53 3.23 0.54 3.29 0.55 3.36 0.56 3.42 0.57 3.48 0.58 3.54 0.59 3.60 0.60 3.66 0.61 3.72 0.62 3.78 0.63 3.84 0.64 3.90 0.65 3.97 0.66 4.03 0.67 4.09 0.68 4.15 0.69 4.21 0.70 4.27 0.71 4.33 0.72 4.39 0.73 4.45 0.74 4.51 0.75 4.58 0.76 4.64 0.77 4.70 0.78 4.76 0.79 4.82 0.80 4.88 0.81 4.94 0.82 5.00 0.83 5.06 0.84 5.12 0.85 5.19 0.86 5.25 0.87 5.31 0.88 5.37 0.89 5.43 0.90 5.49 0.91 5.55 0.92 5.61

(3) R m 0.407 0.415 0.422 0.430 0.437 0.444 0.452 0.459 0.466 0.473 0.480 0.487 0.494 0.501 0.508 0.515 0.522 0.529 0.536 0.543 0.549 0.556 0.563 0.569 0.576 0.582 0.589 0.596 0.602 0.608 0.615 0.621 0.627 0.634 0.640 0.646 0.652 0.659 0.665 0.671 0.677 0.683 0.689 0.695 0.701 0.707

(4) V m/s 5.930 5.806 5.688 5.574 5.464 5.359 5.258 5.161 5.067 4.977 4.889 4.805 4.724 4.645 4.569 4.495 4.424 4.355 4.288 4.223 4.160 4.098 4.039 3.981 3.925 3.871 3.818 3.766 3.716 3.667 3.619 3.573 3.528 3.484 3.441 3.399 3.358 3.318 3.279 3.241 3.203 3.167 3.131 3.097 3.063 3.029

(5) (6) (7) (8) E Se S e bar Del E m 2.2620 2.28E-02 2.1981 2.14E-02 2.21E-02 -6.39E-02 2.1387 2.00E-02 2.07E-02 -5.94E-02 2.0834 1.88E-02 1.94E-02 -5.53E-02 2.0319 1.77E-02 1.82E-02 -5.15E-02 1.9840 1.66E-02 1.71E-02 -4.80E-02 1.9392 1.56E-02 1.61E-02 -4.47E-02 1.8975 1.48E-02 1.52E-02 -4.17E-02 1.8586 1.39E-02 1.43E-02 -3.89E-02 1.8223 1.32E-02 1.35E-02 -3.63E-02 1.7884 1.25E-02 1.28E-02 -3.39E-02 1.7567 1.18E-02 1.21E-02 -3.17E-02 1.7272 1.12E-02 1.15E-02 -2.96E-02 1.6996 1.06E-02 1.09E-02 -2.76E-02 1.6738 1.01E-02 1.04E-02 -2.58E-02 1.6498 9.59E-03 9.84E-03 -2.40E-02 1.6274 9.12E-03 9.35E-03 -2.24E-02 1.6064 8.69E-03 8.90E-03 -2.09E-02 1.5869 8.28E-03 8.48E-03 -1.95E-02 1.5688 7.90E-03 8.09E-03 -1.82E-02 1.5518 7.54E-03 7.72E-03 -1.69E-02 1.5361 7.20E-03 7.37E-03 -1.57E-02 1.5215 6.88E-03 7.04E-03 -1.46E-02 1.5079 6.58E-03 6.73E-03 -1.36E-02 1.4953 6.30E-03 6.44E-03 -1.26E-02 1.4836 6.04E-03 6.17E-03 -1.17E-02 1.4728 5.79E-03 5.91E-03 -1.08E-02 1.4629 5.55E-03 5.67E-03 -9.94E-03 1.4537 5.32E-03 5.44E-03 -9.15E-03 1.4453 5.11E-03 5.22E-03 -8.40E-03 1.4377 4.91E-03 5.01E-03 -7.69E-03 1.4307 4.72E-03 4.82E-03 -7.01E-03 1.4243 4.54E-03 4.63E-03 -6.37E-03 1.4185 4.37E-03 4.45E-03 -5.76E-03 1.4133 4.21E-03 4.29E-03 -5.18E-03 1.4087 4.05E-03 4.13E-03 -4.63E-03 1.4046 3.90E-03 3.98E-03 -4.10E-03 1.4010 3.76E-03 3.83E-03 -3.60E-03 1.3979 3.63E-03 3.70E-03 -3.12E-03 1.3952 3.50E-03 3.57E-03 -2.67E-03 1.3930 3.38E-03 3.44E-03 -2.23E-03 1.3912 3.27E-03 3.33E-03 -1.82E-03 1.3898 3.16E-03 3.21E-03 -1.42E-03 1.3887 3.05E-03 3.11E-03 -1.04E-03 1.3880 2.95E-03 3.00E-03 -6.82E-04 1.3877 2.86E-03 2.90E-03 -3.36E-04

98

(9) (10) (11) Del x Sum Mom. Fn. m Del x, m cu m 0 10.949 3.029 3.029 10.764 3.017 6.046 10.588 3.004 9.050 10.421 2.990 12.04 10.263 2.974 15.01 10.112 2.957 17.97 9.969 2.938 20.91 9.833 2.917 23.83 9.703 2.894 26.72 9.581 2.870 29.59 9.464 2.844 32.43 9.353 2.816 35.25 9.247 2.786 38.04 9.147 2.755 40.79 9.052 2.721 43.51 8.962 2.685 46.20 8.876 2.647 48.84 8.795 2.607 51.45 8.719 2.565 54.02 8.646 2.520 56.54 8.577 2.472 59.01 8.512 2.422 61.43 8.451 2.370 63.80 8.394 2.314 66.11 8.340 2.256 68.37 8.289 2.195 70.57 8.241 2.130 72.70 8.196 2.062 74.76 8.155 1.991 76.75 8.116 1.916 78.66 8.080 1.837 80.50 8.047 1.754 82.25 8.017 1.667 83.92 7.989 1.575 85.50 7.963 1.478 86.97 7.940 1.377 88.35 7.920 1.270 89.62 7.901 1.157 90.78 7.885 1.039 91.82 7.871 0.914 92.73 7.860 0.782 93.51 7.850 0.643 94.16 7.842 0.496 94.65 7.837 0.340 94.99 7.833 0.176 95.17 7.831

Open Channel Hydraulics

CHAPTER 5

The values of the momentum function for the M2 and M3 profiles are plotted next.

Where the curves intersect is the location of the hydraulic jump at x = 56.7 m.

12.0

10.0

(M3)

Momentum function, cu m

(M2) 8.0

6.0

4.0 Hydraulic Jump

2.0

0.0 0

50

100

150

200

250

x, m

The program WSP in Appendix B gives virtually identical results to the spreadsheet. The only difference is in the number of steps taken in the direct step method. The program WSP uses the maximum of 30 steps for the M3 profile, but fewer steps (26) are taken for the M2 profile because the specified channel length is reached before the profile reaches 99.9 percent of normal depth. Nevertheless, the results from WSP for the M2 profile are given next, and the depth at x = 0 is 1.2522 ft compared to 1.2523 ft interpolated from the spreadsheet.

99

300

Open Channel Hydraulics

CHAPTER 5

x, m 300.0 299.8 299.4 298.5 297.3 295.8 293.7 291.3 288.3 284.7 280.5 275.6 270.0 263.5 256.0 247.4 237.6 226.4 213.5 198.7 181.5 161.5 138.1 110.3 76.7 35.5 0.0

y, m V, m/s Mom. F. 0.9260 3.010 7.831 0.9387 2.969 7.832 0.9514 2.929 7.837 0.9641 2.891 7.844 0.9768 2.853 7.854 0.9895 2.816 7.867 1.0022 2.781 7.882 1.0149 2.746 7.900 1.0276 2.712 7.920 1.0403 2.679 7.943 1.0530 2.647 7.968 1.0658 2.615 7.996 1.0785 2.584 8.025 1.0912 2.554 8.057 1.1039 2.525 8.092 1.1166 2.496 8.128 1.1293 2.468 8.166 1.1420 2.440 8.207 1.1547 2.414 8.249 1.1674 2.387 8.293 1.1801 2.362 8.340 1.1928 2.336 8.388 1.2055 2.312 8.439 1.2182 2.288 8.491 1.2309 2.264 8.545 1.2436 2.241 8.600 1.2522 2.226 8.639

The WSP output is obtained from a SAVE event procedure that utilizes the Windows common dialog box (cdb prefix). The output is saved as a text file and imported into a spreadsheet for further processing. The additional code, which is not given in Appendix B, is shown in the following form code segment.

Private Sub cmdSave_Click() Dim strFileOut As String cdbFile.ShowSave strFileOut = cdbFile.FileName If strFileOut = "" Then Exit Sub Open strFileOut For Output As #1 Write #1, " X, m", "Y, m", "V, m/s", "Mom. F." For J = 1 To NP Write #1, XP(J), YP(J), VP(J), MP(J) Next J Close 1 End Sub In addition, the WSP module code from Appendix B is modified to accommodate SI units. These changes are required in the Function and Sub procedures, F and SPENERGY, respectively, in which the values of gravitational acceleration, G, and the Manning equation constant, KN, are specified.

100

Open Channel Hydraulics

CHAPTER 5

Function F(Y, NFUNC, Q, S, b, m, n) As Single Dim A As Single, P As Single, R As Single, T As Single Dim KN As Single, G As Single G = 9.81: KN = 1# A = Y * (b + m * Y) P = b + 2 * Y * Sqr(1 + m ^ 2) R = A / P T = b + 2 * m * Y If NFUNC = 1 Then F = Q - Sqr(G) * A ^ 1.5 / T ^ 0.5 Else F = Q - (KN / n) * A * R ^ (2 / 3) * S ^ (1 / 2) End If End Function Sub SPENERGY(Y, Q, S, b, m, n, E, SEGL, VEL, FM) Dim A As Single, P As Single, R As Single Dim KN As Single, G As Single G = 9.81: KN = 1# A = Y * (b + m * Y) P = b + 2 * Y * Sqr(1 + m ^ 2) R = A / P VEL = Q / A FM = b * Y ^ 2 / 2 + m * Y ^ 3 / 3 + Q ^ 2 / (G * A) E = Y + Q ^ 2 / (2 * G * A ^ 2) SEGL = n ^ 2 * Q ^ 2 / (KN ^ 2 * A ^ 2 * R ^ (4# / 3#)) End Sub 5.6

A very wide rectangular channel carries a discharge of 10.0 m3/s/m on a slope of 0.001 with an n value of 0.026. The channel ends in a free overfall. Compute the distance required for the depth to reach 0.9 y 0 using the direct step method and compare the result with that from the Bresse function.

Solution.

The normal depth for a very wide, rectangular channel is determined from Manning's equation as  nq   yn =   K S 1/ 2    n

0.6

 0.026 × 10  =   1.0 × 0.0011 / 2 

0.6

= 3.540 m

and the critical depth is

 q2  y c =    g 

1/ 3

 10 2   =   9.81 

1/ 3

= 2.168 m

101

Open Channel Hydraulics

CHAPTER 5

The channel is mild, and we are seeking the length of an M2 profile from the free overfall upstream to a depth of 0.9 y 0 = 3.186 m. Using the Bresse function, the length is obtained in one step from Equation 5.36 as   y 3   y   y  S 0 ( x 2 − x1 ) = ( y 2 − y1 ) − y n 1 −  c   φ  2  − φ  1    y n     y n   y n   

in which y 1 = y c and y 2 = 0.9y n . Substituting into this equation with the Bresse function calculated from Equation 5.37, we have

x 2 − x1 =

3 (0.9 × 3.540 − 2.168) 3.540   2.168   − − 1  (0.61385 − 0.04852)   0.001 0.001   3.540  

x 2 − x1 = 1018 − 3540 × 0.7703 × 0.56533 = −523.6 m The length is 524 m and is negative as expected because we are calculating in the upstream direction. The length of the profile is calculated from WSP in Appendix B by assuming a channel width large enough that the normal depth of 3.54 m is obtained while maintaining a discharge per unit width of 10 m2/s. A width of 10,000 m and a discharge of 100,000 m3/s produces a normal depth of 3.541 m, and a profile length of 454 m to reach a depth of 3.186 m as shown in the following output from WSP. The Bresse method gives a 15% larger length due to the assumption of a constant Chezy C. x, m 454.3 453.9 452.8 450.9 448.0 444.1 439.1 432.9 425.4 416.3 405.6 393.0 378.4 361.5 341.9 319.3 293.3 263.3 228.7 188.6 142.0 87.6 23.3 0.0

y, m V, m/s Mom. F. 2.1683 4.61 70520 2.2139 4.52 70551 2.2596 4.43 70642 2.3052 4.34 70790 2.3508 4.25 70994 2.3964 4.17 71251 2.4421 4.09 71561 2.4877 4.02 71920 2.5333 3.95 72327 2.5790 3.88 72781 2.6246 3.81 73281 2.6702 3.75 73826 2.7158 3.68 74413 2.7615 3.62 75043 2.8071 3.56 75713 2.8527 3.51 76423 2.8984 3.45 77173 2.9440 3.40 77961 2.9896 3.34 78786 3.0352 3.29 79648 3.0809 3.25 80546 3.1265 3.20 81479 3.1721 3.15 82447 3.1860 3.14 82748

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Open Channel Hydraulics

5.7

CHAPTER 5

Derive Equations 5.38 and 5.39 using the Bresse function.

Solution.

First note that for a rectangular channel, the ratio of critical depth to normal depth reduces to

y c ( q 2 / g )1 / 3  q 2  = =  3  y0 y0  gy 0 

1/ 3

= F2 / 3

in which F is the Froude number corresponding to normal depth, y 0 . Now apply Equation 5.36 between depths of y 1 = 0.97 y 0 and y 2 = 0.75 y 0 , so that the profile length, L = x 2 – x 1 , is positive, and divide through by y 0 with the result:

  y   y  LS 0  y 2 y1  =  −  − 1 − F 2 φ  2  − φ  1  y0  y0 y0    y 0   y 0 

[

]

[

]

LS 0 = 0.75 − 0.97 − 1 − F 2 [0.252 − 1.040] y0 LS 0 = 0.57 − 0.79F 2 y0 for the length of the M2 profile. Similarly, for the M1 profile, we have

[

]

LS 0 = 1.25 − 1.03 − 1 − F 2 [0.420 − 1.060] y0 LS 0 = 0.86 − 0.64F 2 y0 5.8

For a very wide channel on a steep slope, derive a formula for the length of an S2 profile from critical depth to 1.01 y 0 using the Bresse function. What is this length in meters if the slope is 0.01; the discharge per unit of width is 2.0 m3/s/m; and Manning’s n is 0.025?

Solution.

Let y 1 = y c and y 2 = 1.01y 0 , and substitute into Equation 5.36 with y c /y 0 = F2/3 to yield

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  y   y  LS 0  y 2 y1  =  −  − 1 − F 2 φ  2  − φ  1  y0  y0 y0    y 0   y 0 

[

]

[

][

]

LS 0 = 1.01 − F 2 / 3 − 1 − F 2 1.419 − φ (F 2 / 3 ) y0

LS 0 = −0.409 − F 2 / 3 + 1.419F 2 + (1 − F 2 )φ (F 2 / 3 ) y0 For the specific case of S = 0.01, n = 0.025, and q = 2.0 m2/s, the normal depth from Manning's equation for a very wide channel is given by

 nq   y 0 =  1/ 2   Kn S 

0.6

 0.025 × 2  = 1/ 2   0.01 

0.6

= 0.660 m

and the critical depth, y c = (q2/g)1/3= (22 / 9.81)1/3 = 0.742 m. The Froude number is

 q2  F =  3   gy 0 

1/ 2

  22  =  3   9.81 × 0.660 

1/ 2

= 1.191

Substituting into the expression for the profile length, the result is

LS 0 = −0.409 − (1.191) 2 / 3 + 1.419 × 1.1912 + (1 − 1.1912 ) × φ (1.1236) = 0.222 y0 Then the length, L = 0.222 × 0.660 / 0.01 = 14.6 m.

5.9

A trapezoidal channel of bottom width 20 ft with side slopes of 2:1 is laid on a slope of 0.0005 and has an n value of 0.045. It drains a lake with a constant water surface level of 10 ft above the invert of the channel entrance. If the channel ends in a free overfall, calculate the discharge in the channel for channel lengths of 100 and 10,000 ft using the WSP program.

Solution.

The program WSP is run with trial values of discharge for each length until the specific energy at the channel entrance is equal to the given value of 10.0 ft. For a length of 100 ft, the results from WSP for Q = 2814 cfs are shown next along with a column giving the specific energy, E = y + V 2/2g. The normal depth for this discharge is 16.38 ft, while the critical depth is 6.75 ft. The channel is mild and the profile shown in the following table is an M2 with a specific energy of 10.00 ft at the entrance as required.

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x, ft 100.0 98.4 93.3 84.2 70.3 51.0 25.3 0.0

y, ft 6.7562 7.0763 7.3965 7.7167 8.0368 8.3570 8.6771 8.9222

V, ft/s 12.428 11.644 10.935 10.292 9.706 9.172 8.682 8.334

Mom. F. 1748 1755 1772 1801 1840 1889 1947 1998

E, ft 9.15 9.18 9.25 9.36 9.50 9.66 9.85 10.00

For a channel length of 10,000 ft, the normal depth is 10.11 ft, and the critical depth is 3.78 ft for a discharge, Q = 1018 cfs. This discharge results in a specific energy of 10.00 ft at the channel entrance as shown in the following table of results from WSP. x, ft 10000 9999 9996 9990 9981 9969 9952 9931 9904 9871 9831 9782 9724 9654 9571 9473 9357 9220 9058 8867 8639 8367 8039 7641 7148 6526 5718 4617 2991 163 0

y, ft 3.7833 3.9940 4.2047 4.4154 4.6261 4.8368 5.0475 5.2582 5.4689 5.6796 5.8903 6.1010 6.3117 6.5224 6.7331 6.9438 7.1545 7.3652 7.5759 7.7866 7.9973 8.2080 8.4187 8.6294 8.8401 9.0508 9.2615 9.4722 9.6829 9.8936 9.8977

V, ft/s 9.761 9.107 8.522 7.997 7.523 7.093 6.702 6.344 6.017 5.716 5.438 5.182 4.944 4.723 4.518 4.326 4.147 3.980 3.823 3.675 3.536 3.406 3.283 3.166 3.056 2.952 2.853 2.760 2.671 2.586 2.585

105

Mom. F. 488 490 496 505 518 534 552 574 598 625 655 687 722 760 800 842 887 935 985 1037 1092 1150 1210 1273 1339 1407 1478 1551 1627 1706 1708

E, ft 5.26 5.28 5.33 5.41 5.50 5.62 5.74 5.88 6.03 6.19 6.35 6.52 6.69 6.87 7.05 7.23 7.42 7.61 7.80 8.00 8.19 8.39 8.59 8.79 8.99 9.19 9.39 9.59 9.79 10.00 10.00

Open Channel Hydraulics

5.10

CHAPTER 5

A 3 ft by 3 ft box culvert that is 100 ft long is laid on a slope of 0.001, and has a Manning's n of 0.013. The downstream end of the culvert is a free overfall. For a discharge of 20 cfs, calculate the entrance depth using the WSP program, and the head upstream of the culvert using the energy equation with an entrance loss coefficient of 0.5 for a square-edged entrance. Compare the result with the head calculated from an assumption of a hydraulically long culvert with an entrance depth equal to normal depth.

Solution.

The results from WSP are given in the following table for the given data. The normal depth is 2.03 ft, and the critical depth is 1.11 ft resulting in a mild slope for Q = 20 cfs. x, ft 100.0 99.7 98.6 96.9 94.3 90.8 86.4 80.9 74.2 66.3 56.9 46.0 33.4 18.8 1.9 0.0

y, ft 1.114 1.145 1.176 1.206 1.237 1.267 1.298 1.329 1.359 1.390 1.420 1.451 1.482 1.512 1.543 1.546

V, ft/s Mom. F. 5.982 5.579 5.822 5.583 5.671 5.595 5.527 5.615 5.390 5.643 5.260 5.677 5.136 5.717 5.018 5.764 4.905 5.818 4.797 5.877 4.694 5.941 4.595 6.012 4.500 6.087 4.409 6.168 4.321 6.254 4.313 6.263

Write the energy equation from the upstream pool to the culvert entrance where the depth is 1.546 ft:

H = y ent + (1 + K e )

4.3132 V2 = 1.979 ft = 1.546 + 1.5 × 64.4 2g

If we assume that the channel is very long, then the entrance depth is normal depth and the energy equation becomes

H = y0 + (1 + K e )

Q2 20 2 = + × = 2.281 ft 2 .

03 1 .

5 2 gA02 64.4 × [3 × 2.03]2

The conclusion is that the assumption of normal depth at the entrance of a short culvert can give erroneous results for the head.

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5.11

CHAPTER 5

Using HECRAS, compute the water surface profile in Some Creek for a discharge of 10,000 cfs Begin with a subcritical profile and a downstream water surface slope of 0.0087 as a boundary condition. Then do a mixed flow analysis with an upstream boundary condition of critical depth. The cross-section geometry, reach lengths, roughness values, and subsection breakpoints are shown in the following table. Analyze the results indicating where any hydraulic jumps may occur.

The upstream cross section for Some Creek at River Station 6000 (ft) is given by X (ft)

Elevation (ft)

0 0 23 36 45 55 99 110 119 133 143 150 154 155 160 168 188 193 200 205 210 229 258 266 276 305 344 380 380

465 465 458.8 458 457.8 458.3 458.4 455.9 455.8 455.5 455.3 455.4 454 452 450.3 450.2 450.5 451.5 452.7 454.5 455.3 455.6 455.3 456.3 458 457.8 458 461 465

n 0.055

0.065

0.040

0.065

0.055

The left and right banks are at X = 150 ft and 210 ft, respectively. At subsequent stations downstream, the cross section should be adjusted with a uniform decrease in elevation from the previous section as follows River station (ft)

Decrease in Elev. (ft)

4000 3000 1500 1000

2.0 6.0 2.8 6.0

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Solution.

The cross section at river station 6000 as plotted by HEC-RAS is shown first:

Some Creek Elevation (ft) .055

466

.065

3/12/01

Upstream .04 .065

.055 Legend

464 EG Q10 462

WS Q10

460

Crit Q10

458

Ground

456

Bank Sta

454 452 450 0

100

200

300

400

Station (ft)

The cross-section data are entered in the cross-section editor (Edit-Geometric Data).

First, the river schematic is drawn after clicking on the River Reach button. Then the Cross Section button is clicked, and Add a New Cross Section is chosen from the Options menu, starting from the most upstream cross section. The user is prompted for a river station, and then the transverse station and elevation data are entered in spreadsheet style.

Also required are the distances to the next downstream cross section and the values of Manning's n in the left overbank (LOB), main channel (Channel), and right overbank (ROB). In addition, the transverse stations corresponding to the locations of the right and left banks must be entered. If the Manning's n value varies within the overbank sections, it can be entered directly with the cross section data by choosing Horizontal Variation in n Values under the Options menu. When all the data have been entered for the cross section, the Apply Data button is clicked, although the data are still not saved at this point.

This process is repeated for each succeeding cross section in the downstream direction.

If the cross sections are relatively uniform in the downstream section, the Copy Current Cross Section choice in the Options menu allows the current cross section elevations to be increased or decreased by a fixed amount, as in this exercise, or the channel width can also be adjusted. When all cross section data have been entered, the cross section window is closed and the geometric data are saved under the File menu of the Geometric Data window.

Next, the flow data are entered under the main menu choice of Edit-Steady Flow Data.

The discharge is entered (it can change along the river profile), and the reach boundary conditions are input. For this analysis, choose critical depth for the upstream boundary condition, and normal depth for the downstream condition with the given slope of 0.0087.

Then click on Apply Data, and save the flow data under the File menu. The project itself, which is a collection of geometric, flow, and plan data files, is saved under the main File menu.

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The final step is to run the program under Simulate-Steady Flow Analysis in the main menu. Choose a mixed-flow analysis because this option will compute both the subcritical and supercritical profiles for the reach and choose between the two at each river station as described in the textbook. The program is run by clicking on the Compute button. The results can be viewed in a variety of ways in tables and graphs under the View menu.

The water surface profile for the first run is shown next: Elevation (ft)

Some Creek

3/12/01

Main

465

Legend EG Q10

460

Crit Q10 455 WS Q10 Ground

450 445 440 435 430 1000

2000

3000

4000

5000

6000

Main Channel Distance (ft)

It can be observed that the critical depth profile and water surface profile coincide at stations 1500 and 4000. It is apparent that the profile passes through critical depth at station 1500 and is supercritical downstream of that point. At station 4000, the program could not find a valid subcritical depth and defaulted to critical depth as explained in the View-Summary Errors, Warnings, and Notes menu choice. This could mean that the profile passes through critical depth and then forms a hydraulic jump between river stations 4000 and 3000. However, the warnings indicate that additional cross sections may be needed. These warnings are triggered by a change in velocity head in excess of 0.5 ft; an energy loss that exceeds 1.0 ft between cross sections; or a ratio of conveyances between cross sections that falls outside the range of 0.7 to 1.4. All of these warnings appeared for this first run, so an additional run was made with interpolated cross sections spaced at 100 ft intervals. These can be added under Edit-Geometric Data-Options-XS Interpolation. The resulting water surface profile is shown in the next graph.

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Open Channel Hydraulics

Elevation (ft)

CHAPTER 5

Some Creek

Main

465

3/12/01 Legend EG Q10

460

Crit Q10 455 WS Q10 Ground

450 445 440 435 430 1000

2000

3000

4000

5000

6000

Main Channel Distance (ft)

Upon comparing these results with the previous graph, it appears that between stations 3000 and 4000 there is an M1 profile that approaches normal depth, which is only slightly greater than critical depth. This is indicated in the Profile Table in which S e approaches S 0 from station 3000 to 4000. As a result, there is no supercritical flow in this reach. If the bed elevation at station 4000 is raised sufficiently, the profile does pass through critical depth with supercritical flow occurring downstream followed by a hydraulic jump. It is worthy of note that the conventional Froude number as calculated

by HEC-RAS does not indicate a supercritical flow in the most downstream reach for reasons described in Chapter 2, even though the flow depth is less than critical depth.

5.12

Compute the water surface profile in the Red Fox River for Q = 1000 cfs for which the downstream water surface elevation, WS = 5703.80 (3.80), and for Q = 10,000 cfs with WS = 5715.05 (15.05). The stations for the four cross sections are shown here, and the elevations (Z) and n values are given in the following table (Hoggan 1997).

Cross section 1 2 3 4

Station (ft) 0 500 900 1300

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Open Channel Hydraulics

Cross section 1 X (ft) Z (ft) n 20 25 30 24 45 22 0.100 60 20 110 18 415 17 @ 630 16 0.050 650 14 @ 655 13 660 13 670 2 675 1 0.030 690 0 697 0.1 700 0.8 710 1 710 13 @ 940 13.5 0.050 1020 14 @ 1215 14 1235 12 1575 12 1590 14 0.10 1615 16 1630 20 1635 25

CHAPTER 5

Cross section 2 X (ft) Z (ft) n 30 25 40 24 50 22 0.10 110 20 200 20 295 18 415 17 @ 455 16 505 13 0.05 575 9.5 @ 585 5 596 4.2 615 4.5 0.03 635 16 640 18 @ 940 18.5 1180 18 1195 18 1205 20 0.10 1225 22 1245 24 1250 25

Cross section 3 X (ft) Z (ft) n 40 25 90 24 0.10 260 22 @ 330 20 0.05 370 18.7 @ 420 15 460 11.2 500 7.1 530 7.5 0.03 550 12 560 17.8 580 19 600 20 @ 850 22 865 24 0.05 875 25

Cross section 4 X (ft) Z (ft) n 30 26 75 25 0.10 130 24 @ 0.05 330 23 @ 360 14 370 9.5 400 9.8 0.036 410 13 460 22 @ 0.05 610 22 @ 650 24 675 25 0.10 700 26

@ = subsection breakpoint Solution.

The four cross sections are entered into HEC-RAS as described in Exercise 5.11 starting upstream and moving in the downstream direction from station 1300 to station 0. The steady flow data are entered for two profiles with discharges of 1000 and 10,000 cfs. The downstream boundary conditions are entered as known water surface elevations (subtract 5700 from the values given in the problem statement), and the upstream conditions are entered as critical depth for a mixed flow analysis. The downstream conditions are slightly subcritical. The water surface profiles for the initial computer run are shown in the following figure:

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Open Channel Hydraulics

CHAPTER 5

Red Fox River

Elevation (ft)

Main

25

3/12/01 Legend EG Q = 10000 cfs

20

WS Q = 10000 cfs Crit Q = 10000 cfs EG Q = 1000 cfs

15

WS Q = 1000 cfs Crit Q = 1000 cfs

10

Ground

5

0 0

200

400

600

800

1000

1200

1400

Main Channel Distance (ft)

For Q = 10, 000 cfs, the profile is subcritical except at station 1300 where the program defaulted to critical depth because a subcritical solution could not be found. Accordingly, a hydraulic jump is expected between station 1300 and station 900, where the flow is subcritical. The flow remains in the main channel except at station 0 where there is overbank flooding in the right floodplain. Warning messages indicate that additional cross sections are needed.

The profile for Q = 1000 cfs shows the occurrence of critical depth at station 500 with a possible supercritical flow and hydraulic jump between stations 500 and 0. The water surface at station 1300 is just barely subcritical. The entire profile is within the main channel for this low discharge. Warning messages expressing the need for additional cross sections also occur at this discharge.

The program was run again with interpolated cross sections at 100 ft intervals. The results are shown in the following figure for the same two discharges. This run clears up some uncertainty in the results. Although the energy losses are still greater than 1.0 ft between some cross sections, the ratio of conveyances now falls in the accepted range of 0.7 to 1.4. For Q = 10,000 cfs, it is quite clear in this run that a hydraulic jump occurs between stations 1200 and 1100. Overbank flooding occurs in the right floodplain for stations 0 to 300 ft. For Q = 1000 cfs, with the additional cross sections, critical depth occurs at station 300 with supercritical flow downstream and a hydraulic jump between stations 200 and 100. The flow is definitely subcritical at station 1300 in this run.

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CHAPTER 5

Red Fox River

Elevation (ft)

3/12/01

Main

25

Legend EG Q = 10000 cfs

20

WS Q = 10000 cfs Crit Q = 10000 cfs EG Q = 1000 cfs

15

WS Q = 1000 cfs Crit Q = 1000 cfs

10

Ground

5

0 0

200

400

600

800

1000

1200

1400

Main Channel Distance (ft)

The program used composite n values in the main channel at station 700. Equation 4.27 is used to calculate the composite n value when there is more than one n value in the main channel and the channel side slope is steeper than 5:1. In this example, however, a warning message for Q = 10,000 cfs indicates that the composite value was either greater than the maximum n or less than the minimum n. In such cases, it may be advisable to enter a single Manning's n value for the main channel that is more representative of the composite roughness using one of the other formulas given in Chapter 4.

5.13

The cross-section geometry for Roaring Creek is given below: X (ft) 4 10 20 30 40 42 46 50 54 58 62 66 70 72 76 80 90 100 110 116

Elevation (ft) 10.0 9.5 9.3 9.4 9.2 7.0 6.2 6.0 6.1 6.2 6.0 7.1 6.3 8.3 8.9 9.0 9.5 9.3 9.6 10.0

n .050

.035

.060 .030

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Open Channel Hydraulics

CHAPTER 5

The measured water surface elevation is 9.8 ft. (a) Manually calculate the normal discharge for a slope of 0.0008. (b) Manually calculate the value of α and the specific energy. (c) Is the flow subcritical or supercritical? (d) Verify your manual calculations with the HEC-RAS program.

Solution.

The cross section is plotted in the following figure in which the measured water surface elevation of 9.8 ft is shown. Both the left and right floodplains are inundated. The Manning's n values are also shown for each subsection.

12.0 n = 0.05

11.0

n = 0.035

n = 0.06

n = 0.03

Elevation, ft

10.0 9.0 8.0 7.0 6.0 5.0 4.0 0

20

40

60

80

100

120

140

Station, ft

In the spreadsheet given next are shown the results of computations for area, a, and wetted perimeter, p, of each individual triangle or trapezoid defined by the ground points, as shown for the first few ground points in the figure above. The area and hydraulic radius of each subsection are calculated and used to obtain the conveyance of each subsection from

Ki =

1.49 Ai Ri2 / 3 ni

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Open Channel Hydraulics

CHAPTER 5

in which the subscript i refers to the ith subsection. Now if the total flow area, A = ΣA i , and the total conveyance is given by K = ΣK i , then the kinetic energy flux correction coefficient is calculated in the spreadsheet from

α=

∑ K i3 / Ai2 K 3 / A2

which is shown in the spreadsheet to be α = 1.3350. The discharge comes from Manning's equation written in terms of the conveyance as

Q = KS 1 / 2 = 9879 × 0.00081 / 2 = 279.4 cfs The specific energy is given by

E = y +α

279.4 2 Q2 = 3.881 ft = − + × ( 9 .

8 6 .

0 ) 1 .

3350 64.4 ×141.34 2 2gA 2

Roaring Creek: W.S. Elev., ft = X,ft Z, ft n 4 10 20 30 40 42 46 50 54 58 62 66 70 72 76 80 90 100 110 116

10.0 9.5 9.3 9.4 9.2 7.0 6.2 6.0 6.1 6.2 6.0 7.1 6.3 8.3 8.9 9.0 9.5 9.3 9.6 10.0

0.05

0.035 0.06 0.03

9.8 a, ft2

p, ft

0.54 4.00 4.50 5.00 3.40 12.80 14.80 15.00 14.60 14.80 13.00 12.40 5.00 4.80 3.40 5.50 4.00 3.50 0.30

3.61 10.00 10.00 10.00 2.97 4.08 4.00 4.00 4.00 4.00 4.15 4.08 2.83 4.04 4.00 10.01 10.00 10.00 3.01 sum A i =

A i , ft2

R i , ft

Ki

K i 3/A i 2

14.04

0.418

233.8

64809

114.00

2.704

9418.4 64286919

9.50

0.475

143.6

32778

3.80

0.292

83.1

141.34

sum K i =

9878.8

39711 alpha = 1.3350

If the cross section geometry is placed in a data file that can be read by the computer program Ycomp in Appendix B, the results confirm that a discharge of 279.4 cfs gives a normal depth of 3.8 ft. In addition, the program results indicate that there is only one critical depth and a single limiting discharge of 668 cfs (QU = QL). For the discharge of 279.4 cfs, which is less than the limiting discharge, the single value of critical depth is

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Open Channel Hydraulics

CHAPTER 5

equal to 1.767 ft. This is a lower critical depth in main channel flow, so the normal depth of 3.8 ft is subcritical. The data file necessary for the program Ycomp is given next.

"Roaring Creek", 20, 4 4,10.0 10, 9.5 20, 9.3 30, 9.4 40, 9.2 42, 7.0 46, 6.2 50, 6.0 54, 6.1 58, 6.2 62, 6.0 66, 7.1 70, 6.3 72, 8.3 76, 8.9 80, 9.0 90, 9.5 100, 9.3 110, 9.6 116, 10.0 40, 0.05, 80, 0.035, 100, 0.060, 116, 0.030 The HEC-RAS calculations are made by entering the given cross section at an upstream station and copying it to a downstream station that is arbitrarily 100 ft downstream. Then the upstream elevations are increased by 0.08 ft to give a slope of 0.0008, although this is not absolutely necessary because direct comparisons with the previous results are best made at the downstream section with a boundary condition: Known W. S. = 9.80 ft. Run HEC-RAS for this boundary condition and the discharge of 279.4 cfs to produce the results in the following cross section table at the downstream section: Plan: Plan 01 River: Roaring Creek Reach:Main Riv Sta: 900 Profile: PF 1 E.G. Elev (ft) 9.88 Element Left OB Channel Right OB Vel Head (ft) 0.08 Wt. n-Val. 0.050 0.035 0.049 W.S. Elev (ft) 9.80 Reach Len. (ft) Crit W.S. (ft) 7.76 Flow Area (sq ft) 14.04 114.00 13.30 E.G. Slope (ft/ft) 0.000804 Area (sq ft) 14.04 114.00 13.30 Q Total (cfs) 279.40 Flow (cfs) 6.61 266.38 6.41 Top Width (ft) 106.60 Top Width (ft) 33.60 40.00 33.00 Vel Total (ft/s) 1.98 Avg. Vel. (ft/s) 0.47 2.34 0.48 Max Chl Dpth (ft) 3.80 Hydr. Depth (ft) 0.42 2.85 0.40 Conv. Total (cfs) 9851.8 Conv. (cfs) 233.1 9392.6 226.0 Length Wtd. (ft) Wetted Per. (ft) 33.62 42.17 33.03 Min Ch El (ft) 6.00 Shear (lb/sq ft) 0.02 0.14 0.02 Alpha 1.33 Stream Power (lb/ft s) 0.01 0.32 0.01 Frctn Loss (ft) Cum Volume (acre-ft) C & E Loss (ft) Cum SA (acres)

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Keeping in mind that the minimum channel elevation is 6.00 ft, the earlier spreadsheet calculation of specific energy = 3.88 ft is in agreement with the E.G. Elevation of 9.88 ft given in the HEC-RAS table. Furthermore, the critical depth from Ycomp was 1.767 ft, which agrees closely with the HEC-RAS table value of 7.76 – 6.0 = 1.76 ft. The HECRAS value of α = 1.33 compares well with the spreadsheet value given previously as 1.335. Finally, the total conveyance given by the HEC-RAS output is 9852 in comparison to the spreadsheet value of 9879, but this can be accounted for by the use of K n = 1.49 in the spreadsheet in comparison to 1.486 in HEC-RAS. 5.14

Write a computer program in the language of your choice that computes the water surface profile in a circular culvert using the method of integration by the trapezoidal rule.

Solution.

The Visual Basic code module for computing a water surface profile in a circular channel is given next with bold highlighting of the changes required in WSP, which appears in Appendix B of the textbook. The circular channel poses some difficulties because of the shape of the curve in Figure 4.9 as the crown of the culvert is approached. The maximum in AR2/3 occurs at a relative depth of 0.938d, and there is the possibility of two solutions for normal depth if AR2/3/d 8/3 exceeds 0.3117 (y 0 /d > 0.81). Accordingly, note that the root search for normal depth is limited in the Sub procedure Y0YC with 0.938d as an upper bound. It is assumed that the upper solution for normal depth is unlikely due to filling of the pipe as the crown is approached. Critical depth approaches the diameter asymptotically, so a value of 0.99d is used arbitrarily as the upper limit of the critical depth root search. If no root is found within these bounds for normal and critical depth, they are set equal to the diameter to detect this occurrence. If both are set equal to the diameter, this condition is considered "full flow", which is written in the profile classification variable strPR, and no profile is computed.

The shape of the curve in Figure 4.9 also poses difficulties for specification of the control depth and the computation of a water surface profile in the upper region (y 0 /d > 0.81) as illustrated in the following figure. If a normal depth < 0.938d occurs in this region, and the control depth is specified to be greater than 0.938d such that AR2/3 for the control depth is less than the value for normal depth, then S e > S 0 even though y > y 0 . This is impossible and the program fails to compute a profile. The optimum solution would be to compute the upper "normal" depth and then not allow the specification of a control depth greater than this value. The compromise shown in the next figure is to choose the multiple solutions of 0.97d and 0.90d such that if the normal depth exceeds 0.90d, the control depth, YCONT, is not allowed to exceed 0.938d, and in any case a control depth greater than 0.97d is not allowed. If these control depths are exceeded, then "Submerged" is written in the profile classification variable, strPR. The trapezoidal rule of integration is shown in bold in the Sub procedure PROFILE. The Sub procedure SPENERGY is called to obtain the necessary quantities to compute the integrand g(y) in Equation 5.16, and then the trapezoidal rule is applied as given by Equation 5.17 to advance the solution by one step each time through the loop. The geometric properties for the circular channel replace those for the trapezoidal channel given previously in the program WSP. The geometric formulas appear in the function F and the Sub procedure SPENERGY.

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Open Channel Hydraulics

1.0 0.97 0.938 0.90

CHAPTER 5

o

y0/d

y0/d

0.332

AR2/3/d 8/3

Option Explicit Dim Q As Single, S As Single, b As Single, m As Single Dim n As Single, Y0 As Single, YC As Single Sub Y0YC(Q, S, d, n, Y0, YC) Dim Y1 As Single, Y2 As Single, ER As Single Dim NF As Integer Y1 = 0.0001 Y2 = 0.99 * d ER = 0.0001 NF = 1 Call BISECTION(Y1, Y2, NF, ER, Q, S, d, n, YC) If S = 0 Then Y0 = 1000# Exit Sub End If Y1 = 0.0001 Y2 = 0.938 * d NF = 2 Call BISECTION(Y1, Y2, NF, ER, Q, S, d, n, Y0) End Sub Sub PROFILE(Q, S, d, n, Y0, YC, XL, YCONT, strPR, NP, _ XP(), YP(), VP(), MP()) Dim XSIGN As Single, YLIM As Single, E As Single, _ SEGL As Single, FR As Single Dim DX As Single, DY As Single, Y As Single, INTEGRAND1 As Single Dim INTEGRAND2 As Single, VEL As Single, FM As Single Dim I As Integer, NS As Integer If YC >= d And Y0 >= d Then strPR = "Full flow" Exit Sub End If If YCONT > 0.97 * d Then strPR = "Submerged" Exit Sub End If

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If YCONT = 0 Then YCONT = YC If Y0 > 0.9 * d And Y0 < d And YCONT > 0.938 * d Then strPR = "Submerged" Exit Sub End If If YC < Y0 Then 'classify mild slope profiles If YCONT > Y0 Then strPR = "M1": XSIGN = -1: YLIM = 1.001 * Y0 ElseIf YCONT < YC Then strPR = "M3": XSIGN = 1: YLIM = 0.999 * YC Else strPR = "M2": XSIGN = -1: YLIM = 0.999 * Y0 If YCONT = YC Then YCONT = 1.001 * YC End If Else If YCONT > YC Then 'classify steep slope profiles strPR = "S1": XSIGN = -1: YLIM = 1.001 * YC ElseIf YCONT < Y0 Then strPR = "S3": XSIGN = 1: YLIM = 0.999 * Y0 Else strPR = "S2": XSIGN = 1: YLIM = 1.001 * Y0 If YCONT = YC Then YCONT = 0.999 * YC If YC > 0.99 * d Then YCONT = 0.99 * d End If End If If S = 0 Then 'classify horizontal slope profiles If YCONT >= YC Then strPR = "H1": XSIGN = -1: YLIM = 0.999 * d If YCONT = YC Then YCONT = 1.001 * YC Else strPR = "H2": XSIGN = 1: YLIM = 0.999 * YC End If End If NS = 30 'initialize variables NP = NS + 1 DY = (YLIM - YCONT) / NS If XSIGN = 1 Then XP(1) = 0 Else XP(1) = XL Y = YCONT Call SPENERGY(Y, Q, S, d, n, E, SEGL, VEL, FR, FM) YP(1) = Y: VP(1) = VEL: MP(1) = FM INTEGRAND1 = (1 - FR ^ 2) / (S - SEGL) For I = 2 To NS + 1 'direct integration method loop Y = Y + DY Call SPENERGY(Y, Q, S, d, n, E, SEGL, VEL, FR, FM) YP(I) = Y: VP(I) = VEL: MP(I) = FM INTEGRAND2 = (1 - FR ^ 2) / (S - SEGL) DX = DY * (INTEGRAND1 + INTEGRAND2) / 2 XP(I) = XP(I - 1) + DX INTEGRAND1 = INTEGRAND2 If XP(I) > XL Then 'interpolation for specified length YP(I) = YP(I) - DY * (XP(I) - XL) / DX

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XP(I) = XL NP = I Call SPENERGY(YP(I), Q, S, d, n, E, SEGL, VEL, FR, FM) VP(I) = VEL: MP(I) = FM Exit For End If If XP(I) < 0# Then 'interpolation for specified length YP(I) = YP(I) - DY * XP(I) / DX XP(I) = 0# NP = I Call SPENERGY(YP(I), Q, S, d, n, E, SEGL, VEL, FR, FM) VP(I) = VEL: MP(I) = FM Exit For End If Next I End Sub Sub BISECTION(Y1, Y2, NFUNC, ER, Q, S, d, n, Y3) Dim FY1 As Single, FY2 As Single, FY3 As Single, FZ As Single Dim I As Integer FY1 = F(Y1, NFUNC, Q, S, d, n) FY2 = F(Y2, NFUNC, Q, S, d, n) If FY1 * FY2 > 0 Then Y3 = d Exit Sub End If For I = 1 To 50 Y3 = (Y1 + Y2) / 2 FY3 = F(Y3, NFUNC, Q, S, d, n) FZ = FY1 * FY3 If FZ = 0 Then Exit Sub If FZ < 0 Then Y2 = Y3 Else Y1 = Y3 If Abs((Y2 - Y1) / Y3) < ER Then Exit Sub Next I End Sub Function F(Y, NFUNC, Q, S, d, n) As Single Dim A As Single, P As Single, R As Single, T As Single Dim PI As Double, THETA As Double, X As Double Dim KN As Single, G As Single G = 32.2: KN = 1.486 X = 1 - 2 * Y / d PI = 3.141592654 THETA = 2 * (PI / 2 - Atn(X / (1 - X ^ 2) ^ 0.5)) A = (THETA - Sin(THETA)) * d ^ 2 / 8 P = THETA * d / 2 R = A / P T = d * Sin(THETA / 2) If NFUNC = 1 Then F = Q - Sqr(G) * A ^ 1.5 / T ^ 0.5 Else

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F = Q - (KN / n) * A * R ^ (2 / 3) * S ^ (1 / 2) End If End Function Sub SPENERGY(Y, Q, S, d, n, E, SEGL, VEL, FR, FM) Dim A As Single, P As Single, R As Single, T As Single Dim PI As Double, THETA As Double, X As Double Dim KN As Single, G As Single G = 32.2: KN = 1.486 X = 1 - 2 * Y / d PI = 3.141592654 THETA = 2 * (PI / 2 - Atn(X / (1 - X ^ 2) ^ 0.5)) A = (THETA - Sin(THETA)) * d ^ 2 / 8 P = THETA * d / 2 R = A / P T = d * Sin(THETA / 2) VEL = Q / A FM = 3 * Sin(THETA / 2) - (Sin(THETA / 2)) ^ 3 _ - 3 * (THETA / 2) * Cos(THETA / 2) FM = FM * d ^ 3 / 24 + Q ^ 2 / (G * d ^ 2 _ * (THETA - Sin(THETA)) / 8) E = Y + Q ^ 2 / (2 * G * A ^ 2) FR = Sqr(Q ^ 2 * T / (G * A ^ 3)) SEGL = n ^ 2 * Q ^ 2 / (KN ^ 2 * A ^ 2 * R ^ (4# / 3#)) End Sub 5.15

Write a computer program in a language of your choice that computes a water surface profile in a trapezoidal channel using the fourth order Runge-Kutta method. Test it with the M2 profile of Table 5-1.

Solution.

The code module written in Visual Basic is given next with the necessary changes made to the program WSP in Appendix B to implement the 4th order Runge-Kutta method.

The Sub procedure Y0YC is identical to that in WSP, as is the profile classification algorithm in the Sub procedure PROFILE. Following the profile classification, however, the direct step algorithm is replaced with the Runge-Kutta algorithm according to Equations 5.23. The Sub procedure RKUTTA is used to compute the variables needed to obtain the function f(y) as defined in Equation 5.19. A rudimentary adaptive step size control is introduced in the Runge-Kutta algorithm. The initial spatial step size is set equal to 0.01 m, and then it is doubled whenever the computed change in depth, DY, is less than the uniform increment determined in WSP for 30 equal steps in the total range of depths expected in the profile. With these variable spatial steps, it is necessary to detect the asymptotic approach to normal depth for M1, M2, S2, and S3 profiles, and to detect the approach to critical depth in M3, H2, and S1 profiles with If statements as shown in the code module. In this case, the loop is exited when the approach to normal or critical depth occurs before the specified channel length is reached. If the spatial step size chosen for the next execution of the loop takes the channel length past its specified value, then the step size is adjusted for the last time through the loop and the loop exits because the computed profile length is equal to the given channel length, XL.

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Following the code module, a table of results is given for comparison with Table 5-1.

The value of depth at a distance of 1271 m is 1.744 m in comparison to 1.745 m in Table 5-1 from the direct step method. However, the depth of 1.744 m is in agreement with the result from the corrected Euler method given in Table 5-2. It is essential that the first few spatial steps be very small in the Runge-Kutta method as in the corrected Euler method.

Option Explicit Dim Q As Single, S As Single, b As Single, m As Single Dim n As Single, Y0 As Single, YC As Single Sub Y0YC(Q, S, b, m, n, Y0, YC) Dim Y1 As Single, Y2 As Single, ER As Single Dim NF As Integer Y1 = 0.0001 Y2 = 100 ER = 0.0001 NF = 1 Call BISECTION(Y1, Y2, NF, ER, Q, S, b, m, n, YC) If S = 0 Then Y0 = 1000# Exit Sub End If Y1 = 0.0001 Y2 = 100 NF = 2 Call BISECTION(Y1, Y2, NF, ER, Q, S, b, m, n, Y0) End Sub Sub PROFILE(Q, S, b, m, n, Y0, YC, XL, YCONT, strPR, NP, _ XP(), YP(), VP(), MP()) Dim XSIGN As Single, YLIM As Single, E As Single, SEGL As Single Dim DX As Single, DY As Single, Y As Single, FR As Single Dim VEL As Single, FM As Single, Y1 As Single, Y2 As Single Dim K1 As Single, K2 As Single, K3 As Single, K4 As Single Dim I As Integer, NS As Integer If YCONT = 0 Then YCONT = YC If YC < Y0 Then 'classify mild slope profiles If YCONT > Y0 Then strPR = "M1": XSIGN = -1: YLIM = 1.001 * Y0 ElseIf YCONT < YC Then strPR = "M3": XSIGN = 1: YLIM = 0.999 * YC Else strPR = "M2": XSIGN = -1: YLIM = 0.999 * Y0 If YCONT = YC Then YCONT = 1.001 * YC End If Else If YCONT > YC Then 'classify steep slope profiles strPR = "S1": XSIGN = -1: YLIM = 1.001 * YC ElseIf YCONT < Y0 Then strPR = "S3": XSIGN = 1: YLIM = 0.999 * Y0 Else

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strPR = "S2": XSIGN = 1: YLIM = 1.001 * Y0 If YCONT = YC Then YCONT = 0.999 * YC End If End If If S = 0 Then 'classify horizontal slope profiles If YCONT >= YC Then strPR = "H1": XSIGN = -1: YLIM = YCONT + 3# * YC If YCONT = YC Then YCONT = 1.001 * YC Else strPR = "H2": XSIGN = 1: YLIM = 0.999 * YC End If End If NS = 30 'initialize variables DX = XSIGN * 0.01 If XSIGN = 1 Then XP(1) = 0 Else XP(1) = XL Y1 = YCONT Call RKUTTA(Y1, Q, S, b, m, n, E, SEGL, VEL, FR, FM) YP(1) = Y1: VP(1) = VEL: MP(1) = FM K1 = (S - SEGL) / (1 - FR ^ 2) For I = 2 To 100 'RUNGE-KUTTA method loop Y = Y1 + K1 * DX / 2 Call RKUTTA(Y, Q, S, b, m, n, E, SEGL, VEL, FR, FM) K2 = (S - SEGL) / (1 - FR ^ 2) Y = Y1 + K2 * DX / 2 Call RKUTTA(Y, Q, S, b, m, n, E, SEGL, VEL, FR, FM) K3 = (S - SEGL) / (1 - FR ^ 2) Y = Y1 + K3 * DX Call RKUTTA(Y, Q, S, b, m, n, E, SEGL, VEL, FR, FM) K4 = (S - SEGL) / (1 - FR ^ 2) DY = (K1 + 2 * K2 + 2 * K3 + K4) * DX / 6 Y2 = Y1 + DY Call RKUTTA(Y2, Q, S, b, m, n, E, SEGL, VEL, FR, FM) K1 = (S - SEGL) / (1 - FR ^ 2) YP(I) = Y2: VP(I) = VEL: MP(I) = FM XP(I) = XP(I - 1) + DX If XP(I) = XL Or XP(I) = 0 Then NP = I Exit Sub End If If Abs(YP(I) - YLIM) <= 0.01 Then 'detect asymptotic approach NP = I 'to Y0 Exit Sub End If If Abs(YLIM / YC - 1) <= 0.01 Then 'detect approach to YC If XSIGN * (YP(I) - YLIM) > 0 Then NP = I - 1 Exit Sub End If End If If Abs(DY) < Abs((YCONT - YLIM) / NS) Then DX = DX * 2 If (XP(I) + DX) < 0 Then

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DX = XSIGN * XP(I) ElseIf (XP(I) + DX) > XL Then DX = XL - XP(I) End If Y1 = Y2 Next I End Sub Sub BISECTION(Y1, Y2, NFUNC, ER, Q, S, b, m, n, Y3) Dim FY1 As Single, FY2 As Single, FY3 As Single, FZ As Single Dim I As Integer FY1 = F(Y1, NFUNC, Q, S, b, m, n) FY2 = F(Y2, NFUNC, Q, S, b, m, n) If FY1 * FY2 > 0 Then Exit Sub For I = 1 To 50 Y3 = (Y1 + Y2) / 2 FY3 = F(Y3, NFUNC, Q, S, b, m, n) FZ = FY1 * FY3 If FZ = 0 Then Exit Sub If FZ < 0 Then Y2 = Y3 Else Y1 = Y3 If Abs((Y2 - Y1) / Y3) < ER Then Exit Sub Next I End Sub Function F(Y, NFUNC, Q, S, b, m, n) As Single Dim A As Single, P As Single, R As Single, T As Single Dim KN As Single, G As Single G = 9.81: KN = 1# A = Y * (b + m * Y) P = b + 2 * Y * Sqr(1 + m ^ 2) R = A / P T = b + 2 * m * Y If NFUNC = 1 Then F = Q - Sqr(G) * A ^ 1.5 / T ^ 0.5 Else F = Q - (KN / n) * A * R ^ (2 / 3) * S ^ (1 / 2) End If End Function Sub RKUTTA(Y, Q, S, b, m, n, E, SEGL, VEL, FR, FM) Dim A As Single, P As Single, R As Single, T As Single Dim KN As Single, G As Single G = 9.81: KN = 1# A = Y * (b + m * Y) P = b + 2 * Y * Sqr(1 + m ^ 2) R = A / P T = b + 2 * m * Y VEL = Q / A FM = b * Y ^ 2 / 2 + m * Y ^ 3 / 3 + Q ^ 2 / (G * A) E = Y + Q ^ 2 / (2 * G * A ^ 2) SEGL = n ^ 2 * Q ^ 2 / (KN ^ 2 * A ^ 2 * R ^ (4# / 3#))

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FR = Sqr(Q ^ 2 * T / (G * A ^ 3)) End Sub

x, m 1271.0 1271.0 1271.0 1271.0 1270.9 1270.8 1270.7 1270.4 1269.7 1268.4 1265.9 1263.3 1260.8 1255.6 1250.5 1245.4 1235.2 1224.9 1214.7 1204.4 1184.0 1163.5 1143.0 1122.5 1081.6 1040.6 999.6 917.7 835.8 753.9 590.0 426.2 98.5 0.0

5.16

y, m 1.0300 1.0678 1.0683 1.0691 1.0708 1.0740 1.0796 1.0892 1.1044 1.1272 1.1597 1.1841 1.2042 1.2367 1.2629 1.2852 1.3219 1.3517 1.3770 1.3989 1.4357 1.4658 1.4912 1.5130 1.5490 1.5775 1.6008 1.6366 1.6625 1.6820 1.7084 1.7246 1.7413 1.7441

V, m/s 2.90 2.77 2.77 2.77 2.76 2.75 2.74 2.71 2.66 2.60 2.51 2.44 2.39 2.32 2.26 2.21 2.13 2.07 2.03 1.99 1.92 1.87 1.83 1.80 1.75 1.70 1.67 1.63 1.59 1.57 1.54 1.52 1.50 1.50

Mom. F. 13.83 13.85 13.85 13.85 13.85 13.86 13.87 13.88 13.91 13.97 14.09 14.19 14.28 14.46 14.62 14.78 15.05 15.30 15.52 15.73 16.10 16.42 16.71 16.97 17.41 17.78 18.10 18.61 18.99 19.29 19.70 19.96 20.24 20.28

For the flow over a horizontal bed with constant specific energy and discharge decreasing in the direction of flow, derive the shapes of the subcritical and supercritical profiles for a side discharge weir as shown in Figure 5.17.

Solution.

The shapes of the profiles depend on the energy form of the equation of spatially-varied flow, with the specific energy held constant, as given by Equation 5.44:

q LV dy gA = dx 1 − F 2 125

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in which q L = – dQ/dx; V = Q/A; A = cross-sectional area = by; and F = local value of the Froude number. For y > y c , F < 1 and the numerator is positive so that dy/dx > 0. The depth increases in the downstream direction for the subcritical case as shown in Figure 5.17. If, on the other hand, y < y c , we have a supercritical profile with F > 1 and dy/dx < 0. The depth decreases from critical depth, or from a known supercritical depth at the upstream end of the weir, in the downstream direction. If the tailwater is raised, then a supercritical profile followed by a hydraulic jump to the subcritical profile is possible.

5.17.

Derive the energy equation for spatially varied flow in the form of Equation 5.44, but do not assume that S 0 and S e , the bed slope and slope of the energy grade line, are equal to zero. Compare the result with Equation 5.40 and discuss.

Solution.

Defining the total energy head, H, and differentiating with respect to the flow direction, x, we have

H = z+ y+

Q2 2 gA2

dy Q dQ Q 2 B dy + 2 − dx gA dx gA3 dx qV S0 − Se + L dy gA = 2 dx 1− F

− Se = −S0 +

where S 0 = bed slope; S e = slope of the energy grade line; q L = – dQ/dx; and F = the energy definition of the Froude number. This equation appears to differ from Equation 5.40 only in the third term on the right in the numerator, i.e. the spatially-varied flow or lateral inflow/outflow term. The equation just given is for lateral outflow in contrast to Equation 5.40, which applies to lateral inflow at 90° to the streamwise direction, which accounts for the difference in algebraic signs of the spatially-varied flow terms in the two equations. Furthermore, the lateral inflow term for the energy equation just given should have the kinetic energy flux correction coefficient, α, present, whereas the momentum flux correction coefficient, β, should appear in this term in Equation 5.40. Likewise, the Froude numbers should be defined in terms of α for the energy equation and β for the momentum equation (Equation 5.40), which is a subtle but important distinction. In addition, there is a factor of 2 different in the spatially-varied flow terms in the two equations. Finally, the energy equation involves the slope of the energy grade line, while the momentum equation term includes the friction slope defined as τ 0 /γR. Theoretically, these are different quantities although we evaluate both using Manning's equation. For further discussion of these differences, refer to the reference by Yen(1973) cited in the textbook.

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5.18.

CHAPTER 5

A rectangular side discharge weir has a height of 0.35 m. It is located in a rectangular channel having a width of 0.7 m. If the downstream depth is 0.52 m for a discharge of 0.27 m3/s, how long should the weir be for a lateral discharge of 0.21 m3/s?

Solution.

At the downstream end of the weir, the specific energy can be calculated from

Q2 0.27 2 E = y+ = 0.52 + = 0.548 m 2 gA2 19.62 × (0.7 × 0.52) 2 The downstream velocity, V = 0.27/(0.7 × 0.52) = 0.74 m/s, and the Froude number is 0.74/(9.81 × 0.52)1/2 = 0.33, so a subcritical profile is expected. At the upstream end of the weir, the discharge is (0.27 + 0.21) = 0.48 m3/s, and the depth can be calculated assuming a constant specific energy from

E = 0.548 = y +

0.48 2 19.62 × 0.7 2 × y 2

By trial and error, y = 0.392 m. Then from Equation 5.47 with E = 0.548 m; P = 0.35 m; and C 1 = (2/3) C d = (2/3)(0.59) = 0.393 for a fully-contracted weir from Chapter 2, we have

LC1  2 E − 3P = b  E−P L=

y2

E−y E−y − 3 sin −1  y−P E − P y

1

0.548 − 0.52 0.548 − 0.52  0.7  − 3 sin −1  0.232 0.548 − 0.35  0.52 − 0.35 0.393  −

0.7  0.548 − 0.392 0.548 − 0.392  − 3 sin −1 0.232  0.393  0.392 − 0.35 0.548 − 0.35 

L = 1.78[(−1.06) − (−2.83)] = 3.2 m in which L = x 2 – x 1 ; y 2 = 0.52 m; and y 1 = 0.392 m. Hager (1999) has given an alternative expression for C d to account for the lateral outflow angle and approach velocity:

C d = 0.636

1− P / E 3− 2y / E − P/ E

which reduces to 0.636 as y/E approaches 1. For this exercise, the expression gives C d = 0.39 to 0.56 for P/E = 0.35/0.548 = 0.639, and for y/E varying from 0.71 – 0.95 as the depth varies from 0.39 – 0.52 m. If we take an average value of C d = 0.48 and C 1 = 0.32,

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the weir length, L = 3.9 m, which compares favorably with the value of 4.0 m given by Hager (1999) for this exercise for a continuously varying value of C 1 .

5.19

A concrete (n = 0.013) cooling tower collection channel is rectangular with a length of 45 ft in the flow direction and a width of 31 ft. The addition of flow from above in the form of a continuous stream of droplets is at the rate of 0.63 cfs/ft of length. Find the location of the critical section and compute the water surface profile if the slope is 0.021 ft/ft.

How deep should the collection channel be?

Solution.

The channel slope is 0.021 ft/ft, and the lateral inflow rate, q L = 0.63 cfs/ft of flow length.

Assume initially that critical depth occurs at the downstream end of the collection channel where q = Q/b = 0.63 × 45 / 31 = 0.914 cfs/ft. Then critical depth, y c , is (q2/g)1/3 = (0.9142/32.2)1/3 = 0.30 ft. Substitute into Equation 5.41 to obtain the next trial value of the location of the critical section:

xc =

xc =

8q L2 gP   gB 2  S 0 − 2  C B 

3

8 × 0.632 32.2 × (31 + 2 × 0.3)   32.2 × 31 × 0.021 −  93.8 2 × 31 

3

= 19.9 ft

2

where the Chezy C = (1.49/n)R1/6 = (1.49/0.013)(0.3)1/6 = 93.8. For the resulting value of x c , q = 0.63 × 19.9 / 31 = 0.404 cfs/ft, and y c = 0.172 ft. The next trial value of x c is 22.7 ft. Continuing in this fashion, the final result is y c = 0.185 ft and x c = 22.24 ft. Because the critical section is located upstream of the end of the collector channel, the flow is subcritical upstream of x c and supercritical downstream. The water surface profile is computed from Equation 5.40 beginning at the critical section and proceeding upstream for the subcritical profile and downstream for the supercritical profile. Equation 5.40 is solved by trial in finite difference form as given by

S0 − S f − ∆y =

1− F

2q L V gA 2

∆x

where the overbars indicate mean values from the beginning to the end of the spatial step.

Values of ∆x are fixed, and trial values of y are assumed until the calculated value of ∆y produces the assumed value of depth. These computations are summarized in the spreadsheets and graph given next. The downstream depth is 0.26 ft, which establishes the minimum depth of the collector channel.

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SPATIALLY-VARIED FLOW qL= 0.63 cfs/ft n= 0.013 b= 31 ft S0= 0.021

x, ft 22.2 22.0 21.0 20.0 18.0 16.0 14.0 12.0 10.0 8.0 6.0 4.0 2.0 0.0

x, ft 22.2 23.0 24.0 26.0 28.0 30.0 35.0 40.0 45.0

SUBCRITICAL PROFILE trial y, ft Q, cfs A, ft2 0.18500 0.18427 0.18012 0.17596 0.16729 0.15822 0.14863 0.13844 0.12748 0.11555 0.10229 0.08703 0.06828 0.03999

13.99 13.86 13.23 12.60 11.34 10.08 8.82 7.56 6.30 5.04 3.78 2.52 1.26 0.00

5.735 5.712 5.584 5.455 5.186 4.905 4.608 4.292 3.952 3.582 3.171 2.698 2.117 1.240

SUPERCRITICAL PROFILE trial y, ft Q, cfs A, ft2 0.18500 0.18830 0.19218 0.19989 0.20731 0.21451 0.23164 0.24775 0.26301

13.99 14.49 15.12 16.38 17.64 18.90 22.05 25.20 28.35

5.735 5.837 5.958 6.197 6.426 6.650 7.181 7.680 8.153

∆y, ft

y, ft

0.9992 0.9961 0.9839 0.9704 0.9421 0.9105 0.8750 0.8344 0.7868 0.7294 0.6569 0.5580 0.4015 0.0000

-0.00073 -0.00415 -0.00416 -0.00867 -0.00907 -0.00959 -0.01019 -0.01095 -0.01193 -0.01327 -0.01526 -0.01875 -0.02829

0.18500 0.18427 0.18012 0.17596 0.16729 0.15822 0.14863 0.13844 0.12748 0.11555 0.10229 0.08703 0.06828 0.03999

Sf

F

dy, ft

y, ft

4.363E-03 4.417E-03 4.494E-03 4.630E-03 4.758E-03 4.877E-03 5.146E-03 5.379E-03 5.586E-03

0.9992 1.0081 1.0202 1.0419 1.0624 1.0814 1.1244 1.1617 1.1949

0.003297 0.003885 0.007707 0.007417 0.007205 0.017124 0.016115 0.015256

0.18500 0.18830 0.19218 0.19989 0.20731 0.21451 0.23163 0.24775 0.26301

R, ft

V, ft/s

Sf

F

0.183 0.182 0.178 0.174 0.166 0.157 0.147 0.137 0.126 0.115 0.102 0.087 0.068 0.040

2.439 2.426 2.369 2.310 2.187 2.055 1.914 1.762 1.594 1.407 1.192 0.934 0.595 0.000

4.363E-03 4.342E-03 4.267E-03 4.182E-03 4.006E-03 3.808E-03 3.588E-03 3.338E-03 3.048E-03 2.704E-03 2.281E-03 1.735E-03 9.723E-04 0.000E+00

R, ft

V, ft/s

0.183 0.186 0.190 0.197 0.205 0.212 0.228 0.244 0.259

2.439 2.482 2.538 2.643 2.745 2.842 3.071 3.281 3.477

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Water Surface Profile 0.5

0.4

0.3

xc

y, ft

SUPERCRITICAL

0.2 SUBCRITICAL

0.1

0.0 0

5

10

15

20

25 x, ft

130

30

35

40

45

CHAPTER 6. Hydraulic Structures 6.1

A high overflow spillway with P/H d > 1.33 has a maximum discharge of 10,000 cfs with a maximum head of 20 ft. Determine the design head, spillway crest length, and the minimum pressure on the spillway. Plot the complete spillway crest shape for a compound circular curve in the upstream quadrant of the crest.

Solution: Set the design head equal to 0.75 times the maximum head to give

H d = 0.75H max = 0.75 × 20 = 15 ft Then from Figure 6.2, for P/H d ≥ 1.33 and H e /H d = 1.33, C/C 0 = 1.025 and C = 1.025 × 4.03 = 4.13. The spillway crest length is given by Equation 6.1 as

L=

Q 10,000 = = 27.1 ft 3/ 2 CH max 4.13 × 20 3 / 2

The minimum pressure head is obtained from Figure 6.3 to be –0.43 times the design head with the result

 p   = −0.43 × 15 = −6.45 ft  γ  min and the minimum pressure is –6.45 × 62.4/144 = –2.8 psi.

The coordinates of the key points establishing the tricompound circular curve for the upstream quadrant in Figure 6.4 are shown in the table and figure given next (COE HDC 111-2/1, 1970):

Point No.

X/H d

2

Y/H d 3

1 2 3 4 5

-0.105 -0.175 -0.276 -0.2418 -0.2818

0.219 0.0316 0.1153 0.136 0.136

5

4

X Y 1

0.5 Hd

A set of coordinates sufficient for plotting the upstream quadrant directly is given in the next table:

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Open Channel Hydraulics

CHAPTER 6

X/H d

Y/H d

0.00 -0.0500 -0.1000 -0.1500 -0.1750 -0.2000 -0.2200 -0.2400 -0.2600 -0.2760 -0.2780 -0.2800 -0.2818

0.00 0.0025 0.0101 0.0230 0.0316 0.0430 0.0553 0.0714 0.0926 0.1153 0.1190 0.1241 0.1360

With these coordinates for the upstream quadrant and Equation 6.3 for the downstream quadrant, the resulting spillway shape for an arbitrary crest elevation of 100 ft can be shown as follows:

Ogee Spillway Shape Tricompound Circular Curve Design Head = 15 ft 102 100 98

Elev., ft

96 94 92 90 88 86 84 82 -10.00

-5.00

0.00

5.00

10.00 X , ft

132

15.00

20.00

25.00

Open Channel Hydraulics

6.2

CHAPTER 6

Repeat Example 6.1 for an elliptical approach crest, using a design procedure that guarantees a minimum pressure head of −15 ft. Plot the head-discharge curve.

Solution.

From Example problem 6.1, the discharge, Q = 200,000 cfs at a maximum head of 64 ft, and the spillway height, P = 60 ft. Then using Figure 6.6 (a) with a minimum pressure head of –15 ft, a value of design head, H d , of 53 ft produces a value of H e /H d = 1.22.

The corresponding maximum head is 1.22 × 53 = 65 ft, which is close enough to the actual maximum head of 64 ft considering the graphical error. The value of P/H d becomes 60/53 = 1.13. Figure 6.5 (b) indicates a discharge coefficient, C = 4.05. The resulting spillway crest length is given by

L=

Q max 200,000 = = 96.4 ft 3/ 2 CH e 4.05 × 64 3 / 2

The shape of the downstream quadrant is determined from

X 1.85 = 2.01H d0.85Y = (2.01× 530.85 )Y = 58.73Y in which K sp = 2.01 from Figure 6.5 (a). Also from the same figure, A/H d = 0.269 and B/H d = 0.158 to define the major and minor axes of the upstream elliptical quadrant. It follows that for a design head of 53 ft, A = 14.3 ft and B = 8.4 ft, so the equation for the upstream elliptical quadrant is given by

X2 (8.4 − Y ) 2 + =1 14.32 8.4 2 The spillway shape is given in the following figure for an arbitrary crest elevation of 100 ft. The compound circular curve is shown for comparison with the elliptical curve in the upstream quadrant.

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CHAPTER 6

Ogee Spillway Shape Elliptical Upstream Quadrant Design Head = 53 ft 120 100

Elev., ft

80 60 40 20 0 -20.00

0.00

20.00

40.00

60.00

80.00

100.00

120.00

X , ft DS

6.3

UScirc

Elliptical

An ogee spillway has a crest height of 50 ft and a maximum head of 15 ft. A minimum pressure of –1.5 psi is allowed. The maximum discharge is 16,000 cfs. (a) Determine the crest length of the spillway assuming a compound circular curve for the upstream crest shape. What is the pressure at the crest for the maximum discharge? (b) If the spillway is designed as a stepped spillway, with each step 2 ft high by 1.5 ft long, what is the energy dissipation in feet of water at the maximum discharge?

Solution. (a) From Figure 6.3, the minimum pressure values are shown in the following table for the fixed values of H/H d in the figure. The design head is calculated as 15/(H/H d ).

H/H d 1.00 1.17 1.33 1.50

H d , ft 15.0 12.8 11.3 10.0

(p min /γ)/H d 0.0 –0.2 –0.43 –0.78

p min , psi 0.0 –1.11 –2.11 –3.38

By interpolation, a minimum pressure of –1.5 psi occurs at a design head of 12.2 ft. Then P/ H d = 50/12.2 = 4.1 and H e /H d = 15/12.2 = 1.23. From Figure 6.2, C/C 0 = 1.02 and C = 1.02 × 4.03 = 4.11. The spillway crest length is determined from Equation 6.1 with the result given by

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Open Channel Hydraulics

L=

CHAPTER 6

Q 16,000 = = 67 ft 3/ 2 CH e 4.11 × 153 / 2

Interpolating from Figure 6.3, the pressure at the crest where X = 0 is approximately equal to –0.19 × 12.2 = –2.32 ft (–1.0 psi) for H e /H d = 1.23. (b) The discharge per unit width, q, is equal to Q/L = 16,000/67 = 239 cfs/ft. Then the critical depth, y c , is equal to (q2/g)1/3 = (2392/32.2)1/3 = 12.11 ft. The steps have an aspect ratio, l/h, of 1.5/2 = 0.75 and y c /Nh = 12.11/(25 × 2) = 0.24. Figure 6.9 indicates a value of ∆H/H 0 = 0.22. This corresponds to a friction factor of 0.265 in Equation 6.9 due to the decrease in the friction factor because of aeration. This value of ∆H/H 0 is in good agreement with data from other investigators and Equation 6.9 for aerated flow as shown

by Chanson (1994b). The actual energy loss, ∆H, is 0.22 × (50+15) = 14.3 ft.

6.4

An existing ogee spillway with an elliptical crest has a crest height of 7.0 m and a crest length of 15.2 m. A minimum gage pressure of zero (atmospheric pressure) occurs at a head of 14.0 m. What maximum head and discharge would you recommend for this spillway?

Solution.

Assume a design head of 14.0 m (45.9 ft) since the minimum pressure is atmospheric for this head. Then from Figure 6.6 (a), the value of H e /H d = 1.25 for a minimum pressure head of –15 ft at the maximum spillway head and discharge. It follows that the maximum head is 1.25 × 14.0 = 17.5 m (57.4 ft). The value of the discharge coefficient, C, is 3.98 from Figure 6.5 (b) with P/H d = 7.0/14.0 = 0.5 and H e /H d = 1.25. The maximum discharge for a spillway crest length of 15.2 m (49.9 ft) becomes

Q max = CLH e3 / 2 = 3.98 × 49.9 × 57.4 3 / 2 = 86,400 cfs (2,450 m 3 /s) 6.5

A 0.91 m diameter corrugated metal pipe culvert (n = 0.024) has a length of 90 m and a slope of 0.0067. The entrance has a square edge in a headwall. At the design discharge of 1.2 m3/s, the tailwater is 0.45 m above the outlet invert. Determine the head on the culvert at the design discharge. Repeat the calculation for head if the culvert is concrete.

Solution.

First, assume inlet control with the inlet submerged (Type IC-1) and solve Equation 6.11 for the head assuming a discharge coefficient of 0.46 from Table 6-2 for HW/d = 1.5:

Q2 1.22 HW = = = 0.82 m 2 gCd2 Ao2 19.62 × 0.462 × 0.652 in which A o = π × 0.912/4 = 0.650 m2. At this head, the inlet is unsubmerged, so use Equation 6.12a for unsubmerged inlets in which K = 0.0078 and M = 2.0 for a corrugated metal pipe in a headwall from Table 6-3. This requires the value of minimum specific energy, which can be obtained from Figure 2.14 for Q/(g1/2d 5/2) = 1.2/(9.811/2× 0.915/2) =

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Open Channel Hydraulics

CHAPTER 6

0.48 for which E c /d is approximately equal to 1.05. Substituting into Equation 6.12a, we have M

HW Ec  Q  = + K  0.5  − 0.5S d d  Ad  2

  HW 1.2 × 35.3 = 1.05 + 0.0078 ×  − 0.5 × 0.0067 2 0.5  d  (0.650 × 3.281 ) × (0.91× 3.281)  HW = 1.14 × 0.91 = 1.04 m in which the quantity Q/(Ad 0.5) has been converted to English units to obtain the value 3.5, which is at the upper limit of applicability of Equation 6.12a.

Now consider full flow. The critical depth from Figure 2.13 with Q/(g1/2d 5/2) = 1.2/(9.811/2× 0.915/2) = 0.48 is 0.71 × 0.91 = 0.65 m. The normal depth is obtained from Figure 4.9 with nQ/(S1/2d 8/3) = 0.024 × 1.2/(0.00671/2 × 0.918/3) = 0.45. The resulting normal depth exceeds the pipe diameter. This means that the slope is mild, and the tailwater is below critical depth. Apply Equation 6.15 for full flow in which TW = (0.65+0.91)/2 = 0.78 m, K e = 0.5 from Table 6-5, and fL/4R is obtained from Equation 6.17 as

f

L 2 gn 2 L 19.62 × 0.024 2 × 90 = 2 4/3 = = 7.32 4R Kn R 1.02 × (0.91 / 4) 4 / 3

Sustituting into Equation 6.15, we have

HW = TW − S 0 L + (1 + K e + f

L Q2 ) 4 R 2 gA 2

HW = 0.78 − 0.0067 × 90 + (1 + 0.5 + 7.32) ×

1.2 2 = 1.71 m 19.62 × 0.65 2

Clearly, the full flow headwater is higher, so we would choose 1.71 m as the design headwater for the corrugated metal pipe.

If we change to a concrete pipe of the same diameter, the Manning's n value becomes 0.012. The inlet control headwater is virtually unchanged, but the value of fL/4R is now 1.83 and the outlet control headwater is given by

HW = 0.78 − 0.0067 × 90 + (1 + 0.5 + 1.83) ×

1.2 2 = 0.76 m 19.62 × 0.65 2

This implies that the concrete culvert does not flow full at the design discharge.

Furthermore, there is a change in normal depth. The value of nQ/(S1/2d 8/3) = 0.012 × 1.2/(0.00671/2 × 0.918/3) = 0.23, and Figure 4.9 gives a value of y 0 /d = 0.63, or 0.57 m.

Upon comparison with the critical depth of 0.65 m, it is apparent that the slope has changed from mild to steep, and so the culvert is certain to be in inlet control with the

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CHAPTER 6

low tailwater. Therefore, the headwater for the concrete culvert is given by Equation 6.12a with K = 0.0098 and M = 2 to be 2

  HW 1.2 × 35.3 = 1.05 + 0.0098 ×  − 0.5 × 0.0067 2 0.5  d  (0.650 × 3.281 ) × (0.91× 3.281)  HW = 1.17 × 0.91 = 1.06 m Alternatively, Equation 6.10a gives a value of HW = 1.09 m, which agrees well with the result from Equation 6.12a. By changing from a corrugated metal pipe (CMP) to a concrete pipe culvert, we have switched from outlet control to inlet control, and the design headwater has dropped to 1.06 m for the concrete culvert in comparison to 1.71 m for the CMP culvert.

6.6

Show that Equation 6.10a for a box culvert in inlet control with the entrance unsubmerged can be placed in a form in which Q is proportional to the head, HW, to the 3/2 power.

Solution.

The critical depth, y c , in Equation 6.10a can be expressed in terms of the discharge by the following expression derived for a rectangular channel (or box culvert) of width b: 1/ 3

 (Q / b) 2   yc =   g 

Q2 / 3 = 2 / 3 1/ 3 b g

Substituting into Equation 6.10a, we have

Q2 / 3 Q2 HW = 2 / 3 1 / 3 + (1 + K e ) b g 2 gb 2 [Q 4 / 3 /(b 4 / 3 g 2 / 3 )] Q2 / 3 Q2 / 3 HW = 2 / 3 1 / 3 + (1 + K e ) 2 / 3 1 / 3 b g 2b g 3 Ke Q 2 / 3 HW = ( + ) 2 2 b2 / 3 g 1 / 3 Solving for Q, the result is

 2  Q=   3 + Ke 

3/ 2

g b HW 3 / 2

which is effectively the broad-crested weir equation with a coefficient expressed in terms of the entrance loss coefficient.

137

Open Channel Hydraulics

6.7

CHAPTER 6

A 3-ft by 3-ft concrete (n = 0.012) box culvert has a slope of 0.006 and a length of 250 ft.

The entrance is a square edge in a headwall. Determine the head on the culvert for a discharge of 50 cfs and a discharge of 150 cfs. The downstream tailwater elevation is 0.5 ft above the outlet invert for 50 cfs and 3 ft above the outlet invert at 150 cfs.

Solution.

For the discharge of 50 cfs, the critical depth, y c = (q2/g)1/3 = [(50/3)2/32.2]1/3 = 2.05 ft, and the normal depth is obtained from

A5 / 3 nQ ( 3 y0 ) 5 / 3 0.012 × 50 = = = = 5.20 2/3 2/3 1/ 2 P Kn S ( 3 + 2 y0 ) 1.49 × 0.0061 / 2 with the result, y 0 = 1.94 ft. The slope is steep. Assume inlet control with the inlet unsubmerged and a loss coefficient of 0.5 in Equation 6.10a to yield

Q2 HW = yc + (1 + K e ) 2 gAc2 HW = 2.05 + 1.5 ×

50 2 = 3.59 ft 64.4 × (3 × 2.05) 2

The value of HW/d = 3.59/3.0 = 1.2, so the inlet is likely to be unsubmerged as assumed.

Because the tailwater depth of 0.5 ft is less than critical depth, and the slope is steep, outlet control cannot occur, so the headwater is 3.59 ft for Q = 50 cfs.

For a discharge of 150 cfs, the critical depth, y c = [(150/3)2/32.2]1/3 = 4.27 ft, which exceeds the height of the culvert. Likewise, normal depth exceeds the culvert height of 3 ft. Assuming inlet control with the inlet submerged, the result from Equation 6.13 with c = 0.04 and Y = 0.80 (see Table 6-3) is given by 2

HW  Q  = c  0.5  + Y − 0.5S d  Ad  2

HW  150  = 0.04 ×  + 0.80 − 0.5 × 0.006 = 4.50 0. 5 d  9 × 3  The result for the headwater is HW = 13.5 ft (Equation 6.11 gives HW = 12.7 ft with C d = 0.582). For outlet control, the value of fL/4R is 1.53 from Equation 6.17 and TW = 3.0 ft.

Equation 6.15 for full flow results in

L Q2 HW = TW − S0 L + (1 + K e + f ) 4 R 2 gA2 1502 HW = 3.0 − 0.006 × 250 + (1 + 0.5 + 1.53) × = 14.6 ft 64.4 × 9 2

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Open Channel Hydraulics

CHAPTER 6

The higher headwater is for outlet control with HW = 14.6 ft at 150 cfs.

6.8

Design a box culvert to carry a design discharge of 600 cfs. The culvert invert elevation is 100 ft and the allowable headwater elevation is 114 ft. The paved roadway is 500 ft long and overtops at 115 ft. The culvert length is 200 ft with a slope of 1.0 percent. The tailwater elevations are given below up to the maximum discharge of 1000 cfs: Q, cfs

TW, ft

200 400 600 800 1000

101.4 102.6 103.1 103.8 104.1

Prepare a performance curve for the culvert design by hand and compare with the results of HY8. Also use HY8 to prepare a performance curve if the slope is 0.1 percent.

Solution.

Assuming inlet control at a submerged entrance with a 45° bevel (w/b = 0.042), Equation 6.11 can be used together with Table 6-2 to size the culvert at the design discharge of 600 cfs. Try a 6 ft by 6 ft box culvert with an initial value of the discharge coefficient of 0.6 to yield

HW =

6002 Q2 = = 12.0 ft 2 g Cd2 Ao2 64.4 × 0.62 × 362

which gives a value of HW/d = 12/6 = 2. Then from Table 6-2 with w/b = 0.042, C d = 0.59. The next iteration gives HW/d = 12.4/6 = 2.07 and C d = 0.595. The final iteration produces HW/d = 12.2/6 = 2.03 and C d = 0.593, which is acceptable agreement. This is a satisfactory design because the headwater is less than the allowable headwater of 14 ft.

For comparison, Equation 6.13 with c = 0.0314 and Y = 0.82 from Table 6-3 gives HW = 13.6 ft, which is still less than the allowable headwater. However, we must also check outlet control. The actual tailwater depth is (103.1 – 98.0) = 5.1 ft, but y c = [(600/6)2/32.2]1/3 = 6.77 ft, so use a tailwater depth of 6 ft assuming that the outlet fills to the crown. The value of K e = 0.2 from Table 6-5; and fL/4R = 0.49 from Equation 6.17.

Substituting into Equation 6.15, we have

HW = TW − S0 L + (1 + K e + f

L Q2 ) 4 R 2 gA2

HW = 6.0 − 0.01 × 200 + (1 + 0.2 + 0.49)

6002 = 11.3 ft 64.4 × 362

Because the inlet control headwater is larger, we accept the value of 13.6 ft for a discharge of 600 cfs in the 6 ft by 6ft concrete box culvert with 45° bevels on the entrance.

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CHAPTER 6

Now calculate the performance curve. At 200 cfs, y c = [(200/6)2/32.2]1/3 = 3.26 ft, and from Figure 4.9, y 0 /d = 0.38 for nQ/(1.49S1/2b8/3) = (0.012)(200)/(1.49 × 0.011/2 × 68/3) = 0.136 so that normal depth is 0.38 × 6 = 2.3 ft. The slope is steep and the tailwater depth is (101.4–98.0) = 3.4 ft, which is only slightly greater than critical depth, so that inlet control can be expected. Assume that the inlet is unsubmerged, and Equation 6.10a gives

HW = yc + (1 + K e ) HW = 3.26 + 1.2 ×

Q2 2 gAc2

2002 = 5.2 ft 64.4 × (6 × 3.26) 2

while Equation 6.12b gives HW = 5.1 ft for K = 0.495 and M = 0.667 from Table 6-3.

The headwater value is less than 6 ft, so the inlet is indeed unsubmerged. Similar calculations for Q = 400 cfs also indicate inlet control on a steep slope, although the inlet is likely to be submerged. Equation 6.13 with c = 0.0314 and Y = 0.82 results in 2

HW  Q  = c  0.5  + Y − 0.5S d  Ad  2

HW  400  = 0.0314 ×  + 0.82 − 0.5 × 0.01 = 1.46 0.5 d  36 × 6  and HW = 1.46 × 6 =8.8 ft confirming submergence of the inlet. At 800 cfs the culvert remains in inlet control, which results in a headwater elevation that exceeds the roadway elevation, so iteration must be performed for both weir overflow and inlet control in the culvert until the sum of the two discharges equals 800 cfs. Road overflow is given by Equation 6.18 with the discharge coefficient approximately equal to 3.0 from Figure 6.13.

The overtopping discharge is given by

Qo = 3.0 L HWr3 / 2 = 3.0 × 500 × ( HW − 15)3 / 2 = 1500 × ( HW − 15)3 / 2 The iteration results are shown in the following table with the inlet control discharge, Q IC , computed from Equation 6.13.

HW ft 15.1 15.2 15.3 15.21

Q IC cfs 649 652 656 653

Qo cfs 47 134 246 144

Q TOT cfs 696 786 902 797

The result for Q = 800 cfs is HW = 15.21 ft. Similar calculations for Q = 1000 cfs result in HW = 15.37 ft. The results from HY8 are plotted in the following figure along with the results of the manual calculations. Then in the succeeding figure, HY8 results for a culvert slope of 0.1% are shown.

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CHAPTER 6

Culvert Performance Curve, S = 1% 6 ft by 6 ft concrete box; 45 deg. bevel 18 16 14

HW , ft

12 10 8 6 4 2 0 0

200

400

600

800

1000

1200

1000

1200

Q , cfs Inlet Control

Outlet Control

Manual Calcs.

Culvert Performance Curve, S = 0.1% 6 ft by 6 ft concrete box; 45 deg. bevel 18 16 14

HW , ft

12 10 8 6 4 2 0 0

200

400

600

800

Q , cfs Inlet Control

141

Outlet Control

Open Channel Hydraulics

6.9

CHAPTER 6

A circular concrete culvert has a diameter of 5 ft with a square-edged entrance in a headwall. The culvert is 500 ft long with a slope of 0.005 and an inlet invert elevation of 100.0 ft. The downstream channel is trapezoidal with a bottom width of 10 ft, side slopes of 2:1, slope=0.005, and Manning's n = 0.025. The paved roadway has a constant elevation of 130 ft with a length of 100 ft and a width of 50 ft. The design discharge is 250 cfs and the maximum discharge is 500 cfs. Use HY8 to construct and plot the performance curve for these data, and compare with the performance curve for a 5 ft diameter corrugated steel pipe. Also compare with the performance curve for the 5 ft diameter concrete culvert with a side-tapered inlet.

Solution.

Data entry for HY8 requires discharge, culvert, tailwater, and roadway data. The design and maximum discharge are entered and the full range of discharges is divided into 10 intervals for calculation of the performance curve. Culvert data include the inlet and outlet stations and elevations; culvert shape, diameter, and material; and the type of culvert inlet. Tailwater rating curves can be entered directly, or the program can calculate rating curves using data on the shape, slope, and roughness of the downstream channel. Finally, roadway information includes the crest length for overtopping, the overtopping elevation, the type of roadway surface, and the width of the roadway in the flow direction.

The HY8 (v6.1) performance curve output for the reinforced concrete pipe with a diameter of 5 ft is shown in the following table (the tailwater velocity has been eliminated for brevity). The flow-type column gives the USGS flow type followed by the type of flow profile and outlet depth (n = normal; c = critical; t = tailwater; f = full) if the culvert is flowing partly full over any portion of the length, or by the symbol FF if the culvert is flowing full.

PERFORMANCE CURVE FOR CULVERT 1 - 1(

5.00 (ft) BY

5.00 (ft)) RCP

DISCHARGE FLOW (cfs)

HEAD- INLET OUTLET WATER CONTROL CONTROL FLOW ELEV. DEPTH DEPTH TYPE (ft) (ft) (ft) <F4>

NORMAL DEPTH (ft)

CRIT. OUTLET DEPTH DEPTH (ft) (ft)

TW OUTLET DEPTH VEL. (ft) (fps)

0.00 50.00 100.00 150.00 200.00 250.00 300.00 350.00 400.00 450.00 474.65

100.00 0.00 102.79 2.79 104.29 4.29 105.69 5.69 107.44 7.44 109.73 9.73 113.48 12.59 117.55 15.97 122.16 19.82 127.38 24.20 130.18 26.62

0.00 1.69 2.50 3.24 4.13 5.00 5.00 5.00 5.00 5.00 5.00

0.00 1.98 2.84 3.51 4.03 4.40 4.78 5.00 5.00 5.00 5.00

0.00 1.07 1.58 1.98 2.32 2.61 2.87 3.11 3.34 3.55 3.74

0.00 2.79 4.29 5.69 7.24 9.62 13.48 17.55 22.16 27.38 30.18

El. inlet face invert El. inlet throat invert

0-NF 1-S2n 1-S2n 5-S2n 2-M2c 2-M2c 2-M2c 6-FFc 6-FFc 6-FFc 6-FFc

100.00 ft 0.00 ft

142

0.00 1.58 2.44 3.24 4.03 4.40 4.78 5.00 5.00 5.00 5.00

El. outlet invert El. inlet crest

0.00 9.38 10.52 11.16 11.79 13.69 15.64 17.83 20.37 22.92 24.17

97.50 ft 0.00 ft

Open Channel Hydraulics

CHAPTER 6

For flows up to 150 cfs, the slope is steep, and inlet control prevails (outlet control is not computed for these cases). At 200 through 300 cfs, the channel slope is mild with an M2 profile being computed starting from critical depth, which is higher than the tailwater depth, at the outlet. For outlet control, the inlet submerges for 250 and 300 cfs, but inlet control gives a slightly higher headwater than outlet control at 200 and 250 cfs.

Beginning with a flow of 350 cfs, the culvert flows full for all succeeding discharges.

Note that the final culvert flow is 475 cfs with a headwater elevation of 130.18. This means that slight road overtopping is occurring at the rate of 25 cfs. The performance curve for the 5 ft diameter reinforced concrete pipe (RCP) is shown in the following figure.

Culvert Performance Curve, S = 0.5% 5 ft dia. RCP 35 30

HW , ft

25 20 15 10 5 0 0

100

200

300

400

500

Q , cfs Inlet Control

Outlet Control

For comparison, HY8 can also be executed for a 5 ft diameter corrugated steel pipe (CSP), and for the same RCP in the figure just shown but with a side-tapered inlet instead of a conventional square-edged inlet with a headwall. All three cases are shown in the following figure. The change to a CSP forces outlet control at all discharges because of the greater wall roughness. Consequently, the larger headwater values result in roadway overtopping beginning at a discharge of 300 cfs. The improved inlet on the RCP has a side taper of 4:1, a face height of 5.1 ft, and a face width of 8.0 ft. The face width is determined by trial and error until inlet control is at the throat rather than the face of the inlet, i.e. the face control elevation is below the throat control elevation. The improved inlet lowers the headwater at all discharges in comparison to the conventional inlet, but the culvert remains in outlet control for discharges greater than 150 cfs. As a result, the reduction in the headwater is not as great as it might have been if the culvert were in inlet control at the higher discharges. Improved inlets are best suited to culverts that are in inlet control.

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Culvert Performance Curve Comparisons S = 0.5% 35 30

HW , ft

25 20 15 10 5 0 0

100

200

300

400

500

600

Q , cfs 5 ft RCP Inlet Control 5 ft CSP Outlet Control

6.10

5 ft RCP Outlet Control 5 ft RCP; SDT Inlet

Prove that α 3 = 1/C 2 in the WSPRO methodology where α 3 = kinetic energy flux correction coefficient at section 3 and C = USGS bridge discharge coefficient.

Solution.

Equation 6.30 in the WSPRO methodology can be rearranged and solved for h v3 to obtain

V32 hv 3 = α 3 = h1 − h3 − h f (1−2 ) − h f ( 2−3) + hv1 2g The corresponding energy equation in the USGS width contraction method is Equation 6.28, which is rearranged to yield

V12 Q2 = ∆h − h f (1−3) + α 1 2g 2 gA32 C 2 Now noting the following equivalent terms by definition:

∆h = h1 − h3 h f (1−3) = h f (1−2 ) + h f ( 2−3) V12 hv 1 = α 1 2g it is apparent that α 3 = 1/C 2.

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6.11

CHAPTER 6

Apply the HDS-1 method to the data given in Example 6.3, and compare the backwater to that obtained from WSPRO. Solution.

One of the most important variables needed is the discharge ratio M 0 = Q b /Q in which Q b = the discharge in the approach section for a width equal to the bridge opening width and Q = the total discharge in the approach section, both for the normal water surface elevation. The discharge ratio can be calculated as the corresponding conveyance ratio, which can be obtained by running WSPRO again using the HP record written as HP ihp, secid, elmin, yinc, elmax, q in which ihp = 2 for a table of the cross-section properties of the subareas and ihp = 3 for a table of the velocity and conveyance distributions; secid = APPR for the approach section; elmin = 28.246 for the desired water surface elevation from Table 6-12; yinc = 0 for the elevation increment; and elmax = 28.246 for the maximum water surface elevation since properties are sought at only one elevation. The value of q is the total discharge, which is needed for the computation of the velocity distribution when ihp = 3.

The table given next shows the output resulting from the HP record with ihp = 3. The cross section has been divided into 20 equal conveyance subsections. By interpolation, the bridge section between stations 230 and 430 ft occupies 12.64 conveyance subsections so that M 0 = (12.64/20) = 0.63.

EXAMPLE 6-3. - NORMAL BRIDGE CROSSING BED SLOPE = 0.00052; LSEL = 32.0 FT; NO OVERTOPPING BRIDGE OPENING (TYPE I): X = 230 TO 430 FT;3 PIERS ***

Beginning Velocity Distribution For Header Record APPR SRD Location: 1440.000 Header Record Number 4

***

Water Surface Elevation: 28.246 Element # 1 Flow: 20000.000 Velocity: 3.62 Hydraulic Depth: 7.561 Cross-Section Area: 5518.58 Conveyance: 878235.30 Bank Stations -> Left: .100 Right: 730.000 X STA. A( I ) V( I ) D( I )

.1

195.9 621.6 1.61 3.18

271.4 537.0 1.86 7.11

316.2 371.5 2.69 8.29

329.1 189.8 5.27 14.72

337.8 162.5 6.15 18.81

X STA. A( I ) V( I ) D( I )

337.8

344.9 146.9 6.81 20.66

351.4 144.4 6.92 22.25

357.2 137.2 7.29 23.68

362.7 136.9 7.30 24.68

368.6 139.4 7.17 23.60

X STA. A( I ) V( I ) D( I )

368.6

375.2 145.6 6.87 22.04

383.2 159.9 6.25 20.06

396.1 195.6 5.11 15.21

426.2 271.5 3.68 9.01

466.9 330.3 3.03 8.12

X STA. A( I ) V( I ) D( I )

466.9

494.4 295.4 3.38 10.74

524.2 309.8 3.23 10.40

556.5 320.5 3.12 9.93

598.7 370.2 2.70 8.76

730.0 532.6 1.88 4.06

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From Figure 6.21 with M 0 = 0.63, the bridge contraction coefficient, K b = 0.83 for 90° wingwalls up to 200 ft in length. Next we can calculate the area in the bridge opening for the normal water surface elevation from the WSPRO output given in Table 6-12. The flow area in the bridge for a water surface elevation of 27.435 is 2398.5 ft2. Add to this the quantity 200 × (28.117 – 27.435) to obtain A n2 = 2535 ft2 for the normal water surface elevation in the bridge of 28.117 ft. The pier coefficient depends on the area obstructed by the piers, A p , at the normal water surface elevation given by

Ap = [( 28.12 − 8.0) + ( 28.12 − 10) + ( 28.12 − 20.4)] × 3.0 = 138 ft 2 The pier obstruction ratio, J = A p /A n2 = 138/2535 = 0.0544. Figure 6.22 gives σ = 0.85 and ∆K = 0.09 for continuous round nosed piers so that ∆K p = ∆K × σ = 0.09 × 0.85 = 0.08. Then K* = 0.83 + 0.08 = 0.91. The value of the reference velocity, V n2 = Q/A n2 = 20,000/2535 = 7.89 ft/s. The first iteration of Equation 6.24 gives

h1* = K *

α 2Vn22 2g

= 0.91×

7.89 2 = 0.88 ft 64.4

with α 2 = 1.0. In the second iteration, we need A 4 = 5512 ft2 from the WSPRO output in Table 6-12; A 1 approximately equal to 6338 ft2 and α 1 approximately equal to 1.64, also obtained from Table 6-12. The result is

 A  2  A  2  V 2 + α1  n 2  −  n 2   n 2 h =K 2g  A4   A1   2 g  2535  2  2535  2  7.89 2 * = 0.96 ft h1 = 0.88 + 1.64 ×     −  5512   6338   64.4 * 1

*

α 2Vn22

Because additional changes in A 1 and α 1 will be very small, the final value is taken to be h 1 * = 0.96 ft, which is in fairly good agreement with the WSPRO result of 1.12 ft although it is smaller.

6.12

Using the USGS width-contraction curves in Figures 6.23 and 6.24, verify the value of the bridge discharge coefficient and the discharge ratio m (=M(K)) given in the WSPRO output for Example 6.3.

Solution.

In the WSPRO methodology, the discharge contraction ratio, M 1 = (1 − m), is defined by

M1 =

Kq KT

in which K q = conveyance of the segment of approach flow that is unobstructed by the bridge at the approach water surface elevation; and K T = total conveyance of the

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approach flow at the approach water surface elevation. These conveyances are given in the WSPRO output in Table 6-12 as KQ = 636,363 and K = 1,058,543 for the approach section at the water surface elevation of 29.369 ft so that M 1 = 636,363/1,058,543 = 0.60 and m = 1 − M 1 = 0.40 in agreement with the WSPRO output for M(K) with rounding.

Then from Figure 6.23(a) with L/b = 40/200 = 0.20, in which L = bridge width and b = bridge opening length, and m = 0.40, the uncorrected bridge discharge coefficient, C' = 0.78. The Froude number at section 3, which is the bridge section, is required for the Froude number correction to the bridge discharge coefficient, k F . The Froude number is defined in terms of the average depth, which can be determined from the WSPRO output in Table 6-12 as y 3 = A 3 /b = 2398.5/200 = 12.0 ft. Then the Froude number is (Q/A 3 )/(gy 3 )0.5 = (20,000/2398.5)/(32.2 × 12.0)0.5 = 0.424 in agreement with the WSPRO output in Table 6-12, and k F from Figure 6.23(b) is approximately 0.98. The corner rounding adjustment factor from Figure 6.23(c) is 1.0 for square abutments. The pier adjustment factor, k j , is obtained from Figure 6.24 with j = A j /A 3 = 0.055, which is given as P/A in the WSPRO output in Table 6-12. The result is k j = 0.975. The corrected bridge discharge coefficient, C = 0.98 × 0.975 × 0.78 = 0.745, in agreement with the WSPRO output in Table 6-12.

6.13

Change the bridge type to Type 3 for the WSPRO example (Example 6.3), and determine the backwater for Q values ranging from 5,000 to 25,000 cfs. Plot the results in a graph comparing the Type 1 and Type 3 bridges in this range of discharges for a bridge length of 200 ft.

Solution.

The Type 3 bridge as indicated in Table 6-8 has sloping embankments and sloping abutments. Assume a 2:1 slope for both. The most important change in the input data records is in the CD record which includes the following data for a Type 3 bridge: * CD

bridge type, bridge width, embankment side slope, embankment elevation 3, 40, 2, 35.0

In addition, the AB record must be included to indicate the abutment side slope: * AB

abutment side slope 2

The bridge length record must be modified slightly because the bridge length of 200 ft is measured from the tops of the two abutments rather than the toes of the abutments. The left and right abutment X-stations (XLAB and XRAB) are located nominally at the toes of the abutments, but the bridge will be centered with respect to the midpoint between stations provided. The stations and bridge length cannot be specified such that the toes of the abutments fall inside the interval between the stations. Hence, the stations are specified as 300 ft and 360 ft to maintain the same bridge opening location with its centerline at a station of 330 ft. The WSPRO results are given next for Q = 20,000 cfs.

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EXAMPLE 6-3. - NORMAL BRIDGE CROSSING BED SLOPE = 0.00052; LSEL = 32.0 FT; NO OVERTOPPING BRIDGE OPENING (TYPE 3): CL = 330 FT;3 PIERS

Section: EXIT Header Type: XS SRD: 1000.000 Section: FULV Header Type: FV SRD: 1200.000

WSEL VHD Q AREA SRDL LEW EGEL HF V K FLEN REW CRWS HO FR # SF ALPHA ERR --------- ------ ---------- ---------- --------- --------28.008 .350 20000.000 5512.029 ********* .100 28.358 ****** 3.628 876862.20 ********* 730.000 21.895 ****** .304 ****** 1.708 ****** 28.117 28.466 21.999

.349 .104 .000

20000.000 3.626 .304

5515.302 877547.80 .0005

200.000 200.000 1.708

.100 730.000 .004

<<< The Preceding Data Reflect The "Unconstricted" Profile >>> Section: APPR Header Type: AS SRD: 1440.000

28.246 28.595 22.124

.349 .125 .000

20000.000 3.624 .304

5518.524 878223.40 .0005

240.000 240.000 1.708

.100 730.000 .004

<<< The Preceding Data Reflect The "Unconstricted" Profile >>> <<< The Following Data Reflect The "Constricted" Profile >>> <<< Beginning Bridge/Culvert Hydraulic Computations >>>

Section: BRGE Header Type: BR SRD: 1200.000

WSEL VHD Q AREA SRDL LEW EGEL HF V K FLEN REW CRWS HO FR # SF ALPHA ERR --------- ------ ---------- ---------- --------- --------27.382 2.182 20000.000 2155.826 200.000 239.352 29.564 .210 9.277 463123.00 200.000 420.748 22.418 .995 .606 ****** 1.630 -.001

Specific Bridge Information C P/A PFELEV BLEN XLAB XRAB Bridge Type 3 Flow Type 1 ------ ----- -------- -------- -------- -------Pier/Pile Code 0 .7833 .061 32.000 200.000 253.100 402.700 --------------------------- ------ ----- -------- -------- -------- --------

Section: APPR Header Type: AS SRD: 1440.000

WSEL VHD Q AREA SRDL LEW EGEL HF V K FLEN REW CRWS HO FR # SF ALPHA ERR --------- ------ ---------- ---------- --------- --------29.618 .238 20000.000 6519.709 200.000 .080 29.856 .186 3.068 1100766.00 221.968 730.020 22.124 .107 .231 .0005 1.628 .005

Approach Section APPR Flow Contraction Information M( G ) M( K ) KQ XLKQ XRKQ OTEL -------- -------- --------- -------- -------- -------.747 .436 620395.7 260.992 426.593 29.618

-------- -------- --------- -------- -------- --------- -------- -------- -------- -------The result of this comparison between the two bridge types is that the approach water surface elevation is higher for the Type 3 bridge with a value of 29.618 ft compared to 29.369 ft for the Type 1 bridge. The reason for this is that maintaining the same bridge length at the tops of the abutments actually reduces the bridge opening area due to the sloping abutments and causes a higher backwater.

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A comparison of the backwater for the Type 1 and Type 3 bridges over the discharge range from 5,000 cfs to 25,000 cfs is given in the following figure. The backwater is found at the approach section as the difference between the approach water surface elevation for the constricted water surface profile (with the bridge) and the unconstricted water surface profile (without the bridge). The low steel elevation (LSEL), or low chord elevation, is the same for both bridges and the flow type is Class 1, free surface flow as given in Table 6-7, in both cases for all discharges.

Comparison of Type 3 and Type 1 Bridges LSEL = 32 ft; BL = 200 ft; No Overtopping 2.5

Backwater, ft

2.0

1.5

1.0

0.5

0.0 0

5000

10000

15000

20000

25000

Discharge, cfs Example 6.3 (Type 1 Bridge)

6.14

Type 3 Bridge

For a bridge length of 200 ft and a Type 1 bridge, change the low chord elevation in the WSPRO example to 28 ft with a constant roadway elevation of 31 ft. Allow overtopping to occur and determine the backwater for the same range of discharges as in Exercise 6.13. Plot the results in comparison with Example 6.3. Note: The XR header record is required to locate the centerline of the roadway followed by GR records to give the roadway profile for overtopping analysis.

Solution.

The low chord elevation, or low steel elevation (LSEL), is changed to 28 ft in the BC record as shown by BC

28.0

At the end of the bridge records, the roadway section is added as follows: * XR GR

centerline of roadway 1220 0., 31.0 730., 31.0

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in which the GR record specifies the roadway profile for overtopping analysis. Without the XR and GR records, no overtopping will be considered. The results for the backwater for the discharge range from 5,000 to 25,000 cfs is shown in the following figure. There is no change in backwater for Q ≤ 15,000 cfs. At Q = 20,000 cfs, the flow type is Class 3, submerged orifice flow, while at 25,000 cfs it becomes Class 6, submerged orifice flow with overtopping as defined in Table 6-7. For both these discharges, the backwater increases in comparison to Example 6.3. There is no overtopping at 20,000 cfs, but the change to submerged orifice flow increases the backwater. At 25,000 cfs, there is an increase in backwater for the same reason, but there is also overtopping of the roadway at the rate of 2153 cfs which limits the backwater increase in comparison to the case of no overtopping.

Comparison of Type 1 Bridges BL = 200 ft; With and Without Overtopping 2.5

Backwater, ft

2.0

1.5

1.0

0.5

0.0 0

5000

10000

15000

20000

25000

Discharge, cfs Example 6.3 (LSEL = 32.0 ft)

6.15

LSEL = 28.0 ft

For Example 6.3, reduce the bridge length to 150 ft (with two bridge piers), and introduce a relief bridge with a length of 50 ft at a location of your choice in one of the floodplains.

Plot the results for backwater over the same range of discharges as in Exercise 6.13 in comparison with the results from Example 6.3.

Solution.

First, the case of shortening the main bridge length to 150 ft is considered without the addition of a relief bridge. This requires changing the full valley and bridge section reference distances to 1150 ft to accommodate the new bridge opening length. In addition, the section reference distance for the approach section becomes 1150 + 40 + 150 = 1340 ft, where the bridge width is 40 ft. The BL record is changed to BL 0

150. 280. 430.

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and the PD record is shortened to include only the first two bridge piers. The relief bridge is added after the main bridge, and it is located on the right hand side looking downstream from station 550 to 600 ft with a length of 50 ft for a total bridge opening length of 200 ft as in Example 6.3. The relief bridge requires BR, BL, BC, and CD records similar to those for the main bridge. No piers are included for the relief bridge.

The WSPRO output for the relief bridge for the case of Q = 20,000 cfs is given next to illustrate the iteration process and the order of the output results. Stagnation points are chosen initially and an initial flow distribution between the two bridges is calculated.

Then the water surface profile is calculated for the main bridge (BRGE) followed by that for the relief bridge (ROBR). The relief bridge approach section is located at the section reference distance of 1240 ft because of its shorter opening length. The water surface profile calculation for the relief bridge is carried upstream to the main bridge approach section where a conveyance weighted water surface elevation is computed based on the results for the two bridges. This procedure is iterated until the change in conveyanceweighted water surface elevation at the approach section is small. The final discharge in the relief bridge is 3314 cfs, which is approximately 17 percent of the total Q, and the final approach water surface elevation is 29.048 ft.

EXAMPLE 6-3. - NORMAL BRIDGE CROSSING BED SLOPE = 0.00052; LSEL = 32.0 FT; NO OVERTOPPING BRIDGE OPENING (TYPE I): X = 280 TO 430 FT; + RELIEF BR. Multiple Bridge Opening Calculations Iteration 2 -------------------------------------------------------------Stagnation Point ********* Bridge 1 A3 BOLEW CA3 CDF CDFn Section BRGE QS BOREW CJ CRF QSn ------------- ---------- --------- --------- --------------AS 4835.950 2065.234 280.040 1611.700 .984 .985 KS 1035898.0 16605.81 430.058 .780 2.339 16685.48 Stagnation Point 529.969 Bridge 2 A3 BOLEW CA3 CDF CDFn Section ROBR QS BOREW CJ CRF QSn ------------- ---------- --------- --------- --------------AS 1346.254 414.743 550.033 349.955 1.089 1.084 KS 187036.3 3394.19 600.063 .844 2.140 3314.52 Stagnation Point ********* Total Values A3 BOREW CA3 CDF CDFn All Sections QS BOLEW CJ CRF QSn ------------- ---------- --------- --------- ----- ---------AS 6182.205 ********** ********* 1961.655 ***** ********** KS 1222934.0 ********** ********* ********* ***** ********** --------------------------------------------------------------

Section: BRGE Header Type: BR SRD: 1150.000

WSEL VHD Q AREA SRDL LEW EGEL HF V K FLEN REW CRWS HO FR # SF ALPHA ERR --------- ------ ---------- ---------- --------- --------27.438 1.683 16685.480 2064.730 150.000 280.040 29.121 .121 8.081 486360.50 150.000 430.058 21.183 .627 .384 ****** 1.657 -.001

Specific Bridge Information C P/A PFELEV BLEN XLAB XRAB Bridge Type 1 Flow Type 1 ------ ----- -------- -------- -------- -------Pier/Pile Code 0 .7768 .054 32.000 150.000 280.000 430.000

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Section: SLICE Header Type: AS SRD: 1340.000

WSEL VHD Q AREA SRDL LEW EGEL HF V K FLEN REW CRWS HO FR # SF ALPHA ERR --------- ------ ---------- ---------- --------- --------28.991 .300 16685.480 4776.306 150.000 .088 29.291 .117 3.493 847028.00 163.827 529.969 20.970 .053 .258 .0005 1.580 .003

Approach Section SLICE Flow Contraction Information M( G ) M( K ) KQ XLKQ XRKQ OTEL -------- -------- --------- -------- -------- -------.717 .332 565334.1 283.163 433.182 28.991 -------- -------- --------- -------- -------- -------<<< End of Bridge Hydraulics Computations >>>

Section: ROBR Header Type: BR SRD: 1150.000

WSEL VHD Q AREA SRDL LEW EGEL HF V K FLEN REW CRWS HO FR # SF ALPHA ERR --------- ------ ---------- ---------- --------- --------27.558 1.407 3314.520 415.368 50.000 550.033 28.965 .088 7.980 46495.56 50.000 600.063 24.402 .689 .488 ****** 1.421 -.004

Specific Bridge Information C P/A PFELEV BLEN XLAB XRAB Bridge Type 1 Flow Type 1 ------ ----- -------- -------- -------- -------Pier/Pile Code ** .8389 .000 32.000 50.000 550.000 600.000 --------------------------- ------ ----- -------- -------- -------- --------

Section: SLICE Header Type: AS SRD: 1240.000

WSEL VHD Q AREA SRDL LEW EGEL HF V K FLEN REW CRWS HO FR # SF ALPHA ERR --------- ------ ---------- ---------- --------- --------29.284 .088 3314.520 1392.801 150.000 529.969 29.372 .206 2.380 180863.50 157.052 730.017 22.928 .203 .159 .0005 1.000 .013

Approach Section SLICE Flow Contraction Information M( G ) M( K ) KQ XLKQ XRKQ OTEL -------- -------- --------- -------- -------- -------.750 .549 81228.1 553.253 603.283 29.284 -------- -------- --------- -------- -------- -------<<< End of Bridge Hydraulics Computations >>> Section: APPR Header Type: XS SRD: 1340.000

29.320 29.408 22.980

.088 .034 .000

Section: APPR Header Type: XS SRD: 1340.000

29.048 .273 29.322 ****** 22.072 ******

3314.520 2.385 .160

1389.535 180167.00 .0003

100.000 100.000 1.000

529.969 730.016 .002

20000.000 6142.152 ********* 3.256 1013924.00 ********* .255 ****** 1.657

.088 730.012 .004

The results for bridge backwater are summarized in the following figure. The shorter bridge length of 150 ft has higher backwater than for the 200 ft bridge of Example 6.3, but the shorter bridge with a relief bridge added has lower backwater values than for Example 6.3 even though the total bridge length is the same. In this case, the relief bridge allows for less overall flow contraction and thus less backwater.

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Comparison of Type 1 Bridges Effect of Bridge Length (BL); No Overtopping 2.5

Backwater, ft

2.0

1.5

1.0

0.5

0.0 0

5000

10000

15000

20000

25000

Discharge, cfs Example 6.3 (BL = 200 ft)

6.16

BL = 150 ft

BL = 150 ft + 50 ft Relief BR

Analyze the existing bridge over Duck Creek using WSPRO or HEC-RAS (WSPRO option) and the following table of cross-section data. Use the fixed geometry mode for the bridge section. The bridge is Type 4 with a width of 30 ft, embankment side slopes of 2:1, embankment elevation of 790 ft, and wingwall angle of 30°. The low chord elevation is 788 ft. The design discharge is 6850 cfs with a water surface elevation of 784.66 in the exit cross section. The full valley section should be identical to the bridge section over the bridge opening width and is essentially the same as the exit section in both floodplains. (a) Determine the backwater for the existing bridge. (b) Design a new bridge to replace the old one such that the backwater is < 0.25 ft.

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Duck Creek cross sections Exit, Station 1000 Point

Distance

Elevation

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

–150 –105 –70 –28 –24 –20 18 22 35 50 210 600 860 1005 1050 1112 1260 1310

792 780 780 778 774 773 772 772 780 780 779 779 780 782 784 786 788 798

Subsection

X Subsection Manning's n

1 2 3 4 5 6

–28 35 210 600 860 1310

0.08 0.04 0.08 0.05 0.08 0.05

Bridge, Station 1100 Point

Distance

Elevation

1 2 3 4 5 6 7

–71 –71 –30 –15 16 25 25

788 778.2 776 772.5 772.5 775 788

Subsection

X Subsection Manning's n

1

25

0.04

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Approach, Station 1230 Point

Distance

Elevation

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

–480 –440 –420 –305 –175 –95 –50 –30 –25 2 17 20 28 50 670 990 1070 1120 1260

796 788 786 784 782 780 778 776 774 772 772 774 780 780 780 782 784 786 810

Subsection

X Subsection Manning's n

1 2 3 4

–50 28 670 1260

0.10 0.04 0.08 0.05

Solution.

First, analyze the existing bridge using WSPRO with the fixed geometry option for the bridge. The data file is shown next. *F SI 0 T1 Exercise 6.16 Duck Creek Existing Bridge T2 Q = 6850 cfs; Type 4; LSEL = 788 ft; Embankment ss = 2:1; T3 Embankment elevation = 790 ft; Width = 30 ft; WW = 30 deg. * * Discharge Q 6850. * * Known WS elevation WS 784.66 * * Section reference distance, skew (0), ek (0.5), ck (0.0) XS EXIT 1000.

GR -150,792 -105,780 -70,780 -28,778 -24,774 -20,773 18,772 GR 22,772 35,780 50,780 210,779 600,779 860,780 1005,782 GR 1050,784 1112,786 1260,788 1310,798 *

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* N SA * XS GR GR GR * * BR GR GR * N * CD * * KD * XS GR GR GR GR * N SA * EX * ER

CHAPTER 6

Subsection n values and subsection break points 0.08 0.04 0.08 0.05 0.08 0.05 -28.

35 210 600 860 FULV 1100.

-150,792 25,775 1050,784

-105,780 -71,778.2 -30,776 -15,772.5 16,772.5 35,780 50,780 210,779 600,779 860,780 1005,782 1112,786 1260,788 1310,798

Section reference distance, LSEL for fixed geometry mode BRGE 1100. 788.

-71,788 -71,778.2 -30,776 -15,772.5 16,772.5 25,775 25,788 -71,788 0.04 bridge type, bridge width, ss, embankment elev., wingwall angle 4 30.0 2 790 30 Place Kq section so that main channel is inside it * * * -71 25 APPR 1230.

-480,796 -50,778 28,780 1260,810

-440,788 -30,776 50,780

-420,786 -25,774 670,780

-305,784 2,772 990,782

0.10

0.04

0.08

0.05

-50

28

-175,782 17,772 1070,784

-95,780 20,774 1120,786

670

The specified discharge of 6850 cfs is given in the Q record, but the boundary condition is a known water surface elevation specified by the WS record in contrast to the SK record used in Example 6.3. The actual geometry of the bridge section is entered in GR records following the BR record in which the low chord or low steel elevation (LSEL) of 788 ft must be entered in fixed-geometry mode. The bridge geometry must be "closed" by ending the GR records with the starting point at the low chord of the left abutment.

The CD record provides additional information for the Type 4 bridge (sloping embankment, vertical abutment) including the embankment side slope, the elevation of the top of the embankment, and the wingwall angle). The KD record is entered specifically to force the Kq section to correspond to the bridge opening extended upstream to the approach section. Otherwise, the program centers the Kq section at the center of conveyance for the approach section which results in the Kq section not including the main channel in this asymmetric cross section. The WSPRO results are given in the following table.

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************************* W S P R O *************************** Federal Highway Administration - U. S. Geological Survey Model for Water-Surface Profile Computations.

Input Units: English / Output Units: English *---------------------------------------------------------------* EXERCISE 6.16 DUCK CREEK EXISTING BRIDGE Q = 6850 CFS; TYPE 4; LSEL = 788 FT; EMBANKMENT SS = 2:1; EMBANKMENT ELEVATION = 790 FT; WIDTH = 30 FT; WW = 30 DEG.

Section: EXIT Header Type: XS SRD: 1000.000

WSEL VHD Q AREA SRDL LEW EGEL HF V K FLEN REW CRWS HO FR # SF ALPHA ERR --------- ------ ---------- ---------- --------- --------784.660 .034 6850.000 6202.737 ********* -122.475 784.694 ****** 1.104 517613.00 ********* 1070.459 780.685 ****** .114 ****** 1.781 ******

Section: FULV Header Type: FV SRD: 1100.000

784.679 784.711 780.614

.032 .017 .000

6850.000 1.084 .110

6321.650 528131.70 .0002

100.000 100.000 1.752

-122.545 1071.042 .000

<<< The Preceding Data Reflect The "Unconstricted" Profile >>> Section: APPR Header Type: AS SRD: 1230.000

784.695 784.750 781.428

.054 .029 .011

6850.000 1.155 .162

5930.262 402564.60 .0002

130.000 130.000 2.619

-344.991 1087.387 -.001

<<< The Preceding Data Reflect The "Unconstricted" Profile >>> <<< The Following Data Reflect The "Constricted" Profile >>> <<< Beginning Bridge/Culvert Hydraulic Computations >>>

Section: BRGE Header Type: BR SRD: 1100.000

WSEL VHD Q AREA SRDL LEW EGEL HF V K FLEN REW CRWS HO FR # SF ALPHA ERR --------- ------ ---------- ---------- --------- --------784.448 1.378 6850.000 920.694 100.000 -70.964 785.826 .065 7.440 138988.70 100.000 25.073 780.269 1.067 .536 ****** 1.601 .000

Specific Bridge Information C P/A PFELEV BLEN XLAB XRAB Bridge Type 4 Flow Type 1 ------ ----- -------- -------- -------- -------Pier/Pile Code ** .7903 .000 788.000 ******** ******** ******** --------------------------- ------ ----- -------- -------- -------- --------

Section: APPR Header Type: AS SRD: 1230.000

WSEL VHD Q AREA SRDL LEW EGEL HF V K FLEN REW CRWS HO FR # SF ALPHA ERR --------- ------ ---------- ---------- --------- --------785.999 .027 6850.000 7867.950 100.000 -419.961 786.026 .108 .871 590932.10 188.877 1119.983 781.428 .092 .103 .0002 2.284 .002

Approach Section APPR Flow Contraction Information M( G ) M( K ) KQ XLKQ XRKQ OTEL -------- -------- --------- -------- -------- -------.933 .702 176009.6 -71.000 25.000 785.999 -------- -------- --------- -------- -------- -------<<< End of Bridge Hydraulics Computations >>> ER

157

Open Channel Hydraulics

CHAPTER 6

The velocity in the bridge is quite high at 7.44 ft/s, but choking does not occur and the flow type is Class 1, free surface flow. The bridge discharge coefficient is 0.79. Note that the boundaries of the Kq section are given by XLKQ and XRKQ as specified in the input file. The backwater is given by the difference between the approach water surface elevations for constricted and unconstricted flow and is equal to 786.0 − 784.7 = 1.3 ft, which is considerably higher than the desired backwater of 0.25 ft.

In order to obtain the desired backwater of 0.25 ft, the new bridge obviously requires a larger opening length. However, because of the large floodplain on the right hand side of the cross section, it is more efficient to increase the main bridge length modestly but to also include a relief bridge in the right floodplain. It is assumed that both bridges will be Type 4 with the same design parameters as the existing bridge. The bridge design mode is used in the following input data file to facilitate changing the bridge opening lengths. *F SI 0 T1 Exercise 6.16 Duck Creek New Bridge: Main 200 ft, Relief 150 ft T2 Q = 6850 cfs; Type 4; LSEL = 788 ft; Embankment ss = 2:1; T3 Embankment elevation = 790 ft; Width = 30 ft; WW = 30 deg. * * Discharge Q 6850. * * Known WS elevation WS 784.66 * * Section reference distance, skew (0), ek (0.5), ck (0.0) XS EXIT 1000.

GR -150,792 -105,780 -70,780 -28,778 -24,774 -20,773 18,772 GR 22,772 35,780 50,780 210,779 600,779 860,780 1005,782 GR 1050,784 1112,786 1260,788 1310,798 * * Subsection n values and subsection break points N 0.08 0.04 0.08 0.05 0.08 0.05 SA -28.

35 210 600 860 * XS FULV 1200.

GR -150,792 -105,780 -71,778.2 -30,776 -15,772.5 16,772.5 GR 25,775 35,780 50,780 210,779 600,779 860,780 1005,782 GR 1050,784 1112,786 1260,788 1310,798 * * Section reference distance BR BRGE 1200. * BC 788.

BL 0 200. -71. 129.

PD 0 775.,3.0,1 780.,3.0,1 780.,6.0,2 N 0.04 * bridge type, bridge width, ss, Embankment elev., Wingwall angle CD 4 30.0 2 790 30 * * Place Kq section so that main channel is inside it KD * * * -71 129 * BR ROBR 1200 BC 788.

BL 0 150.

450.

600.

PD 0 779.,3.0,1 788.,3.0,1

158

Open Channel Hydraulics

N CD KD * XS GR GR GR GR * N SA * EX * ER

0.04 4 30. 2 * * *

CHAPTER 6

790.

450

30.

600

APPR 1430.

-480,796 -50,778 28,780 1260,810

-440,788 -30,776 50,780

-420,786 -25,774 670,780

-305,784 2,772 990,782

0.10

0.04

0.08

0.05

-50

28

-175,782 17,772 1070,784

-95,780 20,774 1120,786

670

The FULV, BRGE, and APPR stations are changed to accommodate the new main bridge opening length of 200 ft, which is located from X = -71 ft to X = 129 ft in the cross section. Two bridge piers are added with a width of 3 ft each. The relief opening is 150 ft long and is located from X = 450 ft to X = 600 ft. One bridge pier is proposed in this opening. A KD record is used for both the main bridge and the relief bridge. A portion of the WSPRO output is shown in the following table.

Section: BRGE Header Type: BR SRD: 1200.000

WSEL VHD Q AREA SRDL LEW EGEL HF V K FLEN REW CRWS HO FR # SF ALPHA ERR --------- ------ ---------- ---------- --------- --------784.651 .213 4624.701 1468.299 200.000 -70.966 784.865 .072 3.150 197277.80 200.000 129.061 779.045 .045 .241 ****** 1.381 -.001

Specific Bridge Information C P/A PFELEV BLEN XLAB XRAB Bridge Type 4 Flow Type 1 ------ ----- -------- -------- -------- -------Pier/Pile Code 0 .8509 .029 788.000 200.000 -71.000 129.000 --------------------------- ------ ----- -------- -------- -------- --------

Section: SLICE Header Type: AS SRD: 1430.000

WSEL VHD Q AREA SRDL LEW EGEL HF V K FLEN REW CRWS HO FR # SF ALPHA ERR --------- ------ ---------- ---------- --------- --------784.870 .102 4624.701 3136.499 200.000 -355.018 784.972 .084 1.474 289414.20 217.284 331.718 779.197 .023 .211 .0002 3.007 .007

Approach Section SLICE Flow Contraction Information M( G ) M( K ) KQ XLKQ XRKQ OTEL -------- -------- --------- -------- -------- -------.705 .286 206289.5 -71.000 129.000 784.870 -------- -------- --------- -------- -------- -------<<< End of Bridge Hydraulics Computations >>>

Section: ROBR Header Type: BR SRD: 1200.000

WSEL VHD Q AREA SRDL LEW EGEL HF V K FLEN REW CRWS HO FR # SF ALPHA ERR --------- ------ ---------- ---------- --------- --------784.720 .191 2225.299 857.816 150.000 450.036 784.911 .116 2.594 97078.19 150.000 600.064 780.924 .127 .259 ****** 1.828 -.001

159

Open Channel Hydraulics

CHAPTER 6

Specific Bridge Information C P/A PFELEV BLEN XLAB XRAB Bridge Type 4 Flow Type 1 ------ ----- -------- -------- -------- -------Pier/Pile Code 0 .7396 .020 788.000 150.000 450.000 600.000 --------------------------- ------ ----- -------- -------- -------- --------

Section: SLICE Header Type: AS SRD: 1380.000

WSEL VHD Q AREA SRDL LEW EGEL HF V K FLEN REW CRWS HO FR # SF ALPHA ERR --------- ------ ---------- ---------- --------- --------785.182 .007 2225.299 3283.316 200.000 331.718 785.189 .256 .678 154713.80 264.501 1099.552 781.015 .022 .059 .0002 1.037 -.007

Approach Section SLICE Flow Contraction Information M( G ) M( K ) KQ XLKQ XRKQ OTEL -------- -------- --------- -------- -------- -------.802 .788 32931.2 450.000 600.000 785.182 -------- -------- --------- -------- -------- -------<<< End of Bridge Hydraulics Computations >>> Section: APPR Header Type: XS SRD: 1430.000

785.193 785.200 781.015

.007 .010 .000

2225.299 .676 .059

3291.473 155334.70 .0002

50.000 50.000 1.037

331.718 1099.817 .000

Section: APPR Header Type: XS SRD: 1430.000

784.983 .046 785.028 ****** 781.428 ******

6850.000 1.080 .145

6344.903 ********* 440895.60 ********* ****** 2.529

-361.500 1094.565 .000

The flow type is Class 1, free surface flow, in both bridges. The bridge discharge coefficient is 0.85 for the main bridge and 0.74 for the relief bridge. The velocity in the main bridge has been reduced to 3.15 ft/s which should result in less scour. The final conveyance-weighted water surface elevation in the approach section is 784.983 ft. The unconstricted water surface elevation in the approach section is 784.735 ft (not shown), so that the backwater is given by 784.983 − 784.735 = 0.248 ft.

The WSPRO option in HEC-RAS can also be used to solve this problem. For the existing bridge, cross sections are entered starting upstream at station 1230. In addition to the approach and exit stations, cross sections are entered just upstream of the bridge and just downstream of the bridge, typically at the toe of the embankment. For convenience, station 1100 is taken at the toe of the downstream embankment, and an interpolated cross section is added at station 1170 at the toe of the upstream embankment.

Then a bridge is added in the geometric data editor at station 1135, and the WSPRO option is chosen. Additional information on the bridge type, embankment elevations, wingwall angle, and other data specific to WSPRO are requested when the WSPRO option is selected. The ineffective flow area option is applied to stations 1170 and 1100 just outside the bridge as described in the HEC-RAS Hydraulic Manual. The roadway/deck information must also be entered. HEC-RAS automatically creates two cross sections just inside the bridge, and these are designated BR U and BR D for upstream and downstream locations, respectively. The output profile table, a plot of the water surface profile, and a plot of the cross-section inside the bridge at station BR D are given next. The calculated backwater is 786.0 − 784.7 = 1.3 ft, and the water surface elevation in the bridge at section BR D is 784.52, both of which agree well but not exactly with the WSPRO output. Note that the largest energy loss is the exit expansion loss.

160

Open Channel Hydraulics

CHAPTER 6

RAS Reach

Plan: Plan 02

River: Duck Cr.

River Sta

E.G. W.S. Crit Elev Elev W.S. (ft) (ft) (ft) 1230 786.06 786.04 781.25 1170 785.99 785.51 780.38 1135 785.96 785.20 780.38 BR U 1135 785.88 784.52 780.25 BR D 1100 785.83 785.06 780.19 1000 784.69 784.66 780.59

Main Main Main Main Main Main

Frctn Loss (ft) 0.07 0.04 0.08

C & E Top Vel Loss Width Chnl (ft) (ft) (ft/s) 0.00 1540.60 1.70 0.00 1380.65 5.60 0.00 96.00 6.98

0.05

0.00

96.00

7.39

0.05

1.09 1206.81 1192.93

5.70 1.96

Exercise6.16 Constricted Flow Main

790

Legend EG PF 1 WS PF 1

785

Elevation (ft)

Crit PF 1 Ground 780

775

770 1000

1050

1100

1150

1200

1250

Main Channel Distance (ft)

Exercise6.16 Constricted Flow 800

.08 .

0 4

.08

Bridge Type 4 RS = 1135 .08 .05

BR D .05 Legend

Elevation (ft)

795

EG PF 1 WS PF 1

790

Crit PF 1 785

Ground

780

Ineff Bank Sta

775 770 -200

0

200

400

600 Station (ft)

161

800

1000

1200

1400

Open Channel Hydraulics

6.17.

CHAPTER 6

Apply HEC-RAS to Example 6.3 using the energy, momentum, and Yarnell methods.

Set up the cross sections in the schematic layout starting at the upstream station of 1450.

Use a constant slope of 0.00052 and add the appropriate amount to all elevations given for station 1000 in Example 6.3. Then in the geometric data editor, copy the cross sections downstream adjusting the elevation downward according to the slope and distance between stations. Use values of 0.3 and 0.1 for the expansion and contraction loss coefficients, respectively. Establish stations at 1250, 1200, and 1000 to correspond with the WSPRO sections. Add a bridge at station 1225 using the geometric data editor.

In the bridge/culvert editor, enter the deck and roadway data, pier data, and check the boxes for all three methods of computation as well as the box choosing the highest energy answer in the bridge modeling approach window. Also enter a pier drag coefficient of 2.0 and a Yarnell pier coefficient of 0.9. In the cross section data editor, add ineffective flow areas at stations 1250 and 1200 to the left and right of the bridge opening specified at elevations above the low chord elevation but below the top of the roadway. In the steady flow data menu, enter the discharge of 20,000 cfs and choose normal depth as the downstream control with a slope of 0.00052. Finally, choose steady flow analysis and click the compute button. Compare the results with WSPRO, and then make a second run with the exit section at station 800 instead of 1000. Discuss the results.

Solution.

The input data are entered into HEC-RAS as described in the exercise. The program chooses the momentum method as the highest energy answer at Station 1250, just upstream of the bridge, with an energy grade line elevation of 29.07 ft. The energy and Yarnell methods give energy grade line elevations of 29.04 and 28.92 ft, respectively.

The water surface profile results are shown in the following table.

RAS Plan: 01 River: Example6.3 River Reach River E.G. W.S. Crit Frctn C & E Sta Elev Elev W.S. Loss Loss (ft) (ft) (ft) (ft) (ft) Main 1450 29.30 29.02 0.13 0.10 Main 1250 29.07 27.83 22.13 Main 1225 28.98 27.52 22.56 BR U Main 1225 28.83 27.33 22.53 BR D Main 1200 28.81 27.51 22.10 0.16 0.28 Main 1000 28.36 28.01 21.88

The approach water surface elevation is 29.02 ft giving a backwater of (29.02 − 28.25) = 0.77 ft in comparison to the value of 1.12 ft obtained from the WSPRO results in Table 612. The water surface elevation in the bridge at section BR D is 27.33 ft in comparison to the WSPRO value of 27.44 ft. If the exit section is relocated to station 800, or a distance of 2 bridge opening lengths downstream of the bridge, the momentum method again gives the highest energy answer and the computed approach water surface elevation is 29.16 ft. However, the normal water surface elevation is (1450 − 800) × 0.00052 + 28.008 = 28.35 ft, so the backwater height is (29.16 − 28.35) = 0.81 ft, a slightly higher value than for the exit section located at station 1000. The water surface profile and the downstream bridge cross section (BR D) are given in the following two figures.

162

Open Channel Hydraulics

CHAPTER 6

Exercise 6.17 Plan 01 Main

35

Legend

30

EG PF 1 WS PF 1

25

Elevation (ft)

Crit PF 1 Ground

20 15 10 5 0 1000

1100

1200

1300

1400

1500

Main Channel Distance (ft)

Exercise 6.17 Plan 01 Bridge Type 1 RS = 1225 BR D .07 .035 .045

.045

40

Legend

35

EG PF 1

Elevation (ft)

30

WS PF 1

25

Crit PF 1

20

Ground

15

Ineff

10

Bank Sta

5 0 0

100

200

300

400 Station (ft)

163

500

600

700

800

Open Channel Hydraulics

CHAPTER 6

164

CHAPTER 7. Governing Equations of Unsteady Flow 7.1

Starting with Equation 7.13, derive Equation 7.14.

Solution.

Apply the product rule of differentiation to the first term of Equation 7.13, ∂Q/∂t, and substitute for ∂A/∂t from continuity as given by Equation 7.2:

∂Q ∂V ∂A =A +V ∂t ∂t ∂t ∂V ∂Q  ∂Q  =A + V  qL −  ∂t ∂t ∂x   ∂Q ∂V ∂V ∂A =A + Vq L − VA −V 2 ∂t ∂t ∂x ∂x Also apply the product rule to the second term of Equation 7.13 with the result

∂  Q2  ∂ = β βAV 2 ∂x  A  ∂x

(

)

∂V ∂β ∂A ∂  Q2   = 2 βAV  β + βV 2 + AV 2 ∂x ∂x ∂x ∂x  A  As discussed on p. 272 of the textbook after Equation 7.8, the third term of Equation 7.13 can be written as

∂ ∂y ( ghc A) = g A ∂x ∂x for a prismatic channel. Substituting these expressions into Equation 7.13, dividing by A, and collecting terms, we have

∂V ∂V ( β − 1)V 2 ∂A ∂β ∂y + (2 β − 1)V + +V 2 +g ∂t ∂x A ∂x ∂x ∂x q = g ( S 0 − S f ) + L (v L cos φ − V ) A Note that the third and fourth terms on the left hand side of Equation 7.14 have not been divided by A in the first edition and first printing of the textbook.

165

Open Channel Hydraulics

7.2

CHAPTER 7

Derive Equations 7.18 and 7.19, the characteristic equations.

Solution.

The momentum equation (Equation 7.15) for a prismatic channel without lateral inflow is multiplied first by (D/g)1/2 = c/g, where c is the wave celerity = (gD)1/2 and D = hydraulic depth = A/B, and then by −(D/g)1/2 = −c/g to produce two new equations given by

∂y c ∂V c ∂V +V +c = c( S 0 − S f ) ∂x g ∂t g ∂x ∂y c ∂V c ∂V − −V −c = −c ( S 0 − S f ) ∂x g ∂t g ∂x The continuity equation (Equation 7.5) can be written for a prismatic channel with no lateral inflow as

∂y ∂y ∂V =0 +V +D ∂t ∂x ∂x in which D = c2/g. Adding the first transformed momentum equation to the continuity equation, we obtain

∂y ∂y c ∂V  c c 2  ∂V + (V + c) + +  V +  = c( S 0 − S f ) ∂t ∂x g ∂t  g g  ∂x Collecting terms, we have Equation 7.18, the first characteristic form (C1):

c ∂ ∂ ∂ ∂  ∂t + (V + c) ∂x  y + g  ∂t + (V + c) ∂x V = c( S 0 − S f ) Adding the second transformed momentum equation to the continuity equation, we obtain Equation 7.19 in a similar fashion.

7.3

In the estuary problem given as Example 7.1, determine algebraically (not graphically) from the simple wave method the time in hours required for the depth to drop to 6.50 ft at a distance upstream of x = 25,000 ft. Plot on the same axes the depth hydrographs at x = 0 and x = 25,000 ft.

Solution.

Solve the boundary condition for the time at which y L = 6.5 ft at x = 0:

166

Open Channel Hydraulics

CHAPTER 7

 πt π  6.5 = 8 − 2 cos −   6 2  πt π  −1  1.5   −  = cos   = ±0.7227  2   6 2 t = 1.62 hrs in which the negative value of the inverse cosine has been taken to obtain a solution that satisfies the stated requirement of t ≤ 3 hrs. The characteristic velocity associated with the depth of 6.5 ft is given by the condition that V − 2c = constant to yield

V = 2c + V0 − 2c0 V = 2 × 32.2 × 6.5 − 3 − 2 × 32.2 × 8 = −6.17 ft/s The slope of the characteristic that emanates from the y axis with an associated depth of 6.5 ft at t = 1.62 hrs is given by

dx = V + c = −6.17 + 32.2 × 6.5 = 8.30 ft/s dt Then the time required for the depth of 6.5 ft to reach a distance of 25,000 ft becomes

∆t = 1.62 +

25,000 / 8.30 = 2.46 hrs 3600

The depth hydrographs at x = 0 and x = 25,000 ft are calculated in the following table:

t 1 , hrs 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8

y L ,ft dx/dt, ft/s 8.00 8.00 13.05 7.79 12.42 7.58 11.78 7.38 11.15 7.19 10.54 7.00 9.94 6.82 9.37 6.66 8.84 6.51 8.35 6.38 7.91 6.27 7.52 6.17 7.20 6.10 6.94 6.04 6.75 6.01 6.64

167

t, hrs 0.00 0.53 0.76 0.99 1.22 1.46 1.70 1.94 2.19 2.43 2.68 2.92 3.17 3.40 3.63 3.85

Open Channel Hydraulics

CHAPTER 7

3.0

6.00

6.60

4.05

The second column gives the values of depth at x = 0 and t = t 1 from the stated boundary condition. The value of dx/dt in the third column follows from Equation 7.34a, and the time at which the given depth arrives at x = 25,000 ft is given by t = t 1 + [25,000/(dx/dt)]/3600 in hours. The plot of the depth hydrographs is given in the following figure. Note that there is stretching of the hydrograph in the time domain because the smaller depths travel at a slower speed.

Depth Hydrographs 10.0 9.5 9.0

Depth, ft

8.5 8.0 7.5 7.0 6.5 6.0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

Time, hrs x = 0.0

7.4

x = 25,000 ft

Water is initially at rest upstream and downstream of a sluice gate, which is completely closed in a rectangular channel. The upstream depth is 3.0 m and the downstream depth is 1.0 m. The gate suddenly is opened completely. Determine the speed of the surge, the depth and velocity behind the surge, and the speed of lengthening of the constant depth region. Show your results in both the physical plane and the characteristic plane.

Solution.

With reference to Figure 7.13 and Figure 7.14, the continuity equation as given by Equation 7.38 is rearranged and written as

Vs =

V3 y 3 V y = 3 3 y 3 − y 4 y 3 − 1.0

168

Open Channel Hydraulics

CHAPTER 7

Substituting into the momentum equation as given by Equation 7.39, we have

1  Vs = − 9.81 × 1.0 ×  y 3 ( y 3 + 1) 2  1/ 2 Vs = −2.215 × [ y 3 ( y 3 + 1)]

1/ 2

Finally, the simple wave equation as given by Equation 7.40 becomes

V3 = 2 gy 3 − 2 gy 0 V3 = 2 9.81 y 3 − 2 9.81 × 3.0 V3 = 6.264 y 3 − 10.85 The solution proceeds by assuming a value of y 3 ; solving for V 3 from the simple wave equation; and substituting the results for y 3 and V 3 into both the momentum and continuity equations. Iteration on y 3 continues until the results for V s from the continuity and momentum equations agree with one another. This iterative procedure is illustrated in the following table: y3, m

V 3 , m/s

2.0 1.5 1.85 1.849

−1.99 −3.18 −2.33 −2.332

V s , m/s continuity −3.98 −9.54 −5.07 −5.08

V s , m/s momentum −5.43 −4.29 −5.09 −5.08

The final result is given in the last line of the table. The speed of lengthening of the constant depth region is given by

Vs − (V3 + c3 ) = −5.08 − (−2.33 + 9.81 × 1.85 ) = −7.01 m/s Note that point B in Figure 7.13 is moving to the right in this case at the speed of (V 3 + c 3 ) = 1.93 m/s. The change in Figure 7.13 is that point B is on the right hand side of the y axis so that the characteristics in the constant depth region have a positive slope rather than a negative slope.

7.5

Water is flowing in a rectangular channel under a sluice gate. The upstream depth is 3.0 m and the downstream depth is 1.0 m for steady flow. If the gate is slammed shut, compute the height and speed of the surge upstream of the gate and the depth just downstream of the gate. Under what conditions would the depth downstream of the gate go to zero? Show your results in both the physical plane and the characteristic plane.

Solution.

The physical and characteristic planes are shown in the following figure:

169

Open Channel Hydraulics

CHAPTER 7

Vs y1

y4

y3

y2

t 1 V 3 + c3

1

1

Vs V 4 + c4

x

Calculate the initial steady discharge per unit width under the gate from the energy equation given as

y1 +

q2 q2 y = + 4 2 gy12 2 gy 42 y1 y 4

q= q=

y12 − y 42 3.0 × 1.0 3 −1 2

2

2 g ( y1 − y 4 19.62 × (3 − 1) = 6.64 m 3 /s

The steady-state upstream velocity, V 1 = q/y 1 = 6.64/3 = 2.21 m/s while the downstream velocity, V 4 = 6.64 m/s. Now consider the upstream surge after the gate is slammed shut.

The continuity equation, with the surge made stationary, is given by

(V1 + Vs ) y1 = Vs y 2 Vs =

V1 y1 6.64 = y 2 − y1 y 2 − 3.0

The momentum equation for the stationary control volume becomes

(V1 + Vs ) 2 1 y 2  y  1 + 2  = gy1 y1  2 y1  y   (2.21 + Vs ) 2 = 4.91 y 2 1 + 2   3.0  Solving by trial, we obtain the following table of iterations by assuming a value of y 2 and solving for V s from both the continuity and momentum equations:

170

Open Channel Hydraulics

y2, m

CHAPTER 7

V s , m/s continuity 6.64 4.43 4.99

4.0 4.5 4.33

V s , m/s momentum 4.56 5.22 5.00

The result is a surge height of 4.33 − 3.0 = 1.33 m and a surge speed of 5.0 m/s.

The depth downstream of the gate is obtained using the simple wave solution. If a nonzero depth exists, it must be a region of constant depth with zero velocity such that

V3 − 2c3 = V4 − 2c 4 − 2c3 = 6.64 − 2 × 9.81 × 1.0 = 0.376 In this case, the implication is that c 3 is negative, which is impossible because the depth must be greater than or equal to zero. So the depth is zero downstream of the gate, and the trailing end of the negative wave moves downstream at a speed of dx/dt = V 3 + c 3 = V 3 because c 3 = 0. However, V − 2c = constant for the negative wave so that V 3 = V 4 − 2c 4 , and it follows that dx/dt = 6.64 − 2×(9.81×1.0)1/2 = 0.38 m/s. The leading edge of the negative wave has a speed of V 4 + c 4 = 6.64 + (9.81)1/2 = 9.8 m/s. Thus, the negative wave stretches as it moves downstream. The characteristic plane is similar to that drawn in the previous figure except that no characteristics exist to the left of the characteristic tracing the path of the trailing edge of the negative wave. The limiting case of zero depth downstream of the gate such that V 3 = c 3 = 0, and dx/dt =0, is given by V 4 − 2c 4 = 0.

Stated another way, the limiting case occurs for the initial downstream Froude number, (V 4 /c 4 ), equal to 2.0. In the characteristic plane, the characteristic for the trailing edge of the wave coincides with the time axis in this limiting case. If the downstream Froude number is greater than 2.0 as in this exercise, the negative wave moves away from the gate in the downstream direction, while a constant depth occurs downstream of the gate for a Froude number less than 2.0, but greater than 1.0, of course, because the initial downstream flow must be supercritical due to the gate.

7.6

Water initially flows at steady state under a sluice gate. The upstream flow depth is 3.0 m and the downstream flow depth is 0.3 m. The sluice gate is raised abruptly, completely free of the flowing water. (a) Find the depth and discharge at the gate after the gate has been raised. (b) Determine the height and speed of the surge.

Sketch your results on the physical plane and the characteristic plane. Neglect bed slope and resistance effects.

Solution.

First find the steady-state discharge per unit width from the energy equation as in Exercise 7.5 in which y 1 = depth upstream of the gate, and y 4 = depth downstream of the gate.

171

Open Channel Hydraulics

q= q=

CHAPTER 7

y1 y 4

2 g ( y1 − y 4

y12 − y 42 3 × 0.3 3 − 0.3 2

2

19.62 × (3.0 − 0.3) = 2.194 m 3 /s

Then the steady-state velocities are V 1 = −q/y 1 = −2.194/3.0 = −0.731 m/s, and V 4 =− q/y 4 = −2.194/0.3 = −7.31 m/s, where the negative signs are due to the coordinate system being chosen as shown in Figure 7.13. Applying the continuity equation to the surge made stationary as in Figure 7.14, but with the velocity, V 4 , included, we have

Vs =

V3 y 3 − V4 y 4 V3 y 3 + 2.194 = y3 − y 4 y 3 − 0.3

The momentum equation applied to the same control volume becomes

 (Vs − V4 ) 2 1 y 3  y 3  + 1 = 2 y4  y4 gy 4   y  Vs = −  4.91 y 3  3 + 1  − 7.31  0.3    Finally, the simple wave equation is given by

V3 − 2c3 = V1 − 2c1 = −0.731 − 2 × 9.81 × 3.0 = −11.57 V3 = 2 gy 3 − 11.57 First solve for the surge height and speed by assuming a value of y 3 in the simple wave equation and solving for V 3 . Then solve the continuity and momentum equations for the surge speed and iterate until the results from the two equations are equal. This computation is set up in a spreadsheet as shown in the following table: y3, m

V 3 , m/s

0.5 0.35 0.3764

−7.14 −7.86 −7.727

V s , m/s continuity −6.88 −11.17 −9.351

V s , m/s momentum −9.87 −9.24 −9.351

This is a weak surge with a height of only 0.0764 m because the relative Froude number at the front of the surge is (V s −V4)/(gy4)0.5 = | [−9.35−(−7.31)]/(9.81×0.3)0.5 | = 1.19. The speed of the surge is 9.35 m/s, and note that V3 is in the same direction but with a smaller magnitude than the surge speed.

172

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CHAPTER 7

The depth and discharge at the gate after the gate has been raised is the depth of the simple wave there provided that dx/dt is negative at the boundary between the constant depth region and the simple wave (point B in Figure 7.13). In this case, dx/dt = V3 + c3 = −7.73+2×(9.81×0.3764)0.5 = −5.81 m/s. So at the gate, it must be true that

V g − 2c g = V1 − 2c1 = −0.731 − 2 9.81 × 3.0 = −11.58 m/s dx = 0 = Vg + c g dt Solving simultaneously, we have Vg = −11.58/3 = −3.86 m/s; cg = 3.86 m/s; and yg = 1.52 m all of which remain constant with time. The physical plane and characteristic plane are similar to what is shown in Figure 7.13 with the addition of finite values of velocity upstream and downstream of the gate.

7.7

Prove that the limiting condition of y4/y0 = 0.138 in the simple wave dam-break problem determines whether the constant depth region (point B in Figure 7.13) moves upstream or downstream.

Solution.

In the limiting case we must have dx/dt = V3 + c3 = 0 in addition to satisfying the momentum and continuity equations at x = 0 so that point B in Figure 7.13 is on the y axis. From Equation 7.34a, the first condition requires that c3 = (2/3)c0, or y3 = (4/9)y0.

In addition, V3 = −c3 = −(2/3)c0. The continuity equation (Equation 7.38) becomes

Vs =

− ( 2 / 3) gy 0 y 3 V3 y 3 = y3 − y4 y3 − y4

and the momentum equation (Equation 7.39) is written as

1 y  y  Vs = − gy 4  3  3 + 1   2 y4  y4

1/ 2

Equating the right hand sides of these two expressions for Vs and nondimensionalizing some of the terms, we have 1/ 2

  y3    y3    2 y3      − 1/ 2    y 4   1  y 0    y 0    3 y0 = −   +1  y3 y4 y0   2  y4    y4     −      y0 y0   y 0    y 0      

173

Open Channel Hydraulics

CHAPTER 7

Substituting y3/y0 = 4/9 and squaring both sides of the equation, we obtain

  4   4  9 + 1 = 2  18  y 4    4 y4       −  y 9   y 0   0   4   9

3

With some further algebra, it is easily shown that the solution is y4/y0 = 0.138.

7.8

In the simple wave dam-break problem, derive an expression for the discharge per unit width at x = 0, q0, as a function of the ratio of initial depths, y4/y0. Nondimensionalize q0 as q0/c0y0 and plot the results.

Solution.

For y4/y0 ≤ 0.138, the discharge per unit width has its maximum value as given by Equation 7.36. In dimensionless terms, q0/c0y0 = 8/27. If y4/y0 > 0.138, then Equations 7.38, 7.39, and 7.40 must be solved simultaneously to obtain the discharge per unit width at the dam where x = 0. From Equation 7.38 (continuity), we have

Vs =

V3 y3 y3 − y 4

(1)

while the momentum equation (Equation 7.39) can also be written in terms of the surge speed, Vs, as

1 y  y  Vs = − gy 4  3  3 + 1   2 y4  y4

1/ 2

(2)

The simple wave equation (Equation 7.40) requires that

V3 = 2 gy3 − 2 gy 0

(3)

Equating the right hand sides of Equation 1 and Equation 2, we have

1 y  y  V3 y3 = − gy 4  3  3 + 1 y3 − y 4   2 y4  y4

1/ 2

(4)

Now noting that at x =0, q0 = −V3y3, and nondimensionalizing the preceding equation, the result is

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Open Channel Hydraulics

q0 = c0 y 0

CHAPTER 7

y 4  y3 y 4   1 ( y3 / y 0 )  ( y3 / y 0 )   −   + 1 y 0  y 0 y 0   2 ( y 4 / y 0 )  ( y 4 / y 0 ) 

1/ 2

(5)

Now multiply Equation 3 by y3, substitute q0 = −V3y3 and nondimensionalize to obtain

q0 y  = 2 3 1 − c0 y 0 y0 

y3   y 0 

(6)

Substituting from Equation 6 into Equation 5 and rearranging, we have the following nonlinear algebraic equation in which Y = y4/y0 and η = y3/y0 :

 1 η  η  2η (1 − η ) − Y (η − Y )   + 1  2 Y Y

1/ 2

=0

This equation can be solved for η for a given value of Y using a nonlinear algebraic equation solver and then q0/c0y0 follows from Equation 6. The results are summarized in the following table and graph: Y = y4/y0 0.000 0.138 0.150 0.200 0.250 0.300 0.350 0.400 0.450 0.500 0.550 0.600 0.650 0.700 0.750 0.800 0.850 0.900 0.950 1.000

η = y3/y0 0.4440 0.4576 0.5079 0.5518 0.5914 0.6282 0.6627 0.6955 0.7269 0.7572 0.7866 0.8152 0.8431 0.8704 0.8972 0.9234 0.9493 0.9748 1.0000

175

q/c0y0 0.2963 0.2963 0.2961 0.2919 0.2838 0.2732 0.2606 0.2464 0.2310 0.2143 0.1966 0.1779 0.1583 0.1379 0.1167 0.0947 0.0721 0.0488 0.0247 0.0000

Open Channel Hydraulics

CHAPTER 7

Dimensionless Discharge at Dam (x = 0)

0.35

0.30

q /c 0y 0

0.25

0.20

0.15

0.10

0.05

0.00 0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

y 4/y 0

7.9

Determine the maximum possible height of the surge, (y3 – y4), in Example 7.3 in terms of y0, and the value of y4/y0 for which it occurs.

Solution.

Note that as y4/y0 approaches zero, we recover the dam-break problem of Figure 7.12 for which there is no surge and (y3−y4) approaches zero. Similarly, as y4/y0 approaches 1.0, (y3−y4) must also approach zero. This implies that there is a maximum surge height between these two limits.

Substituting for V3 from Equation 3 in Exercise 7.8 into Equation 1, and equating the right hand sides of Equations 1 and 2, results in

y3 − y 4 =

2 y3 gy3 − 2 y3 gy 0 1 y  y  − gy 4  3  3 + 1   2 y4  y4

1/ 2

Nondimensionalizing this equation and setting Y = y4/y0 and η = y3/y0 as in Exercise 7.8, we have after rearrangement the following nonlinear algebraic equation:

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Open Channel Hydraulics

CHAPTER 7

 1 η  η  (η − Y ) Y   + 1  2 Y Y

1/ 2

+ 2(η 3 / 2 − η ) = 0

which can be solved for η for a given value of Y with a nonlinear algebraic equation solver. The results in the vicinity of the maximum surge height are given in the following table: Y = y4/y0 η = y3/y0 η−Y 0.15 0.4576 0.3076 0.16 0.4683 0.3083 0.17 0.478683 0.308683 0.175 0.483734 0.308734 0.176 0.484735 0.308735 0.177 0.485732 0.308732 0.18 0.488706 0.308706

From the table, it is apparent that the maximum surge height is 0.31y0, and it occurs for a value of y4/y0 = 0.176. The complete functional dependence is shown in the following graph:

Maximum Surge Height

0.35

0.30

(y 3-y 4)/y 0

0.25

0.20

0.15

0.10

0.05

0.00 0.0

0.1

0.2

0.3

0.4

0.5 y 4/y 0

177

0.6

0.7

0.8

0.9

1.0

Open Channel Hydraulics

7.10

CHAPTER 7

The headrace for a turbine is a long rectangular canal that feeds water from a lake to the turbine penstocks. Suppose that the design discharge for a turbine is 68 m3/s. The canal width is 12 m, and the canal slope is very small with an average water depth in the canal of 3.5 m with no flow. If the turbine is brought on line suddenly (load acceptance), what will be the depth of flow at the downstream end of the canal where the penstock inlet is located? How long will it take for the negative wave to reach the reservoir if the canal is 2 km long?

Solution.

By extension of the dam-break problem, the maximum possible discharge that can be supplied by the canal is given by simple-wave theory (Equation 7.36) to be

Qmax =

8 8 by 0 gy 0 = × 12 × 3.5 × 9.81× 3.5 = 72.9 m 3 /s 27 27

Because the turbine is demanding only 68 m3/s, the negative wave can supply this discharge without the necessity for storage. However, the depth at the turbine will be greater than 4/9 y0 because the turbine discharge is less than the maximum. With reference to Figure 7.13, but with point B to the right of the y axis when the turbine is located at x = 0, we must have

V3 − 2c3 = V0 − 2c0 = 0 − 2 × 9.81× 3.5 = −11.72 m/s In addition, we must satisfy continuity at the turbine, which is given by

V3by3 = −68 V3 = −

5.67 y3

with b = 12 m, and with the minus sign employed to preserve the sign convention in Figure 7.13 that has V3 at the turbine in the negative x direction. Solving the two equations simultaneously, we obtain y3 = 2.04 m. The negative wave will move upstream in the canal at the rate c0 = (gy0)0.5 = (9.81×3.5)0.5 = 5.86 m/s. The time required to travel 2 km is 2000/5.86/60 = 5.7 minutes.

7.11

Suppose that the turbine in Exercise 7.10 is operating at steady state at a discharge of 68 m3/s, and the corresponding normal depth of flow in the canal is 3.0 m. If the turbine is shut down suddenly (load rejection), what will be the height of the surge at the downstream end of the canal?

Solution.

Vs

V1

y1

y2

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Open Channel Hydraulics

CHAPTER 7

The velocity V1 = Q/A1 = 68.0/(12×3.0) = 1.89 m/s. The continuity equation applied to the control volume, when it is made stationary, is

(V1 + Vs ) y1 = Vs y 2 Vs =

V1 y1 1.89 × 3.0 5.67 = = y 2 − y1 y 2 − 3.0 y 2 − 3. 0

The momentum equation applied to the stationary control volume becomes

(V1 + Vs ) 2 1 y 2  y  1 + 2  = gy1 y1  2 y1  y   (1.89 + Vs ) 2 = 4.91 y 2 1 + 2   3.0  In both the continuity and momentum equations, the boundary condition, V2 = 0, has been applied. An iteration is set up in which y2 is assumed and the surge speed is computed from both the continuity and momentum equations until the two values are equal. The table of results is given next. y2, m

Vs, m/s continuity 5.67 3.78 5.15 5.044

4.0 4.5 4.1 4.124

Vs, m/s momentum 4.88 5.54 5.01 5.044

The surge has a height of (4.12−3.0) = 1.12 m, and a speed of 5.04 m/s.

7.12

If a dam with a maximum water depth of 50 ft fails abruptly, estimate the time required for the surge to reach a community 5 mi downstream of the dam. What will the surge height be? Initially, the downstream river has a negligible velocity and a depth of 5 ft.

What factors alter your estimates and in what direction?

Solution.

Let y0 = 50 ft and y4 = 5 ft in Figure 7.13. In this case, y4/y0 < 0.138 so that point B will move to the left as shown in Figure 7.13. The continuity equation as given by Equation 7.38 can be rearranged to yield

Vs =

V3 y3 V y = 3 3 y3 − y 4 y3 − 5.0

The momentum equation (Equation 7.39) for this problem becomes

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Open Channel Hydraulics

CHAPTER 7

1 y  y  Vs = − gy 4  3  3 + 1   2 y4  y4

1/ 2

 1 y3  y3  Vs = − 32.2 × 5  + 1   2 5.0  5.0    y  Vs = −4.0125 y3  3 + 1   5.0 

1/ 2

1/ 2

The simple wave equation (Equation 7.40) is given by

V3 = 2 gy3 − 2 gy 0 V3 = 2 32.2 y3 − 2 32.2 × 50 V3 = 11.35 y3 − 80.25 Equating the expressions for the surge velocity, Vs, from the continuity and momentum equations, and substituting for V3 from the simple wave equation, we have the following nonlinear algebraic equation to solve:

 y  (11.35 y3 − 80.25) y3 + 4.0125 ( y3 − 5)  y3  3 + 1    5

1/ 2

=0

The result is y3 = 19.8 ft, and back-substituting, we have V3 = −29.7 ft/s and Vs = −39.8 ft/s. Therefore, the surge height is (19.8−5.0) = 14.8 ft. The time required for the surge to travel 5 miles is (5×5280/39.8/60) = 11.1 minutes. The effect of flow resistance, which is neglected in the simple wave theory, tends to make the surge more dispersive and delays its formation so long as the Froude number of the initial flow is less than 2 (Henderson 1966). If the downstream river bed is dry, then the leading edge of the surge will move at a slower speed than indicated by the simple wave theory, although resistance should have less effect in the case of the submerged downstream river bed. In any case, the simple wave theory applies only in the initial period following the dam break when the wave is governed by inertia rather than slope or resistance effects. We will examine numerical solutions of the surge movement and compare the results with the simple wave theory in Chapter 8.

180

CHAPTER 8. Numerical Solution of the Unsteady Flow Equations 8.1

Write out the complete difference equations for an internal boundary condition of a weir.

What if the weir is in a submerged condition?

Solution. t P1 P2 ∆t

C1

C2

R1

S2

∆x

∆x

x

Using the method of characteristics, the forward characteristic equation is written along C1 upstream of the weir and the backward characteristic equation is applied along C2 downstream of the weir along with the internal boundary conditions given by Equations 8.32 and 8.33. The four equations in terms of the four unknowns V P1 , y P1 , V P2 , y P2 can be written as

VP1 − VR1 +

g ( y P1 − y R1 ) = g ( S 0 − S fR1 ) c R1

2 AP1VP1 = C d 2 g L( y P1 − P) 3 / 2 3 g ( y P 2 − y S 2 ) = g ( S 0 − S fS 2 ) V P 2 − VS 2 − cS 2 AP 2VP 2 = AP1VP1 The interpolation points, R 1 and S 2 , are obtained from values of velocity and depth computed at P 1 and P 2 in the previous time step, which are a negligible distance apart. If the weir is submerged, then the upstream boundary condition given by the weir discharge equation is also dependent on the downstream depth and degree of submergence that it dictates.

8.2.

What causes the diffusive behavior in the Lax diffusive method?

Solution.

With reference to the computational molecule in Figure 8.5b, write the Taylor series expansion for f i+1 and f i−1 at time level k∆t:

181

Open Channel Hydraulics

CHAPTER 8

 ∂ 2 f  (∆x) 2  ∂ 3 f  (∆x) 3  ∂f  +  3  f i +1 = f i +   ∆x +  2   ∂x  i  ∂x  i 2  ∂x  i 6  ∂ 2 f  (∆x) 2  ∂ 3 f  (∆x) 3  ∂f  −  3  f i −1 = f i −   ∆x +  2   ∂x  i  ∂x  i 2  ∂x  i 6 in which all error terms beyond the third order have been dropped, and the superscript k has been omitted for convenience. If these two equations are subtracted and rearranged, we have

f i +1 − f i −1  ∂f   ∂ 3 f  (∆x) 2 =   +  3  2∆x  ∂x  i  ∂x  i 6 which gives the second order finite difference approximation of the first derivative with respect to x in terms of the actual derivative and the error term. Now add the two original equations from the Taylor series expansion and divide by 2 to produce

 ∂ 2 f  (∆x) 2 f i −1 + f i +1 = f i +  2  2  ∂x  i 2 Similarly, we can write the Taylor series expansion with respect to time, t, to give

fi

k +1

 ∂ 2 f  (∆t ) 2  ∂f  = f i +   ∆t +  2   ∂t  i  ∂t  i 2

The finite difference approximation for the time derivative as given by Equation 8.36, with χ = 1 for the Lax diffusive scheme, becomes

f i k +1 −

f i k +1 −

f i −1 + f i +1  ∂ 2 f  (∆t ) 2  ∂ 2 f  (∆x) 2  1   ∂f  2 =  f i +   ∆t +  2  − f i −  2   ∆t ∆t   ∂t  i  ∂t  i 2  ∂x  i 2  f i −1 + f i +1 2 2 2  ∂f   ∂ f  ∆t  ∂ f  (∆x) 2 =   +  2  −  2  ∆t  ∂t  i  ∂t  i 2  ∂x  i 2∆t

The foregoing expressions developed for the finite difference approximations of the x and t derivatives used in the Lax diffusive scheme can be substituted into the difference equation representing continuity for a prismatic rectangular channel to yield

182

Open Channel Hydraulics

CHAPTER 8

∂y ∂ 2 y ∆t ∂ 2 y (∆x) 2  ∂ 2V (∆x) 2  ∂y ∂ 3 y (∆x) 2    + + 2 − 2 + V + 2 ∂t ∂t 2 ∂x 2∆t  2  ∂x ∂x 3 6  ∂x  ∂ 2 y (∆x) 2  ∂V ∂ 3V (∆x) 2  =0  +  y + 2 + 2  ∂x ∂x 3 6  ∂x  Now if ∆x and ∆t are taken to zero at the same rate such that (∆x)2/2∆t = constant = c, we are left with a diffusive term of the form c ∂2y/∂x2 which is responsible for the numerical diffusion in the Lax diffusive scheme. In a similar fashion, the momentum equation yields a diffusive term of the form c ∂2V/∂x2.

8.3.

Derive the Lax-Wendroff method using the Taylor series expansion for U(t+ ∆t) about U(t). Hint: Write the second derivative of U with respect to t in terms of derivatives of F using the governing equations and substitute A ∆U = F. Solution.

The Taylor series expansion of U is given by Equation 8.48 as

U ik +1 = U ik +

∂ 2 U (∆t ) 2 ∂U ∆t + 2 ∂t 2 ∂t

in which terms beyond the second order term are neglected. Now we develop expressions for the derivatives in the Taylor series expansion from the differential equation itself (Equation 8.46) with the source term set equal to zero. See Liggett and Cunge (1975) for the full development for a trapezoidal channel with finite source term.

The first derivative can be expressed by

∂U ∂F(U) =− ∂t ∂x Then after a series of substitutions, an expression for the second derivative is developed:

∂  ∂F  ∂ 2 U ∂  ∂F  = − =−   2 ∂x  ∂t  ∂t  ∂x  ∂t ∂ 2U ∂  ∂U  = − A  2 ∂x  ∂t  ∂t ∂ 2 U ∂  ∂F  = A  ∂t 2 ∂x  ∂x  in which the matrix A is the Jacobian of F(U), ∂F/∂U, as given by Equation 8.57. These expressions for the derivatives are substituted into the Taylor series expansion to obtain the difference equation of the Lax-Wendroff scheme as given by

183

Open Channel Hydraulics

U ik +1 = U ik −

CHAPTER 8

[

∆t (∆t ) 2 Fik+1 − Fik−1 + A ik+1 / 2 Fik+1 − Fik − A ik−1 / 2 Fik − Fik−1 2∆x 2(∆x) 2

(

)

(

)

(

)]

in which

A

k i +1 / 2

 U ik+1 / 2 + U ik   = A 2  

This scheme can be simplified to the two-step scheme given in the textbook as described

by Liggett and Cunge (1975).

8.4.

Verify the definition of A given by (8.57) by taking the Jacobian of F (∂F/∂U).

Solution.

The definition of the vectors F and U are given in (8.47) as

Q    ; F(U) =  Q 2 + ghc A   A 

 A U=  Q 

The Jacobian is defined by

∂Q   ∂Fi ∂A A= = 2 ∂U j  ∂ (Q / A + ghc A)  ∂A

∂Q   ∂Q  ∂ (Q 2 / A + ghc A)   ∂Q

Evaluation of the first row gives (0 1), but the first term in the second row is more involved. In particular, the evaluation of ∂(Ah c )/∂A is given by

∂ ( Ahc ) 1 ∂ ( Ahc ) 1 ∂ y [b( y − η )] dη = = B ∂y B ∂y ∫ 0 ∂A ∂ ( Ahc ) 1 y ∂ ∂0 ∂y [b( y − η )] dη + b( y − y ) − b( y − 0) = ∫ ∂y ∂y B 0 ∂y ∂A ∂ ( Ahc ) 1 y A = ∫ b dη = B ∂A B 0 in which the Leibniz rule has been applied. Now the first term of the second row is given by

184

Open Channel Hydraulics

CHAPTER 8

∂ (Q 2 / A + ghc A) Q 2 gA =− 2 + ∂A B A and the second term of the second row is evaluated as

∂ (Q 2 / A + ghc A) 2Q = ∂Q A The final result for A is given by

0  A =  Q 2 gA − A 2 + B

1  2Q  A 

in agreement with Equation 8.57.

8.5.

Show that the eigenvalues, λ, of A are given by V + c and V − c by taking the determinant of (A − λI).

Solution.

An eigenvalue, λ, of a matrix is defined by

Ax = λx where x is an eigenvector that represents the solution of the foregoing equation unique to a particular eigenvalue. For this equation to have a nontrivial solution, it must be true that det(A−λI) = 0 as discussed on p. 311 of the textbook. Evaluating the determinant, we have

−λ Q2 gA det ( A − λI) = − 2 B A

1 2Q =0 −λ A

2  2Q   gA Q  − λ − λ  −  − 2  = 0  A   B A 

λ2 −

 gA Q 2  2Q − 2  = 0 λ −  A  B A 

The solution comes from the quadratic formula and is given by

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Open Channel Hydraulics

CHAPTER 8

 gA Q 2   1  2Q 4Q 2   4 ± + − 2 A B A 2   A2   

λ= 

λ =V ±

gA =V ±c B

Therefore, these eigenvalues represent the slopes (dx/dt) of the characteristic directions associated with Equation 8.56 since A ∂U/∂x = λ ∂U/∂x.

8.6.

Rederive the Lax diffusive scheme using the method of treating the source term given by (8.45).

Solution.

Equation 8.45 applies to the leapfrog scheme, so it must be modified for the Lax diffusive scheme. Define any variable f with an overbar as

f =

f i −k1 + f i +k1 2

Then the source term can be evaluated in the manner of Equation 8.45 as

φ = g AS 0 −

g A Q  Q k +1 + Q   i  2   2 K  

in which K = channel conveyance. Substituting into the momentum equation in conservation form, Equation 8.41 is modified to produce

k +1 i

Q

k k  1   Q2   ∆t  Q 2 k k  =  (Qi −1 + Qi +1 ) − + gAhc  −  + gAhc   A 2∆x  A    i −1   2 i + 1 

 g A Q Q    g A Q    1 + + ∆t  g AS 0 − ∆t  2 2    K K 2 2    

−1

Note that the difference equation remains explicit because all of the overbar terms are evaluated at the kth time level.

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Open Channel Hydraulics

8.7.

CHAPTER 8

Apply the simple wave method of Chapter 7 to Example 8.1. Calculate the minimum depth at the turbine and compare it with the computer results. What is the maximum discharge that can be supplied to the turbine by the negative wave?

Solution.

Using the methods of previous chapters, the steady-state flow has a normal depth of 7.66 ft from Manning’s equation with n = 0.015, Q = 1000 cfs, S = 0.0002, b = 20 ft, and m = 1.5. The head in the reservoir corresponding to this normal depth is given by

H = y0 +

Q2 1000 2 = + = 7.93 ft 7 .

66 2 gA02 64.4 × [7.66 × (20 + 1.5 × 7.66)]2

When the fluid is at rest prior to the turbine coming on line, the initial downstream depth at the turbine, y i = H + S 0 L = 7.93 + 0.0002×5000 = 8.93 ft. This corresponds to the upstream depth in the dam-break problem. The difference in this exercise is that the channel is trapezoidal instead of rectangular. The simple wave solution that we obtained in Chapter 7 (V−2c = constant) assumed a rectangular channel. It is clear from Equation 7.21 that this simplification does not exist for a nonrectangular channel shape. So for the sake of argument, use a rectangular channel with a width such that the normal depth is the same as in the trapezoidal channel, namely 7.66 ft. The required width is 31.21 ft as can be confirmed from Manning’s equation. It can also be confirmed that y i = H + S 0 L = 8.93 ft for the rectangular channel in agreement with the value for the trapezoidal channel.

The initial wave celerity is c i =(32.2×8.93)1/2 = 16.96 ft/s. The simple wave equation requires that

VT − 2cT = Vi − 2ci = 0 − 2 × 16.96 = −33.9 ft/s in which the subscript T refers to the turbine location at x = 0, and the subscript i represents the initial values before the turbine is started. In addition, continuity must be satisfied at the turbine and it is expressed by

VT AT = −1000 to preserve the sign convention as in Figure 7.13 with the constant depth region existing at the turbine location. We can combine the simple wave equation and continuity equation and solve them simultaneously in the form

−

1000 − 2 gyT = −33.9 31.21 yT

to obtain y T = 6.53 ft. There are two solutions, but the other solution is less than the minimum possible depth at the turbine of (4/9)y i = (4/9) ×8.93 = 3.97 ft. The numerical solution as given in Example 8.1 is y Tmin = 4.94 ft. If the program CHAR is run again for a rectangular channel of width 31.21 ft, the result is y Tmin = 4.83 ft which is very close to the trapezoidal result. It would appear from this example that including friction results in a greater drawdown at the turbine for the same discharge; however, it must be emphasized that the simple wave solution assumes an instantaneous drawdown at the

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turbine that remains constant for subsequent values of time, while the numerical method computes a continuing drawdown with time until the positive wave returns from the reservoir. It is of interest to note that the numerical solution gives a depth at the turbine of 6.45 ft when the discharge reaches 1000 cfs. The maximum possible discharge occurs if dx/dt = 0 at the turbine, which translates into the following expression as shown in Chapter 7 (Equation 7.36):

Qmax =

8 8 byi gyi = × 31.21× 8.93 × 32.2 × 8.93 = 1400 cfs 27 27

in which y i = initial depth of water at rest at the turbine.

8.8.

Apply the momentum and continuity equations in finite volume form (shock compatibility equations) to the surge developed in Example 8.2 and compare the results to the numerical results.

Vs

Solution. y0

y1

V0

Turbine

The normal depth is y 0 = 7.66 ft (A 0 = 241.2 ft2) as determined in Exercise 8.7 for the trapezoidal channel. Applying the continuity equation to a stationary control volume that contains the surge, we have

(Vs + V0 ) A0 = Vs A1 Vs =

V0 A0 1000 = A1 − A0 [ y1 (20 + 1.5 y1 ) − 241.2]

The momentum equation applied to the stationary control volume is given by

M 0 = M1 by 02 my 03 (Vs + V0 ) 2 A02 by12 my13 (Vs + V0 ) 2 A02 + + = + + gy 0 (b + my 0 ) gy1 (b + my1 ) 2 3 2 3 Substituting b = 20, m = 1.5, y 0 = 7.66, A 0 = 241.2, and V 0 = 1000/241.2 = 4.146, we have

811.5 + 7.49(Vs + 4.146) 2 = 10 y12 + 0.5 y13 + (Vs + 4.146) 2 =

0.5 y13 + 10 y12 − 811.5 1807 7.49 − y1 (20 + 1.5 y1 )

188

1807(Vs + 4.146) 2 y1 (20 + 1.5 y1 )

Open Channel Hydraulics

CHAPTER 8

Finally, the expression for V s from the continuity equation can be substituted into the momentum equation and the resulting equation solved by a nonlinear algebraic equation solver to yield y 1 = 9.46 ft and V s = 12.16 ft/s. The numerical solution given in Example 8.2 results in a maximum surge depth of 10.36 ft as it builds up until flow reversal causes it to decrease. It is of interest to note that the numerical solution for the surge depth immediately after shutdown of the turbine compares favorably with the result from the shock compatibility equations.

8.9.

For an earthen dam that is 80 ft in height with a volume of water in storage of 50,000 acft, estimate the time of failure and the average breach width. What will be the height and bottom width of the breach at one third the time to failure?

Solution.

The time of failure is given by Equation 8.78:

∀  g τ =τ = 79 w3  Hd  Hd 

0.47

*

80  50,000 × 43,560  τ= × 79 ×   32.2 80 3  

0.47

= 6321 sec or 1.76 hrs

The average width of the breach comes from Equation 8.77:

b* =

bav = 0.47 k 0 (V * ) 0.25 Hd

 50,000 × 43,560  bav = 80 × 0.47 × 1.4 ×   80 3  

0.25

= 425 ft

in which overtopping has been assumed for the value of k 0 . For t/τ = 1/3 and ρ = 1.0, the height of the breach, h bt , is (2/3)(80) = 53.3 ft from Equation 8.75. The final bottom width of the breach, b f = b av − 2mH d /2 = 425 − 2×1.4×80/2 = 313 ft. Then the bottom width at t/τ = 1/3 and for ρ = 1.0 is 313/3 = 104 ft from Equation 8.76.

8.10.

Use the computer program CHAR (on the website) to route a triangular dam-breach hydrograph with a peak discharge of 100,000 cfs at a time to peak of 1 hour and a base time of 3 hrs in a downstream rectangular prismatic channel having a width of 200 ft, a slope of 0.0003, and Manning's n = 0.025. The initial discharge in the channel is 2500 cfs, and it is 10 mi long. What will be the peak discharge at the downstream boundary and how much time will it take for the peak discharge to arrive?

Solution.

The data entry is organized into two sections: channel data and flood routing data. The values are entered into the following table.

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CHANNEL DATA ENTRY Bed slope Bottom width, ft Side slope Manning’s n Channel length, ft Number of channel reaches, (multiple of 5)

0.0003 200 0 0.025 52,800 50

FLOOD ROUTING DATA ENTRY Base flow discharge, Q B , cfs Peak discharge, Q p , cfs Time to peak, t p , sec Duration of peak, t D , sec Base time, t B , sec Maximum simulation time, t max , sec

Q, cfs

2500 100,000 3600

Qp

0 10800 21600

QB

tp

tD

tB

Time, t

The inflow hydrograph is trapezoidal as defined in the figure, but can also be triangular if t D = 0. The maximum simulation time represents the total time of the routing and should be greater than the base time, in general, to obtain the full outflow hydrograph. The number of channel reaches defines the value of ∆x as the channel length, L, divided by the number of reaches, N. The number of reaches should be divisible by 5 because the velocities and depths are given as output at x/L = 0, 0.2, 0.4, 0.6, 0.8, 1.0. Some experimentation with the number of reaches may be necessary, but generally 50 is sufficient. The maximum allowed by the program is 150 which increases the number of time steps. The maximum number of time steps in the program is 5000. These limits can be increased in the Visual Basic code. The Courant condition is imposed automatically by the program at each time step.

The results indicate a normal depth at base flow of 4.55 ft and a critical depth of 1.69 ft.

The maximum outflow discharge is 60,855 cfs occurring at a time of 119.7 minutes, or approximately 2 hours. The percent reduction in the peak inflow discharge is approximately 39 percent. The plot of the inflow and outflow hydrographs is shown in the following figure.

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FLOOD ROUTING 120000

Discharge, cfs

100000 80000 60000 40000 20000 0 0

100

200

300

400

Time, minutes Inflow

8.11.

Outflow

Apply the computer program CHAR (on the website) for a hydroelectric load acceptance problem in a trapezoidal headrace having a bottom width of 10 m, side slopes of 0.5:1, Manning's n of 0.016, and a bed slope of 0.0001. The steady-state turbine discharge is 40 m3/s, which is brought on line in 2 min. The headrace is 3 km long. Repeat for a slope of 0.0004.

Solution.

The channel data are entered first, and then the load acceptance data are entered in the following tables.

CHANNEL DATA ENTRY Bed slope Bottom width, m Side slope Manning’s n Channel length, m Number of channel reaches (multiple of 5)

0.0001 10 0.5 0.016 3000 50

Q

QT

LOAD ACCEPTANCE DATA ENTRY Turbine discharge, Q T , m3/s Startup time, t s , sec Maximum simulation time, t max , sec

40 120 3600

ts

191

Time, t

Open Channel Hydraulics

CHAPTER 8

The results give a normal depth of 3.26 m and a critical depth of 1.15 m at the steadystate turbine discharge of 40 m3/s. The minimum depth at the turbine is 2.63 m at a time of 16.9 minutes as shown in the next figure which plots the upstream depth at the reservoir and the downstream depth at the turbine. The initial depth at the turbine is 3.62 m, so the depth draws down to a value that is (2.63/3.62)(100) = 73% of its original value.

LOAD ACCEPTANCE (S=0.0001) 4.0

Depth, m

3.0

2.0

1.0

0.0 0

10

20

30

40

50

60

Time, minutes Upstream Depth

Depth at Turbine

If the channel slope is increased to 0.0004, the normal depth becomes 2.12 m, and the critical depth remains the same at 1.15 m. The minimum depth at the turbine is 1.98 m at a time of 18.9 minutes as shown in the next figure which plots the upstream depth at the reservoir and the downstream depth at the turbine. The initial depth at the turbine is 3.47 m, so the depth draws down to a value that is (1.98/3.47)(100) = 57% of its original value.

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LOAD ACCEPTANCE (S=0.0004) 4.0

Depth, m

3.0

2.0

1.0

0.0 0

10

20

30

40

50

60

Time, minutes Upstream Depth

8.12.

Depth at Turbine

Apply the computer program CHAR (on the website) for the same conditions as in Exercise 8.11 except for the load rejection problem.

Solution.

The data input for the load rejection problem is very similar to the load acceptance problem except that the steady-state discharge decreases from 40 m3/s to zero in 120 sec rather than increases.

CHANNEL DATA ENTRY Bed slope Bottom width, m Side slope Manning’s n Channel length, m Number of channel reaches (multiple of 5)

0.0001 10 0.5 0.016 3000 50

Q

QT

LOAD REJECTION DATA ENTRY Turbine discharge, Q T , m3/s Shutdown time, t s , sec Maximum simulation time, t max , sec

40 120 7200

ts

193

Time, t

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CHAPTER 8

The initial state in this exercise is the steady-state discharge of 40 m3/s at the normal depth of 3.26 m as in Exercise 8.11. The depth hydrographs at the turbine and at the upstream reservoir are shown in the following figure. At the turbine, the depth increases rapidly as the turbine is shut down and then more gradually as the water continues to pile up at the turbine. The depth at the turbine reaches a maximum value of 4.12 m at 18.6 minutes before decreasing as the flow reversal begins to take effect at the turbine. The positive wave continues to occur periodically as the flow moves back and forth, but the magnitude of each succeeding positive wave continues to get smaller until the water comes to rest everywhere. The maximum depth at the turbine is 126% greater than the steady-state normal depth of 3.26 m. If the slope is increased to 0.0004, the maximum depth at the turbine is 3.94 m, which is 186% greater than the steady-state depth of 2.12 m. The plot of the depth hydrographs for the slope of 0.0004 is similar to that shown next except that the difference between the upstream and downstream steady-state depths is larger due to the larger slope.

LOAD REJECTION (S=0.0001) 5.0

Depth, m

4.0 3.0 2.0 1.0 0.0 0

20

40

80

60

100

Time, minutes Upstream Depth

194

Depth at Turbine

120

Open Channel Hydraulics

Apply the computer program LAX (on the web site) for the same conditions as in Exercise 8.11 and compare the results with those from CHAR. Solution.

The results of CHAR and LAX for the load acceptance problem of Exercise 8.11 are compared in the figure shown next for a slope of 0.0001. They are virtually indistinguishable even at an exaggerated vertical scale. The minimum depth at the turbine from the LAX results is 2.622 m at 16.6 minutes in comparison to 2.626 m at 16.9 minutes from CHAR. At the larger slope of 0.0004, the minimum depth at the turbine from the LAX results is 1.965 m in comparison to 1.977 m from CHAR.

LOAD ACCEPTANCE (S=0.0001) 4.0

Depth, m

8.13.

CHAPTER 8

3.0

2.0 0

10

20

30

40

50

60

Time, minutes Upstream Depth

Depth at Turbine-CHAR

195

Depth at Turbine-LAX

Open Channel Hydraulics

8.14.

CHAPTER 8

In the load acceptance problem, show that the negative wave cannot supply the steady state turbine discharge if the corresponding Froude number of the uniform flow, F 0 , exceeds 0.319. Set the steady state discharge per unit of channel width, Vy 0 , equal to the unit maximum discharge for the negative wave, which depends on the upstream head, or specific energy, E 0 , if the slope is small. Then solve for F 0 .

Solution.

For a rectangular channel, the upstream head in the reservoir is given by

E0 = H = y 0 +

V02 2g

F2 H = 1+ 0 y0 2 in which y 0 = normal depth of flow and F 0 = Froude number of the uniform flow. Now assuming that the channel slope is small enough that H = initial depth of water at rest at the turbine, Equation 7.36 gives the maximum discharge per unit width supplied by the negative wave at the turbine as

q max =

8 H gH 27

Assuming that the maximum discharge per unit width supplied by the negative wave is equal to the steady-state discharge per unit width, q 0 , and substituting for H/y 0 , we have

8 H =   3/ 2 27  y0  g y0 q0

8  F2  F0 = 1 + 0  27  2 

3/ 2

3/ 2

Solving, the result is F 0 = 0.319.

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Open Channel Hydraulics

8.15.

CHAPTER 8

A steady discharge of 8000 cfs is released by hydroelectric turbines at a dam into the downstream river at full load. The discharge is increased linearly from the minimum release rate of 700 cfs to 8000 cfs in 20 minutes, held steady at 8000 cfs for 2 hr, and brought back down to 700 cfs linearly in 20 minutes. The river cross section is approximately trapezoidal in shape with a bottom width of 300 ft and 1:1 side slopes.

The bed slope is 0.0005 ft/ft, and Manning’s n = 0.035. What will be the time to peak and the peak discharge at a location 25,000 ft downstream of the dam? Use the computer program LAX or CHAR. Solution.

The input data are organized into the following tables:

CHANNEL DATA ENTRY Bed slope Bottom width, ft Side slope Manning’s n Channel length, ft Number of channel reaches (multiple of 5)

0.0005 300 1 0.035 25,000 50 Q, cfs Qp

FLOOD ROUTING DATA ENTRY Base flow discharge, Q B , cfs Peak discharge, Q p , cfs Time to peak, t p , sec Duration of peak, t D , sec Base time, t B , sec Maximum simulation time, t max , sec

700 8000 1200

QB

7200 9600 18000

tp

tD tB

Time, t

The input data represent a controlled release from a dam through the turbines to produce peaking power. The inflow and outflow hydrographs are shown in the following figure.

The peak outflow using CHAR is 7640 cfs, which is nearly 96% of the peak inflow. The trapezoidal inflow hydrograph undergoes a significant change in shape but relatively little attenuation of the peak. Note that at the downstream location, more than 4 hours elapse from the time the turbine release is finished until the discharge returns to the base flow value.

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CHAPTER 8

FLOOD ROUTING 10000

Discharge, cfs

8000 6000 4000 2000 0 0

100

200

300

400

500

Time, minutes Inflow

8.16.

Outflow

In the dam-break problem of Exercise 8.10, repeat the computation using CHAR or LAX, but for a Manning’s n = 0.05 instead of 0.025. Plot on the same axes the maximum stage as a function of distance downstream of the dam for both values of Manning’s n.

Solution.

The input data are identical to that in Exercise 8.10 except that the Manning’s n value is 0.05 instead of 0.025. As shown in the next figure, the peak discharge for the outflow hydrograph is 45,248 cfs at a time of 147 minutes in comparison to the results from Exercise 8.10, which gave a peak discharge of 60,855 cfs at 120 minutes. The larger roughness apparently decreases and delays the peak of the outflow hydrograph as might be expected. Another method of comparison is to plot the maximum depth at various locations downstream of the dam for the two roughness coefficients. The locations are at x/L = 0.0, 0.2, 0.4, 0.6, 0.8, 1.0 as shown in the figure following the hydrographs. There is clearly less decrease in the maximum stage for the channel with the lower roughness coefficient.

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FLOOD ROUTING (n=0.05) 120000

Discharge, cfs

100000 80000 60000 40000 20000 0 0

100

200

300

400

0.8

1

Time, minutes Inflow

Outflow

MAXIMUM FLOOD DEPTHS 60

Depth, ft

50 40 30 20 10 0 0

0.2

0.4

0.6 x/L

n=0.025

199

n=0.05

Open Channel Hydraulics

CHAPTER 8

200

CHAPTER 9. Simplified Methods of Flow Routing 9.1.

A stormwater detention basin has bottom dimensions of 700 ft by 700 ft with interior 4:1 side slopes. The outflow structures consist of a 12-in. diameter pipe laid through the fill on a steep slope with an inlet invert elevation of 100 ft at the bottom of the basin and a broad-crested weir with a crest elevation of 107 ft and a crest length of 10 ft. The top of the berm is at elevation 109 ft. The design inflow hydrograph can be approximated as a triangular shape with a peak discharge of 125 cfs at a time of 8 hrs and with a base time of 24 hr. If the detention basin initially is empty, route the inflow hydrograph through the basin and determine the peak outflow rate and stage. How could these be further reduced?

Explain in detail.

Solution.

The storage-outflow relationship is created first as shown in the following table: Z, ft 0 1 2 3 4 5 6 7 8 9

O, cfs 0 2.1 4.58 6.04 7.21 8.21 9.11 9.92 36.87 85.48

S,cu ft 0 4.956E+05 1.003E+06 1.521E+06 2.051E+06 2.593E+06 3.146E+06 3.712E+06 4.289E+06 4.879E+06

2S/∆t+O, cfs 0 277.5 561.6 851.0 1146.7 1448.6 1757.0 2072.0 2419.9 2796.2

The outflow, O, and the storage, S, depend on the elevation, Z. The first outflow value at Z = 1.0 for the circular pipe is calculated from Equation 6.10a for inlet control with the inlet unsubmerged and an entrance loss coefficient of 0.5 for a square edge in a headwall.

For all remaining Z values, the pipe outflow is obtained from Equation 6.13 for inlet control in a circular concrete pipe having a square edge in a headwall with the entrance submerged. For Z > 7.0, the broad-crested weir discharge is added to the pipe discharge with the weir discharge given by O w = 2.62L(Z−7.0)3/2 in which L = 10 ft. The storage is calculated from the average end area method at each increment of elevation and accumulated. Finally, the 2S/∆t + O values are calculated from the adjacent columns with ∆t = 1 hr = 3600 sec. The routing is executed using Equation 9.8 in the routing table, which is shown next. The outflow values at each time step are obtained by interpolation in the storage-outflow table given previously.

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t, hr 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

I,cfs 0.0 15.6 31.3 46.9 62.5 78.1 93.8 109.4 125.0 117.2 109.4 101.6 93.8 85.9 78.1 70.3 62.5 54.7 46.9 39.1 31.3 23.4 15.6 7.8 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

2S/∆t−O, 2S/∆t+O, cfs cfs 0 0 15.4 15.6 61.3 62.3 137.3 139.4 243.0 246.7 377.5 383.6 540.5 549.4 732.6 743.6 954.0 967.0 1181.4 1196.2 1391.8 1408.0 1585.4 1602.8 1762.4 1780.8 1922.9 1942.1 2064.8 2087.0 2171.5 2213.3 2248.5 2304.3 2300.3 2365.7 2330.9 2401.9 2343.6 2416.9 2341.1 2413.9 2325.7 2395.8 2299.6 2364.8 2264.3 2323.0 2221.2 2272.1 2178.3 2221.2 2142.0 2178.3 2111.3 2142.0 2085.3 2111.3 2063.4 2085.3 2043.6 2063.4 2023.9 2043.6 2004.3 2023.9 1984.9 2004.3 1965.5 1984.9 1946.2 1965.5 1927.0 1946.2 1907.9 1927.0 1888.9 1907.9 1870.0 1888.9 1851.2 1870.0

202

O, cfs 0.00 0.12 0.47 1.06 1.87 3.03 4.47 5.50 6.50 7.37 8.08 8.66 9.17 9.59 11.1 20.9 27.9 32.7 35.5 36.6 36.4 35.0 32.6 29.4 25.4 21.5 18.2 15.3 13.0 11.0 9.9 9.8 9.8 9.7 9.7 9.6 9.6 9.5 9.5 9.4 9.4

Open Channel Hydraulics

CHAPTER 9

The inflow and outflow hydrographs are plotted in the following figure:

Reservoir Routing 140 120

Discharge, cfs

100 80 60 40 20 0 0

5

10

15

20

25

30

35

40

Time, hr Inflow

Outflow

The peak outflow discharge is 36.6 cfs with a peak stage in the detention basin of 7.99 ft. At the peak discharge, the emergency spillway is overflowing with a freeboard of approximately 1 ft with respect to overtopping of the berm. The peak outflow could be reduced further by increasing the storage while using the same outlet structures, or by decreasing the size of the outlet pipe and/or the emergency spillway width.

9.2.

Prove that the temporary storage in reservoir routing is given by the area between the inflow and outflow hydrographs. For triangular inflow and outflow hydrographs, derive a relationship for the detention storage as a function of inflow and outflow peak discharges and the base time of the inflow hydrograph.

Solution.

Q Ip

Inflow, I Storage Outflow, O

Op

t1

tB

Time, t

As illustrated in the figure, the temporary storage occurs only so long as the inflow exceeds the outflow, which occurs up to time t 1 . Integrate Equation 9.1 from zero to t 1 to yield

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Open Channel Hydraulics

t1

t

CHAPTER 9

t

1 1 dS = − d d t I t ∫0 dt ∫0 ∫0 O dt

t1

S t 1 = ∫ ( I − O ) dt 0

in which the temporary storage, S t1 , is obviously equal to the area between the inflow and outflow curves up to time t 1 . This is true regardless of the shape of the hydrographs which are shown to be linear in the figure. For linear hydrographs, we can write the same integration up to the base time, t B , of the inflow hydrograph in algebraic form as

1 S = ( I p − O p )t B 2 9.3.

The inflow hydrograph for a 25,000 mile reach of the Tallapoosa River follows. Route the hydrograph using Muskingum routing with ∆t = 2 hr, ∆x = 25,000 ft, θ = 3.6 hr, and X = 0.2. Plot the inflow and outflow hydrographs and determine the percent reduction in the inflow peak as well as the travel time of the peak. t, hr 0 2 4 6 8 10 12 14 16 18

Q, cfs 100 500 1500 2500 5000 11000 22000 28000 28500 26000

t, hr 20 22 24 26 28 30 32 34 36 38 40

Q, cfs 22000 17500 14000 10000 7000 4500 2500 1500 1000 500 100

Solution.

Calculate the routing coefficients from Equations 9.11, 9.12, and 9.13:

− θX + 0.5∆t − 3.6 × 0.2 + 0.5 × 2 = = 0.0722 θ − θX + 0.5∆t 3.6 − 3.6 × 0.2 + 0.5 × 2 θX + 0.5∆t 3.6 × 0.2 + 0.5 × 2 C1 = = = 0.4433 θ − θX + 0.5∆t 3.6 − 3.6 × 0.2 + 0.5 × 2 θ − θX − 0.5∆t 3.6 − 3.6 × 0.2 − 0.5 × 2 C2 = = = 0.4845 θ − θX + 0.5∆t 3.6 − 3.6 × 0.2 + 0.5 × 2 C0 =

As a check, note that the coefficients have a sum of 1.0 as expected, and that ∆t < 2θX=2×3.6×0.2=1.44 hr. The routing table, in which the outflow hydrograph is determined step by step from Equation 9.10, is given next.

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Time, hr

Inflow, cfs

C 0 *I 2

C 1 *I 1

C 2 *O 1

O 2 , cfs

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50

100 500 1500 2500 5000 11000 22000 28000 28500 26000 22000 17500 14000 10000 7000 4500 2500 1500 1000 500 100 100 100 100 100 100

36 108 181 361 794 1588 2022 2058 1877 1588 1264 1011 722 505 325 181 108 72 36 7 7 7 7 7 7

44 222 665 1108 2217 4876 9753 12412 12634 11526 9753 7758 6206 4433 3103 1995 1108 665 443 222 44 44 44 44 44

48 62 190 502 955 1921 4063 7673 10728 12229 12279 11286 9717 8064 6300 4713 3337 2206 1426 923 558 295 168 106 77

100 129 392 1036 1971 3966 8386 15837 22143 25240 25343 23295 20055 16645 13003 9728 6888 4554 2944 1906 1152 610 347 220 158 128

The peak outflow discharge from the river reach is 25,343 cfs, which is 89 percent of the inflow peak. The travel time of the peak from the routing is 4 hr taken to the nearest 2 hr due to the discretization. This corresponds to the given θ value of 3.6 hr. The plot of the inflow and routed outflow hydrographs is given in the following figure.

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Tallapoosa River Routing

30000

Discharge, cfs

25000 20000 15000 10000 5000 0 0

5

10

15

20

25

30

35

40

Time, hr Inflow , cfs

9.4.

Outflow , cfs

Route the inflow hydrograph of Example 9.6 through the same river, but for a total reach length of 27,000 ft, using the method of characteristics computer program CHAR (on the website). Then use this outflow hydrograph with the inflow hydrograph to derive θ, X, and the Muskingum routing coefficients using the least-squares fitting method. Finally do the Muskingum routing and compare the results with the outflow hydrograph from dynamic routing.

Solution.

The outflow is determined from CHAR for a channel length of 27,000 ft with 50 subreaches. Then the cumulative storage is calculated from Equation 9.17 as shown for the first 9 minutes in the table given next. (There are approximately 900 entries in the full table.)

Time, t minutes

Inflow, I cfs

Outflow, O cfs

I avg cfs

O avg cfs

Storage, S ft3

0.00 0.86 1.72 2.56 3.39 4.20 5.01 5.81 6.60 7.39 8.16 8.93

500 529 557 585 613 640 667 694 720 746 772 798

500 500 500 500 500 500 500 500 500 500 500 500

515 543 571 599 627 654 681 707 733 759 785

500 500 500 500 500 500 500 500 500 500 500

0 7.523E+02 2.959E+03 6.558E+03 1.150E+04 1.772E+04 2.519E+04 3.386E+04 4.368E+04 5.463E+04 6.667E+04 7.976E+04

206

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CHAPTER 9

Equations 9.19, 9.20, and 9.21 are applied for the least squares fitting of the storage vs. weighted flow relationship to obtain A = 2248 sec and B = 3178 sec so that from Equations 9.22, we have θ total = A+ B = (2248 + 3178)/60 = 90.43 minutes and X = A/(A+B) = 2248/(2248+3178) = 0.4143. This value of θ applies to the entire reach with a length of 27,000 ft, but we do the Muskingum routing for three subreaches, each with a length of 9000 ft as determined in Example 9.6, so we take θ = 90.43/3 = 30.14 minutes for each subreach. The routing coefficients for each subreach are calculated from Equations 9.11, 9.12, and 9.13 with ∆t = 30 minutes to yield

− 30.14 × 0.4143 + 0.5 × 30 − θX + 0.5∆t = 0.0770 = θ − θX + 0.5∆t 30.14 − 30.14 × 0.4143 + 0.5 × 30 θX + 0.5∆t 30.14 × 0.4143 + 0.5 × 30 = 0.8418 = C1 = θ − θX + 0.5∆t 30.14 − 30.14 × 0.4143 + 0.5 × 30 θ − θX − 0.5∆t 30.14 − 30.14 × 0.4143 − 0.5 × 30 = 0.0812 = C2 = θ − θX + 0.5∆t 30.14 − 30.14 × 0.4143 + 0.5 × 30 C0 =

Now apply the Muskingum routing procedure as in Example 9.6 but with three reaches as shown in the following routing table.

t, min.

I, cfs

C 0 ×I2

C1×I1

O 2 , cfs O 2 , cfs O 2 , cfs C2×O1 x=9000 ft x=18000 ft x=27000 ft

0 30 60 90 120 150 180 210 240 270 300 330 360 390 420 450

500 1500 2500 3500 4500 4000 3500 3000 2500 2000 1500 1000 500 500 500 500

115 192 269 346 308 269 231 192 154 115 77 38 38 38 38

421 1263 2104 2946 3788 3367 2946 2525 2104 1683 1263 842 421 421 421

41 47 122 203 284 356 325 285 244 204 163 122 82 44 41

500 577 1502 2496 3495 4380 3992 3502 3002 2502 2002 1502 1002 541 503 500

500 506 642 1508 2492 3482 4277 3978 3502 3005 2505 2005 1505 1008 576 509

500 500 516 699 1518 2489 3462 4188 3958 3501 3006 2507 2007 1507 1015 607

The Muskingum routing is shown in the following figure for comparison with the CHAR routing. The peak discharge from the CHAR routing is 3891 cfs, while it is 4188 cfs for the Muskingum routing with some offset from the CHAR value of time to peak.

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M uskingumRoutingfromCHARResults C0=0.077; C1=0.842; C2=0.081 5000

Discharge, cfs

4000 3000 2000 1000 0 0

100

200

300 Time, minutes

Inflow

9.5.

400

CHARRouting

500

600

MuskingumRouting

Derive the kinematic wave celerity for a trapezoidal channel. Plot the ratio of kinematic wave celerity to average channel flow velocity, ck/V, as a function of the aspect ratio b/y for several values of side slope ratio m including m=0 for a rectangular channel and discuss the results.

Solution.

From Equation 9.32, the kinematic wave celerity is given by

ck =

1 dQ B dy

Applying the derivative to the Manning’s equation for discharge Q, we have

dQ K n S 1 / 2 d  A5 / 3    = dy n dy  P 2 / 3  dQ K n S = dy n

1/ 2

2 5/3   5 2/3 A 2 1 + m2   A B   3 2/3 − 3 P5/3   P    

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Open Channel Hydraulics

CHAPTER 9

Now substituting into the expression for kinematic wave celerity, the result is

ck =

K n S 1 / 2  5 2 / 3 4 1 + m 2 R 5 / 3  R −  3 n  3 B 

From Manning’s equation, the flow velocity is V = KnR2/3S1/2/n, and the nondimensional expression for ck as applied to a trapezoidal channel becomes

ck 5 4 1 + m 2 R = − V 3 3 B ck 5 4 y 0 (b + my 0 ) = − 1 + m2 V 3 3 (b + 2 y 0 1 + m 2 )(b + 2my 0 )  b   + m   ck 5 4  y0  = − 1 + m2 V 3 3   b 2  b  y + 2 1 + m  y + 2m   0   0 The final expression is plotted in the figure given next. Note that as the aspect ratio, b/y0, approaches infinity, the value of ck/V approaches the very wide channel value of 5/3 regardless of the value of the sideslope ratio, m. As the aspect ratio approaches zero in a trapezoidal channel, which is just a triangular channel, the value of ck/V approaches the triangular channel value of 4/3. For the rectangular channel, for which m = 0, ck/V approaches 1.0 as the aspect ratio approaches zero.

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Kinematic Wave Celerity 1.8

ck /V

1.6

1.4

1.2

1.0 0.1

1

10

100

1000

b /y 0 m=0

9.6.

m=1

m=2

m=3

m=4

For uniform flow at a depth of 5 cm on a concrete parking lot having a slope of 0.01 and a flow length of 200 m, calculate the kinematic wave number, k. Would kinematic routing apply? Repeat for a very wide river channel with a slope of 0.001, n = 0.035, a depth of 5 m, and the same flow length of 200 m. Discuss the results.

Solution.

Calculate the velocity from Manning’s equation for a very wide channel to give

V=

1.0 2 / 3 1 / 2 1.0 y0 S0 = (0.05) 2 / 3 (0.01)1 / 2 = 0.905 m/s n 0.015

Then the Froude number is F = V/(gy0)1/2 = 0.905/(9.81×0.05)1/2 = 1.29. The kinematic flow number as defined by Equation 9.35 becomes

k=

L0 S 0 200 × 0.01 = = 24.0 2 y0 F 0.05 × 1.29 2

Therefore, for this case of overland flow, k>10 and the kinematic wave approximation is acceptable. For the river case, the velocity is V = (1/0.035)×52/3×0.0011/2 = 2.64 m/s, and the Froude number becomes F = 2.64/(9.81×5)1/2 = 0.377. The kinematic flow number is given by k = (200×0.001/5)/0.3772 = 0.28, for which the kinematic wave approximation is not applicable.

210

Open Channel Hydraulics

9.7.

CHAPTER 9

For the same parking lot of Exercise 9.6, suppose the hydrograph rise time is 15 min.

Would the kinematic wave routing method apply based on Equation 9.37?

Solution.

Substituting into Equation 9.37, we have

TrV0 S 0 (15 × 60) × 0.905 × 0.01 = = 163 y0 0.05 which is considerably greater than the required value of 85 for the kinematic wave approximation to apply, in agreement with the value of the kinematic flow number determined in Exercise 9.6.

9.8.

Route an overland flow hydrograph over a distance of 300 m using the analytical kinematic wave solution with variable wave celerity. The overland flow occurs over a 100 m wide plane strip of constant slope 0.006 and n = 0.015. The inflow hydrograph is triangular with a peak of 1.25 m3/s at a time to peak of 15 min and with a base time of 45 min. The initial base flow is 0.25 m3/s. Plot the inflow and outflow hydrographs and discuss the shape of the outflow hydrograph.

Solution.

The calculations are organized in the following table. The value of the normal depth comes from Manning’s equation for a very wide channel as given by

 nq  y 0 =  1 / 2   S0 

0.6

 0.015 × Q / 100  =  1/ 2  0.006 

0.6

= 0.02356Q 0.6

The velocity is given by the continuity equation as V0 = Q/(by0) = 0.4244Q0.4. The kinematic wave celerity for a very wide channel is (5/3)V0. Then the translation time for a given value of Q is ∆t = L/ck = 300/ck/60 in minutes, and the time of occurrence of that value of Q is t+∆t.

t, min 0 5 10 15 20 25 30 35 40 45 50

Q, m3/s 0.250 0.583 0.916 1.250 1.083 0.917 0.750 0.583 0.417 0.250 0.250

y0, m 0.0103 0.0170 0.0224 0.0269 0.0247 0.0224 0.0198 0.0170 0.0139 0.0103 0.0103

V0, m/s 0.244 0.342 0.410 0.464 0.438 0.410 0.378 0.342 0.299 0.244 0.244

211

ck, m/s 0.406 0.570 0.683 0.773 0.730 0.683 0.631 0.570 0.499 0.406 0.406

∆t, min t+∆t, min 12.31 12.31 8.77 13.77 7.32 17.32 6.46 21.46 6.85 26.85 7.32 32.32 7.93 37.93 8.77 43.77 10.03 50.03 12.31 57.31 12.31 62.31

Open Channel Hydraulics

CHAPTER 9

The results of the routing are shown in the following figure in which it is apparent that using a variable value of the kinematic wave celerity results in steepening of the rising side of the hydrograph and flattening of the receding side.

Analytical Kinematic Wave Routing 1.50

Discharge, m3 /s

1.25 1.00 0.75 0.50 0.25 0.00 0

10

20

30

40

50

60

70

Time, minutes

Inflow

9.9.

Outflow

Derive Equation 9.42 and then show that as yf approaches yi, the monoclinal wave celerity approaches the initial kinematic wave celerity.

Solution.

As shown in the textbook, continuity applied to the monoclinal wave results in the following expression for the monoclinal wave celerity in a very wide channel (Equation 9.41):

cm =

V f y f − Vi y i y f − yi

Now from the Chezy equation, we have V = C(yS)1/2 for a very wide channel so that Vf /Vi = (yf /yi)1/2. The initial kinematic wave celerity based on Chezy’s equation is given by cki = (3/2)Vi. Nondimensionalizing the expression for monoclinal wave celerity and substituting, we have

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Open Channel Hydraulics

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 yf  V f − Vi  yi cm 1   = yf cki (3 / 2)Vi −1 yi  yf    cm 2  y i  = cki 3 y f yi

3/ 2

−1 −1

In the limit as yf /yi approaches 1, both the numerator and denominator approach zero.

Apply l’Hospital’s rule to obtain

3 yf    lim yf →1 2  y i 

c m 2 yi = 3 →1 c ki

lim

yf yi

9.10.

1

1/ 2

=1

Compute and plot the monoclinal wave profile in a very wide river with an initial depth of 1.0 m and a final depth of 4.0 m. The slope of the river is 0.001 and the Manning's n is 0.040. Assume that the depth at x = 0 and t = 0 is 3.99 m. Also calculate the monoclinal wave celerity and compare this to the initial kinematic wave celerity.

Solution.

Calculate Chezy’s C for the initial depth as C = (1.0/n)yi1/6= 25 m1/2/s. Then from Chezy’s equation, the initial and final velocities are given by

Vi = Cy i1 / 2 S 1 / 2 = 25 × 1.01 / 2 × 0.0011 / 2 = 0.7906 m/s V f = Cy 1f / 2 S 1 / 2 = 25 × 4.01 / 2 × 0.0011 / 2 = 1.581 m/s Then from Equation 9.41, the monoclinal wave celerity is given by

cm =

V f y f − Vi y i y f − yi

=

1.581 × 4.0 − 0.7906 × 1.0 = 1.845 m/s 4.0 − 1.0

For comparison, the initial kinematic wave celerity is given by cki = (3/2)Vi = (3/2)(0.7906) = 1.186 m/s. The ratio becomes cm/ cki = 1.845/1.186 = 1.556. The same result could be obtained from Equation 9.42. To obtain the critical depth calculate the overrun discharge as qr = (cm − Vi)yi = (1.845 − 0.7906)×1.0 = 1.054 m2/s. Then the critical depth is given by

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Open Channel Hydraulics

CHAPTER 9

 q2  y c =  r   g 

1/ 3

 1.054 2   =   9.81 

1/ 3

= 0.4838 m

In addition to yc, we need the parameter Y0 calculated from its definition as

Y0 =

1.0 × 4.0

yi y f

( y + y ) ( 1 .0 + 4 .0 ) 2

i

2

= 0.4444

f

Now the constants needed to calculate the monoclinal wave profile can be obtained from Equations 9.48 as

a1 = a2 =

yi3 − y c3 1.0 3 − 0.48383 = = −0.5320 yi ( yi − y f )( yi − Y0 ) 1.0 × (1.0 − 4.0) × (1.0 − 0.4444) y 3f − y c3 yi ( y f − yi )( y f − Y0 )

=

4.0 3 − 0.48383 = 5.9893 1.0 × (4.0 − 1.0) × (4.0 − 0.4444)

Y03 − y c3 0.4444 3 − 0.48383 a3 = = = −0.01290 yi (Y0 − yi )(Y0 − y f ) 1.0 × (0.4444 − 1.0) × (0.4444 − 4.0)

Finally, the constant of integration is obtained from Equation 9.47 by substituting y = 3.99 at x = 0 to yield CI = 24.19. Then the monoclinal wave profile is given by Equation 9.47 as

x=

1.0 [ y − 0.5320 ln( y − 1.0) + 5.9893 ln(4.0 − y) − 0.01290 ln( y − 0.4444) + 24.19] 0.001

The profile is plotted in the following figure.

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Monoclinal Wave Profile 4.0 3.5 3.0

y, m

2.5 2.0 1.5 1.0 0.5 0.0 0

5000

10000

15000

20000

25000

30000

35000

40000

x, m

9.11.

Derive Equation 9.58, the linear diffusion routing equation.

Solution.

By writing the depth and velocity as the sum of an initial or reference value and a small perturbation, i.e. y = yi + y′ and V = Vi + V′, and substituting into the continuity and momentum equations for a wide rectangular channel (without the inertia terms in the momentum equation), we obtain Equations 9.55 and 9.56:

∂y ′ ∂y ′ ∂V ′ + Vi + yi =0 ∂t ∂x ∂x Sf  ∂y ′ + S 0  − 1 = 0 ∂x  S0  From Chezy’s equation, it can be shown that Sf/S0 = (V/Vi)2/(y/yi) which can be expanded as 2

 V + V ′   yi   V ′ V ′ 2  y i      = 1 + 2 + 2  =  i S 0  Vi   y i + y ′   Vi Vi  y i + y ′ 

Sf

215

Open Channel Hydraulics

CHAPTER 9

Neglecting the squared term and rearranging, we have

V′ V ′ y′ 2 − Sf Vi Vi y i V ′ y′ −1 = −1 = ≅2 − ′ ′ y y S0 Vi y i 1+ 1+ yi yi 1+ 2

So the momentum equation becomes

 V ′ y′  ∂y ′ + S 0  2 −  = 0 ∂x  Vi y i  as given by Equation 9.57. Differentiate the momentum equation in this form with respect to x to yield

 ∂V ′ / ∂x 1 ∂y ′  ∂ 2 y′ =0 − + S 0  2 2 Vi y i ∂x  ∂x  But from the continuity equation, we have

∂V ′ 1  ∂y ′ ∂y ′  + Vi =−   ∂x y i  ∂t ∂x  Substituting into the differentiated momentum equation and rearranging, we obtain

Vi y i ∂ 2 y ′ ∂y ′ 3 ∂y ′ = + Vi ∂t 2 ∂x 2 S 0 ∂x 2 which is precisely Equation 9.58, the linear diffusion routing equation, with D = Viyi/2S0; ck = (3/2)Vi; and φ = y′.

9.12.

Using the numerical approximation of Equation 9.60 for the kinematic wave equation with X = 1 and Y = 0, derive the routing coefficients and route the hydrograph of Example 9.4 for Cn = 1.5. With reference to Equation 9.62, what happens to the outflow hydrograph if Cn = 1.0? Explain why the method becomes unstable for Cn < 1.0.

Solution.

Equation 9.60 can be rearranged and the routing coefficients determined for general values of X, Y, and Cn as follows:

216

Open Channel Hydraulics

CHAPTER 9

C0 =

C n (1 − Y ) − X 1 − X + C n (1 − Y )

C1 =

X + C nY 1 − X + C n (1 − Y )

C2 =

1 − X − C nY 1 − X + C n (1 − Y )

Now for the special case of X = 1, Y = 0, and Cn = 1.5, the routing coefficients become

C0 = 1 − C1 =

1 1 = 1− = 0.3333 Cn 1.5

1 1 = = 0.6667 C n 1.5

C2 = 0 The routing table with the solution of the routing equation (Equation 9.14) using these values of the routing coefficients is given next followed by a figure showing the inflow and outflow hydrographs. For this case, Equation 9.62 indicates nonzero values of the coefficients on both the diffusion and dispersion terms of the remainder with Cn = 1.5, and this is reflected in the numerical solution for the outflow hydrograph. If Cn = 1, both remainder terms go to zero, and the numerical solution is identical to the analytical solution. For Cn < 1, the coefficient on the diffusion term is negative, so an unstable numerical solution results.

t, hr 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5

I, m3/s 0 100 200 300 400 500 600 500 400 300 200 100 0 0

C0×I2

C1×I1

33.33 66.67 100.00 133.33 166.67 200.00 166.67 133.33 100.00 66.67 33.33 0.00 0.00

0.00 66.67 133.33 200.00 266.67 333.33 400.00 333.33 266.67 200.00 133.33 66.67 0.00

217

Numerical Analytical O2, m3/s O2, m3/s 0.00 0 0.00 33.33 0 0.00 133.33 100 0.00 233.33 200 0.00 333.33 300 0.00 433.33 400 0.00 533.33 500 0.00 566.67 600 0.00 466.67 500 0.00 366.67 400 0.00 266.67 300 0.00 166.67 200 0.00 66.67 100 0.00 0.00 0

C2×O1

Open Channel Hydraulics

CHAPTER 9

800 X=1; Y=0 Cn=1.5

Q, cu m/s

600

400

200

0 0

1

2

3

Inflow

9.13.

4 5 Tim e, hrs

Outflow

6

7

8

Outflow,Analytical

Using the numerical approximation of Equation 9.60 for the kinematic wave equation with X = 0 and Y = 1, derive the routing coefficients and route the hydrograph of Example 9.4 for Cn = 0.67. With reference to Equation 9.62, what happens to the outflow hydrograph if Cn = 1.0? Explain why the method becomes unstable for Cn > 1.0.

Solution.

Applying the same general solution for the routing coefficients given in Exercise 9.12, the values of the routing coefficients for this exercise (X = 0, Y = 1, Cn = 0.67) become

C0 = 0 C1 = C n = 0.67 C 2 = 1 − C n = 1 − 0.67 = 0.33 The routing table with the solution of the routing equation (Equation 9.14) using these values of the routing coefficients is given next followed by a figure showing the inflow and outflow hydrographs. For this case, Equation 9.62 indicates nonzero values of the coefficients on both the diffusion and dispersion terms of the remainder for Cn = 0.67, and this is reflected in the numerical solution for the outflow hydrograph. If Cn = 1, both remainder terms go to zero, and the numerical solution is identical to the analytical solution as in Exercise 9.12. For Cn > 1, the coefficient on the diffusion term is negative, so an unstable numerical solution results.

218

Open Channel Hydraulics

I, m3/s 0 100 200 300 400 500 600 500 400 300 200 100 0 0

C0×I2

C1×I1

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

0 67 134 201 268 335 402 335 268 201 134 67 0

Numerical Analytical O2, m3/s C2×O1 O2, m3/s 0.00 0 0.00 0.00 0 0.00 67.00 100 22.11 156.11 200 51.52 252.52 300 83.33 351.33 400 115.94 450.94 500 148.81 550.81 600 181.77 516.77 500 170.53 438.53 400 144.72 345.72 300 114.09 248.09 200 81.87 148.87 100 49.13 49.13 0

800 X=0; Y=1 C=0.67 n

600

Q, cu m/s

t, hr 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5

CHAPTER 9

400

200

0 0

1

2

3

Inflow

4 5 Tim e, hrs

Outflow

219

6

7

8

Outflow,Analytical

Open Channel Hydraulics

9.14.

CHAPTER 9

Using the numerical approximation of Equation 9.60 for the kinematic wave equation with X = 0 and Y = 0, derive the routing coefficients and route the hydrograph of Example 9.4 for Cn = 1.0. With reference to Equation 9.62, what happens to the outflow hydrograph if Cn ≠ 1.0? Does the method become unstable for any value of Cn ? Explain.

Solution.

Applying the same general solution for the routing coefficients given in Exercise 9.12, the values of the routing coefficients for this exercise (X = 0, Y = 0, Cn = 1.0) become

C0 =

Cn 1 = 1 + Cn 2

C1 = 0 C2 =

1 1 = 1 + Cn 2

The routing table with the solution of the routing equation (Equation 9.14) using these values of the routing coefficients is given next followed by a figure showing the inflow and outflow hydrographs. For this case, Equation 9.62 indicates a nonzero value of the coefficient on the diffusion term but a zero value on the dispersion term of the remainder, and this is reflected in the numerical solution for the outflow hydrograph. If Cn ≠ 1, both remainder terms are finite so that numerical diffusion and dispersion both occur regardless of the value of Cn. Instability does not occur for any positive value of Cn.

t, hr 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5

I, m3/s 0 100 200 300 400 500 600 500 400 300 200 100 0 0

C0×I2

C1×I1

50.00 100.00 150.00 200.00 250.00 300.00 250.00 200.00 150.00 100.00 50.00 0.00 0.00

0 0 0 0 0 0 0 0 0 0 0 0 0

220

Numerical Analytical O2, m3/s O2, m3/s 0.00 0 0.00 50.00 0 25.00 125.00 100 62.50 212.50 200 106.25 306.25 300 153.13 403.13 400 201.56 501.56 500 250.78 500.78 600 250.39 450.39 500 225.20 375.20 400 187.60 287.60 300 143.80 193.80 200 96.90 96.90 100 48.45 48.45 0 C2×O1

Open Channel Hydraulics

CHAPTER 9

800 X=Y=0

Q, cu m/s

600

400

200

0 0

1

2

3

Inflow

9.15.

4 5 Tim e, hrs

Outflow,Cn=1.0

6

7

8

Outflow,Analytical

Express the routing coefficients of the Muskingum-Cunge method in terms of λ and Cn. Solution.

If we define λ = Q/(BS0ck∆x), then from Equation 9.67, X = 0.5(1 − λ). Substituting for X in Equations 9.64, 9.65, and 9.66 and simplifying, the results for the routing coefficients are

9.16.

C0 =

Cn + λ − 1 Cn + λ + 1

C1 =

Cn − λ + 1 Cn + λ + 1

C2 =

− Cn + λ + 1 Cn + λ + 1

Use the Muskingum-Cunge method to route the hydrograph of Example 9.6 for S0 = 0.001 and compare with the results from the computer program CHAR. Solution.

For an average inflow discharge of 2500 cfs, the normal depth from Manning’s equation, with S0 = 0.001, b = 100 ft, m = 2, and n = 0.025, is y0 = 4.65 ft. The corresponding

221

Open Channel Hydraulics

CHAPTER 9

velocity from continuity is 2500/[4.65×(100+2×4.65)] = 4.92 ft/s. The kinematic wave celerity for a wide channel is given by ck = (5/3) ×4.92 = 8.20 ft/s. Use a time discretization of ∆t = 0.5 hr. Then from (9.73), the spatial discretization is limited by

∆x ≤

Q  1 2500 1   =  8.20 × 0.5 × 3600 +  c k ∆t +  = 8665 ft BS 0 ck  2  118.6 × 0.001 × 8.20  2

Because this is very close to 9000 ft, use ∆x = 9000 with two routing reaches as in Example 9.6. The value of Cn = ck∆t/∆x = 8.20×0.5×3600/9000 = 1.64 and the value of X is given by

X =

1 2500  1 −  = 0.357 2  118.6 × 0.001 × 8.20 × 9000 

The values of the routing coefficients are computed from Equations 9.64, 9.65, and 9.66 with the results given by

C 0 = 0.3165;

C1 = 0.8045;

C 2 = −0.1210

The routing results are shown in the following table and plotted in the figure given after the table. The peak outflow discharge from the method of characteristics routing (CHAR) is 4245 cfs, while it is 4252 cfs from the Muskingum-Cunge method.

t, hr

I, ft3/s

C0×I2

C1×I1

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5

500 1500 2500 3500 4500 4000 3500 3000 2500 2000 1500 1000 500 500 500 500

475 791 1108 1424 1266 1108 950 791 633 475 317 158 158 158 158

402 1207 2011 2816 3620 3218 2816 2414 2011 1609 1207 805 402 402 402

222

x=9000 ft x=18000 ft O2, ft3/s C2×O1 O2, ft3/s -61 -99 -230 -350 -471 -534 -459 -400 -339 -279 -218 -158 -97 -56 -61

500 817 1899 2889 3890 4416 3791 3306 2805 2305 1805 1305 805 463 504 499

500 600 1185 2299 3278 4131 4252 3582 3114 2609 2110 1610 1110 660 452 509

Open Channel Hydraulics

CHAPTER 9

ROUTING COMPARISON S=0.001 (C0=0.3165; C1=0.8045; C2=-0.1210)

5000

4000

Q , cfs

3000

2000

1000

0 0

2

4

6

8

Inflow

CHAR

10

Time, hr Muskingum-Cunge

9.17.

Repeat the Muskingum-Cunge routing of Example 9.6 for a very wide channel using the three-point variable parameter method with ck and λ calculated from three-point averages of Q at each time step. Compare the results with the constant parameter method.

Solution.

At each time step, the average Q is computed from the known values of Q as

Qav = ( I1 + I 2 + O1 ) / 3 = (Qik + Qik +1 + Qik+1 ) / 3 For each value of Qav, the value of y0 is calculated from Manning’s equation for the given trapezoidal channel, and the velocity, V0, is obtained from continuity. Then ck = (5/3)V0; Cn = ck ∆t/∆x; and X = 0.5(1 −λ) from Equation 9.67. Equations 9.64, 9.65, and 9.66 provide the values of the routing coefficients as a function of the current values of Cn and X to be used in the current time step only. The routing equation (Equation 9.14) is solved for O2, or (Qi+1 )k+1, which allows the solution to proceed to the next time step when all calculations are repeated for a new value of Qav. These calculations are summarized in the two tables that follow. The first table shows the routing for the first subreach of 9000 ft, and the second table gives the routing in the second subreach with the outflow from the first reach becoming the inflow for the second reach. The calculation of the normal depth is cumbersome in this procedure if a spreadsheet is used, although the “goal-seek” feature can be used in Excel, for example. A simple computer program could also be written.

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The routing computations are summarized in the figure following the two tables, and it can be seen that the variable-parameter Muskingum-Cunge (VPMC) routing has captured the steepening of the rising side of the hydrograph and flattening of the recession side in contrast to the constant parameter routing. The agreement with the method of characteristics solution is quite good for this case.

Qav, Time, Inflow, hr cfs cfs 0.0 500 833.3 0.5 1500 1537.8 1.0 2500 2531.9 1.5 3500 3574.0 2.0 4500 4105.1 2.5 4000 3907.2 3.0 3500 3438.7 3.5 3000 2946.2 4.0 2500 2452.6 4.5 2000 1960.3 5.0 1500 1470.3 5.5 1000 984.3 6.0 500 672.6 6.5 500 541.9 7.0 500 511.2 7.5 500 336.4

y0, ft 2.980 4.286 5.751 7.037 7.629 7.413 6.880 6.285 5.644 4.947 4.174 3.290 2.623 2.307 2.228 1.736

ck, ft/s 4.398 5.508 6.581 7.420 7.781 7.651 7.322 6.940 6.508 6.010 5.419 4.678 4.060 3.743 3.661 3.122

Qav, Time, Inflow, hr cfs cfs 0.0 500.0 537.8 0.5 613.4 903.8 1.0 1595.7 1678.6 1.5 2722.1 2762.1 2.0 3815.2 3675.4 2.5 4221.6 3976.8 3.0 3816.1 3724.9 3.5 3338.7 3280.2 4.0 2857.7 2809.0 4.5 2380.8 2339.9 5.0 1910.9 1879.7 5.5 1452.9 1435.7 6.0 1017.8 1020.9 6.5 625.8 732.1 7.0 533.6 585.6 7.5 509.2 362.8

y0, ft 2.296 3.127 4.514 6.052 7.153 7.489 7.209 6.693 6.112 5.490 4.826 4.115 3.362 2.759 2.416 1.816

ck, ft/s 3.732 4.533 5.685 6.785 7.492 7.697 7.526 7.204 6.825 6.401 5.921 5.372 4.742 4.191 3.854 3.214

X 0.312 0.235 0.152 0.082 0.051 0.062 0.091 0.123 0.158 0.197 0.242 0.293 0.333 0.353 0.358 0.388

X 0.353 0.303 0.222 0.136 0.076 0.058 0.073 0.101 0.133 0.167 0.204 0.245 0.289 0.325 0.346 0.383

224

Cn

C0

C1

0.880 0.113 1.102 0.240 1.316 0.336 1.484 0.397 1.556 0.421 1.530 0.413 1.464 0.391 1.388 0.363 1.302 0.330 1.202 0.288 1.084 0.231 0.936 0.149 0.812 0.068 0.749 0.021 0.732 0.008 0.624 -0.082

0.666 0.597 0.538 0.497 0.480 0.486 0.501 0.520 0.542 0.569 0.603 0.648 0.689 0.712 0.718 0.758

Cn

C0

C1

0.746 0.019 0.907 0.131 1.137 0.257 1.357 0.352 1.498 0.402 1.539 0.416 1.505 0.405 1.441 0.383 1.365 0.355 1.280 0.321 1.184 0.279 1.074 0.226 0.948 0.156 0.838 0.086 0.771 0.038 0.643 -0.066

0.712 0.658 0.587 0.528 0.493 0.484 0.492 0.507 0.526 0.548 0.574 0.605 0.644 0.680 0.704 0.751

C2 x=9000 O2, cfs 0.220 500 0.163 613 0.126 1596 0.106 2722 0.099 3815 0.101 4222 0.108 3816 0.116 3339 0.128 2858 0.144 2381 0.166 1911 0.203 1453 0.243 1018 0.267 626 0.274 534 0.324 509

C2 x=18000 O2, cfs 0.268 500 0.212 502 0.156 718 0.120 1749 0.104 2989 0.101 3892 0.104 4020 0.110 3644 0.119 3188 0.131 2728 0.147 2275 0.168 1836 0.200 1419 0.234 1037 0.258 714 0.315 579

Open Channel Hydraulics

CHAPTER 9

5000

4000

Q, cfs

3000

2000

1000

0 0

6

4

2

8

10

Tim e, hrs

9.18.

Inflow

M uskingum -CungeO utflow

M OCO utflow

VPM COutflow

Write a computer program for Muskingum-Cunge routing using the four-point variable parameter method and apply it to Example 9.6.

Solution.

This is a programming exercise left to the student, but a flow chart is given next. In the main loop with index K, the values of outflow are determined at each time step for a single subreach. The three-point average Q is used to start an inner iteration on the fourpoint average Q that continues until the change in average Q is smaller than some specified value (ERR). A counter (N) is used in case convergence is not occurring. A subroutine (YNORMAL) is used to calculate the normal depth for each value of QAV.

225

Open Channel Hydraulics

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START

Input slope, geometry, roughness, inflow(t), ERR

Input ∆x and ∆t

NS = 1.5*Base time/∆t O(1) = I(1)

Loop for K = 1 to NS−1 N=1 QAV = (I(K)+I(K+1)+O(K))/3 CALL YNORMAL (QAV, y0)

Compute CK, X, C0, C1, C2

No

O(K+1)=C0*I(K+1)+C1*I(K)+C2*O(K)

N>10?

QAVOLD = QAV QAV = (I(K)+I(K+1)+O(K)+O(K+1))/4

N = N+1 Yes Next K

ABS(QAV− QAVOLD)<ERR?

No

Loop Finished STOP

PRINT

226

STOP – check for nonconvergence

Yes

Open Channel Hydraulics

CHAPTER 9

227

CHAPTER 10. Flow in Alluvial Channels 10.1.

Derive the Engelund-Hansen equation (10.10) for fall velocity of natural sands and gravels from their proposed relationship for C D (= 24/Re + 1.5). Then plot the EngelundHansen equation for fall velocity (cm/s) vs. particle diameter (mm) for sand to gravel size ranges in water at 20oC and compare with the plot of calculated fall velocity of quartz spheres in the same sedimentation fluid (on the same graph).

Solution.

Square both sides of Equation 10.4 and multiply both sides by d 2/ν 2 to yield

 4γ s  − 1 gd 3 4 d 3 * w d 3 γ  3 = = CD ν2 ν 2C D 2 f

2

in which the definition of d * from Equation 10.8 has been substituted. Upon substituting the Engelund-Hansen expression for the drag coefficient and rearranging, we have a quadratic equation in the Reynolds number given by

4 3 d * Re 2 3 Re = 24 + 1.5Re 4 1.5Re 2 + 24Re − d *3 = 0 3 The solution is given by the quadratic formula:

Re =

− 24 + 24 2 + 8d *3 3

= −8 + 8 1 + (8 / 24 2 )d *3

Re = 8[−1 + 1 + 0.0139d *3 ] in agreement with Equation 10.10. The fall velocity is calculated for sand and gravel using this equation, and the results are shown next in a table. In addition, for quartz spheres of the same diameter, the fall velocity is obtained from Equation 10.4 and the coefficient of drag is calculated from Equation 10.5 in an iterative fashion starting with an assumed C D as shown in the table. These fall velocities are compared in the figure following the table to illustrate the influence of particle shape on fall velocity.

227

Open Channel Hydraulics

d s , mm 0.062 0.125 0.25 0.5 1 2 4 8 16 32 64

CHAPTER 10

Sand & Gravel Spheres Equation 10.10 Eqs. 10.4, 10.5 d* Re w f , cm/s C D w f , cm/s 1.57 0.212 0.34 120.9 0.33 3.16 1.598 1.28 19.39 1.18 6.32 9.000 3.60 4.806 3.35 12.65 35.17 7.03 1.870 7.60 25.30 112.3 11.23 1.025 14.51 50.59 331.5 16.57 0.7068 24.71 101 952.0 23.80 0.5632 39.15 202 2707 33.84 0.4909 59.31 405 7672 47.95 0.4520 87.41 809 21714 67.86 0.4302 126.70 1619 61431 95.99 0.4177 181.85

Comparison of Fall Velocities 1000

Fall Velocity, cm/s

100

10

1

0.1 0.01

0.1

1

10

Diameter, mm Sand & Gravel

Quartz spheres

228

100

Re 0.2063 1.474 8.377 37.98 145.1 494.2 1566 4744 13985 40545 116381

CD' 120.9 19.39 4.806 1.870 1.025 0.7068 0.5632 0.4909 0.4520 0.4302 0.4177

Open Channel Hydraulics

10.2

CHAPTER 10

Derive Rubey's equation from his proposed relationship for C D (= 24/Re + 2). Plot the results for fall velocity for sand to gravel size ranges and compare with the results from the Engelund-Hansen equation on the same graph.

Solution.

Manipulate Equation 10.4 as in Exercise 10.1 to obtain

 4γ s  − 1 gd 3 4 d 3 * w d 3 γ  3 = = CD ν2 ν 2C D 2 f

2

Then substitute the Rubey expression for the coefficient of drag and rearrange to produce a quadratic equation:

4 2.0Re 2 + 24Re − d *3 = 0 3 for which the solution is given by the quadratic formula as

− 24 + 242 + (32 / 3)d*3 Re = = = −6 + 36 + (2 / 3)d*3 ν 4 ν  2  w f = − 6 + 36 + d*3  ds  3  wf ds

Then multiply and divide the right-hand side by [d * 3]1/2 to yield the familiar form of the Rubey equation, as given by Equations 10.51 and 10.52, for fall velocity in terms of d * :

 36 36 2  w f = − +  ( SG − 1) gd s + 3 3 3 d d * *  The fall velocities from the Engelund-Hansen equation (Equation 10.10) and the Rubey equation are compared in the figure given next. The Rubey equation is seldom used except in the Einstein bedload equation.

229

Open Channel Hydraulics

CHAPTER 10

Com parison of Fall Velocities

Fall Velocity, cm/s

100

10

1

0 0.01

0.1

1

10

100

Diam eter, m m Engelund-Hansen

10.3

Rubey

A river sand has a sieve diameter of 0.3 mm. Find the fall velocity at 20oC using two methods: (1) use Figures 10.2 and 10.3, (2) use the Engelund-Hansen equation plotted in Exercise 10.1.

Solution.

From Figure 10.3, the fall diameter is essentially the same as the sieve diameter of 0.3 mm for a shape factor of 0.7. The value of d * becomes

1.65 × 9.81× 0.00033  d* =   (1× 10 −6 ) 2  

1/ 3

= 7.6

Then from Figure 10.2, Re ≅ 12.5 and w f = 12.5 × (1 × 10−6)/0.0003 = 0.042 m/s.

In the second method, substitute directly into Equation 10.10 with the same value of d * in this case to produce

wf ds

ν

[

]

= 8 × 1 + 0.0139 × 7.6 3 − 1 = 13.3

from which w f = 13.3 × (1 × 10-6)/0.0003 = 0.044 m/s. Note that for d s = 10 mm, the results for fall velocity are approximately 0.36 m/s by the first method and 0.38 m/s by the second method.

230

Open Channel Hydraulics

10.4

CHAPTER 10

A river sand has a measured fall velocity in water of 10 cm/s at 20oC. Find the sedimentation diameter.

Solution.

From the figure given in Exercise 10.1, it can be surmised that the sedimentation diameter is of the order of 1.0 mm, which is outside the Stokes range (d < 0.1 mm).

Iteration using Equation 10.4 and Figure 10.1 (or Equation 10.5) is required. If we iterate on the coefficient of drag, then we solve Equation 10.4 for the diameter, d, to give

d=

w 2f C D 4 ( SG − 1) g 3

=

0.10 2 C D 4 × 1.65 × 9.81 3

= (4.63 × 10 −4 )C D

The particle Reynolds number for an equivalent sphere having a fall velocity of 0.10 m/s becomes

Re =

wf d

ν

=

0.10d = (1× 10 5 )d −6 1× 10

The iteration is done in the table given next starting with a C D value of 1.0 and applying underrelaxation to speed convergence.

CD 1.0 1.5 1.45 1.42

d, m 0.000463 0.000694 0.000671 0.000657

Re 46.3 69.4 67.1 65.7

C D (Eq. 10.5) 1.7 1.4 1.41 1.42

The result is a sedimentation diameter of 0.66 mm.

10.5

Plot the results of the sieve analysis given below on lognormal paper. Find d 84.1 , d 15.9 , d 50 , d g , and σ g . Derive a formula for d 65 and d 90 in terms of σ g and d g using the theoretical lognormal distribution: Sieve diameter, mm 0.495 0.417 0.351 0.295 0.246 0.208 0.175 0.147 0.124 0.104 0.088 0.074 Pan

Grams retained 0.36 1.52 4.32 7.40 8.80 7.04 4.56 2.88 1.52 0.80 0.44 0.20 0.16

231

Open Channel Hydraulics

CHAPTER 10

Solution.

The cumulative mass of sediment retained and sediment passing a given sieve size are calculated in the table given below. Then the “Percent Passing” is determined as the cumulative mass of sediment passing a given sieve size divided by the total mass of sediment, which is 40 g. The complete results are shown in the table.

Dia.,mm Retained,g Cumulative Cumulative Percent retained,g. passing,g. Passing 0.495 0.36 0.36 39.64 99.1 0.417 1.52 1.88 38.12 95.3 0.351 4.32 6.2 33.8 84.5 0.295 7.4 13.6 26.4 66 0.246 8.8 22.4 17.6 44 0.208 7.04 29.44 10.56 26.4 0.175 4.56 34 6 15 0.147 2.88 36.88 3.12 7.8 0.124 1.52 38.4 1.6 4 0.104 0.8 39.2 0.8 2 0.088 0.44 39.64 0.36 0.9 0.074 0.2 39.84 0.16 0.4 Pan 0.16 40

The results from the table are plotted in the figure given next. The vertical scale is a normal probability scale while the horizontal scale is logarithmic. The vertical grid as well as the ordinates of the data points are plotted using the cumulative distribution function (Eq. 10.12) for the normal probability density function where “Percent Finer” = F(ζ) × 100 and ζ = (log d s −µ)/σ. On the spreadsheet chart created using Excel, the vertical scale is set up as linear in ζ with the value of ζ determined for a given value of F(ζ) from table lookup functions applied to a table of ζ vs. F(ζ). This latter table is calculated in the spreadsheet using the error function (erf):

F (ζ ) = ±0.5 erf (

ζ 2

) + 0.5

In the figure shown next, it can be observed that d 84.1 = 0.35 mm and d 15.9 = 0.18 mm.

Then the geometric mean diameter, d g , and the geometric standard deviation, σ g , are determined as

d g = d84.1d15.9 = 0.35 × 0.18 = 0.25 mm

σg =

0.35 d84.1 = = 1.39 0.18 d15.9

Also from the figure, note that d 50 = 0.26 mm which is close to the value of d g indicating that the sediment size distribution is very close to the theoretical lognormal distribution.

232

Open Channel Hydraulics

CHAPTER 10

Lognormal Size Distribution 99.99 99.95 99.8 99 98 95 90

Percent Finer

84.1 80 70 60 50 40 30

50

20

15.9

10 5

1 0.2 0.1 0.05 0.01

0.001

0.01

0.1

1

10

Sieve Size, mm

Values of d 90 and d 65 are needed for various procedures introduced in this chapter, and they can be determined in terms of d g and σ g from the theoretical lognormal distribution.

For F(ζ) = 0.9, the value of ζ = 1.282 for the standard normal distribution so that we have

log d 90 − log d g log σ g

= 1.282

d 90 = d g σ 1g.282 Similarly, it can be shown that

d 65 = d g σ g0.385

233

Open Channel Hydraulics

10.6

CHAPTER 10

For a slope of 0.002 in a wide stream, calculate the depth at which sediment motion begins for a fine gravel with d 50 = 3.3 mm.

Solution.

Assuming fully-rough turbulent flow so that τ *c = 0.045 from the Shields diagram, Equation 10.22 applies for a very wide channel in which τ c = τ 0 . Solving Equation 10.22 for the depth, we have

y0 =

d 50 3.3 / 305 = = 0.40 ft (0.12 m) 13.5S0 13.5 × 0.002

This is a relatively shallow depth above which sediment motion will be initiated. The corresponding value of the shear stress is γ y 0 S = 62.4×0.4×0.002 = 0.05 lbs/ft2. As a check on fully-rough turbulent flow, first calculate u * = (gy 0 S 0 )1/2 = (32.2×0.4×0.002)1/2 = 0.16 ft/s. Then Re * = u * d 50 /ν = 0.16×(3.3/305)/1.0×10−5 = 173, which is well above the lower limit of the fully-rough turbulent regime on the Shields diagram.

10.7

Bank-full depth for a stream in well-rounded alluvium with d 50 = 12 mm is 1.5 m. The stream slope is 0.0005. The channel shape is approximately trapezoidal with 2:1 side slopes and a bottom width of 10 m. Calculate the bank-full discharge and determine whether the bed and banks are stable at bank-full conditions.

Solution.

From the discussion at the end of section 4.13 in Chapter 4, take c n = 0.04 so that we have 1/ 6 n = 0.04d 50 = 0.04 × (12 / 305)1 / 6 = 0.023

Then from Manning’s equation, bank-full discharge is given by

Q=

1.0 [1.5 × (10 + 2 × 1.5)]5 / 3 1.0 × × 0.00051 / 2 = 21.0 m 3 /s AR 2 / 3 S01 / 2 = 2/3 0.023 [10 + 2 × 1.5 × 5 ] n

Now to check bed stability, first calculate the critical shear stress, which depends on d * :

 ( SG − 1) gd 503  d* =   ν2  

1/ 3

1.65 × 9.81× 0.012 3  =  (1× 10 −6 ) 2  

1/ 3

= 303.6

This is in the fully-rough turbulent region in Figure 10.6 where τ *c = 0.045 so that τ c = 0.045×1.65×9810×0.012 = 8.74 N/m2. Using the Bureau of Reclamation procedure, we calculate the maximum bed shear stress as

(τ 0 ) max = 0.97γy 0 S 0 = 0.97 × 9810 × 1.5 × 0.0005 = 7.14 N/m 2 which is less than the critical shear stress so that the channel bed is stable.

234

Open Channel Hydraulics

CHAPTER 10

To check bank stability, first calculate the tractive force ratio, K r , with the angle of repose, φ ≅ 28° from Figure 4.13 and θ = tan−1(1/2) = 26.6°:

Kr = 1−

sin 2 θ sin 2 (26.6) = − = 0.91 1 sin 2 φ sin 2 (28)

Then the critical wall shear stress, τ c w = K r τc = 0.91×8.74 = 7.95 N/m2, and the maximum wall shear stress is given by

(τ 0w ) max = 0.75γy 0 S 0 = 0.75 × 9810 × 1.5 × 0.0005 = 5.52 N/m 2 which is less than the critical value. Therefore, both the bed and banks are stable for these conditions.

10.8

A roadside drainage ditch has to carry a discharge of 100 cfs. It has a bottom width of 5 ft and side slopes of 2:1. The ditch is to be lined with locally-available gravel which has d50=25 mm. At what maximum slope can the channel be constructed for stability?

Solution.

First, calculate the critical shear stress on the bed and banks. Because this is a coarse gravel, it is clear from Figure 10.6 that the flow is fully-rough turbulent and τ*c = 0.045 (d* = 632). The critical shear stress on the bed is τc = 0.045×1.65×62.4×(25/305) = 0.380 lbs/ft2. On the bank, we have φ ≅ 32° for very rounded gravel from Figure 4.13 and θ = tan−1(1/2) = 26.6° so that the tractive force ratio becomes

Kr = 1 −

sin 2 θ sin 2 (26.6) = 1 − = 0.535 sin 2 φ sin 2 (32)

Then the critical shear stress on the bank is τcw = 0.535×0.380 = 0.203 lbs/ft2. The maximum shear stress depends on the slope and the normal depth, which is also a function of slope from Manning’s equation. Set up an iteration in the table that follows to find the maximum slope for stability. Set Manning’s n = 0.04(d50)1/6 = 0.04×(25/305)1/6 = 0.026 and solve Manning’s equation in the form

[ y 0 (5 + 2 y 0 )]5 / 3 [5 + 2 5 y 0 ]

2/3

=

0.026 × 100 1.745 nQ = = 1/ 2 1/ 2 1.49S 01 / 2 1.49S 0 S0

with a variable slope. Calculate the maximum bed shear stress, (τ0)max, as 0.97γy0S0 and the maximum bank shear stress, (τ0w)max, as 0.75γy0S0. Compare with the corresponding critical shear stresses and iterate on the slope or depth.

235

Open Channel Hydraulics

CHAPTER 10

S0

y0, ft

(τ0)max, lbs/ft2

0.0005 0.0010 0.0020 0.0015

3.75 3.18 2.70 2.89

0.113 0.192 0.327 0.262≤0.380

(τ0w)max, lbs/ft2 0.088 0.148 0.253 0.203≤0.203

Bed Stable?

Yes Yes Yes Yes

Bank Stable?

Yes Yes No Yes

The maximum slope for stability is 0.0015. This exercise illustrates that instability occurs first on the bank for a 2:1 side slope.

10.9

Determine the critical velocity of a uniform flow of water (20oC) with a depth of 3 ft over a bed of very coarse sand with a median size of 2.0 mm using Keulegan's equation.

Repeat for a laboratory uniform flow depth of 0.3 ft. Let ks = 2.5 d50.

Solution.

The critical shear stress depends on d*, which is calculated as

 ( SG − 1) gd 503  d* =   ν2  

1/ 3

1.65 × 32.2 × (2 / 305) 3  =  (1.08 × 10 −5 ) 2  

1/ 3

= 50.4

which gives a value of τ*c = 0.045 from Figure 10.6 so that τc = 0.045×1.65×62.4×(2/305) = 0.0304 lbs/ft2. The critical value of shear velocity becomes u*c = (τc/ρ)1/2 = (0.0304/1.94)1/2 = 0.125 ft/s. Now substituting into Keulegan’s equation for fully-rough turbulent flow in a very wide channel as given by Equation 10.19 with x = 1.0 and R′ = y, we have

 12.2 y  Vc = 5.75u*c log    2.5d 50   12.2 × 3  Vc = 5.75 × 0.125 × log   = 2.41 ft/s  2.5 × (2 / 305)  If the same calculation is repeated for a depth of 0.3 ft, the flow remains fully-rough turbulent for this sediment size, but check the value of x anyway for the purpose of illustration. Calculate the viscous sublayer thickness, δ = 11.6ν/u*′ = 11.6×1.08×10−5/0.125 = 0.001 ft, and then ks/δ = 2.5×(2/305)/0.001 = 16, which gives a value of x = 1.0 from Figure 10.7. Substituting again into Equation 10.19, we have

 12.2 × 0.3  Vc = 5.75 × 0.125 × log   = 1.69 ft/s  2.5 × (2 / 305) 

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10.10

CHAPTER 10

Calculate the critical velocities in Exercise 10.9 using Manning’s equation with the Strickler equation for Manning’s n. For coarse sediments and fully-rough turbulent flow, derive a general relationship for the critical sediment number as a function of d50/y0, where y0 is flow depth, and the Shields parameter using Manning’s equation and Strickler’s equation for n. The critical sediment number is defined as

N sc =

Vc (SG − 1) gd50

in which Vc = critical velocity and SG = specific gravity of the sediment.

Solution.

For a wide channel, substitute into Manning’s equation for the slope in terms of the critical shear stress using the uniform flow formula, and substitute Strickler’s equation for Manning’s n:

Kn 2 / 3  τ c  Vc = y  1/ 6 0  cn d 50  γy0 

1/ 2

=

u*c Kn 1/ 6 cn g (d 50 / y0 )

Then substitute for u*c in terms of the Shields parameter, i.e. u*c = [τ*c(SG−1)gd50]1/2, with the result given by

Vc =

Kn 1/ 3 1/ 6 ( SG − 1)τ *c d50 y0 cn

which is identical to Equation 10.18. Now substitute the values from Exercise 10.9, first for a depth of 3.0 ft and then for a depth of 0.3 ft with cn = 0.034 using English units:

1.49 1.65 × 0.045 × (2 / 305)1 / 3 × 3.01 / 6 = 2.68 ft/s 0.034 1.49 Vc 2 = 1.65 × 0.045 × (2 / 305)1 / 3 × 0.31 / 6 = 1.83 ft/s 0.034 Vc1 =

These critical velocities are slightly higher than those given by Keulegan’s equation.

Now Equation 10.18 can be further generalized in terms of the critical sediment number by dividing both sides of the equation by [(SG−1)gd50]1/2 to yield

K τ d  N sc = n *c  50  cn g  y0 

−1 / 6

The value of Kn/(cng1/2) = 7.72 in English or SI units if cn = 0.034 (0.0414), and taking τ*c = 0.045 for the fully-rough turbulent case (coarse sediment), we have

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d  N sc = 1.64  50   y0 

−1 / 6

This equation is plotted in Figure 10.8 and labeled as “Manning”.

10.11

The Republican River is flowing at a depth of 3.0 ft with a velocity of 7 ft/s for a slope of 0.0017. The sediment sizes are d50 = 0.32 mm, d65 = 0.39 mm, and d90 = 0.59 mm.

Determine the bed form type using the Simons-Richardson diagram and the van Rijn diagram.

Solution.

For the Simons-Richardson diagram, calculate the mean bed shear stress for a wide channel as

τ 0 = γy0 S0 = 62.4 × 3.0 × 0.0017 = 0.318 lb/ft 2 Then the stream power is given by τ0V = 0.318×7.0 = 2.2 ft-lb/s/ft2. The median fall diameter is not significantly different from the median sieve diameter of 0.32 mm according to Figure 10.3, so the Simons-Richardson diagram in Figure 10.12 indicates antidunes and flat bed as the bed forms for these conditions.

The van Rijn diagram (Figure 10.13) classifies bed forms according to the values of d* and the transport parameter, T = τ*′/τ*c − 1 = (u*′)2/(u*c)2 − 1, in which τ*′ = value of Shields’ parameter for the grain shear stress. First the dimensionless particle diameter is given by

 ( SG − 1) gd503  d* =   ν2  

1/ 3

1.65 × 32.2 × (0.32 / 305)3  =  (1.08 × 10− 5 ) 2  

1/ 3

= 8.1

The corresponding value of Shields’ parameter is τ*c = 0.039 from Figure 10.6, and the critical shear velocity is u*c = [τ*c(SG−1)gd50]1/2 = [0.039×1.65×32.2×(0.32/305)]1/2 = 0.0466 ft/s. Next, the grain shear velocity is calculated from Equation 10.37 as

u* ' =

V 12 R 5.75 log 3d90

=

7.0 = 0.321 ft/s  12 × 3.0  5.75 log    3 × (0.59 / 305) 

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10.12

CHAPTER 10

Then the transport parameter, T = (u*′)2/(u*c)2 − 1 = 0.3212/.04662 − 1 = 46.4. With d* = 8.1 and T = 46, Figure 10.13 indicates that the bed forms are upper regime (flat bed and antidunes) in agreement with the Simons-Richardson diagram.

The Missouri River at Omaha flows at a depth of 10.1 ft with a velocity of 4.49 ft/s and a slope of 0.000155 in summer (T = 71°F). At approximately the same discharge of 32,000 cfs, it flows at a depth of 9.1 ft with a velocity of 5.49 ft/s and a slope of 0.00016 in winter (T = 41°F). The measured bed sediment sizes are d50 = 0.199 mm and d90 = 0.286 mm in summer, while in winter, d50 = 0.224 mm and d90 = 0.306 mm. Predict the bed form type for both cases using Figure 10.14 and explain your results. (Use the van Rijn method to estimate u*'.)

Solution.

First, calculate the grain shear velocity for summer and winter conditions from Equation 10.37 to produce

(u* ' )s =

V

5.75 log (u* ' ) w =

12 R 3d90

V

5.75 log

12 R 3d90

=

4.49 = 0.168 ft/s  12 × 10.1  5.75 log    3 × (0.286 / 305) 

=

5.49 = 0.209 ft/s   12 × 9.1 5.75 log    3 × (0.306 / 305) 

in which the subscripts “s” and “w” refer to summer and winter values, respectively. The corresponding values of the viscous sublayer thickness are δs = 11.6ν/u*′ = 11.6×1.05×10−5/0.168 = 7.25×10−4 ft and δw = 11.6×1.64×10−5/0.209 = 9.10×10−4 ft so that (d50/δ)s = (0.199/305)/ 7.25×10−4 = 0.90 and (d50/δ)w = (0.224/305)/ 9.10×10−4 = 0.81.

The reference value Ns* = 1.74/S1/3 = 1.74/0.0001551/3 = 32.4 for summer conditions and 32.1 for winter conditions. The relative sediment numbers become

 Ns  V 4.49  *  = * = = 0.74  N s  s N s (SG − 1) gd 50 32.4 1.65 × 32.2 × 0.199 / 305  Ns  5.49  *  = = 0.87 N 32 .

1 1 .

65 × 32 .

2 × 0 .

224 / 305  s w Either from Figure 2.14 or Equations 10.31 and 10.32, it can be shown that both summer and winter points fall in the transition region, but the summer condition is just slightly above the upper limit of the lower regime (0.62), while the winter condition is just below the lower limit of the upper regime (0.93). In other words, at the same water discharge, the summer conditions tend toward lower regime bed forms while the winter conditions tend toward upper regime bed forms. This has important implications for flow resistance and accounts for the observed higher depth and lower velocity in the summer.

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10.13

CHAPTER 10

A 7 mi reach of the Chattahoochee River downstream of the Buford dam is subject to a maximum discharge of 8,000 cfs due to hydropower releases and a minimum discharge of 550 cfs. The average width of the river is 180 ft, and it can be considered very wide.

The bed sediment is coarse sand with d50 = 1.0 mm and d65 = 1.2 mm. The average slope is 0.00031 ft/ft. (a) Calculate the velocity and depth at low and high flow using the Karim-Kennedy method and the Engelund method. (b) Predict the dominant bed form at low and high flow using the Simons-Richardson diagram and the van Rijn diagram. What are the bed form dimensions at low and high flow?

Solution. (a) Velocity and depth Karim-Kennedy Low Flow: Calculate the value of the Shields parameter for an assumed depth of 2.0 ft to yield

τ* =

2.0 × 0.00031 y0 S = = 0.115 (SG − 1)d50 1.65 × (1.0 / 305)

which is less than 1.2 and therefore in the lower regime. The relative dune height is given by Equation 10.42 into which the value of τ* has been substituted:

∆  0.115   0.115  = 0.08 + 2.24  − 18.13  y0  3   3  3

2

4

 0.115   0.115  + 70.9  − 88.33  = 0.143  3   3  The friction factor follows from Equation 10.41 and is given as

f ∆ = 1.20 + 8.92 = 1.20 + 8.92 × 0.143 = 2.48 f0 y0 The velocity is obtained from Equation 10.43 as

 2.0  V  = 6.683 ×  (SG − 1) gd50  (1.0 / 305) 

0.626

× 0.000310.503 × 2.48− 0.465 = 4.17

so that V = 4.17×[1.65×32.2×(1.0/305)]1/2 = 1.74 ft/s. This gives Q = 180×1.74×2.0 = 626 cfs which is greater than the specified value of 550 cfs. A second iteration with y0 =

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1.84 ft gives τ* = 0.105; ∆/y0 = 0.139; f/f0 =2.44 and V = 1.66 ft/s resulting in a calculated discharge of 550 cfs as desired.

High Flow: After several spreadsheet iterations, assume a depth y0 = 10.3 ft to give τ* = 0.590 (which is still lower regime); ∆/y0 = 0.227; f/f0 =3.22 and V = 4.30 ft/s resulting in a calculated discharge of 7970 cfs which is close enough to the desired value of 8000 cfs.

Engelund Method Low Flow: Assume a value of y0′ = 1.2 ft and then calculate τ*′ as

τ*'=

y0 ' S 1.2 × 0.00031 = = 0.0688 (SG − 1)d 50 1.65 × (1.0 / 305)

The velocity is given by Equation 10.35 as

 y' V = gy0 ' S 6 + 5.75 log 0  2d 65     1.2 V = 32.2 × 1.2 × 0.00031 × 6 + 5.75 log = 2.03 ft/s 2 × (1.2 / 305)   From Figure 10.15, or from τ* = [2.5(τ*′−0.06)]1/2 for the lower regime, estimate τ* = 0.148 so that the depth follows from the definition of τ* to yield

y0 =

τ * (SG − 1)d50 S

=

0.148 × 1.65 × (1.0 / 305) = 2.58 ft 0.00031

The result from continuity is Q = 943 cfs, which is too high. After additional iterations, assume y0′ = 1.105 ft from which τ*′ = 0.0633; V = 1.93 ft/s; τ* = 0.091; and y0 = 1.59 ft.

Then continuity gives Q = 552 cfs, which is close enough.

High Flow: Following the same procedure, assume y0′ = 3.72 ft from which τ*′ = 0.213; V = 4.12 ft/s; τ* = 0.619; and y0 = 10.8 ft. Then continuity gives Q = 8009 cfs, which is close enough to the desired value of 8000 cfs. This solution compares closely with the Karim-Kennedy result; however, if we assume upper regime bed forms rather than lower regime in Figure 10.15 with τ* = τ*′, the result is V = 6.17 ft/s and y0 = 7.2 ft which produces a multi-valued relationship as illustrated for the Rio Grande River in Figure 10.1. (b) Bed forms Simons-Richardson For low flow using the Karim-Kennedy depth of 1.84 ft, τ0 = 62.4×1.84×0.00031 = 0.0356 lb/ft2. The stream power is τ0V = 0.0356×1.66 = 0.06 ft-lb/s/ft2. With a fall diameter of 0.8 mm from Figure 10.3, the Simons-Richardson diagram (Figure 10.12)

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indicates a bed form of dunes. At high flow, τ0 = 62.4×10.3×0.00031 = 0.20 lb/ft2 and the stream power is τ0V = 0.20×4.30 = 0.86 ft-lb/s/ft2. In this case, Figure 10.12 implies a bed form that is on the border between dunes and transition.

Van Rijn Calculate the value of d* to be 25 for d50 = 1.0 mm, and Figure 10.6 gives a value of τ*c = 0.037. The critical shear velocity is u*c = [τ*c(SG−1)gd50]1/2 = [0.037×1.65×32.2×(1.0/305)]1/2 = 0.0803 ft/s. For low flow, τ*′ = 0.0633 and u*′ = 0.105 ft/s from Engelund’s method. Then the transport parameter, T = (u*′)2/(u*c)2 − 1 = 0.1052/.08032 − 1 = 0.71. With d* = 25 and T = 0.71, Figure 10.13 indicates that the bed forms are dunes in agreement with the Simons-Richardson diagram. At high flow, u*′ = 0.193 ft/s from Engelund’s method. Then the transport parameter, T = (u*′)2/(u*c)2 − 1 = 0.1932/.08032 − 1 = 4.8. With d* = 25 and T = 4.8, Figure 10.13 indicates that the bed forms are dunes.

The height of the dunes comes from Equation 10.29 by van Rijn. At low flow, we have 0.3

d  ∆ = 0.11 50  (1 − e − 0.5T )(25 − T ) y0  y0  ∆  1.0 / 305  − 0.5× 0.71 ) × (25 − 0.71) = 0.125 = 0.11×   × (1 − e y0  1.59  0.3

in which the Engelund estimate of depth has been utilized. The resulting value of ∆ = 0.125×1.59 = 0.20 ft while λ = 7.3×1.59 = 11.6 ft. At high flow, the depth is 10.8 ft and T = 4.8, so we have

∆  1.0 / 305  − 0.5× 4.8 ) × (25 − 4.8) = 0.178 = 0.11 ×   × (1 − e y0  10.8  0.3

The dune height is 0.178×10.8 = 1.9 ft and the wave length is 7.3×10.8 = 79 ft.

10.14

Calculate the velocity and flow depth for the Republican River using the measured discharge per unit width of 21 ft2/s as given in Exercise 10.11. Use the van Rijn method and the Karim-Kennedy method.

Solution.

Van Rijn Method From Exercise 10.11, d* =8.1 and the corresponding value of Shields’ parameter is τ*c = 0.039 from Figure 10.6. The critical shear velocity is u*c = [τ*c(SG−1)gd50]1/2 = [0.039×1.65×32.2×(0.32/305)]1/2 = 0.0466 ft/s.

Assume a depth of 4.0 ft, which gives a velocity of V = q/y0 =21/4 =5.25 ft/s. Next, the grain shear velocity is calculated from Equation 10.37 as

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Open Channel Hydraulics

u* ' =

CHAPTER 10

V 5.75 log

12 R 3d90

=

5.25 = 0.233 ft/s  12 × 4.0  5.75 log    3 × (0.59 / 305) 

Then the transport parameter, T = (u*′)2/(u*c)2 − 1 = 0.2332/.04662 − 1 = 24.0, which is less than 25 beyond which van Rijn predicts a flat bed and a bed form height, ∆ = 0.

Calculating the bed form height in this case from Equation 10.29, we have

∆  0.32 / 305  − 0.5× 24 ) × (25 − 24) = 0.00927 = 0.11 ×   × (1 − e y0 4 .

0   0.3

so that ∆ = 0.00927×4.0 = 0.0371 ft. The length of the bed form, λ = 7.3y0 = 7.3×4.0 = 29.2 ft. The equivalent sand grain height for form roughness is given by Equation 10.39:

k s ' ' = 1.1∆(1 − e −25 ∆ / λ ) = 1.1 × 0.0371 × (1 − e −25×0.0371 / 29.2 ) = 0.00128 ft Now substitute into Equation 10.38 to obtain the velocity:

 12 y0   V = 5.75u* log  3d90 + k s ' '  12 × 4.0   V = 5.75 × 32.2 × 4.0 × 0.0017 × log  = 10.3 ft/s  3 × 0.59 / 305 + 0.00128  in which the shear velocity, u* = (gy0S)1/2. From continuity, the value of q = 4.0×10.3 = 41.2 cfs/ft, which is clearly too large. This requires reducing the depth, but u*′ will increase and T will exceed 25 so that ∆ = 0 for subsequent iterations. The final iteration is for y0 = 2.60 ft for which u*′ becomes

u* ' =

V 12 R 5.75 log 3d 90

=

(21 / 2.60) = 0.376 ft/s  12 × 2.60  5.75 log    3 × (0.59 / 305) 

The value of T = 0.3762/0.04662 −1 = 64, which confirms that ∆ = 0 and ks′′ = 0 because T > 25. Equation 10.38 then provides the velocity as

12 × 2.60   V = 5.75 × 32.2 × 2.60 × 0.0017 × log  = 8.09 ft/s  3 × 0.59 / 305 + 0.0 

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and q = 8.09×2.60 = 21.0 cfs/ft as required. The final result is y0 = 2.60 ft and V = 8.09 ft/s.

Karim-Kennedy Method Calculate the value of the Shields parameter for an assumed depth of 2.0 ft to yield

τ* =

y0 S 2.0 × 0.0017 = = 1.96 (SG − 1)d 50 1.65 × (0.32 / 305)

which is greater than 1.5 and therefore in the upper regime where ∆ = 0. Then from Equation 10.41, f/f0 = 1.2 and Equation 10.43 results in:

  2.0 V  = 6.683 ×  (SG − 1) gd50  (0.32 / 305) 

0.626

× 0.0017 0.503 × 1.2− 0.465 = 28.1

The velocity is 28.1×(1.65×32.2×0.32/305)1/2 = 6.63 ft/s and q = 6.63×2.0 = 13.3 cfs/ft.

This means that the depth must be increased, so ∆ will remain zero for the upper regime flat bed. Additional iterations produce a final depth of y0 = 2.65 ft and velocity, V = 7.91 ft/s.

10.15

The Rio Grande River at Bernalillo, New Mexico has the following characteristics: Depth range: Slope: Temperature: Bed Sediment:

y0 = 0.6-5.0 ft S = 0.00086 ft/ft T = 60°F d35 = 0.24 mm d50 = 0.29 mm d65 = 0.35 mm d90 = 0.53 mm

Use a spreadsheet to prepare a depth-velocity curve using three methods: (1) Engelund, (2) van Rijn, and (3) Karim-Kennedy. Plot the results on log-log scales along with the measured data from the table given below. Compare the results and discuss.

Rio Grande River near Bernalillo, New Mexico, Sec. A2 Velocity,ft/s Depth,ft Velocity,ft/s Depth,ft 4.06 2.46 2.09 1.94 6.57 3.63 1.62 0.93 5.96 3.79 .17 0.52 5.09 3.49 1.30 0.58 3.71 2.76 3.73 1.54 2.84 2.69 5.00 3.06 2.65 2.15 5.17 2.19 1.66 1.25 4.31 2.44

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.58 3.11 3.12 2.34 2.23 1.98 1.91 2.05 1.94 2.37 2.51 2.90 1.44 1.70 1.40 2.01 1.97 1.85 1.76 1.89 1.76

2.66 2.56 2.48 2.14 1.44 1.18 1.23 1.29 1.32 1.42 1.50 1.29 1.12 1.48 0.79 1.07 1.70 1.03 1.31 1.10 1.70

5.86 5.56 6.91 6.88 7.82 7.71 6.92 6.27 6.10 5.06 6.50 2.55 3.04 2.71 2.89 3.16 3.99 3.62 6.01 2.88 2.00

3.25 3.03 3.68 4.46 4.11 4.80 4.34 3.43 2.67 2.96 3.40 1.84 2.56 3.44 2.93 2.64 3.12 2.33 3.22 1.54 1.00

Source: Nordin and Beverage (1965).

Solution.

Engelund’s Method A series of values of y′ = depth due to grain roughness are assumed and the calculations are made as shown in the spreadsheet given next. The value of τ*′ = y′S/[(SG−1)d50] and u*′ = (gy′S)1/2. The calculations are divided into lower regime and upper regime as shown in Figure 10.15 with Equations 10.34a, 10.34b, and 10.34c used to determine τ* for each value of τ*′. The values of velocity and depth follow from Equations 10.35 and 10.36, respectively. y',ft 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.1 1.2

τ*′

τ*

0.110 0.164 0.219 0.274 0.329 0.384 0.439 0.493 0.548 0.603 0.603 0.658

0.352 0.511 0.631 0.732 0.820 0.900 0.973 1.041 1.105 1.165 0.603 0.658

u*′, ft/s 0.074 0.091 0.105 0.118 0.129 0.139 0.149 0.158 0.166 0.175 0.175 0.182

245

V,ft/s 1.28 1.66 1.99 2.29 2.57 2.82 3.07 3.30 3.52 3.74 3.74 3.94

y,ft 0.64 0.93 1.15 1.33 1.50 1.64 1.77 1.90 2.02 2.13 1.10 1.20

q, ft2/s 0.82 1.54 2.29 3.05 3.84 4.63 5.45 6.27 7.10 7.94 4.11 4.73

Open Channel Hydraulics

1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0

CHAPTER 10

0.713 0.767 0.822 0.877 0.932 0.987 1.042 1.096 1.151 1.206 1.261 1.316 1.370 1.425 1.480 1.535 1.590 1.645

0.713 0.767 0.822 0.877 0.932 0.987 1.061 1.146 1.238 1.338 1.447 1.569 1.704 1.858 2.034 2.241 2.489 2.797

0.190 0.197 0.204 0.210 0.217 0.223 0.229 0.235 0.241 0.247 0.252 0.258 0.263 0.268 0.273 0.278 0.283 0.288

4.14 4.33 4.52 4.70 4.88 5.06 5.22 5.39 5.55 5.71 5.87 6.02 6.17 6.32 6.47 6.61 6.75 6.89

1.30 1.40 1.50 1.60 1.70 1.80 1.93 2.09 2.26 2.44 2.64 2.86 3.11 3.39 3.71 4.09 4.54 5.10

5.38 6.07 6.78 7.53 8.30 9.10 10.11 11.27 12.54 13.94 15.50 17.23 19.19 21.42 24.00 27.03 30.67 35.18

Van Rijn’s Method Iteration is required for an assumed value of q. First, y is assumed and u*′ is determined from Equation 10.37. The transport parameter, T = (u*′)2/(u*c)2 − 1, in which u*c = 0.0466 ft/s for d* = 6.8 and τ*c = 0.043. The values of ∆ and ks′′ follow from Equations 10.29 and 10.39, respectively. They both reduce to zero when T>25. Finally, the velocity V comes from Equation 10.38 in which u* = (gyS)1/2. The second value of y is obtained from continuity as q/V. Subsequent iterations use the mean value of the two most recent values of V. In the table given next, four iterations are shown for a few selected points.

The fourth value is assumed as the first value in order to repeat the four iterations until convergence is achieved. This process is more easily done with a computer program. q, ft2/s 0.75

1.00

1.25

1.50

y,ft 0.548 0.545 0.547 0.548 0.668 0.665 0.667 0.668 0.777 0.775 0.776 0.777 0.878 0.877 0.878 0.878

u*′,ft/s 0.0768 0.0773 0.0769 0.0768 0.0817 0.0821 0.0819 0.0818 0.0860 0.0862 0.0861 0.0861 0.0899 0.0900 0.0899 0.0899

T 1.712 1.751 1.726 1.717 2.072 2.102 2.085 2.077 2.406 2.422 2.414 2.410 2.719 2.725 2.722 2.721

∆,ft 0.120 0.121 0.120 0.120 0.152 0.153 0.152 0.152 0.181 0.181 0.181 0.181 0.206 0.206 0.206 0.206

246

ks′′,ft 3×d90,ft 0.0696 0.005213 0.0708 0.005213 0.0700 0.005213 0.0697 0.005213 0.0906 0.005213 0.0915 0.005213 0.0910 0.005213 0.0908 0.005213 0.1092 0.005213 0.1095 0.005213 0.1093 0.005213 0.1093 0.005213 0.1254 0.005213 0.1255 0.005213 0.1254 0.005213 0.1254 0.005213

u* ,ft/s 0.123 0.123 0.123 0.123 0.136 0.136 0.136 0.136 0.147 0.147 0.147 0.147 0.156 0.156 0.156 0.156

V,ft/s 1.377 1.366 1.373 1.376 1.503 1.496 1.500 1.502 1.612 1.608 1.610 1.611 1.710 1.708 1.709 1.709

Open Channel Hydraulics

CHAPTER 10

The final values for the van Rijn method are given in the following table: y,ft 0.548 0.668 0.777 0.878 1.062 1.228 1.379 1.647 1.880 2.354 2.694 2.584 3.088 3.546 3.972 4.371 4.750

V,ft/s 1.376 1.502 1.611 1.709 1.883 2.037 2.177 2.429 2.659 3.186 3.713 5.805 6.477 7.049 7.554 8.007 8.422

Karim-Kennedy Method A series of y values are assumed and τ* = yS/[(SG−1)d50]. Values of ∆/y and f/f0 follow from Equations 10.42 and 10.41, respectively. The sediment number, Ns, comes from Equation 10.43 from which the velocity is calculated. No iteration is required. For values of τ* > 1.5, ∆/y = 0 and f/f0 = 1.2 for upper regime bed forms. These calculations are summarized in the following spreadsheet. y, ft 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0

τ* 0.329 0.384 0.439 0.493 0.548 0.603 0.658 0.713 0.767 0.822 0.877 0.932 0.987 1.042 1.096

∆/y 0.188 0.195 0.201 0.209 0.218 0.229 0.243 0.258 0.275 0.293 0.312 0.329 0.344 0.356 0.362

f/f0 2.880 2.936 2.994 3.062 3.145 3.246 3.365 3.503 3.655 3.816 3.980 4.136 4.272 4.376 4.431

247

Ns 6.641 7.249 7.809 8.319 8.777 9.181 9.533 9.838 10.103 10.339 10.558 10.772 10.997 11.250 11.549

V, ft/s 1.493 1.629 1.755 1.870 1.973 2.064 2.143 2.211 2.271 2.324 2.373 2.421 2.472 2.528 2.596

q, ft2/s 0.896 1.140 1.404 1.683 1.973 2.270 2.571 2.875 3.179 3.486 3.797 4.116 4.449 4.804 5.192

Open Channel Hydraulics

2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4.0 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 5.0

CHAPTER 10

1.151 1.206 1.261 1.316 1.370 1.425 1.480 1.535 1.590 1.645 1.699 1.754 1.809 1.864 1.919 1.973 2.028 2.083 2.138 2.193 2.248 2.302 2.357 2.412 2.467 2.522 2.576 2.631 2.686 2.741

0.361 0.350 0.327 0.288 0.232 0.155 0.053 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

4.419 4.320 4.112 3.771 3.271 2.582 1.674 1.200 1.200 1.200 1.200 1.200 1.200 1.200 1.200 1.200 1.200 1.200 1.200 1.200 1.200 1.200 1.200 1.200 1.200 1.200 1.200 1.200 1.200 1.200

11.923 12.405 13.050 13.953 15.294 17.497 21.912 26.172 26.753 27.327 27.894 28.454 29.007 29.554 30.096 30.631 31.161 31.685 32.205 32.719 33.229 33.734 34.235 34.731 35.223 35.711 36.195 36.675 37.152 37.624

2.680 2.788 2.933 3.136 3.438 3.933 4.925 5.882 6.013 6.142 6.269 6.395 6.520 6.643 6.764 6.885 7.004 7.122 7.238 7.354 7.469 7.582 7.695 7.806 7.917 8.026 8.135 8.243 8.350 8.456

5.627 6.134 6.746 7.527 8.594 10.225 13.297 16.471 17.438 18.426 19.435 20.465 21.515 22.585 23.675 24.785 25.914 27.062 28.230 29.416 30.621 31.845 33.087 34.347 35.625 36.921 38.235 39.567 40.916 42.282

The calculated depth and velocity pairs for the three methods are compared with the measured field data in the following figure. The Karim-Kennedy and van Rijn methods produce very similar results, while Engelund’s method gives a slightly different slope and has to be connected smoothly between lower and upper regimes.

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Rio Grande River at Bernalillo, NM

Depth, ft

10

1

0.1 1

10 Velocity, ft/s Data

10.16

Karim-Kennedy

van Rijn

Engelund (upper regime)

Engelund (lower regime)

Calculate the bed load discharge per unit width in a wide gravel-bed stream with a slope of 0.005 and a flow depth of 1.0 m. The median bed sediment size is 20 mm. Use the Meyer-Peter and Mller formula and the Einstein-Brown formula. Give the results in m2/s and in metric tons/day. (1 metric ton = 1000 kg).

Solution.

Calculate the value of τ* = yS/[(SG−1)d50] = 1.0×0.005/(1.65×0.02) = 0.152. Substituting into Equation 10.47, the Meyer-Peter and Mller formula, we have

qb (SG − 1) gd

3 50

= 8.0(τ * − 0.047)3 / 2 = 8.0 × (0.152 − 0.047)3 / 2 = 0.272

qb = 0.272 × 1.65 × 9.81 × 0.023 = 0.00310 m 2 /s Converting to a mass flux, the result is ρsqb = 2650×0.00310 = 8.22 kg/m/s or (8.22/1000)×86400 = 710 metric tons/day/m.

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For the Einstein-Brown formula, evaluate ψ = 1/τ* = 1/0.152 = 6.58, which is greater than 5.5 so that Equation 10.55a applies:

φ EB =

1 e − 0.391ψ = 2.15 × e − 0.391×6.58 = 0.164 0.465

To complete the evaluation of the bed load, we need the value of F*, which depends on d*, from the Rubey formula for fall velocity (see Exercise 10.2 and Equation 10.52). First the value of d* is given by 1/ 3

 (SG − 1) gd503  d* =   ν2  

1.65 × 9.81 × 0.0203  =  (1 × 10− 6 ) 2  

1/ 3

= 506

Then the value of F* follows from Equation 10.52 and is given by

F* =

2 36 36 + 3 − = 3 d* d*3

2 36 36 + − = 0.816 3 3 506 5063

Finally, the bed load transport rate comes from the definition of φEB:

qb = φ EB F* (SG − 1) gd 503 = 0.164 × 0.816 × 1.65 × 9.81 × 0.023 = 0.00152 m 2 /s The corresponding mass flux is 2650×0.00152/1000×86400 = 348 metric tons/day/m, which is about half the value obtained from the Meyer-Peter and Mller formula.

10.17

Calculate the bed load discharge per unit width in a canal constructed in coarse sand (d50 = 1.0 mm) at a slope of 0.001, which is flowing at a depth of 2.0 m and a velocity of 1.5 m/s. Use the van Rijn bed load formula and the Einstein-Brown formula. Discuss the results.

Solution.

The values of the transport parameter, T, and the dimensionless particle diameter, d*, are required in the van Rijn formula. First calculate d* as

 (SG − 1) gd 503  d* =   ν2  

1/ 3

1.65 × 9.81 × 0.0013  =  (1 × 10− 6 ) 2  

1/ 3

= 25.3

for which τ*c = 0.037 from Figure 10.6 and u*c = [τ*c (SG−1)gd50]1/2 = [0.037×1.65×9.81×0.001]1/2 = 0.0245 m/s. Then u*′ follows from Equation 10.37 as

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Open Channel Hydraulics

u* ' =

CHAPTER 10

V 5.75 log

12 R 3d 90

=

1. 5 = 0.0715 m/s  12 × 2.0  5.75 log    3 × 0.0018) 

in which d90 has been taken to be 1.8 mm assuming σg = 1.6 (see Exercise 10.5). Now, T can be calculated as (u*′)2/(u*c)2 − 1 = 0.07152/0.02452 − 1 = 7.52. Substituting into van Rijn’s bed load formula (Equation 10.57), we have

 T 2.1  qb = (SG − 1) gd503 0.053 0.3  d*   7.522.1 qb = 1.65 × 9.81 × 0.001 × 0.053 × = 1.77 × 10− 4 m 2 /s 0.3 25.3 3

Although the Einstein-Brown formula is given in terms of ψ = 1/τ*, it should be expressed in terms of ψ′ in this exercise because the bed consists of dunes as bed forms (see Figure 10.13). Hence, ψ′=1/τ*′=(γs − γ)d50/ρu*′2=1.65×9810×0.001/(1000×0.07152) = 3.166. Because this value is less than 5.5, Equation 10.55b applies as given by 3

1  1  φ EB = 40  = 40  = 1.26  3.166  ψ ′  3

Then for d* = 25.3, Equation 10.52 gives F* = 0.771 and the bed load transport rate is given by

qb = φ EB F* (SG − 1) gd503 = 1.26 × 0.771 × 1.65 × 9.81 × 0.0013 = 1.24 × 10−4 m 2 /s which is in reasonable agreement with the van Rijn formula. This would not have been the case if ψ instead of ψ′ had been used in the Einstein-Brown formula. In fact, the results would have differed by an order of magnitude.

10.18

The following table gives the velocity distribution and the suspended sand concentration distribution for the size fraction between the 0.074 mm and the 0.104 mm sieve sizes for vertical C-3 on the Missouri River at Omaha on November 4, 1952. On this day, the slope of the stream was 0.00012, the depth was 7.8 ft, the width was 800 ft, the water temperature was 45°F, and the flow was approximately uniform. (a) Plot the velocity on semi-log scales and the concentration profile on log-log scales.

Do a regression analysis to obtain the best-fit straight lines. (b) Compute from the data and the graphs the following quantities: u* = shear velocity, V = mean velocity, κ = von Karman constant, f = Darcy-Weisbach friction factor, R0 = Rouse number, gs = suspended sediment discharge per unit width in lbs/ft/sec.

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z, distance above bottom (ft) 0.7 0.9 1.2 1.4 1.7 2.2 2.7 2.9 3.2 3.4 3.7 4.2 4.8 5.8 6.8 7.8

u, velocity (ft/s) 4.30 4.50 4.64 4.77 4.83 5.12 5.30 5.40 5.42 5.42 5.50 5.60 5.60 5.70 5.95 -----

C, concentration (mg/l) 411 380 305 299 277 238 217 -----196 -----184 -----148 130 -----------

Solution.

The velocity profile is plotted with distance above the bottom, z, on the vertical log scale and velocity on the horizontal arithmetic scale as shown in the following figure.

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Velocity Profile 10.0

1

z , ft

1.64 1.0

0.1 4.0

4.3

4.5

4.8

5.0

5.3

5.5

5.8

6.0

u , ft/s Velocity Data

Best Fit

From a least-squares regression of u vs. log(z), the best-fit straight line shown in the figure has an inverse slope N = 1.64, which is equal to du/d(log z). The value of N is related to u* and the von Karman constant, κ, as shown in the following:

N=

du du du = 2.3 = 2.3 z d(log z ) d(ln z ) dz

However, from the theoretical semilogarithmic velocity law, du/dz = u*/κz, as given by Equation 10.63, so we have

u N = 2.3 *

κ

First calculate the value of u* as

u* = gy0 S = 32.2 × 7.8 × 0.00012 = 0.174 ft/s Then the value of the von Karman constant, κ, is given by

u N

κ = 2.3 * = 2.3 ×

0.174 = 0.244 1.64

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which is less than the clear-water value of 0.40. The mean velocity can be obtained by integrating the velocity profile over the depth numerically to produce y

1 0 1 (ui + ui +1 ) 41.11 V= u dz = × ∆zi = = 5.27 ft/s ∑ ∫ y0 0 7.8 2 7.8 in which the trapezoidal rule has been employed. The friction factor follows from Equation 4.14 as

V = u*

8 f

u2 0.1742 f = 8 *2 = 8 × = 0.0087 V 5.27 2 To obtain the Rouse number, we need the log-log plot of (y0−z)/z vs. concentration C as shown in the following figure. The value of R0 is simply the inverse slope of the best-fit straight line between the logs of the two variables. As shown in the figure, R0 = 0.347.

Concentration Profile 100.0

(y 0-z )/z

10.0 1 0.347 1.0

0.1 100

1000 C , mg/L Concentration Data

254

Best Fit

Open Channel Hydraulics

CHAPTER 10

As a matter of interest, we can calculate the value of β = wf /(R0κu*). For a geometric mean size of 0.0877 mm, wf = 0.0159 ft/s from Stokes’ Law (Re = 0.30), so β = 0.0159/(0.347×0.244×0.174) = 1.08.

Finally, the suspended sediment discharge per unit width, gs, can be obtained from integration of the product of velocity and concentration over the depth:

u C + ui +1Cs ,i +1 g s = ∫ uCs dz =∑ i s ,i ∆z = 6357 (ft 2 / s) × (mg/L) 2 a y0

Converting units, we have 6357×(1/1000 mg/g) ×(1/454 g/lb) ×(1/0.0353 ft3/L) = 0.397 lb/ft/s. In reality this represents the measured portion of the suspended sediment discharge.

10.19

The Mississippi River at Arkansas City is very wide and has a bed sediment with d50 = 0.30 mm and d90 = 0.60 mm. For a flow depth of 45 ft, the water surface slope is measured to be 0.0001, and the water temperature is 80°F. The mean velocity is 3.0 ft/s, and the sediment concentration at a location of 5 ft above the channel bottom is 1,000 mg/L. (a) What is the dominant mode of sediment transport? (b) Calculate the middepth suspended sediment concentration. (c) If the water temperature were 40°F while all other factors remained the same, would the suspended sediment discharge change? Explain.

Solution.

The relative contribution of suspended load to the total load is primarily dependent on the value of R0 although Julien proposes that the load is primarily suspended for u*/wf > 2.5 and primarily bed load for u*/wf < 0.4 without consideration of the variation in β and κ (see the discussion on p. 405). The fall velocity follows from the value of d* given by

 (SG − 1) gd 503  d* =   ν2  

1/ 3

1.65 × 32.2 × (0.3 / 305)3  =  (9.26 × 10− 6 ) 2  

1/ 3

= 8.385

Equation 10.10 gives a fall velocity of

 ν  9.26 × 10−6 3   × 8 × [ 1 + 0.0139 × 8.3853 − 1] wf =  × 8[ 1 + 0.0139d* − 1] =  (0.3 / 305)  d50  or wf = 0.153 ft/s. The value of u* = (gy0S)1/2 = (32.2×45×0.0001)1/2 = 0.381 ft/s so that u*/wf = 0.381/0.153 = 2.49, which is indicative of significant suspended load. The value of R0 can be determined from the van Rijn methodology by first calculating β from Equation 10.69 as

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2

w   0.153  β = 1 + 2 f  = 1 + 2 = 1.32  0.381   u*  2

Then R0 = wf / (βκu*) = 0.153/(1.32×0.4×0.381) = 0.76. A more complete analysis using Equations 10.70 and 10.71 shows that ∆R0 = 0.025 which makes a negligible contribution to the total value of R0. Observing Figure 10.19 with R0 = 0.76, it is clear that there is a significant suspended load under these conditions.

For the given concentration of Ca = 1000 mg/L at a value of a = 5 ft, substitute into Equation 10.65 to find the middepth concentration where z = 22.5 ft:

C  ( y0 − z ) a  =  Ca  z ( y0 − a ) 

R0

 (45 − 22.5 5  = × (45 − 5)   22.5

0.76

= 0.206

Then C = 0.206×1000 = 206 mg/L. At a colder temperature, the kinematic viscosity increases and the fall velocity decreases giving a smaller value of R0. This implies that the sediment concentration distribution is more uniform with a resulting larger suspended sediment load---but not so fast. The colder temperature also implies a transition from dunes to flat bed (see Exercise 10.12) with less resistance and a lower depth and smaller value of u*, which implies a larger value of R0. However, the near-bed sediment concentration increases with the transition to flat bed (see Equation 10.71) which generally results in a larger suspended load in the winter regardless of the change in value of R0.

10.20

A reach of Peachtree Creek has a slope of 0.0005, and a sand bed with d50 = 0.5 mm, d65 = 0.6 mm, and d90 = 1.0 mm. The measured water discharge per unit width q = 11.5 ft2/s. (a) Estimate the depth and velocity using the Engelund method. What is the equivalent value of Manning’s n and how does it compare with the Strickler value? Explain. (b) Determine the bed form type, and its height and wavelength from van Rijn's method.

Use the value of τ*′ determined in part (a). (c) What is the mode of sediment transport; that is, is it primarily bed load, mixed load, or suspended load? (d) Calculate the suspended sediment concentration in ppm at a distance of 1 ft above the bed using a reference concentration, Ca, and the methodology of van Rijn. (e) Calculate the total load from Karim’s method, Yang’s method, Engelund’s method, and Karim-Kennedy’s method in ft2/s and ppm by weight. (f) Calculate the bed load transport rate in ft2/s using the Meyer-Peter and Mller formula, first using τ* and then τ*′. Which result is the correct one and why?

Solution. (a) After several iterations using Engelund’s method, the final iteration is for y0′ = 1.37 ft so that τ*′ is given by

τ* '=

y0 ' S 1.37 × 0.0005 = = 0.253 (SG − 1)d 50 1.65 × 0.5 / 305

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Then the velocity is calculated as

 y' V = gy0 ' S 6 + 5.75 log 0  2d 65   1.37   V = 32.2 × 1.37 × 0.0005 6 + 5.75 log = 3.06 ft/s 2 × 0.6 / 305   Assuming lower regime bed forms, τ* follows from Equation 10.34a:

τ * = 2.5(τ * '−0.06) = 2.5 × (0.253 − 0.06) = 0.695 Then the depth comes from the definition of τ*:

y0 =

τ * (SG − 1)d 50 S

=

0.695 × 1.65 × 0.5 / 305 = 3.76 ft 0.0005

The corresponding value of q = 3.06×3.76 = 11.5 ft2/s as given. The equivalent value of Manning’s n is given by

n=

1.49 × 3.765 / 3 × 0.00051 / 2 = 0.026 11.5

The Strickler value of Manning’s n = 0.039d501/6 = 0.039×(0.5/305)1/6 = 0.013. This illustrates the additional resistance offered by bed forms. (b) The value of d* comes from its definition and is given by

 (SG − 1) gd 503  d* =   ν2  

1/ 3

1.65 × 32.2 × (0.5 / 305) 3  =  (1.08 ×10 −5 ) 2  

1/ 3

= 12.6

Then the critical value of Shields’ parameter, τ*c = 0.033 from Figure 10.6. Van Rijn’s transport parameter, T = τ*′/τ*c − 1 = 0.253/0.033 − 1 = 6.67. Figure 10.13 indicates dunes as the bed form. The height of the bed forms is given by Equation 10.29 as

∆  0.5 / 305  −0.5×6.67 = 0.11×  ) × (25 − 6.67) = 0.191  × (1 − e y0 3 .

76   0.3

so that ∆ = 0.191×3.76 = 0.718 ft and λ = 7.3×3.76 = 27.45 ft. (c) For d* = 12.6, the fall velocity is calculated from Equation 10.10 to be wf = 0.230 ft/s.

The shear velocity is given by u* = (gy0S)1/2 = (32.2×3.76×0.0005)1/2 = 0.246 ft/s. The ratio u*/ wf = 0.246/0.230 = 1.07, which falls into the mixed load category according to

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Julien’s criteria given on p. 405 or Laursen’s criteria given at the bottom of p. 429 after Equation 10.86. (d) The reference concentration is evaluated from Equation 10.71 at an elevation a = ∆/2 = 0.718/2 = 0.359 ft:

Ca = 0.015

d50 T 1.5 0.5 / 305 6.671.5 = × × = 0.000552 0 .

015 a d*0.3 0.359 12.60.3

Transforming to ppm, we have from Equation 10.67, Ca = 106×2.65×0.000552/(1+1.65×0.000552) = 1462 ppm. Next, the value of ∆R0 is given by Equation 10.70: 0 .8

 w  C  ∆R 0 = 2.5 f   a   u*   C0 

0 .4

0 .8

 0.230   0.000552  = 2.5 ×   0.246   0.65 

0 .4

= 0.140

while the value of R0 depends on β given by Equation 10.69: 2

w   0.230  β = 1 + 2 f  = 1 + 2 ×  = 2.75  0.246   u*  2

The Rouse number without the influence of sediment concentration is

R0 =

wf

βκu*

=

0.230 = 0.850 2.75 × 0.4 × 0.246

Finally, the corrected value of the Rouse number is R0′ = R0 + ∆R0 = 0.850+0.140 = 0.990. The suspended sediment concentration at an elevation of z = 1.0 ft is given by Equation 10.65 as

 (3.76 − 1.0)  0.359 × C = 1462 ×  1.0 (3.76 − 0.359)   (e) Total sediment discharge Karim (Equation 10.78)

258

0.990

= 431 ppm

Open Channel Hydraulics

Ns =

CHAPTER 10

3.06 V = = 10.37 (SG − 1) gd50 1.65 × 32.2 × 0.5 / 305

φt = 0.00139N s

2.97

 u*     w f 

1.47

= 0.00139 × 10.37

2.97

 0.246  ×  0.230 

1.47

= 1.595

qt = 1.595 × 1.65 × 32.2 × (0.5 / 305)3 = 0.000772 ft 2 /s Ct = 106 (γ s qt ) /(γq) = 106 × (2.65 × 62.4 × 0.000772) /(62.4 × 11.5) = 178 ppm Yang (Equation 10.75)

0.230 × 0.5 / 305 = 34.91 ν 1.08 × 10− 5 u*d 50 0.246 × 0.5 / 305 = = 37.34 ν 1.08 × 10− 5     2.5 Vc = w f  + 0.66  log u*d 50  − 0.06    ν   w f d 50

=

  2.5 Vc = 0.230 ×  + 0.66 = 0.532 ft/s  log(37.34) − 0.06  Vc S 0.532 × 0.0005 = = 0.001156 wf 0.230 VS 3.06 × 0.0005 = = 0.006652 wf 0.230 log Ct = 5.435 − 0.286 × log(34.91) − 0.457 × log(0.246 / 0.230)

+ [1.799 − 0.409 × log(34.91) − 0.314 × log(0.246 / 0.230)] × log(0.006652 − 0.001156) Ct = 102.3616 = 230 ppm Engelund (Equation 10.74)

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τ 0 = γy0 S = 62.4 × 3.76 × 0.0005 = 0.1173 lb/ft 2 2τ 2 × 0.1173 c f = 02 = = 0.0129 ρV 1.94 × 3.062 0.1 0.1 0.1173   × φt = τ *5 / 2 = cf 0.0129 1.65 × 62.4 × 0.5 / 305 

5/ 2

= 3.121

qt = 3.121 × 1.65 × 32.2 × (0.5 / 305)3 = 0.00151 ft 2 /s Ct = 106 (γ s qt ) /(γq) = 106 × (2.65 × 62.4 × 0.00151) /(62.4 × 11.5) = 348 ppm Karim-Kennedy (Equation 10.77)

V 3.06 = = 10.37 (SG − 1) gd 50 1.65 × 32.2 × 0.5 / 305 u* − u*c 0.246 − 0.033 × 1.65 × 32.2 × 0.5 / 305 0.246 − 0.0536 = = = 0.652 0.295 (SG − 1) gd 50 1.65 × 32.2 × 0.5 / 305 y0 3.76 = = 2294 d 50 0.5 / 305 log φt = −2.279 + 2.972 × log(10.37) + 1.06 × log(10.37) × log(0.652) + 0.299 × log(2294) × log(0.652) = 0.3532 qt φt = = 100.3532 = 2.255 3 (SG − 1) gd 50 qt = 2.255 × 1.65 × 32.2 × (.5 / 305)3 = 0.00109 ft 2 /s Ct = 106 (γ s qt ) /(γ q) = 106 × (2.65 × 62.4 × 0.00109) /(62.4 × 11.5) = 251 ppm (f) Bed load discharge First calculate the bed load discharge using Equation 10.47 with the total value of the Shields’ parameter, τ* = 0.695, to obtain

qb (SG − 1) gd

3 50

= 8.0(τ * − 0.047)3 / 2 = 8.0 × (0.695 − 0.047)3 / 2 = 4.173

qb = 4.173 × 1.65 × 32.2 × (0.5 / 305)3 = 0.00202 ft 2 /s Second, substitute the Shields’ parameter based on the grain shear stress, τ*′ = 0.253, into the Meyer-Peter and Mller formula to yield

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Open Channel Hydraulics

qb (SG − 1) gd

3 50

CHAPTER 10

= 8.0(τ * '−0.047)3 / 2 = 8.0 × (0.253 − 0.047)3 / 2 = 0.748

qb = 0.748 × 1.65 × 32.2 × (0.5 / 305)3 = 0.000362 ft 2 /s Because of the existence of the dune bed forms, it is the second answer that is correct.

Using the Karim-Kennedy result for total sediment discharge, the bed load discharge is about 1/3 of the total.

10.21

Calculate the total sediment discharge in the Niobrara River using the Engelund-Hansen formula for the conditions given in Example 10.4.

Solution.

Calculate the shear stress and friction coefficient to substitute into Equation 10.74 for the given depth of 1.60 ft, velocity of 3.57 ft/s, slope of 0.0017, and d50 of 0.27 mm from Example 10.4 for the Niobrara River:

τ 0 = γy0 S = 62.4 × 1.60 × 0.0017 = 0.170 lb/ft 2 2τ 2 × 0.170 c f = 02 = = 0.0138 1.94 × 3.57 2 ρV 0.170 0.1 0.1   φt = τ *5 / 2 = ×  0.0138 1.65 × 62.4 × 0.27 / 305  cf

5/ 2

= 34.43

qt = 34.43 × 1.65 × 32.2 × (0.27 / 305)3 = 0.00661 ft 2 /s Ct = 106 (γ s qt ) /(γq) = 106 × (2.65 × 62.4 × 0.00661) /(62.4 × 1.60 × 3.57) = 3067 ppm 10.22

For the given water discharge in the Niobrara River in Example 10.4, calculate the depth and velocity using the Engelund method and the Karim-Kennedy method. Then determine the sensitivity of the calculated sediment discharge to the depth and velocity using one of the sediment discharge formulas in the example.

Solution.

After several iterations using Engelund’s method, the final iteration is for y0′ = 0.822 ft so that τ*′ is given by

τ* '=

y0 ' S 0.822 × 0.0017 = = 0.957 (SG − 1)d 50 1.65 × 0.27 / 305

Then the velocity is calculated as

261

Open Channel Hydraulics

CHAPTER 10

 y' V = gy0 ' S 6 + 5.75 log 0  2d 65   0.822   V = 32.2 × 0.822 × 0.0017 6 + 5.75 log = 4.435 ft/s 2 × 0.32 / 305   in which d65 has been estimated from d65 = d50σg0.385 = 0.27×1.580.385 = 0.32 mm.

Assuming lower regime bed forms, τ* follows from Equation 10.34a:

τ * = 2.5(τ * '−0.06) = 2.5 × (0.957 − 0.06) = 1.497 Then the depth comes from the definition of τ*:

y0 =

τ * (SG − 1)d50 S

=

1.497 × 1.65 × 0.27 / 305 = 1.286 ft 0.0017

The corresponding value of q = 4.435×1.286 = 5.703 ft2/s which is close enough to the given value of 5.71.

Using the Karim-Kennedy method, calculate the value of Shields’ parameter for an assumed depth of 1.275 ft (after several iterations) to yield

τ* =

y0 S 1.275 × 0.0017 = = 1.484 (SG − 1)d 50 1.65 × (0.27 / 305)

which is just less than 1.5 and therefore in the lower regime where ∆/y0 is given by

∆  1.484   1.484  = 0.08 + 2.24  − 18.13  y0  3   3 

2

4

3

 1.484   1.484  + 70.9  − 88.33  = 0.0448  3   3  Then from Equation 10.41, f/f0 = 1.2 + 8.92×0.0448 = 1.60 and Equation 10.43 results in:

 1.275  V  = 6.683 ×  (SG − 1) gd 50  (0.27 / 305) 

0.626

× 0.0017 0.503 ×1.60 −0.465 = 20.6

The velocity is 20.6×(1.65×32.2×0.27/305)1/2 = 4.47 ft/s and q = 4.47×1.275 = 5.70 cfs/ft.

Now use the Karim-Kennedy formula for sediment discharge with the calculated velocities and depths.

262

Open Channel Hydraulics

CHAPTER 10

Engelund y and V

Karim-Kennedy y and V

y0 , ft =

1.286

1.275

V , ft/s = u* , ft/s =

4.435 0.265

4.47 0.264

V = 20.45 (SG − 1) gd 50

20.61

u* − u*c = 1.01 (SG − 1) gd 50

1.01

y0 = d 50

1453

1440

qt , ft 2 /s =

0.00812

0.00830

These two estimates of sediment discharge are very similar because the calculated velocities and depths are nearly the same for the two methods used. However, the sediment discharge from the Karim-Kennedy formula using the measured velocity and depth in Example 10.4 is 0.00575 ft2/s. The predicted velocities are approximately 25 percent greater than the measured value while the predicted depths are approximately 25 percent smaller than the measured depth.

10.23

The Niobrara River in Nebraska has the following sediment sizes: Bed sediment:

d50 =0.27 mm d65 = 0.34 mm d90 = 0.48 mm

For each of the measured data points in the table given below, apply the van Rijn method, Yang formula, Karim-Kennedy formula, and Karim formula to calculate the total sediment discharge per unit width gt in lbs/sec/ft for comparison with measured sediment discharges per unit width (convert from Ct in ppm). Use a spreadsheet or write a computer program. For each method plot a graph of measured vs. calculated sediment discharges and discuss the results.

Niobrara River Depth, ft 1.44

Velocity, ft/s 3.20

Slope 0.001439

Conc., ppm 1610

Viscosity, ft2/s 1.61E-5

1.53

2.47

0.001344

970

0.99E-5

1.44

2.26

0.001250

1140

1.08E-5

1.60

3.57

0.001704

1890

1.64E-5

1.62

3.61

0.001704

1770

1.55E-5

1.89

4.17

0.001799

1780

1.34E-5

263

Open Channel Hydraulics

CHAPTER 10

1.57

2.45

0.001269

780

1.13E-5

1.43

2.33

0.001288

790

1.00E-5

1.51

2.41

0.001288

910

1.03E-5

1.44

2.52

0.001174

1000

1.16E-5

1.53

3.21

0.001420

1780

1.21E-5

1.56

3.07

0.001401

1490

1.19E-5

1.73

3.70

0.001685

1900

1.25E-5

1.38

3.34

0.001553

1710

1.82E-5

1.67

2.95

0.001250

893

1.76E-5

1.70

3.42

0.001477

1820

1.76E-5

1.56

2.15

0.001250

754

1.08E-5

1.61

2.45

0.001287

934

0.95E-5

1.41

2.35

0.001136

503

1.05E-5

1.38

2.19

0.001250

392

0.88E-5

1.44

2.05

0.001212

429

1.01E-5

1.54

2.20

0.001136

736

1.19E-5

1.42

3.33

0.001609

1520

1.79E-5

1.36

3.11

0.001610

1660

1.72E-5

1.62

3.61

0.001667

2200

1.55E-5

Source: Colby and Hembree (1955); Karim and Kennedy (1981)

Solution.

Van Rijn The van Rijn method is divided into two parts in which the bedload discharge is calculated in the first spreadsheet shown below followed by a calculation of suspended sediment discharge in the second spreadsheet with summation of the two contributions to obtain total sediment discharge.

264

Open Channel Hydraulics

CHAPTER 10

Van Rijn Bedload discharge: d*

τ*c

5.219 0.053 7.218 0.042 6.811 0.044 5.156 0.054 5.353 0.052 5.899 0.049 6.609 0.045 7.170 0.043 7.030 0.043 6.494 0.046 6.314 0.046 6.385 0.046 6.179 0.047 4.810 0.056 4.919 0.055 4.919 0.055 6.811 0.044 7.419 0.042 6.940 0.044 7.808 0.041 7.122 0.043 6.385 0.046 4.863 0.056 4.994 0.055 5.353 0.052

u*, ft/s u*′, ft/s u*c,ft/s 0.258 0.257 0.241 0.296 0.298 0.331 0.253 0.244 0.250 0.233 0.264 0.265 0.306 0.263 0.259 0.284 0.251 0.258 0.227 0.236 0.237 0.237 0.271 0.266 0.295

0.156 0.120 0.110 0.172 0.174 0.197 0.118 0.114 0.117 0.123 0.156 0.148 0.177 0.164 0.141 0.164 0.104 0.118 0.115 0.107 0.100 0.106 0.163 0.153 0.174

265

T

qb , ft2/s

0.050 8.784 0.000594 0.044 6.249 0.000264 0.045 4.879 0.000160 0.050 10.760 0.000913 0.049 11.335 0.001008 0.048 15.837 0.001975 0.046 5.615 0.000216 0.045 5.405 0.000195 0.045 5.762 0.000224 0.047 5.991 0.000249 0.047 10.178 0.000765 0.047 9.177 0.000613 0.047 13.113 0.001310 0.051 9.194 0.000670 0.051 6.731 0.000346 0.051 9.347 0.000689 0.045 4.218 0.000118 0.044 6.045 0.000244 0.045 5.389 0.000196 0.044 4.986 0.000160 0.045 3.949 0.000101 0.047 4.242 0.000121 0.051 9.062 0.000648 0.051 8.031 0.000499 0.049 11.335 0.001008

Open Channel Hydraulics

CHAPTER 10

Van Rijn Suspended Sediment and Total Sediment Discharge: ∆, ft

Ca

0.276 0.00153 0.322 0.00071 0.317 0.00051 0.263 0.00218 0.255 0.00240 0.191 0.00515 0.333 0.00060 0.314 0.00059 0.324 0.00063 0.311 0.00071 0.265 0.00187 0.285 0.00148 0.233 0.00314 0.262 0.00177 0.337 0.00085 0.300 0.00157 0.333 0.00039 0.336 0.00064 0.311 0.00060 0.307 0.00052 0.312 0.00037 0.330 0.00040 0.269 0.00168 0.276 0.00135 0.255 0.00240

wf, ft/s

β

∆R0

R0

R0'

If

qs,ft2/s

qt,ft2/s

0.106 1.334 0.108 0.766 0.874 0.231 0.00162 0.00222 0.134 1.541 0.097 0.844 0.941 0.228 0.00061 0.00088 0.129 1.575 0.087 0.851 0.938 0.236 0.00039 0.00055 0.104 1.248 0.111 0.706 0.817 0.224 0.00279 0.00370 0.108 1.262 0.118 0.717 0.835 0.212 0.00299 0.00399 0.117 1.248 0.157 0.706 0.862 0.150 0.00608 0.00805 0.127 1.499 0.088 0.833 0.921 0.234 0.00054 0.00076 0.133 1.599 0.094 0.856 0.949 0.232 0.00046 0.00065 0.132 1.554 0.093 0.847 0.940 0.231 0.00053 0.00076 0.125 1.574 0.099 0.851 0.950 0.230 0.00060 0.00084 0.123 1.430 0.130 0.811 0.941 0.197 0.00182 0.00258 0.124 1.434 0.119 0.812 0.931 0.208 0.00148 0.00209 0.121 1.310 0.141 0.752 0.892 0.175 0.00353 0.00484 0.098 1.278 0.107 0.730 0.836 0.240 0.00195 0.00262 0.100 1.298 0.082 0.743 0.825 0.253 0.00106 0.00141 0.100 1.248 0.097 0.705 0.802 0.239 0.00218 0.00287 0.129 1.530 0.076 0.841 0.917 0.236 0.00031 0.00043 0.136 1.554 0.094 0.847 0.941 0.226 0.00057 0.00082 0.131 1.661 0.098 0.865 0.963 0.230 0.00046 0.00065 0.140 1.705 0.095 0.871 0.966 0.231 0.00036 0.00052 0.133 1.627 0.079 0.860 0.939 0.233 0.00025 0.00036 0.124 1.542 0.077 0.844 0.921 0.236 0.00032 0.00044 0.099 1.267 0.103 0.721 0.823 0.244 0.00193 0.00258 0.101 1.292 0.098 0.739 0.837 0.250 0.00143 0.00193 0.108 1.267 0.119 0.721 0.840 0.211 0.00296 0.00397

266

Open Channel Hydraulics

CHAPTER 10

Yang: d*

u*, ft/s wf, ft/s u*/wf

5.219 7.218 6.811 5.156 5.353 5.899 6.609 7.170 7.030 6.494 6.314 6.385 6.179 4.810 4.919 4.919 6.811 7.419 6.940 7.808 7.122 6.385 4.863 4.994 5.353

0.258 0.257 0.241 0.296 0.298 0.331 0.253 0.244 0.250 0.233 0.264 0.265 0.306 0.263 0.259 0.284 0.251 0.258 0.227 0.236 0.237 0.237 0.271 0.266 0.295

0.106 2.448 0.134 1.923 0.129 1.866 0.104 2.838 0.108 2.765 0.117 2.837 0.127 2.002 0.133 1.828 0.132 1.901 0.125 1.866 0.123 2.157 0.124 2.147 0.121 2.538 0.098 2.681 0.100 2.591 0.100 2.842 0.129 1.942 0.136 1.900 0.131 1.739 0.140 1.684 0.133 1.786 0.124 1.921 0.099 2.739 0.101 2.617 0.108 2.734

u*d50/ν 14.20 23.01 19.73 15.99 17.03 21.86 19.84 21.56 21.51 17.81 19.35 19.73 21.70 12.78 13.04 14.30 20.54 24.07 19.15 23.71 20.78 17.66 13.41 13.67 16.84

Vc, ft/s 0.311 0.345 0.346 0.297 0.301 0.305 0.339 0.350 0.346 0.345 0.331 0.332 0.316 0.299 0.303 0.294 0.343 0.347 0.353 0.358 0.351 0.342 0.297 0.303 0.302

267

wf d50/ν

VS/wf

VcS/ wf Ct,ppm

5.802 0.0436 0.00424 11.964 0.0248 0.00347 10.577 0.0219 0.00335 5.635 0.0583 0.00485 6.159 0.0570 0.00476 7.704 0.0643 0.00470 9.910 0.0246 0.00340 11.796 0.0225 0.00338 11.315 0.0236 0.00338 9.541 0.0237 0.00324 8.969 0.0372 0.00383 9.191 0.0348 0.00376 8.548 0.0517 0.00441 4.767 0.0529 0.00474 5.033 0.0369 0.00379 5.033 0.0505 0.00435 10.577 0.0208 0.00332 12.672 0.0232 0.00328 11.011 0.0204 0.00307 14.077 0.0196 0.00320 11.632 0.0187 0.00321 9.191 0.0202 0.00314 4.897 0.0541 0.00483 5.222 0.0493 0.00480 6.159 0.0558 0.00467

1325 753 596 1978 1965 2453 694 649 686 648 1212 1113 1872 1654 998 1575 556 705 554 570 492 506 1715 1520 1908

qt,ft2/s 0.00230 0.00107 0.00073 0.00426 0.00434 0.00730 0.00101 0.00082 0.00094 0.00089 0.00224 0.00201 0.00452 0.00288 0.00186 0.00346 0.00070 0.00105 0.00069 0.00065 0.00055 0.00065 0.00306 0.00243 0.00421

Open Channel Hydraulics

CHAPTER 10

Karim-Kennedy: d*

τ*c

5.219 0.053 7.218 0.042 6.811 0.044 5.156 0.054 5.353 0.052 5.899 0.049 6.609 0.045 7.170 0.043 7.030 0.043 6.494 0.046 6.314 0.046 6.385 0.046 6.179 0.047 4.810 0.056 4.919 0.055 4.919 0.055 6.811 0.044 7.419 0.042 6.940 0.044 7.808 0.041 7.122 0.043 6.385 0.046 4.863 0.056 4.994 0.055 5.353 0.052

u*, ft/s u*c,ft/s 0.258 0.257 0.241 0.296 0.298 0.331 0.253 0.244 0.250 0.233 0.264 0.265 0.306 0.263 0.259 0.284 0.251 0.258 0.227 0.236 0.237 0.237 0.271 0.266 0.295

(u* -u*c) y0/d50 [(SG-1)gd50]1/2

Ns

0.0499 14.76 0.0444 11.39 0.0455 10.42 0.0502 16.46 0.0495 16.65 0.0480 19.23 0.0460 11.30 0.0450 10.74 0.0450 11.11 0.0465 11.62 0.0465 14.80 0.0465 14.16 0.0470 17.06 0.0513 15.40 0.0509 13.60 0.0509 15.77 0.0455 9.91 0.0444 11.30 0.0455 10.84 0.0439 10.10 0.0450 9.45 0.0465 10.14 0.0513 15.35 0.0509 14.34 0.0495 16.65

268

0.961 0.982 0.900 1.135 1.147 1.304 0.956 0.916 0.947 0.861 1.005 1.009 1.196 0.975 0.961 1.077 0.946 0.986 0.837 0.884 0.886 0.880 1.014 0.990 1.132

φt

1627 14.35 1728 6.98 1627 4.50 1807 28.89 1830 30.60 2135 64.38 1774 6.45 1615 5.09 1706 6.02 1627 5.64 1728 16.00 1762 14.12 1954 36.34 1559 16.82 1886 11.28 1920 22.55 1762 4.29 1819 6.88 1593 4.35 1559 3.96 1627 3.28 1740 3.97 1604 18.19 1536 14.08 1830 29.70

qt, ft2/s 0.00276 0.00134 0.00086 0.00555 0.00587 0.01236 0.00124 0.00098 0.00116 0.00108 0.00307 0.00271 0.00698 0.00323 0.00217 0.00433 0.00082 0.00132 0.00083 0.00076 0.00063 0.00076 0.00349 0.00270 0.00570

Open Channel Hydraulics

CHAPTER 10

Karim: d*

τ*c

5.219 0.053 7.218 0.042 6.811 0.044 5.156 0.054 5.353 0.052 5.899 0.049 6.609 0.045 7.170 0.043 7.030 0.043 6.494 0.046 6.314 0.046 6.385 0.046 6.179 0.047 4.810 0.056 4.919 0.055 4.919 0.055 6.811 0.044 7.419 0.042 6.940 0.044 7.808 0.041 7.122 0.043 6.385 0.046 4.863 0.056 4.994 0.055 5.353 0.052

u*, ft/s u*c,ft/s 0.258 0.257 0.241 0.296 0.298 0.331 0.253 0.244 0.250 0.233 0.264 0.265 0.306 0.263 0.259 0.284 0.251 0.258 0.227 0.236 0.237 0.237 0.271 0.266 0.295

Ns

0.050 14.76 0.044 11.39 0.045 10.42 0.050 16.46 0.049 16.65 0.048 19.23 0.046 11.30 0.045 10.74 0.045 11.11 0.047 11.62 0.047 14.80 0.047 14.16 0.047 17.06 0.051 15.40 0.051 13.60 0.051 15.77 0.045 9.91 0.044 11.30 0.045 10.84 0.044 10.10 0.045 9.45 0.047 10.14 0.051 15.35 0.051 14.34 0.049 16.65

269

wf, ft/s u* /wf

φt

qt, ft2/s

0.106 2.448 15.36 0.00295 0.134 1.923 4.99 0.00096 0.129 1.866 3.67 0.00070 0.104 2.838 26.42 0.00507 0.108 2.765 26.27 0.00504 0.117 2.837 41.89 0.00804 0.127 2.002 5.17 0.00099 0.133 1.828 3.89 0.00075 0.132 1.901 4.56 0.00088 0.125 1.866 5.07 0.00097 0.123 2.157 12.87 0.00247 0.124 2.147 11.20 0.00215 0.121 2.538 24.93 0.00479 0.098 2.681 19.93 0.00383 0.100 2.591 13.11 0.00252 0.100 2.842 23.30 0.00447 0.129 1.942 3.35 0.00064 0.136 1.900 4.79 0.00092 0.131 1.739 3.71 0.00071 0.140 1.684 2.87 0.00055 0.133 1.786 2.57 0.00049 0.124 1.921 3.53 0.00068 0.099 2.739 20.39 0.00391 0.101 2.617 15.56 0.00299 0.108 2.734 25.85 0.00496

Open Channel Hydraulics

CHAPTER 10

The results of all the total sediment discharge calculations are shown in the figure given below. All methods perform quite well for lower sediment discharges but tend to overpredict at higher sediment discharges. It is left to the student to calculate a quantitative measure of performance for each total sediment discharge formulation.

Niobrara River Total Sediment Discharge 0.010

Calculated qt , ft2/s

0.008

0.006

0.004

0.002

0.000 0.000

0.002

0.004

0.006

0.008

0.010

2

Measured q t , ft /s Van Rijn

Yang

Karim-Kennedy

270

Karim

Perfect Fit

Open Channel Hydraulics

10.24

CHAPTER 10

Discuss the expected channel adjustments in a sand-bed stream that is in equilibrium as a result of the following changes: (a) sand-dredging in a localized reach upstream of a bridge; (b) rapid residential development adjacent to the stream in an urban area; (c) excavation of a wider channel in a bridge cross-section to reduce backwater; (d) cutoff of a meander bend; (e) channel improvement by removal of brush on the banks and deadfalls in the channel.

Solution.

Some insight into the expected consequences can be obtained from Lane’s relationship as given by (10.79):

QS ~ Qt d 50 (a) The dredging will reduce the sediment supply downstream which will cause scouring and reduction of the slope. Upstream of the dredged site, a headcut may begin and progress upstream. (b) Residential development brings greater impervious area and larger values of Q for the same rainfall. The result is an increase in sediment transport in the stream supplied by downcutting and bank erosion. (c) Excavating the channel in the bridge section reduces velocities which may lead to deposition in the bridge section and reduced sediment transport and slope downstream of the bridge caused by scour of the downstream channel. (d) Cutting off a meander causes an increase in stream slope with resultant increase in the sediment discharge. (e) “Channel improvement” reduces the flow resistance and increases velocities for the same discharge with the result that sediment discharge increases in the improved reach at the expense of bank erosion. When delivered to the downstream unimproved reach the result may be deposition and an increase in slope.

10.25

Develop the complete derivation of the equation for live-bed contraction scour given by Equation 10.86.

Solution.

First, show that Equation 10.83 follows from substitution of Manning’s and Strickler’s equations into the definitions of τ0′ and τc:

n 2V 2  τ0' cn2V 2 cn2 g  V2 K n2 y04 / 3 = = 2 = 2  2 / 3 1/ 3 2 / 3 1/ 3  τ c τ *c (γ s − γ )d50 K nτ *c ( SG − 1)d50 y0 K nτ *c  ( SG − 1) gd 50 y0 

γy0

Second, prove that u* is given by Equation 10.85 using Manning’s equation for the slope:

u* = gy0 S = gy0

n gQ nQ = 5/3 K n By07 / 6 K n By0

271

Open Channel Hydraulics

CHAPTER 10

Now, substitute Equations 10.84 and 10.85 into the sediment transport relation given by Equation 10.82. Then assuming that τ0′/τc>>1, and that f(u*/wf) = kp(u*/wf)a, substitute the result into the sediment continuity equation given by (10.81):

 d50     y1 

7/6

a

  n Q  d  Q2  2 7 /c3 2 / 3  71 / 6 c  Qc =  50     y2   B1 y1 d 50  By1 w f 

7 7 7 + + a

Q   y2  6 3 6   =  T   y1   Qc  6

3+ a

 B1     B2 

2+ a

6 (2+ a)

 n2     n1  6

7/6

a

  n Q  Q2  2 7 /T3 2 / 3  72 / 6T  QT    B2 y2 d 50  By2 w f 

a

a

 y2   QT  7  B1  7 (3+ a )  n2  7 (3+ a )     =      n1   y1   Qc   B2 

10.26

Consider the case of a very long contraction with overbank flow in the approach channel and a constant main channel width. The bridge abutments extend to the edge of the main channel. Assume that there is no change in the channel roughness from the approach through the contraction and that the sediment is coarse without bed forms with bed-load transport only in the main channel and no sediment transport in the floodplain. For the live-bed case, derive an expression for the contraction scour using the Meyer-Peter and Mller bed-load formula.

Solution.

Applying the sediment continuity equation as given by Equation 10.81, we have

qb1 (qc Bc ) = qb 2 (qT BT ) qc qT in which qb1 and qb2 are the volumetric sediment discharges per unit width in the approach and contracted sections, respectively; qc and Bc are the water discharge per unit width and main channel width in the approach section; and qT and BT represent the total water discharge per unit width and main channel width in the contracted section. Now because Bc = BT, we must have qb1 = qb2. From the Meyer-Peter and Mller bedload formula, this implies that τ*1 = τ*2, if it is assumed that τ*>>0.047, which can be written as

γy1S1 γy2 S 2 = (γ s − γ )d50 (γ s − γ )d50 Substituting Manning’s equation for the slope, we have

272

Open Channel Hydraulics

y2 y1

=

CHAPTER 10

n 2 Q c2

/3 K n2 B T2 y 10 2

K n2 B c2 y 110 / 3

n 2 Q T2

from which we obtain

y2  QT  =  y1  Qc 

6/7

for Bc = BT. This result is identical to Equation 10.87.

Calculate the live-bed contraction scour depths for bankline abutments in overbank flow as a function of the percent of total flow in the main channel, which varies from 50 to 90 percent. Assume a constant main-channel width and an approach flow depth of 10 ft in the main channel. Plot the results.

Solution.

Assuming that the change in velocity head and the friction loss from the approach to the contracted section are very small, we can substitute y2 = (y1 + dsc) into Equation 10.87 to obtain

 d sc  1  =  y1  (Qc / QT ) 

6/7

−1

in which dsc is the depth of contraction scour. Then as Qc/QT varies from 0.9 to 0.5, the contraction scour depth varies from 0.9 ft to 8 ft for y1 = 10 ft. The results are plotted in dimensionless form in the following figure.

Contraction Scour 1.0 0.9 0.8 0.7 0.6

dsc /y 1

10.27

0.5 0.4 0.3 0.2 0.1 0.0 0.50

0.55

0.60

0.65

0.70 Q c /Q T

273

0.75

0.80

0.85

0.90

Open Channel Hydraulics

10.28

CHAPTER 10

Estimate the maximum scour depth around cylindrical piers in a river bed if the bed sediment has a median size of 1.0 mm, and the bridge piers are 1.2 m in diameter. The approach flow just upstream of the piers has a depth of 3.0 m and a velocity of 1.0 m/s.

Use the following formulas: (1) CSU; (2) Laursen and Toch; (3) Jain; (4) Melville and Sutherland; (5) Froehlich. What value of scour depth would you report to the geotechnical engineer designing the pier foundations?

Solution.

(1) CSU Take Ks = 1.0 for the shape factor (cylindrical) and Kθ = 1.0 for no skew. Further assume that Kb = 1.1 for clear-water scour or live-bed scour with small bedforms. Finally, take the armoring factor Ka = 1.0 because the sediment size is smaller than 60 mm. The resulting scour depth is

y  d s = 2.0bK s Kθ K b K a  1  b

0.35

F10.43

 3.0  d s = 2.0 × 1.2 × 1 × 1 × 1.1 × 1 ×    1.2 

0.35

1.0      9.81 × 3.0 

0.43

= 1.8 m

(2) Laursen and Toch

y  d s = 1.35b 1  b

0.3

 3.0  = 1.35 × 1.2 ×    1.2 

0.3

= 2.1 m

(3) Jain Calculate the critical velocity as derived from Manning’s equation and given by Equation 10.18:

Vc =

1.0 Kn 1/ 3 1/ 6 ( SG − 1)τ *c d50 1.65 × 0.037 × 0.0011 / 3 × 3.01 / 6 = 0.72 m/s y0 = 0.041 cn

in which τ*c = 0.037 for d* = 25. This is slightly outside the fully-rough turbulent region but close enough for this calculation. (Keulegan’s equation gives a critical velocity of 0.60 m/s for ks = 2d50. Also note that the value of cn = 0.041 which is the equivalent of 0.034 in English units. If the value of 0.0475 is used as suggested by Hager, the critical velocity is 0.62 m/s.) The critical Froude number for a critical velocity of 0.72 m/s becomes

Fc =

Vc 0.72 = = 0.133 gy0 9.81 × 3.0

Then from Equation 10.98, we have

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Open Channel Hydraulics

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0.3

0.3

 3.0  y  0.25 d s = 1.84b 1  Fc0.25 = 1.84 × 1.2 ×   × 0.133 = 1.8 m  1 .2  b (4) Melville and Sutherland Take Ks = 1 and Kθ = 1 as in the CSU formula. For the flow intensity factor, set KI = 2.4 for V1/Vc ≥ 1. The depth correction factor becomes Ky = 0.78×(3.0/1.2)0.255 = 0.985.

Finally Kd = 1.0 for b/d50>25. Then Equation 10.100 gives

d s = bK s Kθ K I K y K d Kσ = 1.2 × 1 × 1 × 2.4 × 0.985 × 1 × 1 = 2.8 m (5) Froehlich Substituting into Equation 10.104, we have

y  d s = 0.32bK s Kθ  1  b

0.46

 b   F   d50 

0.08

0.2 1

 3.0  d s = 0.32 × 1.2 × 1 × 1 ×    1.2 

0.46

0.2

1.0    1.2       9.81 × 3.0   .001 

0.08

= 0.7 m

With a factor of safety of 1.0 added to the right hand side of (10.104), the scour depth becomes 1.9 m.

10.29

Find the abutment scour depth and pier scour depth for Example 10.5 using Melville's methodology.

Solution.

Abutment Scour (1) Melville From Example 10.5, the abutment length is 233m and the approach depth and velocity in the floodplain are 0.458 m and 0.366 m/s, respectively, with a critical velocity of 0.69 m/s. Then the length to depth ratio is given by La/yf1 = 233/0.458 = 509, so this qualifies as a long abutment with KyL = 10yf1 = 10×0.458 = 4.58 m. The value of the flow intensity factor KI = Vf1/Vf1c = 0.366/0.69 = 0.53. The sediment size factor Kd = 1.0 because La/d50 > 25. The value of the shape factor, Ks, is 0.45 for the spill-through abutment but it is set to 1.0 because this is a long abutment. Substituting into Equation 10.105 with Kθ = KG = 1.0, we have

d s = K yL K I K d K s Kθ K G = 4.58 × 0.53 × 1.0 × 1.0 × 1.0 × 1.0 = 2.4 m

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Open Channel Hydraulics

CHAPTER 10

(2) Froehlich Substituting into Equation 10.109 with the shape factor Ks = 0.55 for spill-through abutments as recommended in HEC-18, the result is

L  d s = 2.27 y1K s Kθ  a   y1 

0.43

F10.61 + 1

 233  d s = 2.27 × 0.458 × 0.55 × 1.0 ×    0.458 

0.43

0.366      9.81 × 0.458 

0.61

+ 0.458 = 3.3 m

Pier Scour (1) Melville From Example 10.5, the approach velocity and depth are 1.68 m/s and 2.63 m, respectively, and the pier diameter is 1.52 m. The critical velocity for a depth of 2.63 m is 0.86 m/s using Keulegan’s equation as in Example 10.5. Now Ks = Kθ = 1.0 for cylindrical piers. The flow intensity factor KI = 2.4 because V1>Vc. The depth factor is less than one because y1/b is less than 2.6, and it is given by

y  K y = 0.78 1  b

0.255

 2.63  = 0.78 ×    1.52 

0.255

= 0.90

Finally, substituting into Equation 10.100 with Kd = Kσ = 1.0, the pier scour depth is

d s = bK s Kθ K I K y K d Kσ = 1.52 × 1.0 × 1.0 × 2.4 × 0.9 × 1.0 × 1.0 = 3.3 m (2) Froehlich Substituting into Equation 10.104, we obtain

y  d s = 0.32bK s Kθ  1  b

0.46

 b   F   d50 

0.08

0.2 1

 2.63  d s = 0.32 × 1.52 × 1 × 1 ×    1.52 

0.46

0.2

1.68    1.52       9.81 × 2.63   .002 

0.08

= 0.85 m

Adding a factor of safety of 1.0×b brings the final scour depth to 2.4 m.

10.30

The following field data are given for pier diameter, b; approach velocity, V1; approach depth, y1; and median sediment diameter, d50. Calculate the maximum pier scour depth for a cylindrical pier having zero skew using the equations of Froehlich, Melville and Sutherland, Jain, Laursen & Toch; and CSU. Compare with the measured scour depths, ds .

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Open Channel Hydraulics

CHAPTER 10

River

b, ft

y1, ft

V1, ft/s

d50, mm

Red Oahu Knik

26.9 4.9 5.0

14.1 10.2 3.9

2.0 7.8 1.6

0.06 20.0 0.5

Measured ds, ft 14.1 4.3 1.0

Solution.

The results are given in the following tables in which the calculations are summarized for each method. For consistency, factors of safety are not included in any of the formulas.

Red Oahu Knik Froehlich b, ft 26.9 4.9 5 y1, ft 14.1 10.2 3.9 V1, ft/s 2 7.8 1.6 d50, mm 0.06 20 0.5 Ks 1.0 1.0 1.0 Kθ 1.0 1.0 1.0 y1 /b 0.524 2.082 0.780 F1 0.0939 0.4304 0.1428 b/d50 136742 75 3050 ds, ft 10.3 2.6 1.8 % error -27 -40 80

Red Oahu Knik 26.9 4.9 5 14.1 10.2 3.9 2 7.8 1.6 0.06 20 0.5 1.0 1.0 1.0 1.0 1.0 1.0 1.13 378 9.46 τ*c 0.13 0.045 0.036 x 0.5 1 1.58 Vc, ft/s 1.13 6.78 1.40 KI 2.4 2.4 2.4 y1/b 0.524 2.082 0.780 Ky 0.662 0.940 0.732 Kd 1 1 1 1 1 1 Kσ ds, ft 42.7 11.1 8.8 % error 203 158 780

Melville b, ft y1, ft V1, ft/s d50, mm Ks Kθ d* (40°F)

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Open Channel Hydraulics

CHAPTER 10

Jain (live-bed) b, ft y1, ft V1, ft/s d50, mm d* (40°F)

Red Oahu Knik 26.9 4.9 5 14.1 10.2 3.9 2 7.8 1.6 0.06 20 0.5 1.13 378 9.46 τ*c 0.13 0.045 0.036 x 0.5 1 1.58 Vc, ft/s 1.13 6.78 1.40 y1/b 0.524 2.082 0.780 F1 0.0939 0.4304 0.1428 Fc 0.0531 0.3740 0.1253 ds, ft 17.5 6.9 3.2 % error 24 60 220

Laursen Red Oahu Knik & Toch b, ft 26.9 4.9 5 y1, ft 14.1 10.2 3.9 V1, ft/s 2 7.8 1.6 d50, mm 0.06 20 0.5 y1/b 0.524 2.082 0.780 ds, ft 29.9 8.2 6.3 % error 112 91 530

Red Oahu Knik CSU b, ft 26.9 4.9 5 y1, ft 14.1 10.2 3.9 V1, ft/s 2 7.8 1.6 d50, mm 0.06 20 0.5 Ks 1.0 1.0 1.0 Kθ 1.0 1.0 1.0 Kb 1.1 1.1 1.1 Ka 1 1 1 y1/b 0.524 2.082 0.780 F1 0.0939 0.4304 0.1428 ds, ft 17.1 9.7 4.4 % error 21 126 340 In general, all of the formulas overpredict the scour depth for this data except Froehlich’s formula (without the factor of safety). The largest overprediction occurs with the formula of Melville and Sutherland. The next question might be whether the measurements were actually taken after passage of the flood with the occurrence of in-filling of the scour hole.

278

Open Channel Hydraulics

10.31

CHAPTER 10

Write a computer program with interactive input that computes the depth and velocity in an alluvial river for a given discharge per unit width. Implement all methods given in this chapter.

Solution.

This is left as a student programming exercise. Refer to the calculations given in Exercises 10.13, 10.14, and 10.15. The only programming difficulties may lie in the convergence of the iterations for some methods and the selection of the correct bedform regime.

279