Fundamentals of Hydraulic Engineering Systems, 5th edition Robert J. Houghtalen Test Bank

Type:

Test Bank

Resource:

Fundamentals of Hydraulic Engineering Systems

Edition:

5th Edition

Author(s):

Robert J. Houghtalen A. Osman Akan Ned Hwang

TEST QUESTIONS - CHAPTER #1 Short Answer Questions 1. Define specific heat.

Ans. The amount of energy required to raise the temperature of a substance by 1oC. 2. Define cavitation.

Ans. Cavitation is rapid water vaporization that occurs in closed systems (e.g., pipelines or pumps) in regions where the pressure drops below the vapor pressure.

3. The amount of energy required to change water from one phase to another is called a) heat of fusion b) heat of vaporization c) specific heat d) latent energy Ans. (d) 4. (T or F) To change water from one phase to another requires a change in heat or pressure.

Ans. True 5. Two locations in closed hydraulic systems where cavitation is likely are (pressure tanks, pipelines, nozzles, pumps, hydraulic jacks).

Ans. Pipelines and pumps 6. What would be the ramification to our planet if a) water did not have a high heat capacity? b) water did not have a high dissolving capacity? c)water did not have a unique temperaturedensity relationship? (Give one answer to each question.)

Ans. a) Temperature differences between locations on the planet would be much greater. b) Nutrients would not be easily absorbed by plants and animals. c) Lakes that froze completely from the bottom up would kill most life forms each winter.

7. What is the difference between density and specific weight?

Ans. Density is mass per unit volume and specific weight is weight per unit volume.

8. Define specific gravity of a liquid.

Ans. Specific gravity is the ratio of the specific weight of any liquid to that of water at 4˚C. 9. (T or F) Water, like most substances, becomes denser as its temperature decreases.

Ans. False – water becomes less dense at temperatures lower than 4˚C and expands even more when it freezes.

10. (T or F) The density of ice is the same as that of liquid water at the same temperature.

Ans. False 11. Derive the relationship between specific weight and density using Newton’s 2nd Law (F=ma) Ans. F = m∙a Letting a = g results in Equation 1.1: W = m∙g, and dividing both sides of the equation by volume yields γ = ρ∙g 1

12. All of the following are true except: a) The density of water is greatest at 4˚C. b) Specific weight can be found by multiplying density and gravitational acceleration. c) The densities of objects on the moon are the same as they are on earth. d) The specific weights of objects on the moon are the same as they are on earth.

Ans. (d) is not true 13. Dividing the specific weight by the mass density of a liquid yields a) the specific gravity of the liquid. b) the weight of the liquid. c) the gravitational acceleration d) none of the above.

Ans. (c) is true 14. Define absolute viscosity.

Ans. Absolute viscosity is the proportionality constant relating shearing stress to the rate of angular deformation (dθ/dt) in Newtonian fluids.

15. (T or F) Newtonian fluids were consumed in great quantities by Sir Isaac Newtonian.

Ans. False 16. Kinematic viscosity a) is the proportionality constant relating shearing stress to the rate of angular deformation. b) is expressed in units of stress times the time interval (N∙sec/m2). c) is found by dividing the absolute viscosity by the mass density at the same temperature. d) is not related to the density of a fluid.

Ans. c 17. Shear stresses in water are related to all of the following except: a) The surface tension of the water. b) The velocity difference between adjacent layers. c) The distance between adjacent layers. d) The temperature of the water.

Ans. (a) is not true 18. Define surface tension.

Ans. Surface tension is the property of a fluid that causes it to seek a minimum possible surface area. It occurs because fluid molecules seek to bond with other fluid particles, but at the surface are not able to bond in all directions and therefore form stronger bonds with adjacent liquid molecules.

19. (T or F) Determination of the capillary rise in a small tube is based on a force balance.

Ans. True

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20. Give two examples of surface tension at work (i.e., evidence of surface tension).

Ans. A steel needle floating on water, the spherical shape of dewdrops, and the rise or fall of liquid in capillary tubes are the results of surface tension.

21. Capillary rise of water in a small tube is dependent on all of the following except: a) the elasticity of water. b) the nature of the solid surface. c) the temperature of the water. d) the diameter of the tube. e) the angle of contact between the water and the tube.

Ans. a 22. Define bulk modulus of elasticity.

Ans. The bulk modulus of elasticity is the inverse of the compressibility of a fluid. Expressed mathematically, it is the ratio of the pressure change that is exerted on a liquid to the ratio of the change in volume over the original volume.

23. (T or F) Water is less compressible than steel, hence it is commonly assumed to be incompressible.

Ans. False 24. The bulk modulus of elasticity is not dependent upon: a) initial pressure b) initial volume c) pressure change d) volume change Ans. a

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Problems 1. How much energy must be added to ice at -20˚C to produce 250 liters of water at +20˚C? Ans. E1 = energy req’d to bring ice temp. to 0C = (250 L)(1000 g/L)(20C)(0.465 cal/g∙C) E1 = 2.33x106 cal. E2 = energy required to melt ice = (250 L)(1000 g/L)(79.7 cal/g∙C) E2 = 1.99x107 cal. E3 = energy required to raise the water temperature to 20C E3 = (250 L)(1000 g/L)(20C)(1 cal/g∙C) = 5.00x106 cal Etotal = E1 + E2 + E3 = 2.72x107 cal 2. At 0˚C and an absolute pressure of 611 N/m2, 100 g of water, 100 g of water vapor, and 100 g of ice are in equilibrium in a sealed thermal container. Determine how much energy should be removed to freeze all the water and vapor.

Ans. E1 = energy required to change water to ice = (100 g)(79.7 cal/g) = 7.97x103 cal E2 = energy required to change vapor to ice E2 = (100 g)(597 cal/g) + (100 g)(79.7 cal/g) = 6.77x104 cal Total energy removed to freeze water and vapor.

Etotal = E1 + E2 = 7.57x104 cal 3. A container weighs 863 N when filled with water and 49 N when empty. How much water (at 20˚C) does the container hold in cubic meters?

Ans. The weight of water in the container is 863 N – 49 N = 814 N. m = W/g = (814 N)/(9.81 m/sec2) = 83.0 kg.

At 20˚C, = 1 m3 of water = 998 kg (Table 1.2) Therefore, the volume is: Vol = (83.0 kg)(1 m3/998 kg) = 8.32 x 10-2 m3 4. Determine the weight and specific gravity of a gallon of liquid if it has a mass of 0.258 slugs.

Ans. W = mg = (0.258 slug)(32.2 ft/s2) = 8.31 lb. Note: a slug has units of (lb∙sec2)/(ft) By definition, S.G. = liquid/water at 4C and γ = weight/volume Volume of 1 gal = 0.134 ft3.  = (8.31 lb)/(0.134 ft3) = 62.0 lb/ft3 S.G. = (62.0 lb/ft3)/(62.4 lb/ft3) = 0.994

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5. Determine the force required to drag a small, flat-bottom boat (3 m by 1 m) in a shallow canal (8 cm deep) in order to maintain a velocity of 1.5 m/sec. Assume the canal water (20˚C) is not moving and is behaving in a Newtonian fashion.

Ans. Assuming a Newtonian relationship:  = (dv/dy) = (Δv/Δy)  = (1.00x10-3 Nsec/m2)[(1.5 m/sec)/(0.08 m)] = (1.88x10-2 N/m2) F = ∙A = (1.88x10-2 N/m2)(3 m)(1 m) = 5.64 x 10-2 N 6. A piston slides downward with a constant velocity through a vertical pipe. The piston is 5.5 inches in diameter and 9.50 inches in length. An oil film between the piston and pipe wall resists the downward motion. If the film thickness is 0.002 in. and the cylinder weighs 0.5 pounds, estimate the velocity. (Assume the oil viscosity is 0.016 lb∙sec/ft2.)

Ans. ∑Fy = 0 (constant velocity motion); W = Tshear force = A; where A is the surface area of the cylinder in contact with the oil film: A = ()[(5.5/12)ft][(9.5/12)ft] = 1.14 ft2 Now,  = W/A = (0.5 lb)/(1.14 ft2) = 0.439 lb/ft2;  = (dv/dy) = (Δv/Δy), where Δv = v (the velocity of the cylinder). Thus, v = ()(Δy)/ v = [(0.439 lb/ft2){(0.002/12)ft}] / (0.016 lb∙s/ft2) = 4.57 x 10-3 ft/sec 7. A capillary rise test is used to determine the surface tension of a liquid. The liquid is observed to rise to a height of 1.5 cm in a 0.5-mm-diameter glass tube with a contacting angle of 60˚. Determine the surface tension of the liquid in N/m if its density is 980 kg/m3.

Ans. By rearranging Equation 1.3 yields:  = [(h)()(D)] / [(4)(sin)], where γ = ρg = (980 kg/m3)(9.81 m/sec2) = 9,610 N/m3  = [(0.015m)(9,610 N/m3)(0.0005m]/[(4)(sin 60)] = 2.08 x 10-2 N/m 8. A steel tank holds 120 ft3 of water at atmospheric pressure (14.7 lbs/in.2 and 68.4˚F). The water is subjected to a 100-fold increase in pressure. Determine the initial weight and final density of the water if the volume decreases by 0.545 ft3.

Ans. i = 1.94 slugs/ft3 (based on temp. & pressure from the inside jacket of the book). mi = i∙Voli = (1.94 slug/ft3)(120 ft3) = 233 slugs = mf Note: Mass does not change.

W = mg = (233 slugs)(32.2 ft/sec2) = 7,500 lb ρf = i/[1+(ΔVol/Vol)]; ➔Based on Example 1.3 ρf = 1.94 slug/ft3/[1+(-0.545/120)] = 1.95 slug/ft3 Altenatively, f = mf/Volf = 233slugs/(120 – 0.545)ft3 = 1.95 slug/ft3 5

TEST QUESTIONS - CHAPTER #2 Short Answer Questions 1. State Pascal’s law.

Ans. A pressure applied at any point in a liquid at rest is transmitted equally and undiminished in all directions to every other point in the liquid.

2. (T or F) The difference in pressure between any two points in still water is always equal to the product of the density of water and the difference in elevation between the two points.

Ans. False – specific weight, not density.

3. Gage pressure is defined as a) the pressure measured above atmospheric pressure. b) the pressure measured plus atmospheric pressure. c) the difference in pressure between two points. d) pressure expressed in terms of the height of a water column.

Ans. (a) is true 4. Some species of seals dive to depths of 400 m. Determine the pressure at that depth in N/m2 assuming sea water has a specific gravity of 1.03.

Ans. P = γ∙h = (1.03)(9790 N/m3)(400 m) = 4.03∙106 N/m2 5. Pressure below the surface in still water (or hydrostatic pressure) a) is linearly related to depth. b) acts normal (perpendicular) to any solid surface. c) is related to the temperature of the fluid. d) at a given depth, will act equally in any direction. e) all of the above. f) (a) and (b) only.

Ans. (e) 6. (T or F) A single-reading manometer makes use of a reservoir of manometry fluid with a large cross sectional area so that pressure calculations are only based on one reading.

Ans. True.

7. What is an open manometer?

Ans. A manometer is a pressure measurement device that utilizes fluids of known specific gravity and differences in fluid elevations. An open manometer has one end open to the air.

8. (T or F) The total hydrostatic pressure force on any submerged plane surface is equal to the product of the surface area and the pressure acting at the center of pressure of the surface.

Ans. False. The total hydrostatic pressure force on any submerged plane surface is equal to the product of the surface area and the pressure acting at the centroid of the plane surface.

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9. A surface of equal pressure requires all of the following except: a) points of equal pressure must be at the same elevation. b) points of equal pressure must be in the same fluid. c) points of equal pressure must be interconnected. d) points of equal pressure must be at the interface of immiscible fluids.

Ans. (d); points of equal pressure do not need to be at and interface of fluids.

10. The center of pressure on inclined plane surfaces is: a) at the centroid. b) is always above the centroid. c) is always below the centroid. d) is not related to the centroid.

Ans. (c); points of equal pressure do not need to be at and interface of fluids.

11. (T or F) The location of the centroid of a submerged plane area and the location where the resultant pressure force acts on that area are identical.

Ans. False. The resultant force acts at the center of pressure.

12. The equation for the determination of a hydrostatic force on a plane surface and its location are derived using all of the following concepts except a) integration of the pressure equation b) moment of inertia concept c) principle of moments d) Newton’s 2nd Law Ans. Since this deals with hydrostatics (i.e., no acceleration), (d) is the answer.

13. The equation for the righting moment on a submerged body is M = W·GM·sin θ, where GM = MB – GB or MB + GB. Under what conditions is the sum used instead of the difference?

Ans. Use the sum when the center of gravity is below the center of buoyancy.

14. Given the submerged cube with area (A) on each face, derive the buoyant force on the cube if the depth (below the surface of the water) to the top of the cube is x and the depth to the bottom of the cube is y. Show all steps.

Ans. Fbottom= Pavg∙A =γ∙y∙A; Ftop= Pavg∙A =γ∙x∙A; Fbottom-Ftop= γ∙(y-x)∙A=γ∙Vol 15. A 3 ft x 3 ft x 3 ft wooden cube (specific weight of 37 lb/ft3) floats in a tank of water. How much of the cube extends above the water surface? If the tank were pressurized to 2 atm (29.4 psi), how much of the cube would extend above the water surface? Explain.

Ans. ∑Fy=0; W = B; (37 lb/ft3)(3 ft)3 = (62.3 lb/ft3)(3 ft)2(y); y = 1.78 ft Note: The draft does not change with pressure. That is, the added pressure on the top of the cube would be compensated by the increased pressure in the water under the cube.

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16. The derivation of the flotation stability equation utilizes which principles? Note: More than one answer is possible. a) moment of a force couple b) moment of inertia nd c) Newton’s 2 Law d) buoyancy Ans. It utilizes (a), (b), and (d).

17. Rotational stability is a major concern in naval engineering. Draw the cross section of the hull of a ship and label the three important points (i.e., centers) which affect rotational stability.

Ans. See Figure 2.16.

18. A 4 m (length) by 3 m (width) by 2 m (height) homogeneous box floats with a draft 1.4 m.

What is the distance between the center of buoyancy and the center of gravity?

Ans. G is 1 m up from bottom and B is 0.7 m up from bottom. Thus, GB = 0.3 m.

19. Determine the waterline moment of inertia about the width of a barge (i.e., used to assess stability from side to side about its width) if it is 30 m long, 12 m wide, and 8 m high?

Ans. Io =(30m)(12m)3/12 = 4320 m4 20. (T or F) Floatation stability is dependent on the relative positions of the center of gravity and the center of buoyancy.

Ans. True.

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Problems 1. Given the submerged, inclined rod with a top and bottom area (dA), a length of L, and an angle of incline of θ, derive an expression that relates the pressure on the top of the rod to the pressure on the bottom. Show all steps and define all variables.

Ans. Because the prism is at rest, all forces acting upon it must be in equilibrium in all directions. For the force components in the inclined direction, we may write

 F = P dA − P dA + LdAsin  = 0 x

A

B

Note that L∙sinθ = h is the vertical elevation difference between the two points. The above equation reduces to PB − PA = h 2. Collapse depth (or crush depth) is the submerged depth that a submarine can’t exceed without collapsing due to the surrounding water pressure. The collapse depth of modern submarines is not quite a kilometer (730 m). Assuming sea water to be incompressible (S.G. = 1.03), what is the crush depth pressure in N/m2 and psi (lb/in2). Is the pressure you computed absolute or gage pressure?

Ans. P = γ∙h; where γ = (1.03)(9810 N/m3) = 1.01x104 N/m3 (using the specific weight of water at standard conditions since water gets very cold at great depths) P = γ∙h = (1.01x104 N/m3)(730 m) = 7.37x106 N/m2 = 1,070 psi (gage pressure) To get absolute pressure, atmospheric pressure must be added.

3. Mercury (Hg) is a preferred measurement fluid in simple barometers since its vapor pressure is low enough to be ignored. In addition, it is so dense (S.G. = 13.6), the tube height can be shortened considerably. If a barometric pressure reading is 29.9 inches of Hg, determine the pressure (in lb/in.2) and the comparable reading if the measurement fluid was water..

Ans. From Eq’n 2.4: P = (Hg)(h) P = (13.6)(62.3 lb/ft3)(29.9 in.)(1 ft/12 in.)3 =14.7 lb/in.2 Also, h = P/ = [(14.7 lb/in.2) / (62.3 lb/ft3)]·(12 in./1 ft)3 = 408 in. = 34.0 ft

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4. A storage tank (6 m x 6m x 6m) is filled with water. Determine the force on the bottom and on each side.

Ans. The force exerted on the tank bottom is equal to the pressure on the bottom times the area of the bottom. P = γ∙h = (9.79 kN/m3)(6 m) = 58.7 kN/m2 F = P∙A = (58.7 kN/m2)(36 m2) = 2,110 N The force exerted on the sides of the tank may be found in like manner (pressure times the area). However, the pressure is not uniform on the tank sides since P = γ∙h.

Therefore, the average pressure is required. Since the pressure is a linear relationship, the average pressure occurs at half the depth. Now, Pavg = γ∙havg = (9.79 kN/m3)(3 m) = 29.4 kN/m2 F = Pavg ∙A = (29.4 kN/m2)(36 m2) = 1,060 N 5. A water container consists of a 10-ft-high, 1-ft-diameter pipe welded on top of a cube (3ft x 3ft x 3ft). The container is filled with water (20°C). Determine the weight of the water and the pressure forces on the bottom and sides of the bottom cube.

Ans. Wtotal = ·Vol = ()[Volcube + Volpipe] Wtotal = (62.3 lb/ft3)[(3 ft)3 + ()(0.50 ft)2(10 ft)] = 2,170 lb Pbottom = h = (62.3 lb/ft3)(13 ft) = 810 lb/ft2; Fbottom = (810 lb/ft2)(9 ft2) = 7,290 lb Note: The weight of the water is not equal to the force on the bottom. Why? (Hint: Draw a free body diagram of the 3 ft x 3 ft x 3 ft water body labeling all vertical forces acting on it. Don’t forget the pressure from the container top. Now, the side force is: Pavg = γ∙havg = (62.3 lb/ft3)(11.5 ft) = 716 lb/ft2 F = Pavg ∙A = (716 lb/ft2)(9 ft2) = 6,440 lb 6. A weight of 5,400 lbs is to be raised by a hydraulic jack. If the large piston has an area of 120 in.2 and the small piston has an area of 2 in.2, what force must be applied through a lever having a mechanical advantage of 6 to 1?

Ans. From Pascal’s law, the pressure on the small piston is equal to the pressure on the large.

Fsmall/Asmall = Flarge/Alarge Fsmall = [(Flarge)(Asmall)]/(Alarge) = [(5400 lb)(2 in2)]/(120 in2) = 90 lb  The applied force = 90 lb/6 = 15 lb based on the mechanical advantage of the lever.

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7. Carbon tetrachloride (manometer liquid, M, in the figure below, S.G. = 1.6) is poured into a U-tube with both ends open to the atmosphere. Then water is poured into one leg of the Utube until the water column (y) is 20 cm high. Determine the height of the carbon tetrachloride (h) above the water-carbon tetrachloride interface (point 1).

y

Note: y = height of water. Air is above the Hg at point D.

Ans. A surface of equal pressure surface can be drawn at the interface (1-2).

Therefore, P1 = P2; (0.2 m)() = (h)(M) h = [(0.2 m)()]/(M) = [(0.2 m)(γ)]/[(γ)(SGM)] = (0.2 m) / (1.6) h = 0.125 m = 12.5 cm 8. The manometer depicted below is mounted on a city water supply pipe to monitor water pressure. However, the field engineer suspects the manometer reading of h = 3 feet (Hg) may be incorrect. If the pressure in the pipe is measured independently and found to be 16.8 lb/in.2 (psi), determine the correct value of the reading h.

Ans. A surface of equal pressure can be drawn at the mercury-water meniscus. Therefore, Ppipe + (2 ft)() = (h)(Hg) (16.8 lb/in2)(144 in2/ft2) + (2 ft)() = (h)()(S.G.Hg) (2.42 x 103 lb/ft2) + (2 ft)(62.3 lb/ft3) = (h)(62.3 lb/ft3)(13.6) h = 3.00 ft (manometer is correct)

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9. Manometer computations for the figure above yield a pressure of 16.8 lb/in.2 (psi) if h = 3 ft.

If the fluid in the pipe was oil (S.G. = 0.80) under the same pressure, would the manometer measurements (2 ft and 3 ft) still be the same? If not, what would the new measurements be?

Ans. The measurements will not be the same since oil is now in the manometer instead of water. A surface of equal pressure can be drawn at the mercury-oil interface.

Ppipe + (2 ft + Δh)(oil) = (3 ft + 2Δh)(Hg) This is based on volume conservation. If the mercury-oil meniscus goes down Δh on the right, it must climb up Δh on the left making the total difference 2Δh. Now (2.42 x 103 lb/ft2) + (2 ft + Δh)(0.80)(62.3 lb/ft3) = (3 ft + 2Δh)(13.6)(62.3 lb/ft3) Δh = -0.0135 ft New measurements: 2 feet becomes 2 – 0.0135 = 1.99 ft and 3 feet becomes 3 - 2(0.0135) = 2.97 ft 10. Determine the air pressure (in kPa and cm of Hg) in the sealed left tank depicted in the figure below if EA = 32.5 m.

Ans. Using the “swim through” technique, start at the sealed right tank where the pressure is known.

Then “swim through” the tanks and pipes, adding pressure when

“swimming” down and subtracting when “swimming” up until you reach the left tank where the pressure is not known. The computations are as follows: 20 kN/m2 + (4.5 m)(9.79 kN/m3) - (2.5 m)(1.6)(9.79 kN/m3) - (5 m)(0.8)(9.79 kN/m3) = Pleft

Pleft = -14.3 kN/m2 (or -14.3 kPa). Now to determine the pressure in cm of Hg, since P = γh or h = P/γ; h = (-14.3 kN/m2) / [(SGHg)(γ)] h = (-14.3 kN/m2) / [(13.6)(9.79 kN/m3)] = -0.107 m = 10.7 cm (Hg)

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11. A vertical gate keeps water from flowing in a triangular irrigation channel. The channel has a 4-m top width and a 3-m depth. If the channel is full, what is the magnitude of the hydrostatic force on the triangular gate and its location?

Ans. The hydrostatic force and its locations are (based on Table 2.1): F =   h  A = = (9790 N/m3)[(3m)/(3)]∙[6 m2] = 5.87 x 104 N = 58.7 kN yP =





I0 (4m)(3m) 3 / 36 +y= + 1.00m (4m)(3m) / 2(1.00m) Ay

yp = 1.50 m (depth to center of pressure)

12. A 3-ft square (plane) gate is mounted into an inclined wall (45°). The center of the gate is located 4 feet (vertically) below the water surface. Determine the magnitude of the hydrostatic force (in lbs) and its location with respect to the water surface along the incline.

Ans. F =   h  A = (62.3 lb/ft3)(4 ft)(3 ft)2 F = 2,240 lbs; Now noting that y = h / sin 45 = 4ft/(sin 45°) = 5.66 ft yP =





I0 (3 ft )4 / 12 +y= + 5.66 ft Ay (3 ft )2 (5.66 ft )





yp = 5.79 ft (distance from water surface to the center of pressure along incline) 13. A circular gate is installed on a vertical wall as shown in the figure below. Determine the horizontal force, P, necessary to hold the gate closed if the gate diameter is 6 feet and h = 7 feet. Neglect friction at the pivot.

Ans. F =   h  A = (62.3 lb/ft3)[(7 ft)]∙[π(6 ft)2/4] = 1.23 x 104 lbs yP =





I0  (6 ft ) 4 / 64 +y= + 7 ft ; yp = 7.32 ft (depth to the center of pressure) Ay  (6 ft) 2 / 4 (7 ft )





Thus, summing moments: ∑ Mhinge = 0 P(3 ft) – (1.23 x 104 lbs)(7.32 ft – 7 ft) = 0; P = 1.31 x 103 lbs

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14. A vertical plate, composed of a square and a triangle, is submerged so that its upper edge coincides with the water surface (Figure P2.5.5). What is the value of the ratio H/L such that the pressure force on the square is equal to the pressure force on the triangle?

Ans. Fsquare =   h  A = γ(L/2)(L2) = (γ/2)∙L3 Ftri =   h  A = γ(L+H/3)(LH/2) = (γ/2)[L2H + LH2/3];

Setting the two forces equal:

Fsquare = Ytri ; substituting yields; (γ/2)∙L3 = (γ/2)[L2H + LH2/3] L2 - HL - H2/3 = 0; divide by H2 and solve quadratic (L/H)2 – (L/H) - 1/3 = 0; L/H = 1.26 or H/L = 0.791 15. The wicket dam, pictured on the right, is 5 m high and 3 m wide and is pivoted at its center. Determine the reaction force in the supporting member AB.

Ans. F =   h  A = (9790 N/m3)(2.5m)[(5/cos 30˚)(3 m)] = 4.24 x 105 N = 424 kN yp = I 0 + y = Ay

(3m)(5.77m) / 12 + 2.89m = 3.85 m (inclined depth to pressure center) 3

(3m)(5.77m)(2.89m)

Summing moments about the base of the dam; ∑ M = 0 (424 kN)(5.77 m – 3.85 m) – (FAB)(5.77m/2) = 0; FAB = 282 kN 16. Calculate the magnitude and the location of the resultant pressure force on the annular gate shown in the figure to the right if the cross-section of central hub is a square that is 1 m by 1 m.

Ans. F =   h  A = (9790 N/m3)(2.5 m)[(π)(1.5 m)2–(1.0 m)2] = 1.49 x 105 N = 149 kN yp = I 0 + y = { (3m) / 642} − {(1m2)(1m) / 12} + 2.5m = 2.76 m (below the water surface) 4

Ay

3

 (1.5m) − (1m) (2.5m)

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17. Determine the relationship between γ1 and γ2 in the figure to the right if the weightless triangular gate is in equilibrium in the position shown.

(Hint: Use a unit length for the gate.)

Ans. Fincline =   h  A = (γ1)(0.5 m)[(1m/cos30˚)(1m)] = 0.577∙γ1 N; Since (1m/cos 30˚)=1.15 m yP =





I0 (1m)(1.15m)3 / 12 +y= + (1.15m / 2) = 0.767 m (inclined distance to center of pressure) (1m)(1.15m)(1.15m / 2) Ay

Fright =   h  A = (γ2)(0.5 m)[(1m)(1m)] = 0.500∙γ2 N; and yP = 0.667 m

∑ Mhinge = 0; (0.577∙γ1 N)(1.15m - 0.767 m) – (0.500∙γ2 N)(1 m - 0.667 m) = 0; γ2 = 1.33∙γ1 N 18. An inverted hemispherical shell of diameter d = 3 feet as shown in the figure to the right is used to cover a tank filled with water at 20°C. Determine the minimum weight the shell needs to be to hold itself in place (i.e., not be lifted up).

Ans. The vertical component of the total hydrostatic pressure force is equal to the weight of the water column above it to the free water surface. In this case, it is the virtual or displaced weight of water above the shell since the pressure is from below.

FV =   Vol = (62.3 lb/ft3)[(1/2)(4/3)(π)(1.5 ft)3] = 440 lb The weight of the cover must balance this upward pressure; thus W = 3,520 lb 19. The corner plate of a barge’s hull is curved with a radius of 1.75 m. The depth of submergence (draft) is depicted in the figure. However, the barge is leaking and the water on the inside is up to level A producing hydrostatic pressure on the inside as well as the outside.

Determine the resultant horizontal hydrostatic pressure force on plate AB per unit length of the hull.

Ans. FH = (  h  A) right − (  h  A) left = (9790 N/m3) [(1.75 m)(1 m)] (3.875 m – 0.875 m) FH = 5.14 x 104 N = 51.4 kN  (towards the barge) 7

20. The corner plate of a barge’s hull is curved with a radius of 1.75 m. The depth of submergence (draft) is depicted in Figure P2.6.4. The barge is leaking and the water on the inside is up to level A producing hydrostatic pressure on the inside as well as the outside. Determine the resultant vertical hydrostatic force on plate AB per unit length of hull.

Ans. The resultant vertical component of the total hydrostatic force is the weight of the water column above the curved surface (water in the barge) subtracted from the virtual weight of the water column above the curved surface up to the water level outside. Note that this equals the water column above the curved surface between the two water levels.

FV = (  Vol )outside − (  Vol )inside = (9790 N/m3)[(1.75 m)(1 m)(3 m)] FV = 5.14 x 104 N = 51.4 kN ↑ 21. Calculate the horizontal and vertical hydrostatic forces on the curved surface ABC in the figure to the right.

Ans. The horizontal component of the total hydrostatic pressure force equals the total pressure on the vertical projection of the curved surface ABC: FH =   h  A = (γ)(R)[(2R)(1)] = 2(γ)(R)2

The vertical component of the total hydrostatic pressure force equals the weight of the water above the curved surface ABC. The volume of water above this surface is: Vol = (Aquaarter circle + Arectangle – Aquarter circle )(unit length) Vol = (Arectangle)(unit length) = (2R)(R)(1) = 2(R)2; FV =   Vol = γ[2(R)2] = 2(γ)(R)2 22. The tainter gate section shown in figure to the right has a cylindrical surface with a 12-m radius and is supported by a structural frame hinged at O. The gate is 10 m long (in the direction perpendicular to the page). Determine the magnitude, direction, and location of the total hydrostatic force on the gate 7

Ans. First find the height of the vertical projection of area: (R)(sin 45˚) = 8.49 m. Thus, FH =   h  A = (9790 N/m3)(4.25m)[(10 m)(8.49 m)]; FH = 3.53 x 106 N = 3,530 kN

The vertical component of the total pressure force is the weight of the water column above the curved gate. The volume of water above the gate is: Vol = (Arectangle - Atriangle - Aarc)(length) Vol =[(12m)(8.49m) - (1/2)(8.49m)(8.49m) - (π/8)(12m)2](10 m) = 92.9 m3 Now, FV =   Vol = (9790 N/m3)(92.9 m3) = 9.09 x 105 N = 909 kN; and the total force is: F = [(3,530 kN)2 + (909 kN)2]1/2 = 3650 kN; θ = tan-1 (FV/FH) = 14.4˚

.

Since all hydrostatic pressures pass through point O (i.e., they are all normal to the surface upon which they act), then the resultant must also pass through point O. 23. A piece of irregularly shaped metal weighs 301 N. When the metal is completely submerged in water, it weighs 253 N. Determine the specific weight and the specific gravity of the metal.

Ans. The buoyant force equals the weight reduction. Thus, B = 301 N – 253 N = 48.0 N In addition, B = wt. of water displaced = γ∙Vol = (9790 N/m3)(Vol); Thus, Vol = 4.90 x 10-3 m3 and γmetal = W/Vol = 301N/(0.00490 m3) = 6.14 x 104 N/m3 S.G. = (6.14 x 104 N/m3)/(9,810 N/m3) = 6.26 24. A concrete block that has a total volume of 12 ft3 and a specific gravity of 2.67 is tied to one end of a long cylindrical buoy as depicted in the figure below. The buoy is 10 ft long and is 2 ft in diameter. Unfortunately the water level has risen and the buoy is floating away with 1 ft sticking above the water surface. Determine the specific gravity of the buoy. The fluid is brackish bay water (S.G. = 1.02).

Ans. W = 2.67γ(12 ft3) + (SG)γ∙π(1 ft)2(10 ft);

B = 1.02γ(12 ft3) + 1.02γ∙π(1 ft)2(9 ft)

∑Fy = 0; or W = B; 32.0 + 31.4(SG) = 12.2 + 28.8;

SG = 0.287

7

25. A cube of ice measures 10 inches on each side. It has a density of 1.76 slugs/ft3. Determine the weight of the ice and the percentage of the cube below the waterline when it is floating.

Ans. Since ρ = m/Vol; m = (Vol)(ρ) = [(10/12)ft3](1.76 slugs/ft3) = 1.02 slugs Also, W = mg = (1.02 slugs)(32.2 ft/sec2) = 32.8 lbs γice = W/Vol = 32.8 lbs/[(10/12)ft3] = 56.7 lbs/ft3 S.G. = γice/γ = 56.7 lbs/ft3/ 62.4 lbs/ft3 = 0.909 Therefore, 90.9% of the cube will be below the waterline.

26. A rectangular barge is 14 m long, 6 m wide, and 2 m deep. The center of gravity is 1.0 m from the bottom and the barge drafts 1.5 m of seawater (S.G. = 1.03). Find the metacentric height and the righting moments for a 8° angle of heel (or list).

Ans. The center of gravity (G) is given as 1 m up from the bottom of the barge.

The center of buoyancy (B) is 0.75 m up from the bottom since the draft is 1.5 m.

Therefore GB = 0.25 m, and GM is found using GM = MB  GB =

I0  GB ; where Io is the waterline moment of inertia about the tilting Vol

axis. Chopping off the barge at the waterline and looking down we have a rectangle which is 14 m by 6 m. Thus, GM = MB  GB =





(14m)(6m) / 12 − 0.25m = 1.75 m; I0  GB = (14m)(6m)(1.5m) Vol 3

Note: Vol is the submerged volume and a negative sign is used in the equation since G is located above the center of buoyancy.

M = W  GM  sin 

M = [(1.03)(9790 N/m3)(14 m)(6m)(1.5m)](1.75m)(sin 8˚) M = 309 kN∙m (for a heel angle of 8˚)

8

TEST QUESTIONS & PROBLEMS - CHAPTER #3 Short Answer Questions 1. In Osborne Reynolds carefully performed experiments, what observation was apparent when laminar flow was occurring in the glass pipe?

Ans. When colored water was injected into the flow stream, it appeared as a straight line extending to the downstream end, indicating laminar flow in the pipe.

2. (T or F) Turbulent flow in a pipe has a more uniform velocity distribution than laminar flow.

Ans. True 3. What are the dimensions of Reynolds number when calculations are performed in SI units?

Ans. Reynolds number is dimensionless.

4. Sketch the velocity distribution in a pipe for both laminar and turbulent flow.

Ans.

5. (T or F) The transition from laminar to turbulent flow in circular pipes occurs at a Reynolds number of about 1000, but varies somewhat based on differences in experimental conditions Ans. False, it occurs at about 2000.

6. Water loses energy as it flows through a pipe. What causes most of the energy loss?

Ans. (a) friction against pipe walls, and b) viscous dissipation due to internal actions of flow 7. The continuity (conservation of mass) equation has the following limitations: a) steady flow b) incompressible flow c) inviscid (negligible viscosity) flow d) atmospheric pressure e) laminar flow f) turbulent flow Ans. (a) and (b)only 8. Water sprays through a 6-inch-diameter nozzle at a flow rate of 2.75 ft3/sec. Determine the velocity of the water spray.

Ans. V = Q/A= = (2.75 ft3/sec)/[π(0.25 ft)2] = 14.0 ft/sec 9. Which force(s) is (are) not accounted for in the conservation of momentum equation? b) weight of fluid b) anchoring (wall) force c) pressure force d) surface tension force Ans. (d)

18

10. In the derivation of the conservation of momentum equation, ∑F = (MV2 - MV1)/Δt became ∑F = ρQ(V2- V1)/Δt. Explain this step.

Ans. M/Δt = mass flow rate = ρQ 11. Which of the following is true? The friction factor … a) in laminar flow depends on pipe material. b) for turbulent flow depends on fluid pressure. c) in all flow regimes depends on pipe diameter. d) in complete turbulent flow depends on the Reynold’s number.

Ans. (c) 12. What is the significant distinction between the Bernoulli equation and the energy equation?

Ans. The Bernoulli equation does not account for energy losses.

13. (T or F) The velocity of water in a pipe slows down as it goes downstream due to friction and other losses.

Ans. False. The continuity equation (mass balance) would be violated if this happened.

14. Identify the three major forms of energy in pipe flow.

Ans. Potential energy (position or elevation head), pressure energy (pressure head), and kinetic energy (velocity or kinetic head).

15. Which of the following flow regimes are dependent on the Reynolds number in order to determine the friction factor? a) laminar flow b) critical flow c) turbulent flow (smooth pipe) d) turbulent flow (transition zone) e) turbulent flow (complete turbulence, rough pipes) Ans. (a), (c), and (d) 16. Identify the flow assumptions that were made in the derivation of the Bernoulli equation.

Ans. Steady, incompressible, and inviscid (negligible losses) flow.

17. All of these principles are likely to be applied to determine the flow in a pipeline except: a) conservation of momentum b) minor losses c) Darcy-Weisbach d) Hazen-Williams e) energy balance f) conservation of mass (continuity) Ans. (a) 18. (T or F) The Moody diagram in the book is in the SI system of units.

Ans. False. It is dimensionless.

19. (T or F) For laminar flow, the friction factor depends only on the Reynolds number.

Ans. True.

20. Which of the following methods can be used to determine friction loss in pipelines? a) Darcy-Weisbach b) Manning c) Hazen-Williams d) a, b, and c e) a and b only f) a and c only g) b and c only h) none of the above i) there is no i j) same as i Ans. (d)

19

21. A 1 m-diameter corrugated metal storm water pipe (n = 0.024) is flowing full with a discharge of 2.00 m3/sec. Determine the friction head loss over a 100 m length.

Ans. hf = KQm ; m = 2 and K = (10.3∙n2∙L)/(D5.33) = (10.3∙0.0242∙100)/(1.05.33) = 0.593; hf = KQm = (0.593)(2.0)2; hf = 2.37 m 22. A 1 ft-diameter PVC pipe (CHW = 150) is conveying 15.3 ft3/sec. Determine the friction head loss over a 100 ft length.

Ans. hf = KQm ; m =1.85 and K =(4.73L)/(D4.87∙C1.85)= (4.73∙100)/(1.04.87∙1501.85) = 0.0446 hf = KQm = (0.0446)(15.3)1.85; hf = 6.93 ft 23. (T or F) Confusors are used to reduce the head loss at pipe contractions.

Ans. True.

24. Minor losses in pipe systems are a) always less than friction losses b) dependent on temperature c) independent of Q d) assumed to vary linearly with V e) normally found using loss coeffcients f) a and e Ans. (e) 25. Give three examples of minor losses that occur in pipelines.

Ans. Bends, contractions, expansions, valves, entrance losses, exit (discharge) losses.

26. (T or F) Contraction losses exceed expansion losses given the same transition geometry.

Ans. False. Expansion losses are greater. Examine the limiting case of a reservoir entrance loss [(0.5)V2/2g] vs. a reservoir exit (discharge) loss [(1.0)V2/2g].

27. For two pipes in series, which of the following equivalent pipe principles is appropriate? a) hfE = hf1 + hf2 b) Q1 = Q2 = QE c) QE = Q1 + Q2 d) hfE = hf1 = hf2 e) none of the above f) all of the above Ans. (a) and (b) 28. For two pipes in parallel, which of the following equivalent pipe principles is appropriate? a) hfE = hf1 + hf2 b) Q1 = Q2 = QE c) QE = Q1 + Q2 d) hfE = hf1 = hf2 e) none of the above f) all of the above Ans. (c) and (d) 29. (T or F) Equivalent pipe principles can be applied to pipe systems regardless of the friction loss equation used (e.g., Darcy-Weisbach, Manning, or Hazen-Williams).

Ans. True 30. (T or F) In a pipe network that contains multiple junctions, flow from one junction to another produces different head loses depending on the path taken (i.e., based on the length, size, and roughness of the pipes along the route).

Ans. False

20

Problems 1. A jet of water flowing freely in the atmosphere (x-direction) hits a curved vane and shoots straight up (z-direction). The velocity of the water jet is 10.5 ft/sec and has a 2-in. diameter.

If the vane is assumed frictionless, determine the magnitude of the resultant force exerted on the water jet by the vane and its direction.

V

F

V

Ans. The curved vane resists the hydrodynamic force (F) of the jet. From Eq’n. (3.7c),

 F = Q(V x

out , x − Vin , x )

; no pressure forces since water is exposed to atmosphere,

and Vout=Vz = 10.5 ft/s; Vin=Vx= 10.5 ft/s; Q =V∙A=(10.5 ft/s)[(π/4){(2/12)ft}2] = 0.229 ft3/sec; Fx = (1.94 slugs/ft3)(0.229 ft3/s)[0 – 10.5 ft/s] = -4.66 lbs; and Fy = +4.66 lbs; thus, F = [(-4.66 lb)2 + (4.66 lb)2]1/2 = 6.59 lb; θ = tan-1 (FV/FH) = 45˚ 2. The 3-in-diameter, horizontal nozzle depicted below is attached to the 6-in-diameter pipe with flange bolts. The nozzle inlet pressure is 34.7 psi (absolute), and the exit velocity is 42 ft/sec. Compute the force in the flange bolts.

F Vout

Vin

Ans. No exit force exists since the water is exposed to the atmosphere. The entrance pressure force is given as absolute, so atmospheric pressure must be removed. Therefore, Fp = P∙A = [(34.7 – 14.7) lb/in2][π/4(6.0in)2] = 565 lb; Vx,out = 42 ft/s Q = V∙A = (42 ft/s)[π/4(0.25ft)2] = 2.06 cfs; Vx,in= Q/A = (2.06 ft3/s)/[π/4(0.50 ft)2] = 10.5 ft/s

∑Fx = ρQ(Vx,out - Vx,in); 565 lb – F = (1.94 slug/ft3)(2.06 ft3/s)[(42.0 – 10.5) ft/s]; F = 439 lb; which is the force in that must be resisted by the flange bolts.

3. Determine the magnitude and direction of the reaction force (F) at a 90° pipe bend that is passing a mass flow rate of 1,200 kg/sec. The diameter of the pipe is 70 cm and the pressure head is 8.8 m just upstream of the bend and 8.6 m downstream of the bend.

Vout Vin

F

21

Ans. The forces on the control volume: pressure forces and the bend reaction force.

Fin = P∙A= (8.8m)(9.79 kN/m3)[(π/4)(0.7m)2] = 33.2 kN; Fout = P∙A= -32.4 kN Q = m /ρ = (1200 kg/s)/(998 kg/m3) = 1.20 m3/sec V = Q/A = (1.20 m3/sec) / [(π/4)(0.7m)2] = 3.12 m/s; Using equations (3.7a) and (3.7b); ∑Fx = ρQ(Vx,out - Vx,in); (

+), Assume Fx is negative

33.2 kN – Fx = (1200 kg/s)[(0 – 3.12) m/s] (1 kN/1000 N); ∑Fy = ρQ(Vy,out – Vy,in); (

Fx = 36.9 kN

+); Assume Fy is positive

Fy - 32.4 kN = (1200 kg/s)[(3.12 - 0) m/s] (1 kN/1000 N); Fy = 36.1 kN F = [(36.9 kN)2 + (36.1 kN)2]1/2 = 51.6 kN ; θ = tan-1 (Fy/Fx) = 44.4˚ 4. A homeowner is interested in supplying water (20°C) to her wood shop that is about 30 m from her house. However, she is concerned about the lack of water pressure. Calculate the pressure drop she can expect for a flow rate of 10 liters per minute through a 1.5-cm copper pipe. Assume minor losses are negligible.

Ans. Determine the friction head loss and convert it to a pressure drop; Eq’n (3.15a). e/D = (0.0015 mm)/(15 mm) = 0.0001; V = Q/A = (0.01m3/60s)/[(π/4)(0.015 m)2] = 0.943 m/s

NR = DV/ν = [(0.015m)(0.943 m/s)]/(1.00 x 10-6 m2/s) = 1.41 x 104 From Moody diagram; f = 0.028; hf = f (L/D)(V2/2g); for a 30 m length of pipe hf = (0.028)(30m/0.015m)[(0.943 m/s)2/(2∙9.81 m/s2)] = 2.54 m; ΔP = (9,790 N/m3)(2.54 m) = 24.9 kN/m2 = 24.9 kPa 5. A 30-cm diameter, cast-iron pipeline that is 2 km long carries water at 20°C. What is the maximum discharge if a 4.6-m head loss is allowed?

Ans. Use the Darcy-Weisbach equation: hf = f (L/D)(V2/2g); 4.6 m = f (2000m/0.30m)(V2/2∙9.81m/s2); The Moody diagram is required to find f.

However, V is not available so NR cannot be solved. Use (e/D) and the Moody diagram to obtain a trial f value by assuming flow is in the complete turbulence regime. e/D = (0.26 mm)/(300 mm) = 0.000867; f ≈ 0.02, and 4.6 m = (0.02)(2000m/0.30m)(V2/2∙9.81m/s2); V = 0.823 m/sec Now with ν = 1.00 x 10-6 m2/sec; NR = 2.47 x 105 From Moody; f = 0.0205, ok - close 4.6 m = (0.0205)(2000 m/0.30 m)(V2/2∙9.81 m/s2) V = 0.813 m/sec; Q = AV = 0.0575 m3/sec 22

6. A water main made of rough concrete loses 43 psi (lb/in.2) of pressure over a 1-mi horizontal length. Estimate the size of the pipe required to carry 16.5 cfs (ft3/sec) of water at 68°F (20˚C). Assume minor losses are negligible.

Ans. From Eq’n (3.15a) and noting 43 psi = 6190 lb/ft2: (P1 – P2)/γ = hL = hf ; (6190 lb/ft2)/(62.3 lb/ft3) = hf and hf = 99.4 ft; Substituting this into the Darcy-Weisbach equation: hf = f (L/D)(V2/2g) , we have 99.4 ft = f∙(5,280/D)[V2/(2∙32.2ft/s2)]; V = Q/A = 4Q/πD2 Thus; V = 4(16.5 ft3/sec)/[π(D2)] = 21.0/D2 and substituting we have 99.4 ft = f∙(5,280/D)[(21.0/D2)2/(2∙32.2 ft/s2)]; yielding D5 = 364∙f ; Neither D nor V is available so e/D and NR cannot be determined.

Iterate with f = 0.02 as a first trial, which is near midrange of typical f values.

Solving for D: D5 = 364∙(0.02); D = 1.49 ft; Now, e/D = 0.002ft/1.49ft = 0.00134 V = 21.0/D2 = 9.46 ft/s; and w/ν = 1.08 x 10-5 ft2/s NR = 1.31 x 106 From Moody, f = 0.021; the new D: D5 = 364∙(0.021); D = 1.50 ft (Check) 7. Water flows through a 20-cm commercial steel pipe at 20°C. The flow rate is 80 liters per second and pressure is constant throughout the length of the pipe. Determine the slope of the pipe. Assume minor losses are negligible.

Ans. First, apply the energy equation to the pipeline; V12 P1 V2 P + + h1 = 2 + 2 + h2 + hL 2g  2g  ; where hL = hf, V1 = V2; P1 = P2; Thus, h1 - h2 = hf

Now use the Darcy Weisbach equation to find hf . e/D = (0.045 mm)/(200 mm) = 0.000225 V = Q/A = (0.08m3/s)/[(π/4)(0.20 m)2] = 2.55 m/s NR = DV/ν = [(0.20m)(2.55 m/s)]/(1.00 x 10-6 m2/s) NR = 5.10 x 105 From Moody diagram; f = 0.016 Determine the friction head loss and convert to slope hf = f (L/D)(V2/2g); for a 100 m length of pipe hf = (0.016)(100m/0.20m)[(2.55 m/s)2/(2∙9.81 m/s2)] hf = 2.65 m; hf/L = 2.65/100 = 0.0265 or S = 2.65% 23

8. A proposed, 2-mile pipeline (2.5-ft-diameter; new HDPE pipe) is proposed to carry 77.6 cfs (ft3/sec) of water at 4°C (39˚F) between two reservoirs. The receiving reservoir is located near Denver, Colorado with a water surface elevation of 5,280 ft. MSL. The supply reservoir is being constructed in the foothills near the city. Determine the required water surface elevation of the supply reservoir using the (a) Hazen-Williams equation, and the (b) Manning equation? Ignore minor losses.

Ans. First, apply the energy equation to the pipeline; V12 P1 V2 P + + h1 = 2 + 2 + h2 + hL 2g  2g  ; where hL = hf, V1=V2= 0; P1= P2= 0;

Thus, h1 = h2 + hf = 5280 + hf. Both equations can be written in the form: hf = KQm a) Hazen-Williams: K = (4.73L)/(D4.87∙C1.85) = [(4.73)(2∙5280)]/(2.54.87∙1501.85) = 0.0543 hf = KQm = (0.0543)(77.6)1.85 = 170 ft; h1 = 5,450 ft b) Manning: K = (4.64∙n2∙L)/(D5.33) = (4.64∙0.0092∙2∙5280)/(2.55.33) = 0.0300 hf = KQm = (0.0300)(77.6)2 = 181 ft; h1= 5,461 ft 9. The elevation difference between two reservoirs 2000 m apart is 20 m. A 30-cm commercial steel (CHW = 140) pipeline connects the reservoirs. The computed flowrate using the HazenWilliams equation is 0.136 m3/sec. Determine the flowrate using the Manning equation (commercial steel, n = 0.013) and the Darcy-Weisbach equation (e = 0.045 mm).

Ans. The energy equation yields, h1 - h2 = hf = 20 m Both equations can be written in the form: hf = KQm a) Manning: K = (10.3∙n2∙L)/(D5.33); m = 2 K = (10.3∙0.0132∙2000)/(0.305.33) = 2130 hf = 20 = KQm = (2130)(Q)2; Q = 0.0969 m3/sec b) Darcy-Weisbach: K = (0.0826fL)/D5, m = 2 e/D = 0.045/300 = 0.000150; From Moody, try f = 0.014 based on the complete turbulence assumption.

K = (0.0826∙0.014∙2000)/(0.30)5 = 952 hf = 20 = KQm = (952)(Q)2; Q = 0.145 m3/sec Now V = Q/A = (0.145 m3/s)/[π(0.15m)2] = 2.05 m/sec NR = VD/ν = [(2.05)(0.30)]/1.0 x 10-6 = 6.15 x 105 From the Moody diagram, f = 0.0145; K = (0.0826∙0.0145∙2000)/(0.30)5 = 986 hf = 20 = KQm = (986)(Q)2; Q = 0.142 m3/sec 24

10. A check valve (swing) and a globe valve are installed in series in an 8-in. pipe. Determine the flow rate if the pressure drop across the valves is 5.19 psi (lbs/in.2). Assume that the friction losses are negligible due to the short pipe length.

Ans. The headloss is expressed as: hL = [∑Kv] (V)2/2g ΔP/γ = (5.19 lb/in.2)(144 in.2/ft2)/(62.3 lb/ft3) = 12.0 ft = hL 12.0 ft = [2.5 + 10] (V)2/2∙32.2 ft/s2, V = 7.86 ft/sec Q = V∙A = (7.86 ft/s)[π{(1/3)ft}2] = 2.74 ft3/sec 11. Water flows from the bottom of a storage tank through a horizontal, 6-inch pipe (cast iron) and discharges water (20˚C) into the atmosphere at the end. The pipe is 500 feet long, contains two 1-ft-radius bends, a fully open gate valve, and the entrance is square edged.

Calculate the flow rate if the depth of water in the storage tank is 60.2 feet above the inlet (and the outlet) of the pipe.

Ans. Applying the energy equation from the surface of the storage tank (1) to the outlet of the V12 P1 V2 P + + h1 = 2 + 2 + h2 + hL 2g  pipe (2) yields; 2 g  ; where P1 = P2 = 0; hL = hf + [∑K](V2/2g);

V1 = 0 and h2 = 0. Thus h1 = [1 + f(L/D) + ∑K](V2/2g); where V2 = V (pipe V) and Ke = 0.5; Kb = 0.19 (R/D = 2); Kv = 0.15; and assuming complete turbulence for the first trial: e/D = 0.00085ft/0.5ft = 0.00170; thus; f = 0.022, and h1 = 60.2 = [1+0.022(500/0.5)+0.5+2(0.19)+0.15](V2/2g); V = 12.7 ft/sec; NR = DV/ν = [(0.5)(12.7)]/(1.08x10-5) = 5.88 x 105; From Moody; new f = 0.022; (ok) Q = V∙A = (12.7 ft/s)[π(0.25ft)2] = 2.49 ft3/sec 12. A 75-m long cast-iron pipe, 15 cm in diameter, connects two tanks (open to the atmosphere) that have a water (20˚C) surface elevation difference of 5 m. The entrance of the pipe from the supply tank is square-edged, and the pipe contains a 90° bend with a sharp, 15 cm radius.

Determine the flow rate in the pipe.

Ans.

Applying the energy equation from the surface of the supply tank (1) to the

V12 P1 V2 P + + h1 = 2 + 2 + h2 + hL 2g  surface of the receiving tank (2): 2 g  ; where P1 = P2 = 0;

hL = hf + [∑K](V2/2g); and V1 = V2 = 0. Therefore h1 - h2 = [f(L/D) + ∑Kv](V2/2g); where V is pipe V; Ke = 0.5; Kb = 0.35 (R/D = 1); Kd = 1.0 (exit loss); and assuming complete turbulence (first trial): e/D = 0.26mm/150mm = 0.00173; thus; f = 0.022, and h1 - h2 = 5 = [0.022(75/0.15) + 0.5 + 0.35 + 1.0](V2/2g); V = 2.76 m/sec; 25

NR= DV/ν = [(0.15)(2.76)]/(1.0x10-6) = 4.14 x 105; From Moody; new f = 0.0225; Now solving the energy eq’n again yields V = 2.74 m/s and Q = V∙A = (2.74 m/s)[π(0.075m)2] = 0.0484 m3/sec Note: Theoretically, the pipe entrance coefficient of 0.5 should be adjusted according to the pipe velocity and the first column of Table 3.5. However, this usually does not affect the final result and a loss coefficient of 0.5 is often used for square-edged entrances.

13. Pipes AB and CF in the figure below have diameter of 4 feet and carry a discharge of 120 cfs. The length of AB is 1800 feet and that of CF is 1500 feet. Branch 1 is 1800 feet long and has a diameter of 3 feet. Branch 2 has a length of 1500 feet and a diameter of 2 feet. (a) Determine the total head loss due to friction between points A and F, and (b) determine the discharge in each of the two branches (1 and 2). All of the pipes possess a Manning n value of 0.013 and assume a diameter of 4 feet for the equivalent pipe.

Branch 1

Q A

Q

Q1 B

Q2

C

F

Branch 2

Ans. Using the appropriate equivalent pipe equation from Table 3.7 (Manning) yields: [(DE5.33)/(nE2LE)]1/2 = [(D15.33)/(n12L1)]1/2 + [(D25.33)/(n22L2)]1/2 [(45.33)/(0.0132∙LE)]1/2 = [(35.33)/(1800∙0.0132)]1/2 + [(25.33)/(1500∙0.0132)]1/2; LE = 4430 ft.

From Eq’n 3.44; hfAF = hfAB + hfBC + hfCF (and using appropriate equation from Table 3.4) hfAF = [(4.64∙1800∙0.0132)/45.33]1202 + [(4.64∙4430∙0.0132)/ 45.33]1202 + [(4.64∙1500∙0.0132)/45.33]1202 hfAF = 12.6 + 30.9 + 10.5 = 54.0 ft; Flow in pipe branches, hfBC = 30.9 ft = [(4.64∙1800∙0.0132)/35.33]Q12; Q1 = 87.4 cfs hfBC = 30.9 ft = [(4.64∙1500∙0.0132)/25.33]Q22; Q2 = 32.6 cfs

14. Determine the equivalent to replace the pipe system of Example 3.10.

Ans. In Example 3.10, an equivalent pipe was found for the parallel pipes resulting in 3 pipes in series. Replace these 3 pipes with an equivalent pipe with D = 4 ft and f = 0.02 [(fE∙LE)/(DE5)] = [(f1∙L1)/(D15)]+[(f2∙L2)/(D25)]+[(f3∙L3)/(D35)] [(0.02∙LE)/(45)] = [(0.02∙1800)/(45)] + [(0.02∙3310)/(45)] + [(0.02∙1500)/(45)]; LE = 6610 ft.

(Note: The individual pipe lengths could be added directly since all have the same D and f.)

26

TEST QUESTIONS & PROBLEMS - CHAPTER #4 Short Answer Questions 1. Why is an iterative procedure required to determine the size (diameter) of a pipeline needed to pass a given flow from one reservoir to another if you are provided with the reservoir water surface elevations, pipe length, and the pipe material (assuming the Darcy-Weisbach equation is used for friction loss)?

Ans. The difference in reservoir surface elevations is equal to the head loss, and the greatest loss is generally friction loss. The friction factor is unknown and depends on the relative roughness and possibly the Reynold’s number. Both of these parameters require the diameter, which is unknown, and thus an iterative procedure is necessary.

2. (T or F) When a pipeline connects two reservoirs, the difference in water surface elevations is equal to the total head loss (friction and all minor losses).

Ans. True 3. Sketch the energy grade line (EGL) and hydraulic grade line (HGL) for the pipeline below.

Valve

ve

Ans. Sketch the EGL and HGL such that a) both start & end at the reservoir surfaces, b) the minor losses (entrance, valve, expansion, and exit) are accounted for with an abrupt drop in the EGL and HGL; c) the EGL and HGL are parallel lines, but the separation distance (i.e., the velocity head, V2/2g) is less for the larger pipe, and d) the slope of the EGL, which represents the friction loss over length, is greater for the smaller pipe.

4. Sketch the energy grade line (EGL) and hydraulic grade line (HGL) for the pipeline below.

27

Ans. Sketch the EGL and HGL such that a) both start & end at the reservoir surfaces, b) the minor losses (entrance, contraction, expansion, exit) are accounted for with an abrupt drop in the EGL and HGL; c) the EGL and HGL are parallel lines, but the separation distance (i.e., the velocity head, V2/2g) is less for the larger pipes, and d) the slope of the EGL, which represents the friction loss over length, is greater for the smaller pipe.

5. Define cavitation.

Ans. The formation of vapor pockets in pipelines which occurs when the pressure drop falls below the vapor pressure of water. This is often followed by vapor collapse when normal pressure return downstream and is accompanied by damaging vibration and sound waves.

6. What two locations in pipelines are likely places for cavitation to occur?

Ans. The suction side of pumps and the high point in siphons or pipelines.

7. (T or F) Cavitation can occur on the discharge side of pumps or when a pipeline rises above the hydraulic grade line (HGL).

Ans. False – on the suction side of pumps.

8. Do all siphons encounter negative pressure at their summit points? Prove your answer by using an energy grade line (EGL) and hydraulic grade line (HGL) sketch.

Ans. Yes, the pipe center line rises above the HGL as seen in the sketch below.

-P/γ EGL HGL

9. Sketch the energy grade line (EGL) and hydraulic grade line (HGL) for the pipeline below.

Pump

Ans. Sketch the EGL and HGL such that a) both start & end at the reservoir surfaces, b) the minor losses (entrance, bends, exit) are accounted for with an abrupt drop in the EGL and HGL; c) the EGL and HGL are parallel lines, and d) the pump adds a significant boost (abrupt rise) to the EGL. 28

10. (T or F) Pumps add energy to a fluid primarily in the form of increased velocity head.

Ans. False – increased pressure head.

11. (T or F) Negative gage pressure in a pipeline occurs whenever the pipe rises above the hydraulic grade line (HGL).

Ans. True 12. Solving the classical 3-reservoir problem generally does not require the determination of a) friction losses b) minor losses c) mass balance d) energy balance Ans. (b) – minor losses are often ignored because they are small when compared to friction.

13. In the classical 3-reservoir problem, we are typically solving for the three pipe flow rates.

What other unknown enters the analysis and what four equations are used in the solution?

Ans. The fourth unknown is the total head at the junction. The four solution equations include three energy balances between the reservoirs and the junction (which includes a friction loss equation) and mass balance at the junction.

14. Solution of the classical 3-reservoir problem relies on all of the following principles except a) conservation of momentum b) conservation of energy c) conservation of mass d) friction loss equation (e.g., Darcy-Weisbach) Ans. (a) 15. When solving the classical 3-reservoir problem, a trial energy elevation at the junction may be needed. What is a good first estimate and why?

Ans. The middle reservoir’s water surface elevation is a good first estimate; it reduces the number of computations and the result will indicate whether a higher or lower estimate is needed. Thus, the direction of flow (to or from) the middle reservoir is established.

16. All of the following principles are applied to the classical 3-reservoir problem except b) conservation of momentum b) conservation of energy c) Bernoulli principle d) conservation of mass e) friction losses e) minor losses Ans. (a) and (c) are not used, and (e) is generally not used, although it could be.

17. (T or F) Two major assumptions used in the solution of the classical 3-reservoir problem include negligible minor losses in the pipes and a negligible velocity head at the junction.

Ans. True 18. Starting with the Darcy-Weisbach equation, determine the value of K in hf = KQ2.

Ans. hf =f(L/D)(V2/2g) = f(L/D)[Q2/(2g A2)] =[fL/(2g∙D∙A2)] Q2; K=[fL/(2g∙D∙A2)] 19. Give two advantages of having the grid system in pipe networks rather than a dendritic (treelike) system in distributing water to urban customers.

Ans. Fewer people are without service if a line breaks (valves can be closed on either side of the leak). In a dendritic system, everyone downstream of the break is without water until it is fixed. Also, water is less likely to stagnate in a grid system.

29

20. What principles are used to determine flow rates in a water distribution system (network)? c) conservation of momentum b) conservation of energy c) Bernoulli principle d) conservation of mass e) friction losses e) minor losses Ans. (b), (d), and (e) are used; (f) is generally not although it could be..

21. In solving a pipe network problem, one condition which must be satisfied is mass balance at each junction (∑Q = 0). What other condition must be satisfied?

Ans. Between any two junctions, the total head loss is independent of the path taken based on the conservation of energy (loop equation).

22. Given the supply and demands in the pipe network, provide initial estimates of pipe flows.

150 cfs 50 cfs 100 cfs 100 cfs

100 cfs

Ans. Multiple answers exist, but mass balance at junctions must be maintained.

23. Once the flows in a pipe network have been established, how can you determine the location of lowest pressure in the network if all junction elevations are the same?

Ans. The junction of lowest pressure is the one in which all pipe flows move toward. If more than one junction like this exists, head loss computations will determine the lowest.

24. (T or F) The speed of a pressure wave in a pipeline is faster for saltwater than fresh water.

Ans. False; saltwater is more dense and C =(Ec/ρ)1/2 25. Identify two principles used in the derivation of the water hammer equations for pipelines.

Ans. Newton’s 2nd law [Equation (4.25b)], mass balance [Equation (4.23)] , and fluid elasticity [Equation (4.24)].

26. Identify three tactics an engineer can use to eliminate pipeline damage from water hammer.

Ans. Increase pipe wall thickness, slow valve closure, diverter, surge tank, or standpipe.

27. The speed of a pressure wave in a pipeline is dependent on which of the following? a) fluid density b) pipe wall thickness c) pipe diameter d) modulus of elasticity of the fluid e) pipeline anchoring method Ans. The wave is (C) is dependent on all of these factors. See Equations (4.21) and (4.22).

30

Problems 1. In the pipe system shown below the pipe has a diameter of 30 cm, and it is 300 m long. The equivalent roughness height of the pipe, e, is 0.1 mm. The kinematic viscosity is  = 1.13 × 10-6 m2/s. The minor head loss coefficients are: Ke = 0.5 for entrance losses, and Kd = 1.0 for exit losses. The water surface elevations are h1= 68.9 m, and h2= 50 m. The discharge is Q = 0.30 m3/s. Determine the valve head loss coefficient, Kv.

P V2 P V2 Ans. Apply the energy eq’n; h1 + 1 + 1 = h2 + 2 + 2 + hL ; where V1 = V2 = P1 = P2 = 0,  2g  2g 2 hL = hf + [∑K](V) /2g; Ke = 0.5, Kv = ?, Kd = 1.0 (exit coefficient). Thus,

h1 – h2 = [f(L/D) + ∑K](V2/2g) = [f(300/0.30) + 0.5 + Kv + 1.0](V2/2g) V = Q/A = (0.3 m3/sec)/[(π/4)(0.30 m)2] = 4.24 m/sec; e/D = 0.0001m/0.30 m = 0.00033 NR=DV/ν=[(0.30m)(4.24 m/s)]/(1.13 x 10-6 m2/s)= 1.13 x 106; From Moody; f = 0.016; thus, h1 – h2 = 68.9 m – 50 m = [0.016(300/0.30) + 0.5 + Kv + 1.0]∙[(4.24)2/2∙9.81] and, Kv = 3.13 2. At a water treatment plant, water (68˚F) flows from tank A to tank B at a rate of 0.50 cfs through a 3-in.-diameter, cast-iron pipe that is 200 feet long. Determine the difference in water surface elevation between tanks (open to the atmosphere) if there are two bends (R/D = 2.0) and a fully open gate valve in the pipeline.

P V2 P V2 Ans. Apply the energy eq’n; hA + A + A = hB + B + B + hL ; where VA = VB = PA = PB = 0,  2g  2g 2 hL = hf + [∑K](V) /2g; Ke = 0.5, Kb = 0.19, Kv = 0.15, and Kd = 1.0 (exit coefficient).

Thus, hA - hB = [f(L/D) + ∑K](V2/2g) = [f(200/0.25) + 0.5 + 2(0.19) + 0.15 + 1.0](V2/2g) V = Q/A = (0.50 ft3/s)/[(π/4)(0.25 ft)2] = 10.2 ft/sec; e/D = 0.00085ft/0.25ft = 0.0034 NR = DV/ν = [(0.25 ft)(10.2 ft/s)]/(1.08 x 10-5 ft2/s) = 2.36 x 105; From Moody; f = 0.027; and, hA - hB = [0.027(200/0.25) + 0.5 + 2(0.19) + 0.15 + 1.0]∙[(10.2)2/2∙32.2] and, hA - hB = 38.2 ft

31

3. If the pressure at point 1 in the figure below is 50.9 kPa and the water flow in the 4-inch pipe (0.102 m) is 10.1 litres/sec (20°C),, determine the pressure P0 at reservoir A.

VA2 P V2 = h1 + 1 + 1 + hL ; VA = 0, V1 = V (pipeline V),  2g  2g 2 hL = hf + [∑K](V) /2g; hA= 10 m, h1= 7 m, PA= P0, and P1/γ = 50.9/9.79 = 5.20m. Thus,

Ans. Apply the energy eq’n; hA +

PA

+

P0/γ = h1 - hA + P1/γ + [1+f(L/D)+∑K](V2/2g) = -3m + 5.2m - [1+f(L/D)+∑K](V2/2g) V = Q/A = (0.0101 m3/s)/[(π/4)(0.102 m)2] = 1.24 m/s; e/D = 0.045/102 = 0.000441 NR = DV/ν = [(0.102m)(1.24 m/s)]/(1.00 x 10-6 m2/s) = 1.26 x 105; hence f = 0.02, and, P0/γ = 2.2 m - [1+0.02(10/0.102)+0.5+12.0][(1.24)2/2∙9.81] = 0.988m; P0 = 9.67 kPa 4. The 40-m-long, 4-in. diameter commercial steel pipe connects reservoirs A and B as shown in the Figure from Problem #3 above. The water temperature is 20°C. Determine the flow rate (liters/sec) if reservoir A is subjected to atmospheric pressure, the entrance is square-edged, and bend losses are negligible.

2

2

Ans. Apply the energy eq’n between reservoir surfaces; hA + PA + VA = hB + PB + VB + hL ; and  2g  2g VA=VB=PA=PB= 0, hL = hf+[∑K](V)2/2g; Ke= 0.5, Kv1=12, Kv2= 10, Kd= 1.0 (exit coef) Thus, hA – hB = 2.5 m = [f(40m/0.102m) + 23.5](V2/2g); where V is pipe V. Assume complete turbulence for the first trial: For commercial steel: e/D = 0.045mm/102mm = 0.000441; thus; f = 0.0165 and solving energy eq’n 2.5 m = [0.0165(40m/0.102m) + 23.5](V2/2g); V = 1.28 m/sec.

Solving for the Reynolds number, NR = DV/ν = [(0.102)(1.28)]/(1.00x10-6) = 1.31 x 105; From Moody; new f = 0.02; thus; 2.5 m = [0.02(40m/0.102m) + 23.5](V2/2g); V = 1.25 m/sec. NR = DV/ν = 1.28 x 105; f = 0.02; OK Thus, Q = V∙A = (1.25 m/s)[(π/4)(0.102m)2] = 0.0102 m3/sec = 10.2 L/sec

32

5. An irrigation company must transport 5.71 x 10-2 m3/sec of water (20˚C) from reservoir A to reservoir B. The reservoirs are 600 m apart with an elevation difference of 18.4 m. Determine the required pipe diameter if the roughness (e) of the pipe material is 0.36 mm. (Include minor losses.) Would the solution differ if minor losses were neglected?

2

2

Ans. a) Apply the energy eq’n; hA + PA + VA = hB + PB + VB + hL ; where VA = VB = PA = PB = 0, 

2g



2g

and hL = hf + [∑K](V) /2g; with Ke = 0.5, Kd = 1.0 (exit coef.), and hA – hB = 18.4 m 2

Now, V = Q/A = 0.0727/D2. Thus, 18.4 m = [f(600m/D) + 1.5]∙[(0.0727/D2)2/2g].

Assume D = 0.2 m, thus V = 1.82 m/s, e/D = 0.36/200 = 0.0018; and NR = DV/ν = [(0.2)(1.82)]/(1.00x10-6) = 3.64 x 105; From the Moody diagram, f = 0.023; Solve the energy e’qn for new D; 18.4 m = [0.023(600m/D) + 1.5]∙[(0.0727/D2)2/2g]; D = 0.183 m. Now, V = 2.17 m/s, e/D = 0.00197; and NR = 3.97 x 105; new f = 0.0235 From the energy eq’n: 18.4 m = [0.0235(600m/D)+1.5]∙[(0.0727/D2)2/2g]; D = 0.184 m; close enough so D = 0.184 m; Without minor losses from the energy equation; 18.4 m = [0.0235(600m/D)]∙[(0.0727/D2)2/2g]; D = 0.183 m; essentially the same.

Note: The solution process is easier without minor losses; it is often a good idea to solve this way first for an initial trial size and then solve again with minor losses included.

6. The siphon spillway shown below is 100 feet long with a 1.0-ft-diameter and conveys water to a downstream reservoir. The friction losses in the rough concrete siphon are distributed evenly throughout its length. If the siphon’s summit (high point) is 7 feet above the surface of the upper reservoir and 18 feet (in pipe length) from the entrance, is cavitation (i.e., a pressure less than two thirds of atmospheric pressure or -9.8 lb/in2) a concern?

Ans. Balancing energy between the upstream reservoir and the summit, where cavitation is most likely to occur, would yield two unknowns, the velocity and the pressure at the summit. Therefore, balance energy between the upstream reservoir (A) and the downstream reservoir (B) to determine the velocity. Thus, hA +

PA



+

VA2 P V2 = hB + B + B + hL ; where VA= VB = PA = PB = 0; hA – hB = 53 ft; & head 2g  2g

losses are hL = hf + [∑K](V)2/2g; where Ke = 0.5, Kd = 1.0 (exit coef.). Therefore, 53 ft = [f(100/1.0) + 1.5](V2/2g); where V is pipe V. Assume complete turbulence: 33

For rough concrete: e/D = 0.002ft/1.0ft = 0.002; thus; f ≈ 0.0235 & solving energy eq’n 53ft = [0.0235(100/1.0) + 1.5](V2/2g); V = 29.8 ft/sec. Now solve for Reynolds number, NR = DV/ν = [(1.0)(29.8)]/(1.08x10-5) = 2.76 x 106; From Moody; new f = 0.0235; OK Now balancing energy from A to the summit yields; hA +

PA



+

P V2 VA2 = hS + S + S + hL ; where VA= PA= 0; VS= V; hA–hs= -7ft, Ke = 0.5; thus 2g  2g

-7 ft = PS/γ+ (29.8)2/2g + [0.5 + 0.0235(18/1.0)](29.8)2/2g; solving yields PS/γ = -33.5 ft Ps = (-33.5ft)(62.3 lbs/ft3)(1 ft2/144 in2) = -14.5 lbs/in2 Hence, cavitation is a concern.

7. Water flows in a new 20-cm, 300-m-long ductile iron pipe between reservoirs A and B, as shown in the figure below. The pipe is elevated at S, which is 150 m downstream from reservoir A. The water surface in reservoir B is 25 m below the water surface in reservoir A. If Δs = 7.0 m, is cavitation a concern?

Ans. Balancing energy between the upstream reservoir and the summit, where cavitation is most likely to occur, would yield two unknowns, the pipe velocity and the pressure at the summit. Therefore, balance energy between the upstream reservoir (A) and the downstream reservoir (B) to determine the velocity. Thus, hA +

PA



+

VA2 P V2 = hB + B + B + hL ; where VA= VB = 0; PA = PB = 0; hA – hB = 25 m; and 2g  2g

hL = hf + [∑K](V)2/2g; where Ke = 0.5, Kv1 = 0.15, Kd = 1.0 (exit coef.). Therefore, 25 m = [f(300m/0.20m) + 1.65](V2/2g); where V is pipe V. For complete turbulence: Ductile iron: e/D = 0.12mm/200mm = 0.0006; thus; f ≈ 0.018 and solving energy eq’n 25 m = [0.018(300m/0.20m) + 1.65](V2/2g); V = 4.14 m/sec. Thus, Reynolds number, NR = DV/ν = [(0.20)(4.14)]/(1.00x10-6) = 8.28 x 105; From Moody; new f = 0.018; OK 2

2

Balancing energy (A to summit); hA + PA + VA = hS + PS + VS + hL ; VA= PA= 0; VS= V; 

2g



2g

hA=Δs=7.0 m (datum at hs); 7.0m = PS/γ+(4.14)2/2g+[0.5+ 0.018(150/0.20)](4.14)2/2g PS/γ = -6.10 m (> -7.0 m; no cavitation; see paragraph prior to Ex. 4.4 in book)

34

8. Water flows in a new 8-in.-diameter, 1000-ft-long ductile iron pipe between reservoirs A and B, as shown in the figure below. The pipe is elevated at S (location of the summit) which is 500 ft downstream from reservoir A. The water surface difference between the two reservoirs is 82 ft. Determine the head that should be provided by a booster pump 330 feet downstream from reservoir A if Δs is 10 feet and the pressure head at that location (summit) must be kept above -20 feet due to cavitation concerns. Sketch the EGL and the HGL for the pipeline.

Ans. Balancing energy between the summit and the downstream reservoir (B), yields, P V2 P V2 hS + S + S = hB + B + B + hL ; where VB = PB = 0; hS – hB = 72 ft; PS/γ = -20 ft; and  2g  2g hL = hf + [∑K](V)2/2g; w/Kv = 0.15; Kd = 1.0 (exit). Substituting yields: 72ft + (-20ft) = [f(500ft/0.667ft)+1.15](V2/2g); Assuming complete turbulence: e/D = 0.0004ft/0.667ft = 0.0006; thus; f ≈ 0.0175 and solving energy eq’n 52ft = [0.0175(500ft/0.667ft) + 1.15](V2/2g); V = 15.3 ft/sec. Therefore, NR = DV/ν = [(0.667)(15.3)]/(1.08x10-5) = 9.45 x 105; From Moody; new f = 0.0175; OK Balance energy (A to B); H A + H p = H B + hL ; HA- HB = 82ft; Ke= 0.5; Kv= 0.15; Kd= 1.0 82 ft + HP = [0.0175(1000ft/0.667ft) + 1.65]∙[(15.3ft/sec)2/2g); HP = 19.4 ft Start drawing your EGL and HGL from the reservoir surface (on the left) and move to the right accounting for head losses along the way. Note that a) the EGL and HGL both start and end at the reservoir surfaces, b) the minor losses (entrance, valve, and exit) are accounted for with an abrupt drop in the EGL and HGL; c) the separation distance between the EGL and HGL accounts for the velocity head, V 2/2g, so they are parallel lines, d) the slope of the EGL represents the friction loss over length, and e) the pump adds energy shown as an abrupt rise in the EGL and HGL.

EGL A HGL P B

35

9. Determine the flow in each pipe in the figure below if all of the pipes are made of the same material with e = 0.05 mm and water temperature at 20°C (ν = 1.00 x 10-6). Try an initial energy level at the junction of 82 m, and go through one complete iteration of the solution including a comment on your second trial energy level at the junction.

Ans. Assume a total energy elevation at the junction of HJ = 82 m. Then balance energy between all reservoirs and the junction. (Use Darcy-Weisbach eq’n for friction loss.) h1 = 100 – 82 = f1(L1/D1)[(V1)2/2g] = f1(3000/0.8)[(V1)2/2∙9.81]. For full turbulence e/D = 0.05mm/800mm = 0.0000625; thus; f ≈ 0.011 & solving the energy eq’n; V1 = 2.93 m/sec. Now NR = DV/ν = [(0.8)(2.93)]/(1.00x10-6) = 2.34 x 106; From Moody; new f = 0.0125; New V1 = 2.74 m/sec. Now NR = 2.19 x 106 (f is OK) Q1 = 1.38 m3/sec h2 = 82 - 80 = f2(L2/D2)[(V2)2/2g] = f2(4000/1.2)[(V2)2/2∙9.81]. For full turbulence e/D = 0.05mm/1200mm = 0.0000417; thus; f ≈ 0.010 & solving the energy eq’n; V2 = 1.08 m/sec. Now NR = DV/ν = [(1.2)(1.08)]/(1.00x10-6) = 1.30 x 106; From Moody; new f = 0.0125; New V2 = 0.970 m/sec. Now NR = 1.16 x 106 (f is OK) Q2 = 1.10 m3/sec h3 = 82 - 70 = f3(L3/D3)[(V3)2/2g] = f3(5000/0.6)[(V3)2/2∙9.81]. For full turbulence e/D = 0.05mm/600mm = 0.0000833; thus; f ≈ 0.0115 & solving the energy eq’n; V3 = 1.57 m/sec. Now NR = DV/ν = [(0.6)(1.57)]/(1.00x10-6) = 9.42 x 105; From Moody; new f = 0.0135; New V3 = 1.47 m/sec. Now NR = 8.82 x 105 (f is OK) Q3= 0.416 m3/sec Hence, the summation of flows at the junction: ∑Q = 1.38 - (1.10+0.42) = - 0.14m3/sec There is a little too much water leaving the junction: Try HJ = 81.7 m

36

10. Determine the flow into or out of each reservoir in the figure if the connecting pipes all have a Hazen-Williams coefficient of 120. (Hint: Try a junction energy elevation of 81.8 m.)

Ans. Assume a total energy elevation of 81.8 m at the junction J Reservoir Water

Total Head at Junction (m) HJ = 81.8 Trial until ∑Qs Balance

Surface Elevations

WS1 = WS2 = WS3 =

100 80 70

m m m

Pipe Lengths L1 = 3000 m L2 = 4000 m L3 = 5000 m Pipe#

CHW

hf

D1 = D2 = D3 = S*

(m) 1 2 3

120 120 120

18.2 1.8 11.8

0.00607 0.00045 0.00236

* S = hf/L (friction slope or EGL slope in this case)

Pipe Diameters 0.80 m 1.20 m 0.60 m

Rh**

V***

Q

(m)

(m/sec)

(m3/sec)

0.20 0.30 0.15

2.35 0.74 1.18

1.18 0.84 0.33

If HJ < WS2; ∑Q =

1.69

If HJ > WS2; ∑Q =

0.01

** Rh = D/4 (Equation 3.26) *** V = 0.849CRh0.63S0.54 (Equation 3.27)

11. A small branching pipe system distributes water to three holding tanks. Pipe 1 contains a valve that is partly closed that limits the flow rate to 1.0 cfs with a valve head loss of 3.60 feet. Estimate the length of Pipe 2 given the following water surface (WS) elevation and pipe data (lengths, diameters, and Hazen-Williams coefficients): WS1 = 4020 ft L1 = 1000 ft D1 = 1.0 ft C1 = 80 WS2 = 4018 ft L2 = ?

D2 = 10 in.

C2 = 100 WS3 = 4000 ft L3 = 5000 ft D3 = 1.0 ft C3 = 120 The total energy elevation at the junction (HJ) is 4015 ft.

37

WS1 Reservoir 1

WS2 HJ Pipe 1 1 WS3

Reservoir 2

Pipe 2 1

J Pipe 3 1

Reservoir 3

Ans. The flow to the junction from reservoir 1 (in Pipe 1) is 1.0 cfs. Using the HazenWilliams equation, determine the flow in Pipe 3.

Q3 = VA = (1.318CHWRh0.63S0.54 )(A) = (1.318)(120)(1ft/4)0.63(15/5000)0.54 (π/4)(1ft)2 Q3 = 2.25 cfs; Therefore, the flow in Pipe 2 is 2.25 – 1.0 = 1.25 cfs toward J. Hence, Q2 = 1.25 cfs = (1.318)(100)[(10/12)ft/4]0.63[3/(L2)]0.54 (π/4)[(10/12)ft]2; L2 = 873 ft 12. The total discharge from A to B in the figure below is 50.0 cfs (ft3/sec). Pipe 1 is 4000 ft long with a diameter of 1.5 ft, and pipe 2 is 3000 ft long with a diameter of 2 ft. Using HardyCross principles, determine the head loss between A and B and the flow rate in each pipe.

Assume f1 = 0.019, f2 = 0.018, ν = 1.08 x 10-5 ft2/sec, and ignore minor losses.

1 A

B 2

Ans. Use the principles of mass balance and energy balance to determine flows and heads: hL = f(L/D)(V2/2g); and based on conservation of energy, hL1 = hL2. Therefore, (0.019)(4000/1.5)[(V1)2/2g] = (0.018)(3000/2.0)[(V2)2/2g] V1 = Q1/A1 = Q1/[(π/4)(1.5)2] = 0.566∙Q1; V2 = Q2/A2 = Q2/[(π/4)(2.0)2] = 0.318∙Q2; thus (0.019)(4000/1.5)[(0.566∙Q1)2/2g] = (0.018)(3000/2.0)[(0.318∙Q2)2/2g] Q1 = 0.410∙Q2; from mass balance, Q1 + Q2 = 50.0 ft3/s; substituting yields Q2 = 35.5 ft3/s; and therefore Q1 = 14.5 ft3/s. The head loss (using pipe 1) is hL = f(L/D)(V2/2g) = (0.019)(4000/1.5)[{(0.566)(14.5)}2/2g] = 53.0 ft ➔Using pipe 1 38

13. The total discharge from A to B in the figure below is 12 liters/sec. Pipe 1 is 25 m long with a diameter of 4 cm, and pipe 2 is 30 m long with a diameter of 5 cm. Using the method of equivalent pipes, determine the head loss between A and B and the flow rate in each pipe.

Assume f1 = 0.033, f2 = 0.030, ν = 1.00 x 10-6 m2/sec, ignore bend losses, and for the equivalent pipe, let DE = 0.06 m and fE = 0.029.

1 A

B 2

Ans. Find the equivalent pipe to replace 1 and 2, letting D = 0.06 m and f = 0.029; [(DE5)/(fE∙LE)]1/2 = [(D15)/(f1∙L1)]1/2 + [(D25)/(f2∙L2)]1/2; from Equation (3.47) [(0.065)/(0.029∙LE)]1/2 = [(0.045)/(0.033∙25)]1/2 + [(0.055)/(0.030∙30)]1/2; therefore, LE = 30.2 m. and to determine the head loss (from Table 3.4) hfAB = [(0.0826∙f∙L)/D5]Q2 = [(0.0826∙0.029∙30.2)/(0.06)5](0.012)2 = 13.4 m Q in pipe branches, hf1= 13.4 m =[(0.0826∙0.033∙25)/(0.04)5]Q12; Q1 = 0.00449 m3/s = 4.49 L/sec. Also, hf2 = 13.4 m = [(0.0826∙0.030∙30)/(0.05)5]Q22; Q2 = 0.00751 m3/s = 7.51 L/sec. ;The same result can be found from mass balance (Q1 + Q2 = 12 L/sec)

14. Fill in the blanks for the partially completed pipe network solution table given below.

Loop

Pipe

Q (m /sec) ?

0.025 (0.095) 0.153

K (sec2/m5) ?

4880 775 131

3

C

8 9 10 4

hf (m) 1.30 3.05 ?

3.07

hf/Q (sec/m2) 31.7 122.0 ?

20.1

New Q (m3/sec) ? ? ? ?

Ans. The table can be filled in using the equation hf = KQ2 and the flow adjustment equation: Q =

Loop

(

h fc − h fcc

2 h fc / Qc + h fcc / Qcc

Pipe

Q (m /sec) 0.041 0.025 (0.095) 0.153 3

C

8 9 10 4

)

=

(1.30 + 3.05 + 3.07 ) − (6.99) = 0.001 m3 / sec 2((31.7 + 122.0 + 20.1) + (73.6)) K (sec2/m5) 773 4880 775 131

hf (m) 1.30 3.05 (6.99) 3.07

hf/Q (sec/m2) 31.7 122.0 (73.6) 20.1

New Q (m3/sec) 0.040 0.024 (0.096) 0.152 39

15. From the pipe network data below, determine the flow in pipes CE and DE, and the pressure head HE. Also determine the pressure (in psi) at E if the ground elevation is 650 ft.

Pipe

Flow (ft3/sec) 15.35 5.76 9.55 2.29 ? ?

AB BC BD DC CE DE

Length (ft) 1000 1500 1200 1500 2000 1000

Diameter (in.)

12 8 12 6 12 8

K (sec2/ft5) 0.424 1.510 1.100 9.530 1.540 0.949

HA = 1000 ft A C

B

QE = ?

HE = ?

E D

Ans. Use the principles of mass balance and energy balance to determine flows and heads: QDE = QBD - QDC = 9.55 – 2.29 = 7.26 cfs; QCE = QBC + QDC = 5.76 + 2.29 = 8.05 cfs HE = HA – hf(AB) – hf(BD) - hf(DE) ; hf(AB) = (kQ2)AB = (0.424)(15.35)2 = 100 ft hf(BD) = (kQ2)BD = (1.100)(9.55)2 = 100 ft; hf(DE) = (kQ2)DE = (0.949)(7.26)2 = 50.0 ft HE = 1000 – 100 – 100 – 50 = 750 ft; (P/γ)E = HE - ZE = 750 – 650 = 100 ft. (Assuming V2/2g = 0); and PE = (100 ft)(62.3 lbs/ft3)(1 ft2/144 in.2); PE = 43.3 psi 16. Pipe flows are estimated in the pipe network below based on mass balance principles. These flow rates are given in the table with the direction specified (e.g., AB indicates flow from A to B). Proceed through one iteration of the Hardy Cross algorithm for each loop. Also, if the total head at A is 50 feet, determine the pressure (psi) at E (elev = 0) using your results.

B

1.8 cfs

A 0.4 cfs

1

D

0.6 cfs

2 0.8 cfs

C

E

40

Ans. The original data provided is in black; the information requested is in bold red.

Loop

Pipe

1

AB BC AC

Q (cfs) 1.00 0.20 (0.8)

K 14.25 7.00 14.25

hf (ft) 14.25 0.28 (9.12)

hf /Q 14.25 1.40 (11.40)

New Q (cfs) 0.90 0.10 0.90

ΔQ = [∑(hfc) - ∑(hfcc)] / [2(∑(hfc/Qc) + ∑(hfcc/Qcc))] = [(14.25+0.28)-(9.12)] / [2(14.25+1.40)+(11.40))] = +0.10

Loop

Pipe

2

BD DE BC CE

Q (cfs) 0.80 0.20 (0.10) (0.60)

K 14.25 7.00 7.00 14.25

hf (ft) 9.12 0.28 (0.07) (5.13)

hf /Q 11.40 1.40 (0.70) (8.55)

New Q (cfs) 0.70 0.10 0.20 0.70

ΔQ = [(9.12+0.28)-(0.07+5.13)] / [2(11.40+1.40)+(0.70+8.55))] = +0.10

HE = HA – hf(AC) – hf(CE) = 50 – 9.12 – 5.13 = 35.8 ft; (P/γ)E = HE - ZE = 35.8 ft.

(Assuming V2/2g = 0); and PE = (35.8 ft)(62.3 lbs/ft3)(1 ft2/144 in.2); PE = 15.5 psi

17. In the pipe network shown below, all of the pipes have a Darcy-Weisbach friction factor of 0.02. Pipe AB is has a length of 5000 ft and a diameter of 3 ft. Each of the pipes BC, BD, CE, and DE have a length of 1000 ft and a diameter of 2 ft. Pipe DC is 1410 ft long and has a diameter of 2 ft. The final solution obtained for this network indicates that the flows in pipes BC, CE, and DE are respectively 20 cfs, 10 cfs, and 10 cfs as shown in the figure. Also, the energy head at D is 100 feet. Determine the water surface elevation in Reservoir A, the energy loss in pipe DC, and the energy loss in DC if the diameter were increased to 3 feet.

41

Ans. From the mass balance equation based on system outflows: QAB = 20 + 20 + 10 + 10 + 5 = 45 cfs.

Also, from the mass balance at junction B: QBD = 45 - 20 - 5 = 20 cfs.

Now determine energy losses in pipes AB and BD: VAB = Q/A = (45 ft3/sec)/[(π/4)(3.0 ft)2] = 6.37 ft/sec; hfAB = f(L/D)(V2/2g) = (0.02)(5000/3)[(6.37)2/(2·32.2)] = 21.0 ft VBD = Q/A = (20 ft3/sec)/[(π/4)(2.0 ft)2] = 6.37 ft/sec; hfBD = f(L/D)(V2/2g) = (0.02)(1000/2)[(6.37)2/(2·32.2)] = 6.30 ft Balancing energy between reservoir A and junction D yields: HA - hfAB - hfBD = HD

Therefore, HA = 100 + 21.0 + 6.3 = 127.3 ft

Also, from the mass balance at junction D: QDC = 20 - 10 - 10 = 0 cfs.

Therefore, hfDC = 0 ft. If the pipe is enlarged, the Q and energy loss are still zero.

18. A 2,400-ft-long, 2-ft-diameter pipeline conveys water from a hill-top reservoir to an industrial site. The pipe is made of ductile iron, has an outside diameter of 2.25 feet, and has expansion joints. If the flow rate is 30 cfs, determine the maximum water hammer pressure (in psi) that is likely to occur if the downstream flow valve is closed in 0.95 seconds. Also determine the water hammer reduction (in psi) if a diverter is added that reduces the flow rate from 30 cfs to 10 cfs almost instantly upon valve closure.

Ans. For rapid valve closure; ΔP = ρ∙C∙V0; where C = (Ec/ρ)1/2; 1/Ec = 1/Eb + (Dk)/(Epe) and for pipelines with expansion joints: k = 1 – 0.5∙0.25 = 0.875). Therefore, 1/Ec = 1/(3.2 x 105 psi) + [(24 in.)(0.875)]/[(2.3 x 107 psi)(1.5 in.)]; Ec = 2.68 x 105 psi C = (Ec/ρ)1/2 = [(2.68 x 105 lb/in.2)(144 in.2/ft2)/(1.94 slug/ft3)]1/2 = 4460 ft/sec. Now, 2L/C = [2(2,400 ft)]/(4460 ft/sec) = 1.08 sec > 0.95 sec; thus it a is rapid valve closure.

Now we can use ΔP = ρ∙C∙V0; where V0 = Q/A = 30 cfs/[π∙(1 ft)2] = 9.55 ft/sec; thus ΔP = (1.94 slug/ft3)(4,460 ft/sec)(9.55 ft/sec) = 8.26 x 104 lb/ft2 (574 psi) If the flow rate was reduced by 20 cfs, V0 = Q/A = 10 cfs/[π∙(1 ft)2] = 3.18 ft/sec; thus ΔP = 2.75 x 104 lb/ft2 (191 psi); and the pressure reduction is 574 – 191 = 383 psi

42

19. A pipeline conveys water from a reservoir to a point 100 m below its water surface and discharges into the air. The commercial steel pipe is 30 cm in diameter, 420 m long, has a wall thickness of 1 cm, and is free to move longitudinally. A rotary valve is installed at the downstream end. Calculate the maximum water hammer pressure that can be expected on the valve if it closes in a 0.5-sec period. Also, determine the total (maximum) pressure the pipeline will be exposed to during the water hammer phenomenon.

Ans. Applying the energy equation from the reservoir surface (1) to the pipe outlet (2) yields; h1 +

P1



+

V12 P V2 2 = h2 + 2 + 2 + hL ; where P1 = P2 = V1 = h2 = 0; hL= hf+ [∑K](V /2g). Thus, 2g  2g

h1 = [1 + f(L/D) + ∑K](V2/2g); where V2 = V (pipe V); Ke = 0.5 (assume); Kv = 10.0; and assuming complete turbulence for the first trial: e/D = 0.045mm/300mm = 0.00015; thus; f = 0.013, and h1 = 100 m = [1+0.013(420/0.3)+0.5+10.0](V2/2g); V = 8.13 m/sec; Now, NR = DV/ν = [(0.3)(8.13)]/(1.00x10-6) = 2.44 x 106; From Moody; new f = 0.0135 Thus, new V = 8.03 m/sec; and Q = V∙A = 0.568 m3/sec For rapid valve closure; ΔP = ρ∙C∙V0; where C = (Ec/ρ)1/2; 1/Ec = 1/Eb + (Dk)/(Epe) 1/Ec = 1/Eb + (Dk)/(Epe) = 1/(2.2 x 109 N/m2)+[(0.3 m)(1)]/[(1.9 x 1011 N/m2)(0.01 m)]; Ec = 1.63 x 109 N/m2; C = (Ec/ρ)1/2 = [(1.63 x 109 N/m2)/(998 kg/m3)]1/2 = 1280 m/sec; Now, 2L/C = [2(420 m)]/(1280 m/sec) = 0.656 sec > 0.50 sec; ➔ rapid valve closure. ΔP = ρ∙C∙V0; = (998 kg/m3)(1280 m/sec)(8.03 m/sec) = 1.03 x 107 N/m2 (10.3 MPa) Pmax = γH0 + ΔP = (9790 N/m3)(100 m) + 1.03 x 107 N/m2 = 1.13 x 107 N/m2 (11.3 MPa) 20. A pipeline is being designed to withstand a total maximum pressure of 2.13 x 106 N/m2. The 20-cm pipeline is ductile iron and conveys water at 40 liters/sec. Determine the required thickness of the pipe wall if the operational head on the pipeline is 40 m and it is also subject to water hammer if the flow control valve on the downstream end is closed suddenly.

Assume that the longitudinal stresses will be negligible when the pipe is installed.

Ans. Pmax = γH0 + ΔP; 2.13x106 N/m2 = (9790 N/m3)(40m)+ΔP; thus, ΔP = 1.74x106 N/m2 Also, ΔP = ρCV0; where V0 = Q/A = (0.04 m3/s)/[(π/4)(0.20 m)2] = 1.27 m/sec; and C = ΔP/(ρ∙V0) = (1.74 x 106 N/m2)/[(998 kg/m3)(1.27 m/sec)] = 1370 m/sec.

Also, C = (Ec/ρ)1/2; therefore C = 1370 m/sec = (Ec/ρ)1/2 = (Ec/998 kg/m3)1/2; and thus Ec = 1.87 x 109 N/m2. Now determine the required pipe thickness noting that (k = 1.0): 1/Ec = 1/Eb + (Dk)/(Epe); 1/(1.87 x 109 N/m2) = 1/(2.2 x 109 N/m2)+[(0.20 m)(1.0)]/[(1.6 x 1011 N/m2)(e)]; e = 0.0156 m = 15.6 mm (roughly 16 mm) 43

21. A 425-m-long, commercial steel pipe with a 0.90 m diameter carries irrigation water between a reservoir and a distribution junction. The maximum flow is 2.81 m3/sec. A simple surge tank is installed just upstream from the control valve to protect the pipeline from water hammer damage. Determine the minimum diameter of the surge tank if the allowable water surface rise is 5 m over the supply reservoir water level. Ignore minor losses.

Ans. Neglecting minor losses, determine the head loss and the pipeline friction factor. hL = hf = f(L/D)(V2/2g); where V = Q/A = (2.81 m3/sec)/[(π/4)(0.90m)2] = 4.42 m/sec; e/D = 0.045mm/900mm = 0.00005; NR = DV/ν = [(0.90)(4.42)]/(1.00x10-6) = 3.98 x 106 f = 0.0115; and, hL = f(L/D)(V2/2g) = [0.0115(425/0.90)[(4.42)2/2g] = 5.41 m. Thus, Kf= hL/V2= (5.41 m)/[(4.42 m/sec)2]= 0.277 sec2/m & the damping factor is found using (ymax + hL)/(β) = ln [(β)/(β - ymax)]; & substituting, (5 m + 5.41m)/(β) = ln [(β)/(β – 5 m)] β ≈ 6.13 m. Now solving for As from β = (LA)/(2gKfAs) yields 6.13m= [(425m)(0.636m2)]/[2(9.81m/sec2)(0.277 sec2/m)(As)]; As= 8.11m2; Ds = 3.21 m

22. Water flows from a supply reservoir (surface elevation, 450 ft, MSL) through a 2500-ft pipeline at a rate of 350 cfs. A surge tank is required to be installed in the pipeline prior to a gate valve. The pipeline is made of smooth concrete and has a 6-ft diameter. Determine the required height of a 20-ft diameter surge tank if the pipeline is 50 ft lower than the water surface of the supply reservoir. Neglect minor losses.

Ans. Neglecting minor losses, determine the head loss and the pipeline friction factor. hL = hf = f(L/D)(V2/2g); where V = Q/A = (350 ft3/sec)/[(π)(3.0 ft)2] = 12.4 ft/sec; e/D = 0.0006/6.0 = 0.0001; NR= DV/ν= [(6.0)(12.4)]/(1.08x10-5)= 6.89x106; f = 0.0125 hL = hf = f(L/D)(V2/2g) = [0.0125(2500/6.0)[(12.4)2/2g] = 12.4 ft. Therefore, Kf = hL/V2 = (12.4 ft)/[(12.4 m/sec)2] = 0.0806 sec2/ft; and the damping factor is β =(LA)/(2gKfAs)= [(2500ft)(28.3ft2)]/[2(32.2ft/sec2)(0.0806 sec2/ft)(π)(10 ft)2]= 43.4 ft Now to determine ymax using: (ymax + hL)/(β) = ln [(β)/(β - ymax)]; and by substituting, (ymax + 12.4 ft)/(43.4 ft) = ln [(43.4 ft)/(43.4 ft – ymax)]; ymax ≈ 25.0 ft and the surge tank height: HS = HB + ymax = 50.0 ft + 25.0 ft = 75.0 ft

44

TEST QUESTIONS & PROBLEMS - CHAPTER #5 Short Answer Questions 1. Define the physical meaning of Vo, βo, and υo in the centrifugal pump vector diagram.

Ans. Vo is the absolute velocity of the water leaving the impellor, βo is the angle of the impeller at exit, and υo is the velocity of the water if the impellor was not rotating (or the relative velocity of the water with respect to the vanes).

2. The fundamental principle of the centrifugal pump was first demonstrated by: a) Manning b) Chezy c) Darcy-Weisbach d) Hazen-Williams e) Demour f) Wojak-Vieck Ans. (e) - Demour is the correct answer.

3. What is the difference between a turbo hydraulic pump and a positive displacement pump?

Ans. Turbo-hydraulic pumps move fluids with a rotating vane or another moving fluid.

Positive-displacement pumps move fluids strictly by precise machine displacements 4. Which variables represent the impeller speed, the exit velocity, and the angle of the impeller in the vector diagram below?

Ans. ω represents the impellor speed, Vo represents the exit velocity of the water, and βo is the angle of the impeller.

5. Using the equations for motor efficiency and pump efficiency, derive the expression for overall efficiency of a pump system (sometimes called the wire to water efficiency).

Ans. ep = Po/Pi and em = Pi/Pm. Therefore, e = ep∙em = (Po/Pi)(Pi/Pm) = Po/Pm. 6. Which of the following represents units of power? a) work/time b) hp (horsepower) c) kW (kilowatts) d) dynes Ans. (a), (b), and (c) all represent appropriate units of power.

7. (T or F) Pumps add energy to a fluid primarily in the form of increased pressure.

Ans. True 42

8. (T or F) Pumps add energy to a pipeline; thus the outflow from a pump exceeds the inflow.

Ans. False – this would violate the mass balance principle.

9. Propeller blades may be mounted on the same axis of rotation in a common housing to form a multistage propeller pump. What is the advantage of this configuration?

Ans. Propeller pumps are appropriate for high capacity flow, but they cannot overcome large heads. By staging propellers, they can pump water over a greater elevation difference.

10. (T or F) Propeller pumps move a lot of water, but they do not overcome a lot of head.

Ans. True.

11. Describe two differences between an propeller (axial-flow) pump and a centrifugal pump. Is the theory of both based on the principle of impulse-momentum?

Ans. Centrifugal flow is radial and axial flow is not. Centrifugal flow is based on angular momentum and axial flow is based on linear momentum.

12. (T or F) Jet pumps are often used in combination with centrifugal pumps and have a high efficiency.

Ans. False – they have very low efficiencies.

13. Centrifugal pump curves (Hp vs. Q) start at a high head, low flow rate and trend downward with increasing flow. What is the reason for this?

Ans. Pumps add energy to overcome a position head difference and pipe losses. As the flow rate increases, so do the losses so less position head can be overcome. Alternatively, as the position head difference decreases, more energy is available for flow.

14. On centrifugal pump curves (Hp vs. Q), the point where Q = 0 and Hp = Hp(max) is called the a) shut-off head b) rated capacity c) brake horsepower d) match point e) no idea!!

Ans. (a) is the correct answer.

15. Draw a typical pump characteristic curve and label the axes. Then draw a typical system curve on the same graph. Label the match point and the shut-off head. Now draw a typical efficiency curve on the same graph.

Ans. See Figure (5.10) in the book. The pump head at zero discharge is the shut-off head.

16. Centrifugal pump characteristic curves are a plot of Q vs. which of the following a) pump head b) rated capacity c) brake horsepower d) efficiency e) friction loss Ans. (a), (c), and (d).

17. What is the difference between the match point and the shut-off head?

Ans. The match point is the intersection of the pump characteristic curve (Hp vs. Q) and the system curve. The pump head at zero discharge on the pump characteristic curve is the shut-off head.

43

18. Which of the following statements about pumps are false? a) Pumps operate under maximum efficiency over a very broad range of discharges. b) The shut-off head is the height water can be raised but produce no flow. c) Pump discharge increases with decreasing pump head. d) Two identical pumps in parallel on one discharge line will double the flow rate. e) Pump selection involves superimposing a system curve on a pump characteristic curve.

Ans. (a) and (d) are false.

19. Sketch the energy grade line (EGL) and hydraulic grade line (HGL) for the pipeline below.

Pump

Ans. Sketch the EGL and HGL such that a) both start & end at the reservoir surfaces, b) the minor losses (entrance, bends, exit) are accounted for with an abrupt drop in the EGL and HGL; c) the EGL and HGL are parallel lines, and d) the pump adds a significant boost (abrupt rise) to the EGL. 20. Draw a typical pump characteristic curve (Hp vs. Q) and label the axes. On the same graph, sketch the characteristic curve for the system of identical pumps depicted below if valve A is closed and valves B and C are open.

Ans. By closing valve A you have pumps in parallel, and flows should be doubled on the characteristic curve. See Figure (5.13) in the book.

21. Draw a typical pump characteristic curve (Hp vs. Q) and label the axes. On the same graph, sketch the characteristic curve for the system of identical pumps depicted in Question 20 if valve A is open and valves B and C are closed.

Ans. By closing valves B and C you have pumps in series, and pump heads should be doubled on the characteristic curve. See Figure (5.13) in the book.

44

22. Draw a typical pump characteristic curve (Hp vs. Q) and label the axes. On the same graph, sketch the characteristic curve for the system of identical pumps depicted in Question 20 if valve A is open and valves B and C are replaced by two more identical pumps making a four pump arrangement.

Ans. This arrangement produces two pumps in series on each line, producing parallel lines.

Both pump heads and pump flows should be doubled on the characteristic curve.

23. (T or F) By placing two identical pumps in a pipeline in a parallel configuration, you can expect to double the discharge.

Ans. False – losses will increase for the increased flow and use some of the pumps energy.

24. (T or F) A pump is placed in a branching pipeline and is required to pump water from a supply reservoir to two receiving reservoirs. The total discharge will be split between the two pipes such that an equal amount of head is added to each of the two pipelines.

Ans. True 25. How is a pipe network analysis modified to incorporate a pump?

Ans. The energy equation for the loop containing the pump is modified to incorporate the energy added by the pump.

26. Many factors contribute to the loss of pressure on the suction side of pumps. Of those listed here, which does the designer of pumps have the most control over to avoid cavitation. a) position of the pump b) screen loss c) friction loss d) entrance loss e) “b” and “d” Ans. (a) is the correct answer.

27. Give three examples of head losses that can be encountered on the suction side of pumps that may contribute to the loss of pressure and cavitation problems.

Ans. Entrance loss, screen loss, friction loss, other minor losses (i.e., bends, valves, etc.).

28. (T or F) To avoid cavitation in a pump, one of the few parameters that the designer has much control over is the position of the pump.

Ans. True.

29. To avoid cavitation in a pump, one of the few items that a designer has control over is the a) vapor pressure b) velocity head c) position of the pump d) suction line head losses Ans. (c) is the correct answer.

30. Where is cavitation most likely to occur in a pump installation?

Ans. A common site of cavitation is near the tips of the impeller vanes where the velocity is very high and pressure energy is converted to kinetic energy.

31. The maximum velocity near the tip of impeller vanes is an important parameter in assessing pump cavitation potential. It is normally supplied by the pump manufacturers using the term: a) total suction head b) tip velocity c) cavitation parameter d) net positive suction head Ans. (e) is the correct answer.

45

32. Sketch the energy grade line (EGL) and the hydraulic grade line (HGL) for the pump installation depicted below. Include all factors that impact the cavitation potential of the pump.

Pump

Ans. See Figure 5.20 in the textbook. The sketch should include the EGL with an entrance loss, friction losses, and the net positive suction head. The HGL should be parallel to the EGL and (one velocity head below)..

33. Pump specific speed is directly or indirectly dependent on all of the following except a) friction loss b) pump head c) vapor pressure d) discharge e) rotational speed Ans. (c) is the correct answer. Friction loss affects the pump head.

34. (T or F) The dimensionless shape number is commonly used in engineering practice to select or design pumps.

Ans. False, specific speed is more commonly used.

35. If a pump is turning at the rate of 1000 revolutions per minute (rpm), what is its rotational speed in radians per second?

Ans. (1000 rev/min)∙[(2π rad)/(1 rev)]∙[(1 min)/(60 sec)] = 105 rad/sec 36. (T or F) Supplying water to high rise buildings requires pumps with low specific speeds.

Ans. True, small discharges at high heads; Ns = [ω(Q)1/2]/(Hp)3/4 37. Which pump is the most appropriate for high capacity-low head installations? a) positive displacement b) centrifugal c) jet (mixed flow) d) propeller e) aquarium pump Ans. (d) is the correct answer. See Figure 5.22 in the book..

38. (T or F) Positive displacement pumps move a lot of water, but can’t overcome much head.

Ans. False – it is just the opposite.

39. Which pump is the most appropriate for low capacity-high head installations? a) positive displacement b) centrifugal c) jet (mixed flow) d) propeller e) aquarium pump Ans. (a) is the correct answer. See Figure 5.22 in the book..

46

Problems 1. A centrifugal pump is installed in a pipeline between two reservoirs. The pump is required to produce a flow rate of 2500 gpm (gallons per minute) in moving water from the lower reservoir to the upper reservoir. If the water surface elevations of the two reservoirs are separated by 104 feet, and the pump operates at an overall efficiency of 78.5%, determine the required power input (in kW) to the motor. (Assume that pipeline losses are negligible.)

Ans. Convert the pump discharge to cfs (ft3/sec); (2500 gpm)(1 cfs/449 gpm) = 5.57 cfs Po = γQHP = (62.3 lb/ft3)(5.57 ft3/s)(104 ft) = 36,100 ft-lb/s(1 hp/550 ft∙lb/s) = 65.6 hp Pm = Po/e = (65.6 hp)/(0.785) = 83.6 hp; Pm = 83.6 hp(1 kW/1.341 hp) = 62.3 kW 2. A 26-kW motor with 78% efficiency drives a centrifugal pump delivery 68 ℓ/sec against a head of 24 m. Determine the pump efficiency.

Ans. Po = γQHP = (9.79 kN/m3)(0.068 m3/sec)(24 m) = 16.0 kN∙m/sec = 16.0 kW e = Po/Pm = (16.0 kW)/(26 kW) = 0.615 (61.5%) and since e = ep∙em ep = e/em = (0.615)/(0.78) = 0.788 (78.8 %) 3. A centrifugal pump is being tested in a laboratory. The inside and outside radii of the impeller are 7.5 cm and 15.0 cm respectively. The width of the impeller vanes (or flow opening height) varies from 5.0 cm at intake to 3.0 cm at outflow. If the measured flow rate is 55 liters/sec, compute the pump speed (in rpm) and the power input to the pump (in kW).

Assume a shockless entry (αi = 90˚), an exit flow angle (αo) of 22.4˚, and βi = 150°.

Ans. Since αi = 90˚; Vi = Vri = υri; and Vti = 0 (from Fig. 5.3).

Also, υri = Q/Ai = (0.055 m3/s)/[(2π)(0.075m)(0.05m)] = 2.33m/s; υti = υri/tan βi = (2.33)/tan 150˚ = -4.04 m/sec ui = -υti = 4.04 m/s; and ui = ωri Thus, ω = ui/ri = (4.04 m/s)/(0.075m) = 53.9 rad/s ω = (53.9 rad/sec)(1 rev/2π rad)(60 sec/1 min) = 515 rpm Vro= Q/Ao= (0.055 m3/s)/[(2π)(0.15m)(0.03m)] = 1.95 m/s Vo = Vro/sin αo = (1.95)/sin 22.4˚ = 5.12 m/sec.

Finally, Pi = ρQω[roVocos(αo)- riVicos(αi)] Pi = (998 kg/m3)(0.055 m3/s)(53.9rad/s)[(0.15m)(5.12 m/s)cos22.4˚- (0.075m)(2.33)cos 90˚] Pi = 2.10 x 103 N-m/s(1 kW/1000 N-m/s) = 0.210 kW

47

4. A centrifugal pump runs at 1800 rpm and has an outside radius of 12 in. (βo = 170º), an inside radius of 4 in. (βi = 160º), and an impeller thickness (width) of 2 in. at r = ri and 0.75 in. at r = ro. Determine the pump flow rate for a shockless entry (i.e., αi = 90º), and the exit angle αo.

Ans. Impeller speed: ui = ωri = (1800 rev/min)(1 min/60s)(2π rad/rev)(0.333 ft) = 62.8 ft/sec & uo = (1800 rev/min)(1 min/60s)(2π rad/rev)(1.0 ft) = 188 ft/sec. From Fig 5.3 w/αi = 90˚; Vi = Vri = (ui){tan (180˚- βi)} thus Vri = 62.8 ft/sec (tan 20˚) = 22.9 ft/sec = υri Q = Aυri = 2π(0.333ft)(0.167ft)(22.9 ft/sec) = 8.00 cfs. Likewise, υro = Q/Ao υro = (8 cfs)/[(2π)(1 ft)(0.0625 ft)] = 20.4 ft/sec; From Fig 5.3, υto = υro/tan 170˚ = -116 ft/sec Vto = uo + υto = 188 – 116 = 72.0 ft/sec; and αo = tan-1 (Vro/Vto) = tan-1 (20.4/72.0) = 15.8˚ 5. Consider a pump-pipeline system that delivers flow from reservoir A to B with EA = 100 ft and EB = 120 ft. The pipe has a length of L = 2,800 ft, diameter of D = 2 ft, and a Hazen Williams coefficient of CHW = 100. The pump characteristics are tabulated below and minor losses (in feet) accumulate at the rate of 0.01*Q2 where Q is the flow in cfs. Estimate the flow rate in the pipeline by filling in the solution table below.

Q

Hp

hf

hminor

Hs

HSH

(cfs)

Pump (ft)

(ft)

(ft)

(ft)

(ft)

0 10 20 30 40

55 53 47 38 22

Ans. HSH = Hs+ hL; where hL = hf + hminor and Hs = EB - EA= 120 - 100 = 20 ft.

Using the Hazen Williams friction equation (Table 3.4); h f = KQ 1.85 , and K = (4.73∙L)/[(D)4.87(C)1.85] = (4.73∙2,800)/[(2.0)4.87(100)1.85] = 0.0904. The solution table is below with the last four columns filled in. From the table, the pipeline flow is: Q = 20 cfs and Hp ≈ 47 ft. (based on the match point).

Q

Hp

hf

hminor

Hs

HSH

(cfs)

Pump (ft)

(ft)

(ft)

(ft)

(ft)

0 10 20 30 40

55 53 47 38 22

0.0 6.4 23.1 48.8 83.1

0.0 1.0 4.0 9.0 16.0

20.0 20.0 20.0 20.0 20.0

20.0 27.4 47.1 77.8 119.1 48

6. As field engineer on a construction project, you need to pump water from a river (water surface elevation of 382.5 meters) to a reservoir (water surface elevation 400 meters). The pump available to you has the pump characteristics tabulated below. The pipeline will be 400 meters long, and you would like to use a 4-cm-diameter ductile iron pipe (f = 0.020). What flow rate can be expected if minor losses are ignored.

Q

Hp

(m3/hr)

Pump(m)

0 4 8 12 16

52.2 51.3 49.5 46.8 43.1

Ans. HSH = Hs+ hf = Hs + f(L/D)[(V)2/2g]; where Hs = EB - EA= 400 – 382.5 = 17.5 m hf = f(L/D)[(V)2/2g] = (0.02)(400/0.04)[(V)2/2∙9.81] = 10.2(V)2. The solution table is below with appropriate designations. From the table, the pipeline flow is determined to be Q = 8 m3/hr (2.22 x 10-3 m3/s) and Hp = 49.5 m.

Q

Hp

Q

V

hf

HSH

(m3/hr)

Pump(m)

(x10-3 m3/s)

(m/s)

(m)

(m)

0 4 8 12 16

52.2 51.3 49.5 46.8 43.1

0.00 1.11 2.22 3.33 4.44

0.00 0.88 1.77 2.65 3.54

0.0 7.9 32.0 71.6 127.8

17.5 25.4 49.5 89.1 145.3

7. As field engineer on a construction project, you need to pump water from a river (water surface elevation of 362 meters) to a supply reservoir (water surface elevation 400 meters).

The available pump has the performance characteristics tabulated below. The discharge pipeline is a 5-cm-diameter ductile iron pipe (f = 0.018) and is 400 meters long. You have performed the calculations in the table below and you note that the likely flow rate is 8 m3/hr, which is too low. To increase the flow rate, you may have access to an identical pump and can hook it up in parallel to the existing pump, sending the increased flow through the same pipeline. What is the flow rate that can be expected if minor losses are ignored?

Q

Hp

Q

V

hf

HSH

(m3/hr)

Pump(m)

(x10-3 m3/s)

(m/s)

(m)

(m)

0 4 8 12 16

51.6 50.4 47.5 40.8 30.1

0.00 1.11 2.22 3.33 4.44

0.00 0.57 1.13 1.70 2.26

0.0 2.4 9.4 21.2 37.4

38.0 40.4 47.4 59.2 75.4 49

Ans. HSH = Hs+ hf = Hs + f(L/D)[(V)2/2g]; where Hs = EB - EA= 400 – 362 = 38 m hf = f(L/D)[(V)2/2g] = (0.018)(400/0.05)[(V)2/2∙9.81] = 7.34(V)2. For identical pumps in parallel, double the flow rate for each value of pump head. The solution table is below, which shows that the flow rate has increased, but not significantly because friction losses are accumulating rapidly in the common discharge line. By interpolation, we can estimate that Q ≈ 9 m3/hr (2.50 x 10-3 m3/s) and Hp ≈ 50.0 m (from eq’n above).

Q

Hp

2Q

2Q

3

(m /hr)

Pump(m)

3

(m /hr)

-3

3

V

hf

HSH

(x10 m /s)

(m/s)

(m)

(m)

0 4 8 12 16

51.6 50.4 47.5 40.8 30.1

0 8 16 24 32

0.00 2.22 4.44 6.66 8.88

0.00 1.13 2.26 3.40 4.53

0.0 9.4 37.5 84.9 151

38.0 47.4 75.5 123 189

8. Each of the three pipe flows is required in the branching pipe system depicted below. Fill in the blanks of the solution table provided which eventually leads to the graphical solution.

The reservoir water elevations are: EA = 110 ft, EB = 120 ft, and EC = 140 ft.

Pipe 1 2 3

Pipe Characteristics L (ft) D (ft) K 10000 2 0.0512 15000 2 0.0768 5000 2.5 0.0103

m 2 2 2

EC

2 pe Pi

EB Hs2

EA

Pipe 1

e3 Pip

Hs1

P

Q (cfs) 0.0 10.0 20.0 30.0

Hp, ft (pump) 80.0 78.5 74.0 66.5

hf3 (ft)

Net Hp* (ft)

HSH1 (ft)

HSH2 (ft)

50

Ans. This is a partial solution table to Example 5.6 in the book. Friction losses (KQ2) for pipe 3 are determined for the given flows and subtracted from the heads of the pump curve. This yields the “Net Hp” which is available to push flow from the junction to the reservoirs through pipes 1 and 2. The system head curves for the two pipes are found by adding the elevation rise and the friction losses (HSH = Hs + hf = Hs + KQ2) where Hs1 = EB - EA= 10 ft and Hs2 = EC - EA= 30 ft. The completed solution table is provided below.

Q (cfs) 0.0 10.0 20.0 30.0

Hp, ft (pump) 80.0 78.5 74.0 66.5

hf3 (ft) 0.0 1.0 4.1 9.3

Net Hp* (ft) 80.0 77.5 69.9 57.2

HSH1 (ft) 10.0 15.1 30.5 56.1

HSH2 (ft) 30.0 37.7 60.7 99.1

9. The branching pipeline shown below contains a pump and three pipes. The characteristics of the pipes and pump are tabulated below. Pipes 1 and 2 contain partially closed valves. If a flow of 10 cfs is required to flow into Reservoir B and 10 cfs into Reservoir C, determine the head loss in each valve. The reservoir water elevations are: EA = 100 ft, EB = 128 ft, and EC = 131 ft.

Pipe 1 2 3

Pipe Characteristics L (ft) D (ft) f 1000 1.5 0.02 1000 1.5 0.02 1000 1.5 0.02 EC

Q

Hp

(cfs)

Pump (ft)

0 5 10 15 20 25 30

100 96 90 80 66 48 24

2 pe Pi

EA e3 Pip

EB Hs2 Pipe 1

Hs1

P

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Ans. Based on mass balance, the flow in Pipe 3 is Q3 = 10 + 10 = 20 cfs Based on an energy balance, the energy head at the junction is: EA + Hp = EJ + hf3 where Hp = 66 ft (energy head added by the pump when Q = 20 cfs hf3 = f(L/D)(V2/2g) and V3 = Q/A = (20 ft3/s)/[(π/4)(1.5 ft)2] = 11.3 ft/s; Thus; hf3 = (0.02)(1000 ft/1.5 ft)[(11.3 ft/sec)2/(2·32.2 ft/sec2) = 26.4 ft and EJ = EA + Hp - hf3 = 100 ft + 66 ft – 26.4 ft = 139.6 ft Now determine valve losses based on an energy balance from the junction to the reservoirs.

EJ - hf1 - hV1 = EB and EJ - hf2 - hV2 = EC where hf1 = f(L/D)(V2/2g) and V1 = Q/A = (10 ft3/s)/[(π/4)(1.5 ft)2] = 5.66 ft/s = V2 Therefore, hf1 = (0.02)(1000 ft/1.5 ft)[(5.66 ft/sec)2/(2·32.2 ft/sec2) = 6.63 ft = hf2 Now, hV1 = EJ - hf1 - EB = 139.6 ft - 6.63 ft – 128 ft = 4.97 ft And

hV2 = EJ - hf2 – EC = 139.6 ft - 6.63 ft – 131 ft = 1.97 ft

10. A pump is installed in a 400-ft-long, 1.0-ft-diameter (ductile-iron) pipeline to deliver water (68°F) from a supply reservoir to an elevated tank. The design discharge is 4.0 ft3/sec. The pump overcomes an elevation difference between the reservoir and the tank of 150 ft. The suction line is 35 ft of the 400-ft length. Minor loss coefficients in the suction line add up to 3.7. If the pump is installed 3 to 10 ft above the supply reservoir (depending on fluctuations in the water level), will the installation be susceptible to cavitation? The net positive suction head of the pump is rated at 20 ft.

Ans. hp ≤ [(Patm – Pvapor)/γ] – [H’s + V2/2g + hL] where H’s= 20 ft (given), V = Q/A = 5.09 ft/sec, and Patm= 14.7 psi = 2117 psf and Pvapor= 0.344 psi = 49.5 psf from front book jacket. Therefore, (Patm – Pvapor)/γ = (2117 – 49.5)/(62.3) = 33.2 ft hL (suction side) is hL = hf + [∑K](V)2/2g = [f(L/D) + ΣK]∙[(V)2/2g] and e/D = 0.0004/1.0 = 0.0004; T = 68.4˚F, ν = 1.08 x 10-5 ft2/s VD/ν = (5.09 ft/s)(1.0ft/(1.08 x 10-5 ft2/s) = 4.71 x 105 And from the Moody diagram, f = 0.017; thus hL= [0.017(35/1.0)+3.7]∙[(5.09)2/2g] = 1.73 ft, and hp ≤ [(Patm – Pvapor)/γ] – [H’s + V2/2g + hL] hp ≤ [33.2ft] – [20ft + (5.09)2/2g + 1.73ft] = 11.1 ft hp = 3 to 10 ft ≤ 11.1 ft. Cavitation is not a problem.

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11. A pump is installed in a 100-m pipeline to raise water (20° C) from reservoir A with a water surface elevation of 145m to reservoir B with a water surface elevation of 165m. The 80-cmdiameter, concrete (rough) pipe has a design discharge of 2.06 m3/sec. The suction line is 15 m of the 100-m length with a square-edged entrance condition. If the pump has a cavitation parameter of 0.10, determine the allowable height the pump can be placed above the supply reservoir to avoid cavitation. Assume completely turbulent flow in the pipeline.

Ans. To find HP, balance energy from reservoir A to B: HA + HP = HB + hL; (Eq’n 4.2), where HB – HA = 20 m; and hL = hf + [∑K](V)2/2g; Ke = 0.5; Kd = 1.0 (exit coef.)

V = Q/A = 4.10 m/s; e/D = 0.60mm/800mm = 0.00075. From Moody; f = 0.0185 Solving the energy eq’n; HP = 20m+[0.0185(100/0.80) +1.5]∙[(4.1)2/2g] = 23.3m (Patm – Pvapor)/γ = (101,400 – 2370)/(9790) = 10.1 m for 20.2˚C. Now, σ = 0.10 (given), hL (suction side) is hL = hf + hminor= [f(L/D) + ∑K] (V)2/2g hL= [0.0185(15/0.80)+0.5](4.10)2/2g = 0.73 m, and hp = [(Patm – Pvapor)/γ] – Vi2/2g - hL- σHP hp= 10.1m–[(4.10)2/2g]–0.73m–(0.10)(23.3m)= 6.18m 12. Two geometrically-similar, centrifugal pumps are operated at the same efficiency. Pump A operates at 450 rpm, and delivers 2.4 m3/sec against a 22-m head. The scale model (pump B) is one fourth the size with corresponding dimensions and operates at four times the speed.

What is the discharge of pump B when it is delivering water against the same head? If the efficiency of pump A is 0.80, determine the power input required of pump B. Ans. Ns(A) = [ω(Q)1/2]/(Hp)3/4 = [(450)(2.4)1/2]/(22)3/4 = 68.6; for pump B Ns(B) = 68.6 = [1800(Q)1/2]/(22)3/4; Q = 0.150 m3/sec Pi(A) = γQHp/e = [(9790 N/m3)(2.4 m3/sec)(22 m)]/(0.80) = 646 kW Pi(B) = γQHp/e = [(9790 N/m3)(0.150 m3/sec)(22 m)]/(0.80) = 40.4 kW Alternatively, with the power input for pump A, the specific speed (power) could be determined and used for pump B to obtain its power input.

13. A pump with a specific speed based on unit discharge of 50 has an impeller diameter of 72 cm, an operational speed of 1720 rpm, and an efficiency of 80%. Determine the pump head and shaft power required to operate the pump if the discharge is 12.7 m3/sec.

Ans. Note that 1720 rev/min∙[(2π rad)/(1rev)]∙[(1min)/(60sec)] = 180 rad/sec Now: Ns = [ω(Q)1/2]/(Hp)3/4; 50 = [(180)(12.7)1/2]/(Hp)3/4; Hp = 30.0 m and, Pi = γQHp/e = (9.79 kN/m3)(12.7 m3/sec)(30 m)/0.80 = 4,660 kW

53

14. A pump is required to deliver 0.125 m3/sec of water (20˚C) from reservoir A to reservoir B (water surface elevations of 385.7m and 402.5m, respectively). The pipeline (avg. concrete) is 300 m long with a diameter of 0.20 m and contains five bends (R/D = 6) and two gate valves. Determine the appropriate pump from Figure 5.23 and 5.24 and the pump’s working conditions. Assume complete turbulence.

Ans. HSH = Hs + hL; where hL = hf +∑K(V)2/2g; and Ke = 0.5, Kv1 = 5(0.22) = 1.10, Kv2 = 2(0.15) = 0.30, Kd = 1.0, and thus, ∑K = 2.9. Also, e/D = 0.36/200 = 0.0018 and from Moody diagram, f = 0.023. Also, V = 4Q/(π(0.2m)2) = 31.8Q, and from Table 3.4, hf = KQ2 where K = (0.0826∙f∙L)/(D5) = (0.0826∙0.023∙300)/[(0.20)5] = 1,780 Thus, HSH = Hs + hf + [∑K](V)2/2g = (402.5m – 385.7m) + 1,780Q2 + (2.9)V2/2g: Using this equation, plug in a range of Q values, determine HSH, and superimpose this system curve on the pump characteristic curve. This will yield the following: Q (m3/sec) V (m/sec) hf hminor HSH 0.100 3.18 17.8 1.5 36.1 0.125 3.98 27.8 2.3 46.9 0.150 4.77 40.1 3.4 60.3 From Figure 5.23, with Q = 125 L/sec and Hp = 46.9 m (design conditions; table above) Pump IV will work. Plotting Q vs HSH (system curve) on Figure 5.24 (Pump IV): ω = 4350 rpm, e ≈ 62%, Q ≈ 130 L/s, Hp ≈ 48 m 15. A pump, whose characteristic curve is depicted is placed in a 6-in. dia. pipeline, 850 ft long (f = 0.020) to raise water from one reservoir to another. The water surface elevation difference between the reservoirs fluctuates from 136 ft to 58 ft based on water demands. Determine the water surface elevation difference between the supply and receiving reservoirs when the pump is delivering flow rates of 400 gpm, 600 gpm, and 800 gpm.

Ans. HSH = Hs + hL; where hL = hf + [∑K](V)2/2g; and Ke = 0.5 and Kd = 1.0.

Table 3.4, Q = KQ2; K = (0.0252∙f∙L)/(D5) = (0.0252∙0.02∙850)/[(0.5)5] = 13.7 Thus, HSH = Hs + hf + [∑K](V)2/2g = Hs + 13.7Q2 + (1.5)V2/2g: Since, V = Q/A = Q/(πD2/4) = Q/[π(0.5)2/4] = 5.09∙Q and therefore; 54

HSH = Hs + 13.7Q2 + 1.5[(5.09∙Q)2/2g] = Hs + 13.7Q2 + 0.60Q2 = Hs + 14.3Q2.

Therfore, Hs = HSH - 14.3Q2.

For each Q, find Hp from the pump characteristic curve. Then substitute the Q and Hp for Q and HSH in the above energy expression to find the water surface elevation differences (Hs).

For Q = 400 gpm (0.891 cfs), Hp ≈ 147 ft, and Hs = 147 - 14.3(0.891)2 ≈ 136 ft For Q = 600 gpm (1.34 cfs), Hp ≈ 130 ft, and Hs = 130 - 14.3(1.34)2 ≈ 104 ft For Q = 800 gpm (1.78 cfs), Hp ≈ 103 ft, and Hs = 103 - 14.3(1.78)2 ≈ 57.7 ft

43

TEST QUESTIONS & PROBLEMS - CHAPTER #6 Short Answer Questions 1. The flow of water in a small stream during a rainfall event can be described as: a) steady and uniform b) steady and rapidly varied c) steady and gradually varied d) unsteady and uniform e) unsteady and rapidly varied f) unsteady and gradually varied Ans. (f) Flood waves are not considered rapidly varied flow phenomenon.

2. What is the difference between steady and uniform flow?

Ans. In uniform flow, the discharge and depth of flow remain constant over distance.

However, in steady flow, the discharge and depth of flow remain constant over time.

3. (T or F) In gradually varied flow, the channel bottom, water surface, and EGL are parallel.

Ans. False This condition is true only in uniform flow.

4. What is the difference between normal depth and uniform depth?

Ans. There is no difference.

5. The flow of water in a prismatic irrigation canal during dry weather can be described as: a) steady and uniform b) steady and rapidly varied c) steady and gradually varied e) unsteady and uniform e) unsteady and rapidly varied f) unsteady and gradually varied Ans. (a) Most irrigation canals are prismatic and steady, uniform flow is common.

6. (T or F) Manning’s equation is derived from an energy balance.

Ans. False It is derived from a force balance. .

7. What type of flow regime is appropriate for the application of Manning’s equation?

Ans. Uniform flow.

8. The derivation of Manning’s equation is based upon: a) Newton’s 2nd Law b) impulse-momentum principles c) Bernoulli principle d) continuity principle e) Darcy-Wesibach Ans. (a) Newton’s 2nd Law, with no acceleration, and thus reduces to a force balance..

9. Sketch a profile view of uniform open channel flow and label the channel bottom, the HGL, and the EGL. Also identify the three major components of energy on the figure.

Ans. See Figure 6.1 in the textbook.

54

10. What is the definition of wetted perimeter in open channel flow terminology?

Ans. It is the contact length of the water and the channel at a cross section.

11. What is the definition of hydraulic radius in open channel flow terminology?

Ans. It is the flow area divided by the wetted perimeter, Rh = A/P 12. What are the three forces that are considered in the force balance on a control volume that leads to the development of Manning’s equation for uniform flow?

Ans. Hydrostatic pressure forces, the weight of the water body in the reach, and the resistance force exerted by the channel (bottom and sides) on the flow.

13. Which force term leads to the introduction of hydraulic radius in Manning’s equation? a) hydrostatic pressure b) the weight of the water c) impulse-momentum d) the resistance force exerted by the channel on the flow Ans. (d) 14. Derive the relationship between depth of flow and width for a rectangular channel that is the most hydraulically efficient (i.e., best hydraulic section). Show all steps.

Ans. P = b + 2y = A/y + 2y (since A = by; taking the first derivative of P with respect to y: dP/dy = -A/y2 + 2 = 0; A/y2 = 2; or (b∙y)/y2 = 2; b = 2y 15. If a semicircle represents the best hydraulic section, give two reasons why this shape isn’t used for constructed water conveyance channels.

Ans. Excavation is difficult, forming is expensive, and bank stability is a problem.

16. What is the width to depth relationship for the best rectangular hydraulic section?

Ans. b = 2y 17. (T or F) In constructed channels, the vertical distance from the designed water surface to the top of the channel banks is known as the freeboard of the channel.

Ans. True.

18. In the derivation of critical depth, the equation (Q2T)/(gA3) = 1 became V2/(gD) = 1. Show the steps that led to the final equation and define T and D. Ans. Note that Q/A = V and hydraulic depth, D = A/T, where T is the top width. Substituting these into the first equation and simplifying leads to the second equation.

19. A Froude number of 0.8 indicates: a) subcritical flow b) critical depth c) same as “e” d) normal depth f) supercritical flow g) none of the above h) “a” and “d” Ans. (a)

e) there is no “e”

20. (T or F) Normal depth is the depth of flow when the specific energy is at a minimum.

Ans. False, critical depth is the minimum energy depth..

55

21. Starting with the specific energy equation for open channel flow, E = V2/2g + y, show that you will arrive at (Q2T)/(gA3) = 1 for critical depth (i.e., for minimum energy). Show and explain all steps in the derivation.

Ans. E = V2/2g + y = Q2/(2gA2) + y; since V = Q/A. To minimize energy with respect to depth, take the first derivative of this expression and set it equal to zero. dE/dy = -(Q2/gA3)dA/dy + 1 = 0, since dA/dy = T; Q2T/gA3 = 1 22. Which statement below is incorrect? (a) In a given channel the flow depth is higher than the critical depth for the same discharge when the flow is subcritical (b) In a given channel the specific energy for a given discharge is always higher for supercritical flow than for subcritical flow. (c) The critical depth in a channel for a given discharge does not depend on the longitudinal bottom slope of the channel (d) None of the above Ans. (b), examine the specific energy curve to see the fallacy of this.

23. Draw a specific energy curve for a given flow rate, label axes, identify critical depth, subcritical flow, and supercritical flow.

Ans. The critical depth is labeled as yc in the graph below, subcritical flow occurs above that depth and supercritical flow occurs below that depth for the Q specified.

Depth, y subcritical yc

supercritical Specific energy, E

24. Which of the following statements is not true when describing the Froude number? a) The Froude number is the ratio of the mean flow velocity to the speed of a surface wave. b) The Froude number is derived from the specific energy equation. c) The Froude number is used to determine normal depth. d) If Nf < 1, flow is subcritical, and if Nf > 1, flow supercritical. e) The concept of top width is introduced in the derivation of the Froude number.

Ans. (c) Manning’s equation is used to determine normal depth.

25. Write out the equation for the Froude number. What is it a ratio of and what is it used for?

Ans. Nf = V2/gy = V/(gy)1/2. It is the ratio between the mean flow velocity, V, and the speed of a small gravity (surface) wave traveling over the water surface. It is used to distinguish between subcritical (Nf < 1) and supercritical (Nf > 1) flow.

26. (T or F) The difference between energy and specific energy in open channel flow is that the specific energy uses the channel bottom as a datum (reference).

Ans. True.

27. Derive the equation for critical depth in a rectangular channel starting with the Froude number and setting it equal to unity: Nf = V2/gy = 1.

56

Ans. V2/gy = (Q/A)2/(gy) =1; based on V = Q/A. For a rectangular channel, A = by and [Q/(by)]2/(gy) = 1 which leads to y = [Q2/(gb2)]1/3 28. A rectangular channel has a bottom width of 10 ft, longitudinal bottom slope of 0.0001, and a Manning roughness factor of 0.016. If the critical depth in the channel is 1.91 ft (a) The discharge in the channel is nearest 150 cfs (b) The discharge in the channel is nearest 22 cfs (c) The discharge in the channel is nearest 623 cfs (d) Not enough information is given to determine the discharge.

Ans. (a), using the explicit equation for critical depth in a rectangular channel.

29. Channels A and B have identical characteristics except A has a steeper longitudinal slope.

Determine which of the statements below is correct for the same discharge in both channels (a) The critical depth in channel A is larger than the critical depth in channel B (b) The critical depth in channel A is smaller than the critical depth in channel B (c) The critical depth in channel A is same as the critical depth in channel B (d) Critical flow does not occur in the steeper channel Ans. (c), critical depth is not affected by channel slope, normal (uniform) depth is.

30. The appropriate equations for analysis of a hydraulic jump are based on which underlying concepts? a) energy principles b) momentum principles c) friction loss principles d) uniform flow principles e) continuity principles Ans. (a), (b), and (e) 31. (T or F) Hydraulic jumps occur in open channels when the flow changes from subcritical to supercritical.

Ans. False – when supercritical flow changes to subcritical flow.

32. What is the difference between alternate depth and sequent depth?

Ans. Alternate depths have the same specific energy, one depth is subcritical and the other is supercritical. Sequent depths occur before and after a hydraulic jump and do not have the same specific energy.

33. The depth of flow downstream of a hydraulic jump is known as a) normal depth b) critical depth c) alternate depth d) sequent depth Ans. (d)

e) “a” and “d”

34. (T or F) The depth of flow prior to a hydraulic jump (initial depth) and the depth after the jump (sequent depth) have the same specific energy.

Ans. False. A lot of energy is lost in the jump.

35. The flow in a mild-sloped stream during a flood event can be described as a) steady b) unsteady c) uniform d) rapidly varied e) gradually varied Ans. (b) and (e)

57

36. What is the relationship between critical depth and normal depth in channels that are classified as steep.

Ans. Critical depth is greater than normal depth in steep channels.

37. (T or F) In gradually varied flow, the channel bottom, HGL, and EGL are all parallel.

Ans. False, this is only true in uniform flow.

38. What type of gradually varied flow profile would you expect upstream of a dam that is located on a mountain stream? a) S-1 b) M-1 c) S-2 d) M-2 e) S-3 f) M-3 Ans. (a), S because the slope is steep and 1 because the water will be raised considerably.

39. The figure to the left has a (M, C, or S) channel classification and a (1, 2, or 3) flow classification. The figure to the right is a (M, C, or S) channel and a (1, 2, or 3) flow.

Ans. To the left: M-1 and to the right: S-2.

40. Which of the following descriptors are appropriate in describing the flow phenomenon of a flood wave moving down the Mississippi River or the Nile River? a) steady b) unsteady c) uniform d) rapidly varied e) gradually varied Ans. (b) and (e) 41. A trapezoidal channel carrying 300 cfs has a bottom width of 12 ft, side slope of 2 and a longitudinal bottom slope of 0.02. The critical depth is 2.34 ft and the normal depth is 1.25. If the flow depth at a given section is 3.5 ft, the flow profile in the channel is of type (a) M-2 (b) S-1 (c) S-2 (d) None of the above Ans. (b), the channel is steep because yn < yc and flow is S-1 because y is higher than both.

42. What type of gradually varied flow profile do we have if yn < y < yc? If y > yn < yc?

Ans. S-2 and M-1 43. (T or F) A control section in an open channel is a location where the relationship between depth and discharge is known.

Ans. True.

44. A rectangular channel carrying 200 cfs has a bottom width of 12 ft and a longitudinal bottom slope of 0.001. It has a critical depth of 2.32 ft and a normal depth of 4.0. If the flow depth at a given section is 3.5 ft determine which statement below is incorrect (a) The flow profile is of type M-2 (b) The flow depth increases in the flow direction (c) The depth decreases in the flow direction (d) The Froude number is less than one.

Ans. (b), in an M-2 profile, the depth decreases in the direction of flow.

58

45. What is the principle upon which water surface profiles are determined?

Ans. Energy balance.

46. Which principle is not used in the solution of water surface profile computations? a) mass balance b) energy balance c) momentum balance d) Manning’s equation Ans. (c) 47. A trapezoidal channel carrying 280 cfs has a bottom width of 12 ft, side slope of 2 and a longitudinal bottom slope of 0.001. The critical depth is 2.25 ft and the normal depth is 2.78.

The flow depth at the downstream end of this channel is 3.12 ft. Determine which of the statements below is correct (a) A flow depth of 2.55 ft will not occur in this channel. (b) The flow depth decreases in the flow direction. (c) If the channel is very long, a hydraulic jump will occur. (d) None of the above Ans. (a), it is an M-1 profile and will reach normal depth (2.78 ft) but not 2.55 ft.

48. A very long trapezoidal channel has a bottom width of 18 ft, side slope of 2, longitudinal bottom slope of 0.01, Manning roughness factor of 0.020, and it carries 800 cfs. The normal depth is 9.25 ft and the critical depth is 3.45 ft. The flow depth at a section along this channel is 3.70 ft. The flow depth 10 ft upstream is most likely to be (a) 3.79 ft (b) 3.61 ft (c) 5.20 ft (d) 3.45 ft Ans. (a), it is an M-2 profile and the depth will increase as you move upstream.

49. What must be true for a channel to be classified as a critical channel ?

Ans. yn = yc 50. What type of gradually varied flow profile is likely to occur when water flows from a reservoir into a spillway? a) S-1 b) M-1 c) S-2 d) M-2 e) S-3 f) M-3 Ans. (c) 51. (T or F) Open channels are generally designed for uniform flow conditions.

Ans. True.

52. Give three examples of flexible channel liners.

Ans. Grass, riprap, gabions, and grass.

53. Give three examples of rigid channel liners.

Ans. Concrete, asphaltic concrete, soil cement, and grouted riprap.

59

Problems 1. A triangular roadside channel has side slopes of 3:1 (H:V) and a longitudinal slope of 0.01.

Determine the flow rate in the concrete channel if uniform flow is assumed when the top width of the flowing water is 2 meters.

Ans. From Table 6.1; T = 2my; substituting 2m = 2(3)(y); thus, y = 0.333m (flow depth) Now, A = my2 = (3)(0.333m)2 = 0.333 m2 and P = 2y(1 + m2)1/2 P = 2(0.333m)(1 + 32)1/2 = 2.11m; and Rh = A/P = (0.333m2)/(2.11m) = 0.158m Using Manning’s eq’n (6.5a): Q = (1.00/n)(A)(Rh)2/3(S)1/2 with n = 0.013 (Table 6.2) Q = (1.00/0.013)(0.333)(0.158)2/3(0.01)1/2 = 0.748 m3/s 2. Water flows at a depth of 1.83 m in a trapezoidal, riprap lined channel with a bottom width of 3m, side slopes of 2:1 (H:V), and a bottom slope of 0.005. Determine the discharge in the channel assuming uniform flow. Check your solution using Figure 6.4.

Ans. From Table 6.1; A = (b + my)y = [3 + 2(1.83)](1.83) = 12.2 m2 P = b + 2y(1 + m2)1/2 = 3 + 2(1.83)(1 + 22)1/2 = 11.2 m; and Rh = A/P = 1.09 m.

Applying Manning’s equation; Q = (1.0/n)(A)(Rh)2/3(S)1/2; with n = 0.04 (Table 6.2) Q = (1.0/0.04)(12.2)(1.09)2/3(0.005)1/2 = 22.8 m3/s Fig 6.4, (yn/b) = 0.61; nQ/[S1/2b8/3] = 0.71; therefore Q = 23.5 m3/s 3. A discharge of 2,200 cfs flows in a trapezoidal channel at a normal depth of 8.0 ft. The channel has a slope of 0.01, a 12 ft bottom width, and side slopes of 1:1 (H:V). Determine the Manning coefficient of channel roughness. Also determine the sensitivity range of the roughness coefficient estimate if the flow rate can only be estimated to within ±200 cfs?

Ans. For a trapezoidal channel, the area, wetted perimeter and hydraulic radius (Table 6.1) are found to be: A = (b + my)y = [12 ft + 1(8.0 ft)](8.0 ft) = 160 ft2 P = b + 2y(1 + m2)1/2 = 12 + 2(8.0)(1 + 12)1/2 = 34.6 ft Rh = A/P = (160 ft2)/(34.6 ft) = 4.62 ft Now applying Manning’s equation (Eq’n 6.5b): Q = (1.49/n)(A)(Rh)2/3(S)1/2 2,200 = (1.49/n)(160)(4.62)2/3(0.01)1/2; Solving, n = 0.0300 For a flow range of 2,000cfs < Q < 2,400 cfs; the roughness coefficient has a range of : 0.0276 < n < 0.0330

60

4. Determine normal depth for a 4-m-wide trapezoidal channel that is carrying a discharge of 49.7 m3/sec. The excavated earth channel is a well-maintained (clean) with a flow slope of 0.2% and side slopes of 4:1 (H:V). Solve using successive substitution and Figure 6.4.

Ans. For a trapezoidal channel, from Table 6.1: A = (b + my)y = [4 + 4(y)](y) = 4y(1 + y) P = b + 2y(1 + m2)1/2 = 4 + 2y(1 + 42)1/2 = 4 + 8.25y; From Manning’s eq’n (6.5a), Q = (1/n)(A)(Rh)2/3(S)1/2 or alternatively: Q∙n/(S)1/2 = (A)5/3(P)-2/3; (49.7)(0.024)/(0.002)1/2 = 26.7 = [4y(1+y)]5/3(4+8.25y)-2/3 By successive substitution: yn = 2.00 m; From Fig 6.4a with nQ/[(S1/2)(b8/3)] = 0.66; yn/b = 0.48; yn = 1.92 m 5. A grass-lined roadside channel (n = 0.02) is triangular in shape with 30° side slopes and a bottom slope of 0.006. Determine the normal depth of flow when the discharge is 4 cfs (ft3/sec). Check your solution using appropriate computer software.

Ans. Since m = 1/(tan 30˚) = 1.73; referring to Table 6.1: A = my2 = 1.73y2; and P = 2y(1 + m2)1/2 = 4y; Apply Manning’s eq’n: Q = (1.49/n)(A)(R)2/3(S)1/2 or Q = (1.49/n)(A)5/3(P)-2/3(S)1/2; thus, 4 = (1.49/0.02)(1.73y2)5/3(4y)-2/3(0.006)1/2 0.693 = (1.73y2)5/3/(4y)2/3 = 0.989y8/3: yn = 0.875 ft 6. Find the best hydraulic (rectangular) channel cut in rock (n = 0.035) to pass 25m3/sec on a slope of 0.003 m/m. Determine the excavation volume of the 100-m-long channel Ans. The best hydraulic section for a rectangle is a half square (b = 2y): The flow area is A = by = 2y2; the wetted perimeter is: P = b + 2y = 4y; and Rh = A/P = y/2 Now using Manning’s eq’n: Q = (1/n)(A)(Rh)2/3(S)1/2 25 = (1/0.035)(2y2)(y/2)2/3(0.003)1/2; y8/3 = 12.7; yn = 2.59 m; and the cross section is A= by = (5.18 m)(2.59 m) = 13.4 m2; Vol = (area)(length) = (13.4)(100) = 1,340 m3 7. A 12-ft-wide rectangular channel conveys 3,030 cfs. Determine the normal and critical depths for the concrete channel (1% slope). Also determine the flow classification (critical, supercritical, or subcritical) when the channel flows at normal depth.

Ans. From Table 6.1: A = by = 12y; and P = b + 2y = 12 + 2y and using Manning’s eq’n: Q∙n/[1.49(S)1/2] = (A)5/3(P)-2/3; (3030)(0.013)/[1.49(0.01)1/2] = 264 = (12y)5/3(12+2y)-2/3 Successive substitution, yn = 9.30 ft. Eq’n 6:14, yc = [Q2/(gb2)]1/3 = [(3030)2/(32.2∙122)]1/3 yc = 12.5 ft since yn < yc, flow is supercritical at normal depth.

61

8. A 60-cm-diameter concrete pipe on a 1:400 slope carries water at a depth of 30 cm.

Determine the flow rate and flow classification (subcritical, critical, or supercritical).

Ans. Use the Manning eq’n (6.5a) and Froude number (Eq’n 6.12): From Table 6.1: A = (π/8)do2 = (π/8)(0.6)2 = 0.141 m2; P = (π/2)do = (π/2)(0.6) = 0.942 m; Rh= A/P Rh= 0.150m. Q = (1/n)ARh2/3S1/2 = (1/0.013)(0.141)(0.150)2/3(0.0025)1/2 = 0.153 m3/s V = Q/A = 0.153/0.141 = 1.09 m/s. Nf = V/(gD)1/2; D = A/T = 0.141/0.6 = 0.235 m; Nf = (1.09)/(9.81∙0.235)1/2 = 0.718 (<1, subcritical)

9. Water flows in a long, uniform trapezoidal channel lined with brick (n = 0.013) at a depth of 4.5 feet. The base width of the channel is 16.0 ft., the side slopes are 3:1 (H:V), and the bottom drops a foot for every 1000 feet of length. Determine the velocity of flow and the flow classification (subcritical, critical, or supercritical).

Ans. From Table 6.1): A = (b + my)y = [16 + 3(4.5)](4.5) = 133 ft2 and P = b + 2y(1 + m2)1/2 = 16 + 2(4.5)(1 + 32)1/2 = 44.5 ft; Rh = A/P = 2.99 ft Use Manning eq’n: V = (1.49/n)(Rh)2/3(S)1/2 V = (1.49/0.013)(2.99)2/3(0.001)1/2 = 7.53 ft/sec Nf = V/(gD)1/2 where D = A/T; Now from Table 6.1 T = b + 2my = 16 + 2(3)(4.5) = 43 ft D = 133/43 = 3.09 ft; Nf = 7.53/(32.2∙3.09)1/2 = 0.755 (subcritical) 10. A hydraulic jump occurs in a rectangular channel that is 7 m wide. The initial depth is 0.8 m and the sequent depth if 3.1 m. Determine the energy loss and the discharge in the channel.

Also determine the Froude number before and after the jump.

Ans. Using Eq’n 6.20: ΔE = (y2 – y1)3/(4y1y2) = (3.1-0.8)3/(4∙0.8∙3.1) = 1.23 m From Eq’n 6.16: q2/g = y1∙y2[(y1 + y2)/2]; q2/(9.81) = 0.8∙3.1[(0.8 + 3.1)/2]; Solving, q = 6.89 m3/sec-m and Q = b∙q = (7 m)(6.89 m3/sec-m) = 48.2 m3/sec The Froude numbers are determined as: NF1 = V1/(gy1)1/2; where V1 = q/y1 = 6.89/0.8 = 8.61 m/s; NF1 = 8.61/(9.81∙0.8)1/2 = 3.07 (supercritical) NF2 = V2/(gy2)1/2; and V2 = q/y2 = 6.89/3.1 = 2.22 m/s NF2 = 2.22/(9.81∙3.1)1/2 = 0.403 (subcritical)

62

11. A hydraulic jump occurs in 12-ft wide rectangular channel. If the flow rate is 403 cfs and the downstream depth is 5 feet, determine the upstream depth, the loss of energy in the jump, and the Froude number before and after the jump.

Ans. From Eq’n 6.16: q2/g = y1∙y2[(y1+y2)/2]; where q = Q/b = 403/12 = 33.6 cfs/ft Substituting yields: (33.6)2/(32.2) = y1∙5[(y1+5)/2]; y1 = 2.00 ft From Eq’n 6.20: ΔE = (y2 – y1)3/(4y1y2) = (5.0 – 2.0)3/(4∙5.0∙2.0) = 0.675 ft NF1 = V1/(gy1)1/2; and V1 = q/y1 = 33.6/2.0 = 16.8 ft/s NF1 = 16.8/(32.2∙2.0)1/2 = 2.09 (supercritical) NF2 = V2/(gy2)1/2; and V2 = q/y2 = 33.6/5.0 = 6.72 ft/s NF1 = 6.72/(32.2∙5.0)1/2 = 0.530 (subcritical) 12. Water flows at a rate of 22.8 m3/sec in a trapezoidal channel (n = 0.04) with a bottom width of 3 m, side slopes of 2:1 (H:V), and a bottom slope of 0.005. At a certain location, the bottom slope increases to 0.022 with no change in channel dimensions. Normal depths are 1.83 m (upstream) and 1.27 m (downstream). Determine the channel and flow classification (e.g., M-3, S-2, etc.) just upstream of the break in slope. Explain your supporting logic.

Ans. From Table 6.1): A = (b + my)y = [3 + 2y]y = 3y + 2y2 and T = b + 2my = 3 + 4y From Eq’n 6.13; Q2/g = A3/T; (22.8)2/9.81 = 53.0 = (3y + 2y2)3/(3 + 4y) By successive substitution of Figure 6.9: yc = 1.34 m. Since upstream yn > yc; the upstream channel is mild and flows to a steep channel. Flow will go through critical depth near the slope break (see Fig. 6.12); so the classification is M-2.

13. Water flows at the rate of 50 m3/sec in a 5-m-wide, rectangular channel (n = 0.011) at a depth of 1.90 m where the bottom elevation is 150 m above MSL. Upstream 100 m, the depth is 2.00 m where the bottom elevation is 151 m above MSL. What is the slope of the energy grade line in this section of channel? If critical depth is 2.17 m, determine the channel and flow classification (e.g., M-3, S-2, etc.). (Note: MSL stands for Mean Sea Level, the datum.)

Ans. The slope of the energy grade line (EGL) may be found from an energy balance. Thus V12/2g + y1 + z1 = V22/2g + y2 + z2 + ΔL∙Se; V1 = Q/A = 5.00 m/s; V2 = 5.26 m/s (5.00)2/2g + 2.00 + 151 = (5.26)2/2g + 1.90 + 150 + (100)∙Se; thus, Se = 0.00964 m/m For a rectangular channel, A = by = 5y; P = b + 2y = 5 + 2y; from Manning eq’n: Q∙n/(So)1/2 = (A)5/3(P)-2/3; (50)(0.011)/(0.01)1/2 = 5.50 = (5y)5/3(5+2y)-2/3 By successive substitution: yn = 1.25 m. Since yn < yc; the channel is steep. Since the water depth is between yc and yn; the flow is Type 2; Classification is S-2 63

14. A trapezoidal, concrete channel (n = 0.013) has an 8-ft bottom width and z = 1.0 side slope. It discharges 650 ft3/sec on a 0.001 channel slope. The channel is flowing at normal depth (5.85 ft) when it encounters an obstruction. The obstruction causes the depth to rise to 7.00 ft. If critical depth is 4.80 ft, determine the channel and flow classification (e.g., M-3, S-2, etc.).

Also, verify that the depth of flow at a point 100 feet upstream is 6.93 ft by filling in the standard step table shown below.

Section

1 2

z

A

V

V2/2g

Rh

(ft)

(ft)

(ft2)

(ft/s)

(ft)

(ft)

7.00 6.93

0.000

y

Se

Se(avg)

ΔL∙Se(avg)

Total Energy

(ft)

(ft)

Ans. Since yn > yc; the channel is mild. Since the water depth is above yc and yn; the flow is Type 1; Classification is M-1 Section

1 2

y

V2/2g

Rh

z

A

V

(ft)

(ft)

(ft2)

(ft/sec)

(ft)

(ft)

7.00 6.93

0.000 0.100

105.00

6.190 6.282

0.595 0.613

3.777 3.749

103.46

Se

Se(avg)

ΔL∙Se(avg) (ft)

(ft)

4.96E-04

5.06E-04 ΔL =

0.051 100

7.646 7.643

5.16E-04

Total Energy

15. Water flowing at normal depth (2.95 m) in a rectangular concrete channel (n = 0.013) that is 12 m wide encounters an obstruction. The water level rises to 4.55 m to flow over the obstruction. The flow rate is 126 m3/sec and the bottom slope is 0.00086. Determine the channel and flow classification (e.g., M-3, S-2, etc.). Also, verify that the depth of flow at a point 236 m upstream is 4.39 m by filling in the standard step table shown below.

Section

1 2

z

A

V

V2/2g

Rh

(m)

(m)

(m2)

(m/sec)

(m)

(m)

4.55 4.39

10.000

y

Se

Se(avg)

ΔL∙Se(avg)

Total Energy

(m)

(m)

Ans. yc = [Q2/(gb2)]1/3 = [1262/(9.81∙122)]1/3 = 2.24 m. Since yn > yc; the channel is mild.

Since the water depth is above yc and yn; the flow is Type 1; Classification is M-1 Section

1 2

y

V2/2g

Rh

z

A

V

(m)

(m)

(m2)

(m/sec)

(m)

(m)

4.55 4.39

10.000 10.203

54.6 52.7

2.31 2.39

0.271 0.292

2.59 2.54

Se

Se(avg)

ΔL∙Se(avg) (m)

(m)

2.53E-04

2.67E-04 ΔL =

0.063 236

14.884 14.885

2.80E-04

Total Energy

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16. A 10-m-wide rectangular channel (S0 = 0.001, n = 0.013) with a discharge of 100 m3/sec flows from this mild channel into a steep channel (S0 = 0.05) passing through critical depth at the transition between the two bottom slopes. The water surface profile in the mild channel (M-2 classification) has been computed using the standard step method in the table below by solving for the flow depths at points 5 m, 25 m, and 100 m upstream of the transition point (critical depth). Fill in the missing values (designated with question marks) in the table. ΔL∙Se(avg)

Total Energy

(m)

(m)

0.010

3.262 3.261

y

z

A

V

V2/2g

Rh

(m)

(m)

(m2)

(m/sec)

(m)

(m)

1 2

?

2.24

0.000 0.005

?

22.40

4.613 4.464

1.084 1.016

1.512 1.547

2.07E-03 1.88E-03

1.98E-03 ΔL =

2 3

2.24 2.34

0.005 0.025

22.40 23.40

4.464 4.274

1.016 0.931

1.547 ?

? ?

? ΔL =

3 4

2.34 2.48

0.025 0.100

23.40 24.80

4.274 4.032

0.931 0.829

?

1.658

1.66E-03 1.40E-03

1.53E-03 ΔL =

0.115 75

3.411 3.409

Se

Se(avg)

ΔL∙Se(avg)

Total Energy

Section

Se

Se(avg)

5 ? ?

? ?

Ans. Critical depth is the control section (starting point); yc = [Q2/(gb2)]1/3 = [1262/(9.81∙102)]1/3 = 2.17 m.

Section

V2/2g

Rh

y

z

A

V

(m)

(m)

(m2)

(m/sec)

(m)

(m)

(m)

(m)

1 2

2.17 2.24

0.000 0.005

21.68 22.40

4.613 4.464

1.084 1.016

1.512 1.547

2.07E-03 1.88E-03

1.98E-03 ΔL =

0.010 5

3.262 3.261

2 3

2.24 2.34

0.005 0.025

22.40 23.40

4.464 4.274

1.016 0.931

1.547 1.594

1.88E-03 1.66E-03

1.77E-03 ΔL =

0.035 20

3.296 3.296

3 4

2.34 2.48

0.025 0.100

23.40 24.80

4.274 4.032

0.931 0.829

1.594 1.658

1.66E-03 1.40E-03

1.53E-03 ΔL =

0.115 75

3.411 3.409

17. Flow enters a long, rectangular flume at its upstream end from under a sluice gate. The flume has a bottom width of 3 ft, n = 0.013, and a bottom slope of So = 0.02. The flow depth at the entrance is 1.30 ft and the discharge is 30 cfs. The normal and critical depths have already been determined as yn = 0.91 ft and yc = 1.46 ft. a) Determine the channel and flow classification in the flume (e.g., M-2, S-1, etc.) and explain your supporting logic. b) Will depths of 0.80 ft and 0.60 ft ft occur in this channel? If your answer is YES calculate how far downstream from the sluice gate. If your answer is NO explain why. c) Will depths of 1.20 ft and 1.10 ft occur in this channel? If your answer is YES calculate how far downstream from the sluice gate. If your answer is NO explain.(20 points).

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Ans. Use information on critical depth and normal depth to the water is the control section a) The channel is steep since yn is smaller than yc. The depths vary between 1.30 ft at the sluice gate to yn = 0.91 ft further downstream. An S2 profile will occur. b) The depths of 0.80 ft and 0.60 ft are lower than the normal depth. They belong to the S3 zone and will not occur. c) The following table is used to determine the locations downstream where the depths; 1.20 ft and 1.10 ft, will occur.

Sec

U/D

#

y

A

P

Rh

V

V2/2g

E

(ft)

(ft2)

(ft)

(ft)

(ft/sec)

(ft)

(ft)

Se

ΔL (ft)

1

U

1.30

3.90

5.600

0.696

7.692

0.9188

2.2188

0.00730

2

D

1.20

3.60

5.400

0.667

8.333

1.0783

2.2783

0.00908

2

U

1.20

3.60

5.400

0.667

8.333

1.0783

2.2783

0.00908

3

D

1.10

3.30

5.200

0.635

9.091

1.2833

2.3833

0.01154

Distance from Sluice Gate (ft)

0 5.04

5.04 5.04

10.83

15.87

18. An irrigation channel is to be constructed in silt clay (earth excavation, clean) with side slopes of m = 1.5, a bottom slope of S0 = 0.002 m/m, and a bottom width of b = 2.0 m. Can this channel convey 2.4 m3/sec without being eroded?

Ans. This is an analysis problem, not a design problem. For the given channel, determine the normal depth using the Manning equation (and Table 6.1) or Fig 6.4 with n = 0.024 from Table 6.2; Solving, yn = 0.70 m, A = 2.13 m2, and thus V = Q/A = 1.13 m/sec.

The velocity is larger than VMAX = 1.0 m/sec (Table 6.7). Therefore, the channel will likely erode when the design discharge of 2.4 m3/sec.

19. The best hydraulic section for a channel that passes 342 cfs (n = 0.013, m = 1.5, So = 0.001) yields b = 3.0 ft and y = 4.95 ft. (prior to freeboard). However, the depth is limited to 3.5 ft due to a high water table at the site. Resize the channel section considering the depth limitation.

Ans. Due to the limitation on the flow depth, the section will no longer be the most efficient hydraulically. Applying the Manning equations while limiting the depth yields: Q = (1.49/n)ARh2/3S1/2 = (1.49/n)(A)5/3(P)-2/3 (S)1/2 342 = (1.49/0.013)[{b+(1.5)(3.5)}3.5]5/3[b+2(3.5)∙ (1+1.52)1/2]-2/3(0.001)1/2 By successive substitution or by using Figure 6.4 (or computer software); b = 9.97 ft (use 10 ft). The corresponding Froude number is 0.70 and is acceptable. The freeboard is still 2.7 feet based on Figure 6.15.

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TEST QUESTIONS & PROBLEMS - CHAPTER #7 Short Answer Questions 1. (T or F) Aquifer permeability is affected by groundwater temperature.

Ans. True. It affects the specific weight and viscosity in Equation (7.3).

2. Why is the velocity head component of hydraulic head not considered in Darcy’s equation?

Ans. Groundwater moves very slowly, so the velocity head term is generally very small compared to the position head and the pressure head (for confined flow) terms. Thus, it can be ignored without a loss in accuracy.

3. The three different types of groundwater velocity include all of the following except: a) pore velocity b) apparent velocity c) actual velocity d) seepage velocity Ans. (a) Pore velocity was not a term used in this book.

4. (T or F) The higher the porosity, the easier it is to extract water from an aquifer.

Ans. False. Other factors enter in including size and connectivity of the pore spaces.

5. Name the three different types of groundwater velocity and define each.

Ans. Apparent - used in Darcy’s equation V = k(dh/dL) or V = Q/A; A = total area Seepage - average flow speed based on pore area = ΔL/t = V/α; ΔL = straight distance Actual - based on real (tortuous path) distance traveled by molecules (>Vs).

6. (T or F) The theoretical upper limit of porosity is 100%.

Ans. True.

7. Explain the difference between a confined and an unconfined aquifer.

Ans. A confined aquifer contains water that is under pressure and has a relatively impermeable boundary above the saturated layer.

8. Withdrawals from aquifers can result from which of the following mechanisms? a) springs b) wells c) influent streams d) infiltration of rainwater e) transpiration Ans. (a), (b), and (e); the rest contribute water to the aquifer.

9. Define two limitations to laboratory determined permeabilities using core samples.

Ans. Representativeness, wall effects, entrained air, sample disturbance, anisotropy.

10. The ratio of volume of voids to total volume in porous media is called: a) void ratio b) specific yield c) water ratio d) porosity Ans. (d) porosity.

11. (T or F) In general, the porosity of gravel is higher than that of clay.

Ans. False. See Table 7.1.

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12. What is the difference between seepage velocity and actual velocity?

Ans. Seepage velocity relates to average straight line velocity between two locations, or the average flow based on pore area; Vs = ΔL/t = V/α; ΔL = straight distance Actual velocity is based on real (tortuous path) distance traveled by molecules (>Vs). .

13. Show all steps in the proof of the well equation for unconfined aquifer under state flow conditions starting with Darcy’s expression: Q = 2πrhK(dh/dr).

Ans. (Q/r)dr=2πKh(dh); Q and K are constant, and integrating both sides yields Q(ln r) = 2πK(h2/2)+ C; w/boundary conditions (r = rw, h = hw) and (r = ro, h = ho): Q∙ln (ro/rw)= πK(ho2 – hw2); thus Q = πK∙ [(ho2 – hw2)/ ln (ro/rw)].

14. (T or F) A homogeneous aquifer implies the permeability is independent of flow direction.

Ans. False. This is a characteristic of an isotropic aquifer.

15. Define the cone of depression in groundwater flow.

Ans. In a homogeneous, isotropic aquifer, it is the axisymmetric drawdown curve with a conic-shaped geometry produced by a pumping well.

16. When the permeability of an aquifer is dependent upon direction, the appropriate term is: a) homogeneous b) heterogeneous c) isotropic d) anisotropic e) don’t know Ans. (d); anisotropic 17. Describe how you would determine the drawdown in a confined aquifer at a location that was impacted by two pumped wells in the vicinity. What principle is being invoked?

Ans. Determine the drawdowns separately and add them together, called superposition .

18. A 50 cm3 sand sample is collected from an aquifer. When the sample is oven dried and poured into a graduated cylinder, it displaces 30 ml of water. Determine the porosity.

Ans. α = Volv/Vol = (50 – 30)/50 = 0.40 (Note: 30 ml = 30 cm3) 19. (T or F) In confined aquifers, the value of the storage coefficient is much lower than in unconfined aquifers.

Ans. True. In confined aquifers, the pores do not drain. Water is removed by compression of the saturated layer and expansion of the ground water.

20. Why is the storage coefficient larger in confined aquifers than in unconfined aquifers?

Ans. The storage coefficient in unconfined aquifers may approach the porosity since water is drained from the pores. However, in confined aquifers the pores do not drain. Water is removed by compression of the saturated layer and expansion of the ground water.

21. Which of the following terms is a dimensionless number defined as the water yield from a column of aquifer (of unit area) that results from lowering the water table or piezometric surface by a unit height. a) porosity b) permeability c) hydraulic conductivity d) transmissibility e) storage coefficient Ans. (e); storage coefficient

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22. Describe the difference between steady and unsteady radial flow in an unconfined aquifer.

Ans. For unsteady flow, the radius of influence of the water table has not achieved a steady state (i.e., the radius of influence continues to expand with time).

23. (T or F) For unsteady radial flow in confined aquifers, the principal of superposition may be used to solve for the drawdown of multiple wells and for unsteady well pumping rates.

Ans. True.

24. For unsteady radial flow in confined aquifers, which of the following parameters affect the drawdown at a specific location within the radius of influence? a) time since pumping began b) the radial distance from the well to the point of interest c) transmissibility d) storage coefficient e) the well pumping rate Ans. All of these parameters affect the drawdown 25. Define the term aquifer recovery in unconfined aquifers.

Ans. Aquifer recovery is the rise of the water table once pumping of the well has stopped.

26. (T or F) The basic idea (premise) in analyzing field pumping test data to determine an aquifer’s hydraulic characteristics is to fit the observed drawdowns to available analytical solutions.

Ans. True.

27. Why are field pumping tests more reliable in determining an aquifer’s hydraulic paramenters than laboratory tests done on aquifer samples.

Ans. There are many reasons that lab tests are not reliable, but one of the biggest reasons is that it is rare to get a sample or samples that is representative of the entire aquifer.

28. (T or F) An aquifer’s storage coefficient can be obtained from a field pumping test under equilibrium (steady aquifer flow) conditions.

Ans. False. It requires a non-equilibrium test (unsteady flow in the aquifer).

29. All of the following items are required for an equilibrium test in a confined aquifer except: a) the aquifer storage coefficient b) drawdown measurements in at least two wells c) drawdown measurements taken after aquifer equilibrium is achieved d) radial distances to the wells where drawdowns are being measured Ans. (a) the aquifer storage coefficient is not required 30. All of the following are required for an equilibrium test in an unconfined aquifer except: a) a completely penetrating pumped well b) pump flow equilibrium must be achieved c) drawdown measurements in at least two wells d) radial distances to the observed wells Ans. (b) aquifer flow equilibrium must be achieved, not pump flow equilibrium 31. (T or F) Analytical solutions are not available for unsteady flow in unconfined aquifers to determine the storage coefficient.

Ans. True. However, the confined aquifer procedures can be applied to unconfined aquifers if the drawdowns are very small compared to the thickness of the aquifer.

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32. Identify two different types of aquifer boundaries that will upset the radially-symmetric drawdown pattern of a pumping well if the boundary is within the radius of influence.

Ans. Aquifer boundaries include impermeable strata (no flow boundary) and water bodies, like lakes and rivers (constant head boundary).

33. In order to use the method of images to solve aquifer boundary problems, which of the following limitations must be met: a) the aquifer must be confined b) the aquifer must be unconfined c) the boundary must fully penetrate the aquifer d) the aquifer flow must be steady Ans. (a) and (c) are required limitations of the method of superposition 34. (T or F) An impermeable boundary next to a well will increase the drawdown between the boundary and the well when compared to radially-symmetric drawdown conditions.

Ans. True.

35. Define an image well.

Ans. Hydraulic image wells are imaginary sources or sinks, with the same strength (i.e., flow rate) as the original well, placed on the opposite side of a boundary to represent the effect of the boundary.

36. (T or F) Drawdowns for unconfined wells are linear; thus the drawdown of a well and its image can be added to obtain the drawdown created when a well operates near a boundary.

Ans. False. Unconfined aquifers are not linear in drawdowns, rather they are linear in differences in h2.

37. Draw the system of images to replace the boundary conditions depicted (recharge/river & barrier/impermeable).

Show all distances, and whether the image well is a discharge well or a recharge well.

Ans.

38. Which of the following methods are not used to determine the presence of groundwater? a) electrical resistivity b) fluid capacitance c) hydraulic spectrophotometry d) seismic wave propagation e) infrared telemetry f) devination Ans. (b), (c), and (e)

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39. (T or F) The process of saltwater penetrating freshwater aquifers near the coast is called sea water intrusion.

Ans. True.

40. Sketch the saltwater-freshwater interface of an unconfined coastal aquifer.

Ans.

41. Define equipotential lines.

Ans. Equipotential lines connect all points in the flow field that have equal velocity potential (or equal head). In a properly constructed flow net, the drop in head (∆h) between adjacent equipotential lines typically remains constant.

42. Flow nets drawn under a dam allow you to determine all of the following except: a) the amount of seepage b) the hydrostatic pressure anywhere in the net c) the seepage velocity anywhere in the net d) the porosity anywhere in the net Ans. (d). The porosity can not be determined by the flow net.

43. Identify two methods used to control sea water intrusion.

Ans. Pumping troughs, pressure ridges, and subsurface barriers.

44. (T or F) Divination is an ancient art used to locate groundwater.

Ans. True..

45. Excessive seepage through an earth dam could lead to which of the following: a) piping b) cavitation c) scaling d) sloughing e) global warming Ans. (a) and (c). .

46. Define piping.

Ans. The removal of soil by exiting seepage water, often on the downstream side of a dam).

47. (T or F) The upper surface of the seepage flow through an earth dam is known as the surface of saturation or phreatic surface.

Ans. True..

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Problems 1. A permeameter experiment is being set up to determine the permeability of some very fine sands. The cylindrical test samples used are 30 cm long and have a diameter of 10 cm. Also, 50 ml of water will be collected using a head difference of 40 cm. Estimate how much time each experiment will take in minutes? Also, estimate how much time (on average in hours) will it take a conservative tracer to pass through the test sample?

Ans. Using Equation 7.4, try K = 10-6 m/sec = 10-4 cm/sec (from Table 7.2 for very fine sands, low end of range); Q = KA(dh/dL); where A = π(5 cm)2 = 78.5 cm2; thus Q = (10-4cm/s)(78.5cm2)(40/30) = 0.0105cm3/s (ml/s). Also, Q = Vol/t; or t = Vol/Q = (50 ml)/(0.0105 ml/s) = 4760 sec = 79.3 min ≈ 80 min per test Tracer time requires the seepage velocity. From Table 7.1, use a porosity of 0.30.

Vs = V/α = Q/(A∙α) = (0.0105 cm3/s)/[(78.5 cm2)(0.30) = 4.46 x 10-4 cm/sec t = ΔL/Vs = (30 cm)/(4.46 x 10-4 cm/sec)(1 hr/3600 sec) = 18.7 hrs.

2. A high-water table exists behind a retaining wall shown in the figure below. Weep holes are drilled at the bottom of the retaining wall every 3.0 m to alleviate the hydrostatic pressure. If the permeability is 10-5 m/s, determine the seepage (in cm3/sec) through each weep hole.

Ans. The apparent (Darcy) velocity can be computed as: V = K(dh/dL) = (1.00 x 10-5 m/s)(1/50) = 2.00 x 10-7 m/sec The slope of the water table is the hydraulic gradient; not the slope of the confining bedrock layer. The flow rate through each weep hole requires the continuity equation (Q = AV) where the area is in the aquifer where the Darcy V was found, not the weep hole area. Therefore, using the aquifer area contributing to each weep hole yields Q = AV = [(1m)(3m)](2.00 x 10-7 m/sec) = 6.00 x 10-7 m3/sec = 0.600 cm3/sec

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3. A chemical spill occurs at a manufacturing plant. Sampling reveals the conservative pollutant is now in the groundwater directly under the location of the spill. If the aquifer is composed of sand and gravel, how many hours will it take the pollutant to travel 82 feet to the property boundary assuming the aquifer is homogeneous and the water table has a slope of 2%.

Ans. Solve for apparent (Darcy) velocity (K = 5.0 x 10-3 m/sec = 0.0164 ft/sec; Table 7.2).

V = K(dh/dL) = (0.0164 ft/sec)(0.02) = 3.28 x 10-4 ft/sec. Now find seepage velocity; Vs = V/α = (3.28 x 10-4 ft/sec)/(0.275) = 1.19 x 10-3 ft/sec (Note: α from Table 7.1) And finally, t = ΔL/Vs = [(82 ft)/(1.19 x 10-3 ft/sec)](1 hr/3600 sec) = 19.1 hours Note: Groundwater moves very slowly, even in course soils.

4. A fully penetrating well in a 33 m thick confined aquifer pumps at a constant rate of 2000 m3/day for a long time. In two observation wells located 20 m and 160 m from the well, the difference in hydraulic head is 2.0 m. If the undisturbed head is 250 m, and the head in the furthest observation well is 249 m, calculate the radius of influence of the well.

Ans. T and K are not given. Thus use Eq’n (7.11) with the observation well data to obtain T. s = sob + (Q/2πT)[ln(rob/r)]; Data for the closer observation well ➔ (s, r). s - sob = 2 m = [2000/(2πT)]∙[ln(160/20)]; T = 331 m3/day-m; Now use Eq’n (7.11) again with the furthest observation well and the radius of influence. s = sob + (Q/2πT)[ln(rob/r)]; Use (s, r) for radius of influence.

0.0 = 1.0 + [2000/(2π ∙331)]∙[ln(160/ro)];

ro = 453 m

Alternative solution: Use Equation (7.8) with both of the observation wells; use the inner one for (rw, hw): h - hw = (ho - hw)[ln(r/rw)/ln(ro/rw)]; 249-247(250-247)[ln(160/20)/ln(ro/20)];

ro = 453 m

5. After pumping a 12-in.-diameter well for a long time, equilibrium conditions have been reached in a confined aquifer. The coefficient of permeability (based on lab tests) is 7.55  10 −4 ft/sec. At a location of about 90 ft from the well, two closely spaced observation wells indicate the slope of the piezometric surface in the radial direction is 0.0222. If the aquifer thickness is 50 ft, compute the well discharge in gallons per minute.

Ans. Since the slope is known, apply Equation 7.5 directly: Q = 2πrbK(dh/dr) = 2π(90)(50)(7.55 x 10-4)(0.0222) Q = 0.474 ft3/sec (449 gpm/ 1cfs) = 213 gpm

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6. A pharmaceutical industry owns a discharge well that completely penetrates a confined aquifer and is pumped at a constant rate of 2,150 m3/day. The aquifer transmissivity is 880 m2/day. The steady state drawdown measured at a distance 80 m from the pumped well is 2.72 m. What is the drawdown impact of the industrial well on a domestic well that is 100 m away? If a second industrial well is installed with the same diameter and pumping rate at a distance of 140 m from the domestic well, what is the drawdown impact from both wells?

Ans. Using Eq’n 7.11; s1 = sob + (Q/2πT)[ln(rob/r)] = 2.72 + [2150/(2π ∙880)][ln(80/100)] s1 = 2.63 m, for the first well acting alone. For the second well acting alone: s2 = 2.72 + [2150/(2π ∙880)][ln(80/140)] = 2.50 m; Thus, sTotal = s1 + s2 = 5.13 m 7. An industrial manufacturer owns a 12-in-diameter well that completely penetrates a 130-ftthick, unconfined aquifer with a coefficient of permeability of 0.00055 ft/sec. The well pumping rate is 3.5 cfs and the radius of influence is 500 feet. Another industry plans to move into an adjacent property and install a well with the same characteristics. If the new well is 250 feet away, what is the impact in additional drawdown at the existing well? (Hint: Consider using the radius of influence as an observation well with zero drawdown.)

Ans. Apply Eq’n (7.15) to determine the existing drawdown prior to adding the new well: hw2 = ho2 – [Q/(π ∙K)]∙[ln(ro/rw)]; hw2 = 1302 – [3.5/(π ∙0.00055)]∙[ln(500/0.5)] = 2910 ft2 hw = 53.9 ft; thus, sw1 = 130 – 53.9 = 76.1 ft. Now apply Eq’n (7.17) using the radius of influence as the observation well with zero drawdown; hob2 = 1302, rio= 500. Therefore, h2 = hob2 – Σ[Qi/(π ∙K)]∙[ln(rio/ri)] = 1302 – [3.5/(π ∙0.00055)]∙[ln(500/0.5)] – [3.5/(π ∙0.00055)]∙[ln(500/250)] = 1504; h = 38.8 ft; The drawdown for both wells is sw(1+2) = 130 – 38.8 = 91.2 ft; and the addition drawdown caused by the new well is: Δsw = 91.2 – 76.1 = 15.1 ft of additional drawdown 8. A well is located in the middle of a circular island, as depicted below. If the well is pumped at the rate of 7.5 gpm, the water table height in the well above the impervious layer is 25 ft as shown. What would the water table height in the well be for a pump flow of 9.0 gpm?

Assume K = 1.02  10 −5 ft / sec . Also determine the water table height 150 ft from the well.

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Ans. Applying Eq’n (7.14): Q = πK(ho2-hw2)/ln(ro/rw), noting that 449 gpm = 1.0 cfs yields: (9.0/449) ft3/sec = π(1.02 x 10-5)[(602-hw2)/ln(300/1)]; hw = 5.67 ft Now using Eq’n 7.16 with the radius of influence as the observation well yields: h2 = ho2 – [Q/(π ∙K)]∙[ln(ro/r)]; h2 = 602 – [(9/449)/(π ∙(1.02 x 10-5)]∙[ln(300/150)]; h = 56.3 ft

9. A well penetrating a confined aquifer will be pumped at a constant rate of 0.579 cfs. How long can the well be pumped at this rate if the drawdown at a point 300 ft from the well can’t exceed 3.66 ft. The aquifer transmissivity is 12,000 ft2/day and storage coefficient is 0.0003.

Ans. Applying Eq’n (7.21) yields: s = [Qw/(4πT)]W(u) where 0.579 cfs = 50,000 ft3/day 3.66 = [50,000/(4π∙12,000)]W(u); thus , W(u) = 11.0 From Table 7.3, u = 9.0 x 10-6, and substituting into Eq’n (7.20) yields: u = (r2S)/(4Tt);

9.0 x 10-6 = (3002∙0.0003)/(4∙12000∙t); solving: t = 62.5 days.

10. An industrial well that fully penetrates a confined aquifer is pumped at a constant rate of 300 m3/hr when needed. A second well is needed with a pumping rate of 200 m3/hr. However, the drawdown of the aquifer at a nearby residential well (a distance of 100 m from the first well) cannot exceed 10.5 m. If both wells are turned on at the same time and run for four days, how close can the second well be placed to the residential well before the drawdown limitation is violated? The aquifer transmissivity is 25.0 m2/hr and the storage coefficient is 0.00025.

Ans. Use superposition, unsteady flow, multiple wells. For the first well, Eq’n (7.27) yields: u1 = (r12S)/(4Tt) = (1002∙0.00025)/(4∙25∙96) = 2.60 x 10-4; From Table 7.3, W(u1) = 7.697. Now applying Eq’n (7.26): s = [1/(4πT)]{Q1∙W(u1) + Q2∙W(u2)} 10.5 = [1/(4π∙25)]{300∙7.697 + 200∙W(u2)}; W(u2) = 4.948.

From Table 7.3, u2 = 4.0 x 10-3. Now Eq’n (7.27) yields: u2 = (r22S)/(4Tt) 4.0 x 10-3 = (r22∙0.00025)/(4∙25∙96); r2 = 392 m 11. A confined aquifer has a storage coefficient of 0.0005 and a transmissivity of 400 m2/day.

Wells 1 and 2 completely penetrate this aquifer and are pumped at constant rates of 600 m3/day and 1400 m3/day respectively. An observation well is located 400 m from well 1 and 490 m from well 2. Determine the drawdown in the observation well after 2.5 days of pumping if the pumps in both wells start at the same time..

Ans. Use superposition, unsteady flow, multiple wells: Eq’ns (7.27), (7.26), & Table 7.3 Well 1: u1 = (r12S)/(4Tt) = (4002∙0.0005)/(4∙400∙2.5) = 2.00 x 10-2; W(u1) = 3.355 Well 2: u2 = (r22S)/(4Tt) = (4902∙0.0005)/(4∙400∙2.5) = 3.00 x 10-2; W(u2) = 2.959 s = [1/(4πT)]{Q1∙W(u1) + Q2∙W(u2)}= [1/(4π·400)]{600∙3.355 + 1400∙2.959} = 1.22 m 74

12. A 30-m-thick confined aquifer has a piezometric surface 75 m above the top of the confining layer. A 40-cm-diameter well draws water from the aquifer at the rate of 0.1 m 3/sec, and the drawdowns have stabilized. If the drawdown at the well is 30 m and the drawdown in an observation well 50 m away is 10 m, determine the transmissivity and the radius of influence.

Ans. Using Equation (7.34) with observation well #1 represented by the pumped well yields T = (Qw/2π)[ln(r1/r2)/(s2-s1)] = (0.1/2π)[ln(0.2/50)/(10-30)] = 4.39 x 10-3 m2/sec Now using Equation (7.6) noting that Kb = T;

Q = 2πT(ho-hw)/ln(ro/rw)

0.1 = 2π(4.39 x 10-3)(30)/ln(ro/0.2); ro = 785 m 13. A field test is conducted in an unconfined aquifer by pumping 1,300 ft3/hr from an 8-inchdiameter well penetrating the aquifer. The undisturbed aquifer thickness is 46 ft. The drawdowns, measured at steady state at various locations, are tabulated below along with the associated semi-log plot of h2 vs.log r. Determine the coefficient of permeability and how far away from the well you must be before the drawdown becomes less than 2.5 ft. r (ft) 0.33 40 125 350 s (ft) 6.05 4.05 3.57 3.05 Unconfined Aquifer Test Data 1900

1850

1800

h2 (ft2)

1750

1700

1650

1600

1550 0.10

1.00

10.00

100.00

1000.00

Radial Distance, r (ft)

Ans. From the best fit line, ∆* h2 = 1710 – 1630 = 80 ft2 (r = 10 ft to 1 ft).

Then Eq’n (7.39) yields: K = [2.30∙Qw/(π ∙Δ*h2)] = [2.30∙1300/(π ∙80)] = 11.9 ft/hr Now use Eq’n (7.16) w/any observation well: s = 2.5 ft; h = ho – s = 46 – 2.5 = 43.5 ft.

Let rob = 40 ft, hob= 46 - 4.05 = 41.95 ft:

h2 = hob2 – (Q/πK)[ln (rob/r)];

43.52 = 41.952 – (1300/π ∙11.9)[ln (40/r)]; r = 1800 ft (or approximate from the plot)

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14. A pumping test was conducted in a 40-m-thick confined aquifer. Drawdown data were collected in an observation well 20 m away, and is plotted below (drawdown = s = h o – h).

The test well was pumped at the constant rate of 8.50 m3/hr. Determine the aquifer transmissivity, the storage coefficient, and the drawdown after a year if pumping continues.

Ans. From the best fit line, ∆os = 0.90 m. Then Equation (7.42) yields the transmissivity.

T = [2.30∙Qw/(4π ∙Δos)] = [2.30∙8.50/(4π ∙0.90)] = 1.73 m2/hr. Then from Eq’n (7.43): S = (2.25∙T∙to)/(r2) = (2.25∙1.73∙1.72 x 10-2)/(202) = 1.67 x 10-4. To determine the drawdown after a year, use Eq’n (7.40): s = [2.30∙Qw/(4πT)]log[2.25Tt/(r2S)] s = [2.30∙8.50/(4π∙1.73)]log[2.25∙1.73∙(365∙24)/(202∙1.67 x 10-4)] = 5.13 m 15. A 10-m-thick confined aquifer has a transmissivity of 1.30 x 10-3 m2/sec. When a 30-cmdiameter well is pumped at the rate of 30 m3/hr, the drawdown in the piezometric surface at an observation well 30 m away is 9.60 m when it is not impacted by an aquifer boundary. If an identical well is installed in the same aquifer, but located 30 m from a fully-penetrating impermeable boundary, determine the drawdown at the boundary and at the well.

Ans. The equivalent hydraulic system is a pumped (image) well placed on the other side of the boundary the same distance (30 m). Using superposition principles, the drawdown at the boundary (30 m away) is: s30 = sreal + simage = 9.60 m + 9.60 m = 19.2 m The drawdown caused by the pumped (real) well at the well itself can be solved using Eq’n 7.11 with the original observation well data (Q = 30 m3/hr = 8.33 x 10-3 m3/sec): s = sob + (Q/2πT)[ln(rob/r)] = 9.60 + [8.33 x 10-3/(2π ∙1.3 x 10-3)][ln(30/0.15)] = 15.0 m The image well drawdown at the pumped (real) well is: s = 9.60 + [8.33 x 10-3/(2π ∙1.3 x 10-3)][ln(30/60)] = 8.9 m sw = sreal + simage = 15.0 m + 8.9 m = 23.9 m

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16. An industrial well taps into an 80-ft-thick confined aquifer that has a transmissivity of 0.0455 ft2/sec. The well is located 600 feet from a completely penetrating stream. A farmer’s irrigation well is located halfway between the stream and the industrial well. What is the maximum flow rate that can be pumped from the industrial well to limit the drawdown impact on the irrigation well to 5 feet.

Ans. The equivalent hydraulic system is a recharge (image) well 600 m on the other side of the boundary. The drawdown caused by the pumped (real) well at the irrigation well using Eq’n 7.11 (assume observation well at boundary): sreal = sob-r + (Q/2πT)[ln(rob/r)] sreal= sob-r + [Q/(2π ∙0.0455)][ln(600/300)]; The drawdown caused by the image well at the irrigation well: simage = sob-i + [-Q/(2π ∙0.0455)][ln(600/900)]; and by superposition; sTotal = sreal + simage Noting that: sob-r + sob-i = 0.0 (buildup=drawdown) and sTotal= 5.0 ft; sTotal= 5.0 = [Q/(2π ∙0.0455)][ln(600/300) - ln(600/900)]; Q = 1.30 cfs 17. A homeowner has a well that has been used for their domestic water supply for years. The well taps into a shallow, confined aquifer that has enough pressure to deliver their water without a pump. An industry purchases an adjacent property and installs a high capacity well in the same aquifer. To protect the homeowner’s domestic well from being impacted, industrial representatives propose to drive sheet piling into the ground all along the border of the two properties and deep enough to reach the aquifer’s underlying impermeable bottom.

Will the pressure (piezometric surface) in the homeowner’s well be affected by any of these activities. How would go about analyzing the impacts?

Ans. It appears that the industry has taken proper measures to insure the homeowner will be protected from their high capacity well. However, the sheet piles may prove harmful. If the sheet pile boundary (a new impermeable boundary in the homeowner’s aquifer) is within the radius of influence of the homeowner’s well, the domestic well will be impacted. An equivalent hydraulic system contains an image well that is on the other side of the boundary the same distance from the boundary as the domestic well.

18. A fully-penetrating, 12-in. diameter well pumps groundwater from a 25-ft thick confined aquifer. An impermeable rock stratum is located in the aquifer 105 meters away. Will the impermeable boundary impact the performance of the well when equilibrium conditions are achieved if the pump rate is 20,000 gpd? The aquifer permeability is 20 gpd/ft2 and the drawdown at the well is 30 feet. Prove your answer.

Ans. The key to this problem is finding the well’s radius of influence. Eq’n 7.6 yields: Q = 2πKb(ho-hw)/ln(ro/rw) ➔ 20,000 = 2π(20)(25)(30)/ln(ro/0.5); ro = 55.7 ft Thus, the boundary won’t impact well performance.

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19. A permeable stratum (coefficient of permeability of 0.12 m/hr, porosity of 0.45, and void ratio of 0.82) lies under a dam and over an impermeable rock layer. A sheet pile is driven into the stratum near the heal of the dam to reduce seepage, but doesn’t extend to the impermeable rock strata. Based on the flow net below, determine the rate of seepage under the dam (m3/day per meter width of dam), the pressure head at point “3” in meters, and the seepage velocity at point “3” in m/day. Use the impervious rock layer as the datum for all energy head determinations. The upstream water surface is 30 m above the impervious rock layer (datum). The upstream water depth is 10 m, and this dimension can be used as a scale.

Ans. The seepage under the dam: q = K(m/n)H = (0.12)(5/16)(10) = 0.375 m3/hr per meter (9.00 m3/day per meter of dam width). The head drop between equipotential lines is: Δh = H/n = (10 m)/16 = 0.625 m. Head at point 3: H3 = 30 – (10.5/16)(10) = 23.4 m Pressure head at “3”: (P/γ)3 = H3 – h3 = 23.44 – 2 = 21.44 m (h3 is distance above rock strata estimated from the map scale) Apparent velocity at #3, V3= K(Δh/Δs)3 = (0.12)[(3.75 – 3.125)/10] = 0.0075 m/hr The seepage velocity is: VS3 = V3/α = (0.0075)/0.45 = 0.0167 m/hr (0.401 m/day) 20. Determine the quantity of seepage (in gallons per minute) that can be expected to flow under 1,000 feet of sheetpile depicted in the figure below. The permeability of the soil under the sheetpile is 1.50 x 10-5 ft/sec and the porosity is 0.35. Also determine the approximate exit velocity of the water next to the sheetpile in ft/sec. (Note: The drawing below is roughly to scale.)

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Ans. A flow net must be constructed. If three flow channels are drawn, there are approximately six equipotential drops. However, the (m/n) ratio is roughly 3:6, regardless of the number of streamlines used. Now applying Eq’n 7.56: q = K(m/n)H = (1.5 x 10-5)(3/6)(5) = 3.75 x 10-5 cfs/ft Q = q∙L = (3.75 x 10-5 cfs/ft)(1000 ft) = 0.0375 cfs = 16.8 gpm Δh = H/n = (5 ft)/6 = 0.833 ft. The velocity at the exit next to the sheetpile: VA= K(Δh/Δs)A= (1.5 x 10-5)[(0.833)/(1.5)] = 8.33 x 10-6 ft/sec, where Δs is the distance measured across the last cell next to the sheet pile. The seepage velocity is VSA = VA/α = (8.33 x 10-6)/0.35 = 2.38 x 10-5 ft/sec 21. Determine the seepage in m3/day through the homogeneous earth dam depicted in the figure below. Also determine the seepage velocity when the water begins to surface on the downstream embankment at point “D.” The upstream headwater is 7.0 m, the dam is 80 m long, and soil used to construct the dam is silt. Assume this is a scale drawing with the upstream depth of 7.0 m being the scale.

Ans. From Table 7.2, silt has a mid-range permeability of 5.0 x 10-8 m/sec. The flow net provides the (m/n) ratio with n ≈ 17 (half way from D to F). Applying Eq’n 7.56: q = K(m/n)H = (5.0 x 10-8)(4/17)(7.0) = 8.24 x 10-8 m3/s-m Q = (8.24 x 10-8 m3/s-m)(80m)(86,400s/day) = 0.570 m3/day The head drop (loss) between equipotential lines is: Δh = H/n = (7.0 m)/17 = 0.412 m.

At point “D”: H = 7.0 – 14.5(0.412) = 1.03 m. Using Darcy’s eq’n: V= K(Δh/Δs)= (5.0 x 10-8)[(0.412)/1.0] = 2.06 x 10-8 m/s where Δh is the equipotential drop from the 14th to the 15th equipotential line and the distance (Δs = 1.0 m) is measured from the scale drawing.

Thus, VS = V/α = 4.58 x 10-8 m/s where the porosity is 0.45 from Table 7.1.

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TEST QUESTIONS & PROBLEMS - CHAPTER #8 Short Answer Questions 1. (T or F) Buttress dams are generally built in steep rock canyons.

Ans. False – arch dams rely on rock canyon walls for stability.

2. Identify at least four functions of some common hydraulic structures.

Ans. Storage, conveyance, transportation, measurement, energy conversion, sediment and fish control, energy dissipation, and collection.

3. Which of the dams listed relies upon the mass of water on the upstream face for stability? a) gravity b) arch c) buttress Ans. (c).

4. (T or F) The most common type of dam is the concrete gravity dam.

Ans. False. The earth dam is by far the most common type of dam built. .

5. Identify at least four functions of dams.

Ans. Hydroelectric power, irrigation water, downstream flood control, municipal and industrial water supply, cooling water, and recreation.

6. Identify at least four different types of forces that must be resisted by a gravity dam.

Ans. Hydrostatic forces (upstream and downstream), weight of the dam, uplifting force on the base of the dam, sedimentation (silt deposit) pressure force, earthquake force on the dam and, earthquake force due to the water mass behind the dam.

7. (T or F) Two failure modes that must be accounted for in the design of gravity dams are overturning and sliding.

Ans. True.

8. Why are so many small earth dams built each year? Identify 3 purposes for these structures.

Ans. Small earth dams are built for storm water management ponds, tailing ponds (mining applications), farm ponds (irrigation/stock water), constructed wetlands, and flood protection levees.

9. Which of the following terms represent components of a small earth dam? a) riser and barrel b) silt core c) antivortex collar d) spillway e) buttress f) key/keyway Ans. (a), (d), and (f). For some of the other answers, it is a clay core, not a silt core and an antiseep collar, not an antivortex collar. Buttress is a particular type of dam.

10. (T or F) A common outlet device for small earth dams is a riser and barrel assembly, often called a drop inlet.

Ans. True.

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11. Identify two critical elevations that need checking after an earth dam has been constructed?

Ans. The elevations of the service spillway, the emergency spillway, and the top of the dam.

12. Place numbers on the following tasks in earth dam construction to designate the normal order in which they are undertaken (e.g. clear site - 1). a) clear site ___ b) place embankment ___ c) install drawdown pipe ___ d) scrape/compact ___ e) excavate keyway ___ f) check critical elevations ___ Ans. The proper order is: a) clear site - 1 b) place embankment - 5 c) install drawdown pipe - 4 d) scrape/compact - 2 e) excavate keyway - 3 e) check critical elevations - 6 13. (T or F) The purpose of an emergency spillway is to lower the reservoir level when the earth embankment is showing signs of leaking.

Ans. False. A low-level outlet is used to drain a reservoir if necessary.

14. Identify two purposes for weirs constructed in streams and rivers.

Ans. Flow measurement, flow diversion (irrigation etc.), and minimum energy bridges.

15. The flow equation for weirs; q = 3.09H s3 / 2 , was derived using all of the following principles and assumptions except: a) critical depth b) frictionless weir c) normal depth d) specific energy Ans. (c) 16. Identify the three greatest drowning dangers of poorly designed low-head dams.

Ans. The reverse roller, reduced buoyancy, and the force of the plunging overflow.

17. (T or F) Negative pressure is a design concern in overflow spillways.

Ans. True.

18. Overflow spillways fulfill many purposes. Which of the following purposes is not typically associated with overflow spillways? a) flow measurement b) maintain target water levels in a reservoir c) emergency spillway d) maximize flow per unit of head Ans. (a) 19. What is the ideal shape of an overflow spillway?

Ans. The ideal shape of an overflow spillway should closely match the underside of the freefalling water nappe of a sharp-crested weir as depicted in Figure 8.11.

20. (T or F) A side-channel spillway carries water away from an overflow spillway in a channel that is perpendicular to the spillway crest.

Ans. False. It is parallel to the spillway crest.

21. What fundamental principle is used to determine depths of flow in a side-channel spillway?

Ans. The momentum principle.

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22. Which of the following statements is not true about siphon spillways when they are functioning as true siphons? a) Siphon spillways function like pipes under pressure flow. b) The head on a siphon spillway does not depend solely on the upstream reservoir level. c) The crest section of siphon spillways can encounter sub-atmospheric pressure. d) Siphon spillways require a circular cross sectional flow area, like a pipe.

Ans. (d)is false. See Example Problem (8.4).

23. What happens when a siphon spillway primes?

Ans. Free surface (gravity) flow changes to pressure flow.

24. (T or F) Water that is passing through a closed conduit will experience negative pressure when the conduit is elevated above the energy grade line.

Ans. False. The statement if true if hydraulic grade line is inserted for energy grade line. .

25. Is the culvert depicted below operating under inlet or outlet control?

Ans. Outlet control; hydraulic operation category (b).

26. Is the culvert depicted below operating under inlet or outlet control?

Ans. Inlet control; hydraulic operation category (c).

27. (T or F) Although culverts appear to be simple structures, the hydraulics can be complex and involve the principles of pressure pipe flow, orifice flow, and open channel flow.

Ans. True.

28. What is a stilling basin and what purpose does it fulfill?

Ans. A stilling basin is a type of energy dissipator that is used at the outlet of a hydraulic structure to reduce the energy in the flowing water and avoid damage to the hydraulic structure and the downstream channel.

29. The relationship between depth of flow in a stilling basin and the energy to be dissipated is contained in which of the flowing relationships? a) Froude number b) critical depth c) normal depth d) specific energy Ans. (a) 30. (T or F) A stilling basin relies primarily upon baffle blocks to dissipate unwanted energy.

Ans. False. It relies primarily upon a controlled hydraulic jump.

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Problems 1. A gravity dam is depicted in the figure below. If a force ratio against sliding of 1.3 is required, determine whether the 33 m high dam is safe. Assume the coefficient of friction between the dam base and the foundation is 0.6, the specific gravity of concrete is 2.5, and full uplift forces exist on the base of the dam. Neglect earthquake and sedimentation forces.

Ans. Find all forces on dam per unit width: W = γ(Vol) = (2.5)(9.79)[(33)(5) + ½(33)(22)] W = 12,900 kN/m. Full uplift pressure at the heel: Pheel = γH = (9.79)(30) = 294 kN/m2 With no tailwater, Ptoe = 0; thus the uplift force is: Fu = ½(294)(5 + 22) = 3,970 kN/m The hydrostatic force on the dam face, Eq’n (2.12) is: FHS = γ h A = (9.79)(15)(30) = 4410 kN/m. The force ratio against sliding [Eq’n (8.1)]: FRslide= [(0.6)(12,900-3970)]/(4410) = 1.21 < 1.3 (Not safe) 2. A gravity dam is depicted in Problem #1 above. If a force ratio against overturning of 2.0 is required, determine whether the dam is safe if the water level rises to the top of the 33 m high dam during a flood event. Assume the specific gravity of concrete is 2.5, and full uplift forces exist on the base of the dam. Neglect earthquake and sedimentation forces.

Ans. Weight in sections per unit width [rectangle (Wr) & triangle (Wt)]: W = γ(Vol) Wr = (2.5)(9.79)[(33)(5)] = 4,040 kN/m; Wt = (2.5)(9.79)[½(33)(22)] = 8,880 kN/m The full uplift pressure at the heel is: Pheel = γH = (9.79)(33) = 323 kN/m2 With no tailwater, Ptoe = 0; thus the uplift force is: Fu = ½(323)(5 + 22) = 4,360 kN/m The hydrostatic force on the dam face, Eq’n (2.12) is: FHS = γ h A =(9.79)(16.5)(33) = 5,330 kN/m Thus, the force ratio against overturning, Eq’n (8.2), is FRover = [(4040∙24.5)+(8880∙14.7)]/[(4360∙18)+(5330∙11)] FRover = 1.67 (< 2.0, not safe)

83

3. A 15-ft-high gravity dam is depicted in the figure below. Determine the foundation pressure at the heel and toe of the dam if the water behind the dam rises to a depth of 15 ft. Assume the coefficient of friction between the 12-ft-long dam base and the foundation is 0.6, the specific gravity of concrete is 2.65, and full uplift forces exist on the base of the dam.

H

15 ft

12 ft

Ans. W = γ(Vol) = (2.65)(62.3)[½(12)(15)] = 14,900 lbs/ft. Full uplift pressure at the heel: Pheel = γ(15) = 935 lb/ft2, Ptoe = 0; thus the uplift force: Fu = ½(935)(12) = 5,610 lbs/ft Thus, the resultant of all vertical forces is: Rv = 14,900 – 5,610 = 9,290 lbs/ft Hydrostatic force on dam face, Eq’n (2.12): FHS = γ h A = (62.3)(7.5)(15) = 7,010 lbs/ft Use the principle of moments to find “e” (Fig 8.3): Rv∙e = ΣMCL; clockwise positive 9,290∙e =[(5,610∙2)-(14,900∙2)+(7,010)(7.5)]; e = 3.66 m; Since clockwise moments produced by the component forces exceed counter-clockwise moments, the location of Rv is on the downstream side of the centerline. Therefore, PV will be greater than PH. From Eq’n (8.3): PT = (Rv/B)(1 + 6e/B) = (9,290/12)(1 + 6(3.66)/12) = 2,190 lbs/ft2 From Eq’n (8.4): PH = (Rv/B)(1 - 6e/B) = (9,290/12)(1 - 6(3.66)/12) = -643 lbs/ft2 Must re-design: PH is not positive and Rv is not in the middle third of base (e > B/6).

4. A weir 0.78 m in height is built across the floor of a 4-m-wide rectangular channel. If the water depth just upstream of the weir is 2.00 m, determine the flow rate in the channel.

Neglect friction loss and the velocity head upstream. Also, determine the flow rate if the velocity head is not neglected.

Ans. Applying Equation (8.7) yields: H = (3/2)yc + x;

2.00 = (3/2)yc + 0.78 m;

yc = 0.813 m. Now applying Eq’n (8.8a), q = [g(yc)3]1/2 = [9.81(0.813)3]1/2 q = 2.30 m3/sec per meter width. Therefore, Q = bq = (2.30)(4) = 9.20 m3/sec.

Including the upstream velocity head in Eq’n (8.8c): q = 1.70 Hs3/2; where Hs = H + V2/2g - x = H + (q/H)2/2g – x (See Fig 8.7) q = 1.70[(2.0 + {(q/2)2/2∙9.81} - 0.78]3/2 Solving the implicit equation yields: q = 2.52 m3/sec-m; Q = bq = 10.1 m3/sec

84

5. The elevation of the crest of a weir is needed for a court case. The weir is 24.5 feet long with a design flow of 155 cfs when the water surface elevation upstream is 9.28 feet above mean sea level (MSL). Assume the weir friction loss and the velocity head upstream are negligible.

Ans. Apply Eq’n (8.8b): q = 3.09 Hs3/2; where q = Q/b = 155/24.5 = 6.33 cfs per ft of width.

Thus, 6.33 = 3.09 Hs3/2; Hs = 1.61 ft. Now, Elev(crest) = 9.28 - 1.61 = 7.67 ft MSL 6. Water is flowing over a 75-m-wide spillway (C = 2.15) at the rate of 375 m3/sec. The channel leading to the spillway is 6 meters deep at normal lake level (i.e., with no static head included). Compute the static head with and without the approach velocity being considered.

Ans. From Equation (8.9): Ha = [Q/(CL)]2/3 = [375/(2.15∙75)]2/3 = 1.76 m From Equation (8.10): Ha = Hs + Va2/2g, where Va = Q/A = 375/[(75)(6+Hs)] Va = 5/(6+Hs); thus 1.76 = Hs + [5/(6+Hs)]2/2∙9.81; Solving yields: Hs = 1.74 m 7. Determine the maximum approach velocity (as a function of Hs) that can occur before an error of 2% enters into the calculation for spillway discharge. (Use BG units.)

Ans. Ignoring the approach velocity, we obtain the ratio: Reducing yields:

Qignore CLH s3 / 2 = = 0.98 ; Qinclude CLH a3 / 2

Hs H s3 / 2 = 0.987 ; = 0.98 ; 2 3/ 2 ( H s + Va2 / 2 g ) ( H s + Va / 2 g )

Hs = 0.987(Hs + Va2/2g); Hs = 76.9[0.987∙(Va2/2∙32.2)]; Va = 0.921∙Hs1/2 8. An overflow spillway (C = 2.0), under a head of 1.5 m, discharges water into a side-channel spillway that is 15 m long. At the end of the rectangular side channel, the depth of flow is 3.67 m. If the side channel has a bottom width of 2.5 meters (n = 0.015) and a bottom slope of 1.00 %, determine the depth 5 meters upstream by filling in the table below. Δx

Δy

y

A

Q

(m2)

(m3/s)

(m)

(m)

(m)

5

-2.21

3.67

V

Q1+Q2

(m/s)

(m3/s)

V1+V2

ΔQ

ΔV

Rh

Sf

Δy

(m/s)

(m3/s)

(m/s)

(m)

(m)

(m)

Ans. From Equation (8.9) with a negligible approach velocity: Q = CLHs3/2. Therefore, Q = CLHs3/2 = (2.00)(15)(1.5)3/2 = 55.1 m3/sec. Filling out the table yields a depth of flow of 5.88 m at a point 5 m upstream from the end of the side channel. Δx

Δy

y

A

Q

V

Q1+Q2

V1+V2

ΔQ

ΔV

Rh

Sf

Δy

(m3/s)

(m/s)

(m3/s)

(m/s)

(m3/s)

(m/s)

(m)

(m)

(m)

55.1 36.7

6.00 2.50

91.9

8.50

18.4

3.50

1.03

-

-

0.0013

-2.21

(m)

(m)

(m)

(m2)

5

-2.21

3.67 5.88

9.18 14.7

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9. A 300-ft-long, rectangular, side-channel spillway (n = 0.015) receives a flow of 28.5 ft3/sec per foot of channel length from an overflow spillway (C = 3.73). The channel is 15 ft wide and has a bottom slope of 0.001. The depth at the end of the channel is 21.6 feet. Define the water surface profile (at 100-ft intervals) by filling in the blanks of the table below. Δx

Δy

y

A

Q

V

Q1+Q2

V1+V2

ΔQ

ΔV

Rh

Sf

Δy

(ft3/s)

(ft/s)

(ft3/s)

(ft/s)

(ft3/s)

(ft/s)

(ft)

(ft)

(ft)

?

5700 2850 0

26.4 10.9 ?

0.00

?

8550 2850

37.3 15.9 5.1

2850 2850 ?

15.50 5.83 5.06

6.17 ?

6.27

?

?

0.0002

-2.68

0.0000

-0.69

(ft)

(ft)

(ft)

(ft2)

100 100 100

-13.29 ?

-0.69

21.6 34.9 37.6 ?

324 523 ?

574

Ans. The total channel flow is: Q = q∙L = (28.5)(300) = 8550 cfs. Filling out the table yields: Δx

Δy

y

A

Q

V

Q1+Q2

V1+V2

ΔQ

ΔV

Rh

Sf

Δy

(ft3/s)

(ft/s)

(ft3/s)

(ft/s)

(ft3/s)

(ft/s)

(ft)

(ft)

(ft)

8550 5700 2850 0

26.4 10.9 5.06 0.00

14250 8550 2850

37.3 15.9 5.1

2850 2850 2850

15.50 5.83 5.06

6.17 6.25 6.27

-

-

0.0011

-13.29

0.0002

-2.68

0.0000

-0.69

(ft)

(ft)

(ft)

(ft2)

100 100 100

-13.29 -2.68 -0.69

21.6 34.9 37.6 38.3

324 523 564 574

10. A rectangular siphon (3 ft by 6 ft) discharges into a pool with an elevation of 335 ft, mean sea level (MSL). Determine the discharge when the siphon primes at an upstream pool elevation of 368 ft, MSL if losses are 10.5 ft excluding the exit loss. Also, determine the pressure head at the siphon crown under these conditions if 3.5 ft of losses accrue prior to the crown. The siphon crown is at an elevation of 366.5 ft.

Ans. Balancing energy from the reservoir (1) to downstream pool (2): z1 +

P1



+

V12 P V2 = z2 + 2 + 2 + hL : 2g  2g

368 + 0 + 0 = 335 + 0 + 0 + 10.5 + V2/2g;

V2/2g = 22.5 ft., V = 38.1 ft/sec,

Q = AV = (38.1)(3x6) = 686 cfs

Balancing energy from reservoir (1) to siphon crown (C): z1 +

P1



+

P V2 V12 = zC + C + C + hL ; hL = hf + he 2g  2g

368 + 0 + 0 = 366.5 + Pc/γ + 22.5 + 3.5; Pc/γ = -24.5 ft (< -23 ft; cavitation concern)

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11. A 60-m-long siphon spillway discharges water at the rate of 0.32 m3/sec. The siphon crown is 1.2 m above the water surface elevation of the reservoir and a length of 10 m away from the siphon entrance. If the siphon diameter is 30 cm, the friction factor is 0.02, and the head loss coefficients are 0.2 (inlet) and 1.0 (exit), determine the elevation difference between the siphon crown and the downstream pool. What is the pressure at the crown in kN/m2?

Ans. Balancing energy from reservoir to downstream pool: z1 +

P1



+

V12 P V2 = z2 + 2 + 2 + hL ; hL = hent + hf + hexit; 2g  2g

V = Q/A = (0.32)/[π(0.15)2] = 4.53 m/s z1 + 0 + 0 = z2 + 0 + 0 + (0.2+0.02(60/0.3)+1.0)[4.532/2g] z1 - z2 = 5.44 m; thus hc - h2 = 1.2 + 5.44 = 6.64 m (i.e., the elev. difference between crown and downstream pool). Balancing energy from reservoir to siphon crown: z1 +

P1



+

2 P V V12 = zC + C + + h ; h L = hf + he 2g  2g L

0 + 0 + 0 = 1.2 + Pc/γ + [4.532/2g] + (0.2+0.02(10/0.3))[4.532/2g]; Pc/γ = -3.15 m.

Pc = (-3.15m)(9.79 kN/m3) = -30.8 kN/m2

12. Stage recorders register headwater and tailwater depths (4.05 m and 3.98 m) of a culvert during a flood event. The 2 m by 2 m concrete culvert (square-edged entrance) is 15 m long and has a slope of 3.0%. Based on this information, determine the flood flow rate that passed through the culvert. The stream depths are measured above the culvert inverts (bottom).

Ans. Since the outlet is submerged, the hydraulic operation category is (a) and the energy balance is: Hup + SoL = Hdown + hL; where H = stream depths 4.05 + (0.03)(15) = 3.98 + hL; hL = 0.52 m The head loss from hydraulic category (a) is: hL = [Ke + {n2L/Rh4/3}(2g) + 1]{Q2/(2g∙A2)}; where Rh = A/P = (4 m2)/(8 m) = 0.5 m. Thus, 0.52 = [0.5+{0.0132(15)/(0.5)4/3}(2g)+1]{Q2/(2g(42)}; Q = 10.0 m3/sec

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13. A concrete culvert is required to convey a design flow of 654 cfs over a length of 330 ft. and a slope of 0.015 ft/ft. Since there is a large lake downstream, the tail water elevation is constant at 526.4 ft, MSL (mean sea level), which will submerge the culvert outlet. The head water elevation cannot exceed 544.4 ft., MSL without overtopping the roadway embankment.

Find the required diameter for a single barrel culvert assuming a square-edged entrance.

Ans. Since the outlet is submerged, the hydraulic operation category is (a) and the energy balance is: HW = TW + hL; hL = 544.4 – 526.4 = 18.0 ft The head loss from hydraulic category (a) is: hL = [Ke + {n2L/2.22Rh4/3}(2g) + 1]{8Q2/(π2∙g∙D4)}; 18 = [0.5 + {0.0132(330)/2.22(D/4)4/3}(2g) + 1]{8∙6542/(π2∙g∙D4)} 18 = [1.5 + 10.3/D4/3]{10,800/D4} Solving the implicit equation: D = 6.16 ft

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TEST QUESTIONS & PROBLEMS - CHAPTER #9 Short Answer Questions 1. Define liquid pressure.

Ans. The normal force exerted by a liquid on a unit surface area.

2. Which of the following devices are used to measure liquid pressure? a) Pitot tube b) Venturi meter c) propeller-type meter d) orifice meter e) manometer f) Parshall flume g) Bourdon tube gauge h) transducer Ans. (a),(e),(g), and (h).

3. If the flow of a liquid is from left to right above the following (five) pressure openings, identify whether the pressure will register too high (+) or too low (-).

Ans. From left to right: +, - , +, - , - .

4. (T or F) Pitot tubes measure the average velocity in a conduit.

Ans. False. They measure the pressure at a point, which can be converted to a point velocity.

5. Define stagnation pressure.

Ans. Stagnation pressure is the stagnant point pressure at the tip of a Pitot tube which combines static pressure and dynamic pressure.

6. Which of the following devices is not normally used to measure open-channel velocities? a) Pitot tube b) cup-type meter c) propeller-type meter d) acoustic velocity meter Ans. (a).

7. (T or F) Although pipe flow measurements can be accomplished by several different methods, the simplest and most reliable measurement is the volumetric (or weight) method.

Ans. True.

8. What two hydraulic principles are applied when determining flow rates using Venturi, nozzle, and orifice meters?

Ans. Bernoulli’s equation and the continuity equation.

9. Which of the following devices are not used to measure discharge in pressure pipes? a) Pitot tube b) Venturi meter c) propeller-type meter d) orifice meter e) bend meter f) Parshall flume g) Bourdon tube gauge h) nozzle meter Ans. (a),(c),(f), and (g).

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10. (T or F) Energy losses in Venturi meters is greater than orifice meters.

Ans. False. Nozzle meters and orifice meters produce a significant amount of head loss since most of the pressure energy that converts to kinetic energy (to speed the fluid through the narrow opening) cannot be recovered.

11. What hydraulic principle forms the basis for determining flow rates in bend meters?

Ans. The centrifugal force developed at a pipe bend creates a difference in pressure between the inside and outside of the bend, which increases as the flow rate increases.

12. Which of the following devices are not used to measure discharge in open channels? a) Pitot tube b) Venturi meter c) propeller-type meter d) sharp-crested weir e) bend meter f) Parshall flume g) Bourdon tube gauge h) broad-crested weir Ans. (d),(f), and (h).

13. (T or F) The basic discharge equation for a standard horizontal weir contains a discharge coefficient (C) which is dimensionless.

Ans. False. The discharge coefficient has units of length0.5/time.

14. There are four basic types of sharp-crested weirs. Name three of them.

Ans. Horizontal weirs without end contractions, horizontal weirs with end contractions, Vnotch weirs, and trapezoidal weirs.

15. The pertinent equation for broad-crested weirs is derived from what fundamental principle? a) momentum balance b) energy balance c) Manning equation d) Bernoulli equation Ans. (a).

16. (T or F) The basic discharge equation for a standard horizontal weir contains a discharge coefficient (C) whose value varies depending on the unit system being used.

Ans. True.

17. Which of the following statements is not true about Parshall flumes? a) Parshall flumes produce a critical flow depth within the measurement structure. b) Parshall flume equations were developed using experimental data. c) Parshall flume equations were developed for the SI system of units. d) Parshall flume equations can be corrected for high tailwater (submerged flow conditions).

Ans. (c).

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Problems 1. An inclined oil manometer (shown below) is used to measure pressure in a wind tunnel.

Determine the specific gravity of the oil (shaded color) if a pressure change of 323 Pascals occurs causing a 15-cm reading change on the 15° incline.

Ans. Since P = γ∙h; pressure can be expressed as the height of any fluid. In this case, P = (γOil)Δh = (S.G.)Oil(γ)(Δh); where Δh = (Δl)(sin θ) Δh = (0.15 m)(sin 15°) = 0.0388 m. Therefore, P = (S.G.)Oil(γ)(Δh); 323 N/m2 = (S.G.)Oil(9,790 N/m3)(0.0388 m); (S.G.)Oil = 0.850 2. A piezometer and a pitot tube are attached to a pipe as shown if the figure below. Which one will record the higher water level? Explain. Also determine the flow velocity in the pipe if the difference in the water levels between the piezometer and the pitot tube is 34.4 cm.

Piezometer

Pitot tube

Ans. The Pitot tube will register the higher water level because it captures the pressuer head and the velocity head. The piezometer only captures the pressure head. Applying Equation (9.1b) yields: V = [2g(ΔP/γ)]1/2 where ΔP = γ∙Δh; thus V = [2g(Δh)]1/2 = [(2∙9.81)(0.334)]1/2 ; V = 2.56 m/s 3. Pitot tube measurements (Figure 9.4b) in a 1-ft-diameter water pipe indicate that the stagnation pressure and the static pressure differ by a mercury column height of 5.65 inches.

Estimate the discharge coming through the pipe. Why is it just an estimate of discharge?

Ans. Based on Example 9.2, we see that: V = [2gΔh(SG – 1)]1/2

Substituting,

V = [(2∙32.2)(5.65/12)(13.6 – 1)]1/2 = 19.5 ft/sec; Note: This is the point velocity at tip of the Pitot tube. Q = AV = [π(1)2/4](19.5) = 15.3 cfs.

This is a flow estimate since the velocity is not an average velocity.

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4. A maintenance crew discovers an orifice meter in an old water line. The engineer in charge would like to know the size of the orifice plate inside without dismantling the delicate meter.

A specification plate on the side of the meter reads, Cv = 0.605. For a flow rate of 0.00578 m3/sec, the attached mercury-water manometer registers a pressure difference of 9.0 cm. If the horizontal pipeline has an inside diameter of 10 cm, what is the size of the orifice plate?

Ans. Apply Eq’n (9.6b): Q = CvCdA1[2g(ΔP/γ)]1/2; From manometry principles: ΔP = Δh(γHg - γ) or ΔP/γ = Δh(SG – 1) = 0.09(13.6 – 1) = 1.13 m 0.00578 = (0.605)Cd[(π/4)(0.10)2][2∙9.81(1.13)]1/2 Cd = 0.258 Applying Equation (9.5b) yields Cd = 1/[(A1/A2)2 – 1]1/2 = 1/[(D1/D2)4 – 1]1/2 0.258 = 1/[(0.10/D2)4 – 1]1/2;

D2 = 0.0500 m (5.00 cm)

5. A discharge of 0.0820 m3/sec pass through a horizontal, 20-cm waterline containing a 10-cm ASME flow nozzle. Determine the pressure difference that will register across the flow meter in Pascals.

Ans. Apply Eq’n (9.6b): Q = CvCdA1[2g(ΔP/γ)]1/2; where A1 = π(0.1)2 = 0.0314 m2; A2 = π(0.05)2 = 0.00785 m2 Eq’n (9.5b) yields: Cd = 1/[(A1/A2)2 – 1]1/2 = 1/[(0.0314/0.00785)2 – 1]1/2 = 0.258 To obtain Cv: V2 = Q/A2 = 0.082/0.00785 = 10.4 m/sec NR2 = V2d2/ν = (10.4)(0.10)/(1.00 x 10-6) = 1.04 x 106 From Fig. 9.8 with d2/d1 = 0.5; Cv = 0.991; Now Q = CvCdA1[2g(ΔP/γ)]1/2; 0.082 = (0.991)(0.258)(0.0314)[2g(ΔP/9,790)]1/2; ΔP = 52.1 x 103 Pa = 52.1 kPa 6. A Venturi meter with a 4-in. throat is installed in an 8-in. diameter (vertical) waterline. The pressure difference reading between the throat and entry section in a mercury-water manometer is 2.01 feet. Determine the flow rate if the flow is moving downwards and the distance between the pressure taps is 0.80 feet. SG (mercury) = 13.6.

Ans. From manometry principles: ΔP = Δh(γHg - γ); or ΔP/γ = Δh(SG – 1); substituting ΔP/γ = 2.01(13.6 – 1) = 25.3 ft. Therefore, Δ(P/γ + z) = 25.3 + 0.80 = 26.1 ft Equation (9.5b) and Equation (9.5a) yield: Cd = 1/[(A1/A2)2 – 1]1/2 = 1/[(8 in/4 in)4 – 1]1/2 = 0.258 Q = CdA1[2g{Δ(P/γ + z)]1/2 = (0.258)[π(8/12)2/4][2∙32.2(26.1)]1/2 = 3.69 cfs 91

7. The crest of a standard, U.S.B.R. 90°-V-notch weir is 3 feet above the bottom of an irrigation channel. It measures 25.6 cfs of flow when the upstream water depth is at the top of the channel bank. An increased flow rate (33.3 cfs) is needed without flooding the upstream channel. Determine the required length of a replacement weir (USBR standard contracted horizontal weir with the same crest height) to accommodate the new flow demand.

Ans. Applying Eq’n (9.14) yields: Q = 2.49H2.48; 25.6 = 2.49H2.48; H = 2.56 ft Applying Eq’n (9.12a) yields: Q = 3.33(L – 0.2H)H3/2 33.3= 3.33[L – 0.2(2.56)](2.56)3/2; L = 2.95 ft 8. What is the length of the Cipolletti weir (U.S.B.R. standard trapezoidal weir) required to accommodate a flow up to 0.793 m3/sec if the maximum head is limited to 0.259 m?

Ans. Applying Eq’n (9.15): Q = 3.367LH3/2 requires BG units. Thus, H = 0.259 m = 0.850 ft and Q = 0.793 m3/sec (35.3 cfs/1 cms) = 28.0 cfs Substituting into Eq’n (9.15) yields: 28.0 = 3.367(L)(0.850)3/2; L = 10.6 ft = 3.23 m 9. Determine the discharge in cubic meters per second measured by a 15-ft Parshall flume if the gage reading Ha is 1 m and the gage reading Hb is 0.5 m.

Ans. Hb/Ha = 0.5/1.0 = 0.5, thus flow is not submerged.

Applying Eq’n (9.24), which requires BG units (i.e., Ha = 1 m = 3.28 feet): Q = (3.6875W + 2.5)Ha1.6 = [3.6875(15 ft) + 2.5)(3.28 ft)1.6 = 387 cfs = 11.0 m3/s 10. A contracted, horizontal weir with a discharge coefficient of 3.30 and end contractions has a 6.5-ft-long crest. Determine the weir’s height (p) to maintain an upstream depth of 7.40 ft for a discharge of 34.0 ft3/sec?

Ans. Applying Eq’n (9.11) yields: Q = C[L – (n∙H/10)]H3/2; and noting that p + H = 7.4 Q = 3.30[6.5 – {2(7.4 – p)/10)](7.4 - p)3/2 = 34.0 cfs Solving the implicit equation (by calculator, iteration, MathCad or other software) yields: p = 6.00 ft

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TEST QUESTIONS & PROBLEMS - CHAPTER #10 Short Answer Questions 1. Define dimensional analysis.

Ans. Assessment of the appropriate physical quantities and fundamental hydraulic relationships (both static and dynamic) involved in the performance of the structure.

2. (T or F) The reduced version of a hydraulic structure used to study complex hydrostatic and hydrodynamic phenomena is called a prototype.

Ans. False. It is called a scaled model or simply a model.

3. Name three hydraulic structures that are designed or analyzed with the help of model studies.

Ans. Spillways, weirs, energy dissipaters, intake structures, transition sections, cooling ponds, docks, locks, breakwaters, transportation facilities, offshore structures, etc..

4. Which of the following is not one of the three basic forms of hydraulic similitude? a) geometric b) kinetic c) kinematic d) dynamic Ans. (b).

5. Identify the three basic forms of hydraulic similitude.

Ans. Geometric, kinematic, and dynamic.

6. Which of the following is not one of the parameter ratios applicable to geometric similarity? a) velocity b) length c) area d) volume Ans. (a).

7. (T or F) Kinematic similarity between a model and prototype is attained if the homologous moving particles have the same velocity ratio along geometrically similar paths.

Ans. True.

8. Which of the following is not one of the parameter ratios applicable to kinematic similarity? a) angular acceleration b) discharge c) velocity d) mass e) time Ans. (d).

9. (T or F) Dynamic similarity between a model and its prototype is attained if the ratio of homologous forces (prototype to model) is kept at a constant value.

Ans. True.

10. When inertial forces and viscous forces govern fluid motion, which dimensionless ratio must be maintained between the model and the prototype?? a) Reynolds number b) Froude number c) Weber number d) Social Security number Ans. (a).

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11. Based on the Reynolds number law (i.e., ratio), determine the dimensions for the velocity ratio if the same fluid is used in the model and in the prototype.

Ans. The Reynolds number law (for ρr = 1, μr = 1) is (ρrLrVr)/μr = LrVr = 1. Therefore, Vr = 1/Lr = Lr-1 12. Based on the Reynolds number law (i.e., ratio), determine the dimensions for the time ratio if the same fluid is used in the model and in the prototype.

Ans. The Reynolds number law (for ρr = 1, μr = 1) is (ρrLrVr)/μr = LrVr = 1. Therefore, Vr = 1/Lr = Lr-1 and from Eq’n (10.6): Vr = Lr/Tr thus Lr-1 = Lr/Tr; Tr = Lr2 13. When inertial forces and gravity forces govern fluid motion, which dimensionless ratio must be maintained between the model and the prototype?? a) Reynolds number b) Froude number c) Weber number d) Social Security number Ans. (b).

14. Based on the Froude number law (i.e., ratio), determine the dimensions for the velocity ratio if the same fluid is used in the model and in the prototype.

Ans. The Froude number law (for gr = 1, ρr = 1) is (Vr)/(gr1/2Lr1/2) = (Vr)/(Lr1/2) = 1.

Therefore, Vr = Lr1/2 15. Based on the Froude number law (i.e., ratio), determine the dimensions for the time ratio if the same fluid is used in the model and in the prototype.

Ans. The Froude number law (for gr = 1, ρr = 1) is (Vr)/(gr1/2Lr1/2) = (Vr)/(Lr1/2) = 1.

Therefore, Vr = Lr1/2 and from Eq’n (10.6): Vr = Lr/Tr; Lr1/2 = Lr/Tr; Tr = Lr1/2 16. When surface tension forces are prominent, which dimensionless ratio must be maintained between the model and the prototype?? a) Reynolds number b) Froude number c) Weber number d) Social Security number Ans. (c).

17. Based on the Weber number law (i.e., ratio), determine the dimensions for the velocity ratio if the same fluid is used in the model and in the prototype.

Ans. The Weber number law (for ρr = 1, σr = 1) is (ρrVr2Lr)/σr = Vr2Lr = 1.

Therefore, Vr = 1/Lr1/2 18. Based on the Weber number law (i.e., ratio), determine the dimensions for the time ratio if the same fluid is used in the model and in the prototype.

Ans. The Weber number law (for ρr = 1, σr = 1) is (ρrVr2Lr)/σr = Vr2Lr = 1.

Therefore, Vr = 1/Lr1/2 and from Eq’n (10.6): Vr = Lr/Tr; 1/Lr1/2 = Lr/Tr; Tr = Lr3/2 19. When both gravity and viscous forces are prominent, which dimensionless ratio(s) must be maintained between the model and the prototype?? a) Reynolds number b) Froude number c) Weber number d) Social Security number Ans. (a) and (b).

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20. (T or F) To determine the friction drag along the boundary of a floating or submerged vessel, the Reynolds number ratio must be maintained between the model and the prototype.

Ans. True.

21. To determine the force expended in the generation of gravity waves on a floating vessel, which dimensionless ratio(s) must be maintained between the model and the prototype?? a) Reynolds number b) Froude number c) Weber number d) Social Security number Ans. (b).

22. Identify two different kinds of problems that can be analyzed using open channel models.

Ans. Velocity-slope relationships and channel bed movement due to flow patterns.

23. (T or F) Open channel models generally use the same vertical and horizontal scales.

Ans. False.

24. Show that the Manning velocity ratio for an undistorted open channel model equals Lr2/3/nr.

Ans. From the Manning equation: Vr = (1/nr)Rhr2/3Sr1/2; but Sr= 1 (undistorted) and Rhr = Lr; Therefore, Vr = Lr2/3/nr 25. Identify the equation that relates mass in the MLT system to force in the FLT system.

Ans. Newton’s 2nd law, F = ma.

26. Dimensional analysis produces many advantages in solving complex hydraulic problems.

These advantages include all of the following except: a) provides an appropriate solution to the complex problem, b) reduces the number of pertinent variables by forming dimensionless groupings, c) provides a guide to point out applicable relationships among the parameters, and d) reduces the complexity of the model experiments necessary to identify the solution.

Ans. (a).

27. Define repeating variable as the term applies to dimensional analysis using the Pi-theorem?

Ans. Pertinent variables that will appear in each of the dimensionless Π–groups.

28. Identify two dimensionless parameters that commonly appear in hydraulic model studies.

Ans. The Reynolds number, the Froude number, and the Weber number.

29. (T or F) The Buckingham Pi-theorem states that if a physical phenomenon involves n dimensional variables in a dimensionally homogenous equation described by m fundamental dimensions, the variables may be combined into (n – m) dimensionless groups for analysis.

Ans. True.

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Problems 1. A 1:15 scale model will be used to study circulation patterns in a rectangular detention pond.

The pond has a bottom length of 40 m and bottom width of 10 m. All sides of the pond have 3:1 (H:V) slopes. If the design depth is 5 m, determine the depth, surface area, and storage volume of the model. (Note: The storage volume expression is: Vol = LWd + (L+W)zd2 + (4/3)z2d3; where L, W, d, and z are the bottom length, bottom width, depth, and side slope.)

Ans. For a pond depth of 5 m, determine the surface area of the pond (prototype): Width: Wtop = 10m + 2(3)(5m) = 40 m Length: Ltop = 40m + 2(3)(5m) = 70 m; Surface Area: Ap-top = (40)(70) = 2,800 m2 Now determine the volume of the detention pond (prototype): Volp = LWd + (L+W)zd2 + (4/3)z2d3 Volp = (40)(10)(5) + (40+10)(3)(5)2 + (4/3)(3)2(5)3 = 7,250 m3; Based on geometric similarity and using Equations (10.2), (10.3), and (10.4): dp/dm = Lr; therefore, dm = dp/Lr = 5/15 = 0.333 m; Ap/Am = Lr2; Am = Ap/Lr2 = 2,800/(15)2 = 12.4 m2 Volp/Volm = Lr3; Volm = Volp/Lr3 = 7,250/(15)3 = 2.15 m3 2. A centrifugal pump discharges 36 cfs and overcomes a 90-ft head when rotating at 400 rpm.

A 1:6 scale model operating at 1200 rpm is used to study the prototype. Find the model discharge and head. (Hint: For hydraulic machines, head has the units of Lr2/Tr2 since it incorporates the gravitational acceleration which is the same for model and prototype.)

Ans. Kinematic similarity seems appropriate since time is involved, but not force.

Using Eq’n (10.10): Np/Nm = 1/Tr; 400/1200 = 1/Tr; Tr = 3; From Eq’n (10.8): Qp/Qm = Lr3/Tr = (6)3/3 = 72.

Thus, Qp/Qm = Qr; (36)/Qm = 72; Qm = 0.5 cfs Based on the hint (and Example 10.4): Hr = Lr2/Tr2 = (6)2/(3)2 = 4 and Hr = Hp/Hm; 4 = 90/Hm; Hm = 22.5 ft 3. Laboratory models are commonly used to determine the effectiveness of energy dissipation structures. Consider a 1:20 scale model of an energy dissipator that is constructed to study force distribution and water depths. A velocity ratio of 7.75 is used. Determine the force ratio and prototype discharge if the model discharge is 10.6 cfs.

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Ans. Dynamic similarity is used since force is involved. From Equation (10.6): Vr = Lr/Tr; 7.75 = 20/Tr; Tr = 2.58. Then from Equation (10.13): Fr = ρrLr4Tr-2 Fr = (1.00)(20)4(2.58)-2 = 2.40 x 104. Then from Equation (10.8): Qr = Lr3/Tr; Qr = (20)3/(2.58)r = 3,100. Since, Qp/Qm = Qr; Qp = (10.6)(3,100) = 3.29 x 104 cfs 4. The operation of the prototype spillway is studied on a 1:50 scale model in a hydraulic laboratory. The prototype has a 100-m-long crest and conveys a design discharge of 1150 m3/sec under a permitted maximum head of 3.00 m. The flow rate in the model is 0.0650 m3/sec. If the force on a bucket energy dissipater at the toe of the spillway model is measured to be 37.5 N, determine the corresponding force on the prototype.

Ans. Dynamic similarity is appropriate since force is involved. From Equation (10.13): Fr = Fp/Fm = ρrLr4Tr-2. Therefore, Tr must be obtained from Equation (10.8): Qp/Qm = Lr3/Tr; 1150/0.0650 = 503/Tr; Tr = 7.07. Therefore, assuming ρr = 1.0, Fp/Fm = ρrLr4Tr-2; Fp/37.5 = (1.0)(50)4(7.07)-2; Fp = 4,690 kN

5. For a particular model study, the inertial forces and viscous forces are known to govern the motion. If the model and prototype both use water verify the scale ratios for: a) time (Tr = Lr2), b) acceleration (ar = Lr-3), and c) discharge (Qr = Lr).

Ans. The Reynolds Number Law (for ρr = 1, μr = 1) is (ρrLrVr)/μr = LrVr = 1. Rearranging, a) Vr = 1/Lr = Lr-1; and since from Eq’n (10.6): Vr = Lr/Tr; Substituting for Vr yields: Lr-1 = Lr/Tr; therefore, Tr = Lr2 b) Substituting for Tr into Eq’n (10.7): ar = Lr/Tr2 = Lr/(Lr2)2; ar = Lr-3 c) Substituting for Tr into Eq’n (10.8): Qr = Lr3/Tr = Lr3/Lr2: Qr = Lr 6. A Reynolds number 1:10 scale model of a water supply piping system is to be tested at 20°C to find the head loss in the prototype that carries water at 85°C (ρp = 969 kg/m3; μ p = 3.35 x 10-4 N∙s/m2). The prototype is designed to carry 5.0 m3/sec. Determine the model discharge.

Ans. Given: Lr = 10, μr = μp/μm = (3.35 x 10-4)/(1.00 x 10-3) = 0.335; and ρr = ρp/ρm = 969/998 = 0.971 From Equation (10.19): (ρrLrVr)/μr = 1 (0.971)(10)(Vr)/(0.335) = 1; Vr = 0.0345 = Vp/Vm; Since Q = AV; dimensionally, Qr = (Lr)2(Vr) or Qr = (10)2(0.0345) = 3.45. Thus, Qr = Qp/Qm or Qm = Qp/Qr = 5.00/3.45 = 1.45 m3/sec

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7. An overflow spillway with a 100-ft crest is designed to discharge 12,700 cfs. A 1:20 model of a portion of the cross section of the dam is built in a laboratory flume 1 ft wide. Calculate the required laboratory flow rate assuming viscosity and surface tension effects are negligible (i.e., inertial and gravity forces dominate).

Ans. The unit flow rate for the prototype is: qp = Qp/b = 12,700/100 = 127 cfs/ft. Also, qr = Lr2/Tr = Lr2/Lr1/2 = Lr3/2, since Tr = Lr1/2 from Table 10.3. Also, qm = qp/qr qm = qp/Lr3/2 = 127/(20)3/2 = 1.42 cfs/ft. (i.e., the flow rate for the 1 ft wide flume).

8. A 50-m-wide overflow spillway will carry a design discharge of 1,200 m3/sec. The spillway operation is studied on a 1:50 scale model in a hydraulic laboratory assuming inertial and gravity forces are dominant. Determine the following prototype characteristics: (a) The velocity if the spillway toe velocity in the model is 3.54 m/sec. (b) The depth at the bottom (toe) of the prototype spillway. (c) The Froude number at the bottom (toe) of the prototype spillway. (d) The energy dissipation (in kilowatts) if a hydraulic jump occurs with y2 = 10.6 m. (e) The energy removal efficiency of the hydraulic jump (energy dissipater)?

Ans. a) From Eq’n 10.6 and Table 10.3: Vp = Vm∙Vr = Vm∙Lr1/2 = 3.54(50)1/2 = 25.0 m/sec b) Vp = Qp/Ap; 25.0 = 1200/[(50)(yp)]; yp = 0.960 m c) Nf = (Vp)/(gp∙yp)1/2 = (25.0)/(9.81∙0.960)1/2 = 8.15 d) From Eq’ns (6.20) and (5.4), ΔE = (y2 – y1)3/(4y1y2) = (10.6 – 0.96)3/(4∙0.96∙10.6) ΔE = 22.0 m (headloss). P = γQHP = (9.79)(1200)(22.0) = 2.58 x 105 kW e) Based on Eq’n (6.8), the approach energy: E = V2/2g + y = (25.0)2/2(9.81) + 0.960; E = 32.8 m. The energy removal efficiency: e = ΔE/E = 22.0/32.8 = 0.671 (67.1%) 9. Determine the scale ratios based on the Weber Number Law for a) velocity, b) time, c) acceleration, d) discharge, e) force, and f) power. Assume the same liquid is used in the model and the prototype.

Ans. The Weber Number Law (for ρr = 1, σr = 1) is (ρrVr2Lr)/σr = Vr2Lr = 1. Therefore, a) Vr = 1/Lr1/2 b) Eq’n (10.6): Vr = Lr/Tr; 1/Lr1/2 = Lr/Tr; Tr = Lr3/2 c) Eq’n (10.7): ar = Lr/Tr2 = Lr/(Lr3/2)2; ar = 1/(Lr)2 d) Eq’n (10.8): Qr = Lr3/Tr = Lr3/Lr3/2 = Lr3/2 e) Eq’n (10.13): Fr = ρrLr4Tr-2 = (1)(Lr4)(Lr3/2)-2 = Lr f) Eq’n (10.16): Pr = FrLrTr-1 = (Lr)(Lr)(Lr3/2)-1 = Lr1/2

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10. A 1:300 scale model is constructed to study the discharge-depth relationship in a river reach with Manning’s coefficient n = 0.031. If the model discharges 52ℓ/sec and has Manning’s coefficient n = 0.033, determine the appropriate vertical scale ratio and the flowrate for the prototype. (Assume the channel has a large width-to-depth ratio, and the hydraulic radius is roughly equal to the water depth.)

Ans. For a large width to depth ratio, Rhr = Yr. Also, since gravitational forces dominate, we have Nf = Vr/[gr1/2∙Rhr1/2] = 1.00; Vr = Rhr1/2 = Yr1/2.

Using Manning’s eq’n: Vr = Vp/Vm = (1/nr)Rhr2/3Sr1/2; but since Sr = Yr/Xr, Vr = Yr1/2, and Rhr = Yr; Yr1/2 = (1/nr)Yr2/3(Yr/Xr)1/2. Rearranging and substituting (w/nr = 0.031/0.033 = 0.939) yields: Yr = (nr∙Xr1/2)3/2 = [(0.939)(300)1/2]3/2 = 65.6 Since Qr = Ar∙Vr = (Xr∙Yr)(Yr1/2) = (Xr)(Yr3/2) Qr = (300)(65.6)3/2 = 1.59 x 105. Therefore, Qp = Qm∙Qr = (0.052)(1.59 x 105) = 8.27 x 103 m3/sec 11. A model study is proposed for a barge channel (np = 0.03) that has a sedimentation problem.

The channel conveys a flow rate of 10,600 cfs. The following model parameters have been established: a vertical scale ratio of 1:65 and a roughness coefficient of nm = 0.02. Determine the horizontal scale ratio, the time ratio, and the model flow. (Assume the channel has a large width-to-depth ratio, and the hydraulic radius is roughly equal to the water depth.)

Ans. For a large width to depth ratio, Rhr = Yr = 65. Since gravitational forces dominate: Nf = Vr/[gr1/2∙Rhr1/2] = 1.00;

Vr = Rhr1/2 = Yr1/2 = (65)1/2.

Using Manning’s eq’n: Vr = Vp/Vm = (1/nr)Rhr2/3Sr1/2; but since Sr = Yr/Xr, Vr = Yr1/2, and Rhr = Yr; substituting yields: Yr1/2 = (1/nr)Yr2/3(Yr/Xr)1/2.

Rearranging and substituting (w/nr = 0.03/0.02 = 1.5) yields: Xr = Yr4/3/nr2 = (65)4/3/(1.5)2 = 116. Also, since Vr = Rhr1/2 = Yr1/2 = (65)1/2 And dimensionally: Vr = Xr/Tr; substituting, Tr = Xr/Vr = 116/(65)1/2 = 14.4.

Also, Qr = Ar∙Vr = (Xr∙Yr)(Yr1/2) = (Xr)(Yr3/2) Qr = (116)(65)3/2 = 60,800. And since, Qr = Qp/Qm;

Qm = 10,600/60,800 = 0.174 cfs

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12. A model is used to study the drag force (Fd) on a sphere as if moves through a viscous fluid.

The drag force is a function of sphere diameter (D), the sphere velocity (V), the viscosity of the fluid (μ), and the fluid density (ρ). Use the Buckingham Pi-theorem (with diameter, velocity, and density as repeating variables) to derive an expression for the drag force.

Ans. Based on the problem statement: Fd = f (D, V, ρ, μ) or f′ (Fd, D, V, ρ, μ) = 0 with n = 5, m = 3. Thus, there are n – m = 2 dimensionless groups, and Ø(Π1,Π2) = 0.

Taking D, V, and ρ as the basic repeating variables, we have Π1 = DaVbρcμd ; Π2 = DwVxρyFdz. From the Π1 group, we have M0L0T0 = La(L/T)b(M/L3)c[M/(L∙T)]d.

Thus, M: 0 = c + d;

L: 0 = a + b - 3c - d;

and T: 0 = -b – d. Hence, c = -d, b = -d, and a = -d. Therefore, Π1 = D-dV-dρ-dμd = [μ/(DVρ)]d = [(DVρ)/μ]-d. From the Π2 group, noting that the units for force from Newton’s 2nd law are M∙L/T2, thus M0L0T0 = Lw(L/T)x(M/L3)y(M∙L/T2)z; yielding, M: 0 = y + z; L: 0 = w + x – 3y + z; and T: 0 = -x – 2z.

Hence, y = -z, x = -2z, and w = -2z. Therefore, Π2 = D-2zV-2zρ-zFdz = [Fd/(D2V2ρ)]z.

The resulting function is Ø(Π1,Π2) = Ø[(DVρ)/μ, Fd/(D2V2ρ)] = 0; which can be restated as: Fd = D2V2ρ ∙ Ø´[(DVρ)/μ] = D2V2ρ ∙ Ø´[NR] 13. Use the Buckingham Pi theorem to develop two Pi terms for weir unit discharge (q) using the following variables: spillway height (h), gravitational acceleration (g), viscosity (μ), head (H), and density (ρ). Use h, g, and μ as repeating and the FLT system of units.

Ans. Based on the problem statement: q = f (h, g, μ, H, ρ) or f′(q, h, g, μ, H, ρ) = 0 which results in n = 6, m = 3. There are n – m = 3 dimensionless groups, and therefore Ø(Π1, Π2, Π3) = 0. Taking h, g, and μ as the repeating variables, we have Π1 = hagbμcqd

Π2 = hegfμgHh

Π3 = higjμkρs Therefore, from the Π1 group,

F0L0T0 = La(L/T2)b[FT/L2]c[L3/TL]d For F: 0 = c; for L: 0 = a + b – 2c + 2d ; and for T: 0 = -2b + c - d;

Hence, c = 0; d = -2b; and a = 3b.

yielding, Π1 = h3bgbμ0q-2b

or Π1 = [gh3/q2]

Π2 group: F0L0T0 = Le(L/T2)f[FT/L2]g[L]h For F: 0 = g; for L: 0 = e + f – 2g + h; and for T: 0 = -2f + g;

Hence, g = 0; f = 0; and e = -h.

yielding, Π2 = h-hg0μ0Hh or Π2 = [H/h]

100

Bonus Points for Students who want to complete other Pi term: For the Π3 group, we need density in FLT units. Based on Newton’s 2nd Law (F = ma), we have, F = (M)(L/T2) or M = F∙T2/L making the density units ρ = (F∙T2/L)/ L3 = F∙T2/L4. Therefore, Π3 group: F0L0T0 = Li(L/T2)j[FT/L2]k[F∙T2/L4]s For F: 0 = k + s; for L: 0 = i + j – 2k – 4s; and for T: 0 = -2j + k + 2s; Hence, k = -s; j = (½)s; and i = (3/2)s; yielding, Π3 = h(3/2)sg(1/2)sμ-sρs or Π3 = [ρg1/2h3/2/μ] which may also be expressed as: Π3 = [ρ(gh3)1/2/μ]

101

TEST QUESTIONS & PROBLEMS - CHAPTER #11 Short Answer Questions 1. Identify the difference between hydrology and hydraulics.

Ans. Hydraulics - a branch of engineering that applies fluid mechanics principles to problems of collection, storage, control, transport, regulation, measurement, and use of water.

Hydrology - a science dealing with properties, distribution, & circulation of the earth’s water.

2. (T or F) Hydraulics is a science dealing with the properties, distribution, and circulation of the earth's water.

Ans. False. This is the definition of hydrology.

3. All of the following are primarily transport phases (rather than storage phases) of the hydrologic cycle except: a) precipitation b) interception c) rivers d) evaporation e) surface runoff Ans. (b) Interception is a storage phase of the hydrologic cycle.

4. Define interception.

Ans. Precipitation that is caught by buildings, trees, or other vegetation.

5. All of the following are primarily storage phases (as opposed to transport phases) of the hydrologic cycle except: a) evaporation b) interception c) oceans d) aquifer e) depression storage Ans. (a) Evaporation is a transport phase of the hydrologic cycle.

6. What happens to precipitation that is infiltrated?

Ans. This water is either held in the soil pores or moves downward to the water table.

7. The two phases of the hydrologic cycle that are easiest to quantify are: a) evaporation b) interception c) precipitation d) infiltration e) runoff Ans. (c) and (e) Precipitation and runoff.

8. Define transpiration.

Ans. Infiltrated water held in soil pores, used by plants, and released back to the atmosphere.

9. (T or F) Because the continuous simulation approach for generating a design runoff is timeconsuming and costly, the single-event design-storm method is widely used in the practice.

Ans. True.

10. Two fundamentally different approaches are used to generate the design runoff from rainfall when runoff records are unavailable. Which of the following are the two approaches used? a) statistical b) single event c) measurement d) continuous simulation e) rainfall-runoff Ans. (b) Single event and (d) continuous simulation.

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11. (T or F) Other terms used for a watershed include catchment and drainage basin.

Ans. True.

12. Define watershed.

Ans. The land area that contributes flow to a hydraulic structure.

13. (T or F) Watershed boundaries on topographic maps will pass through contour lines in a perpendicular fashion.

Ans. True.

14. Identify four of the five components of a design storm.

Ans. Return period, storm duration, total depth (or average intensity), time distribution, and spatial distribution.

15. A design storm is characterized in terms of: a) return period b) total rainfall depth or average rainfall intensity c) storm duration b) time distribution of rainfall e) spatial distribution of rainfall e) all of the above Ans. All of the above or (e) 16. What is the probability of a 20-year storm event being equaled or exceeded next year?

Ans. The probability is 1/20 = 0.05 = 5% chance of occurrence next year or any year.

17. Define the term “return period.”

Ans. Return period is the average number of years between occurrences of a hydrologic event with a certain magnitude or greater.

18. Hyetographs are plots of time vs.: b) streamflow b) rainfall c) infiltration Ans. (e) Rainfall Intensity.

d) runoff

e) rainfall intensity

19. A 10-yr, 30-minute storm has a rainfall intensity of 3.4 in./hr. What rainfall depth is produced?

Ans. Depth = (3.4 in./hr)(0.5 hr) = 1.7 inches 20. Identify two factors that impact the selection of the design return period (i.e., 2-yr, 5-yr, 10-yr, etc.) when designing urban hydraulic structures. .

Ans. The importance of the structure, the cost, the level of protection it provides, and the consequences of its failure.

21. Which of the following hydraulic structures are sized to accommodate a peak volume? a) storm sewers b) culverts c) detention basins d) street gutters e) street inlets Ans. (c) Detention basins.

22. Which of the following hydraulic structures are sized to accommodate a peak flow? a) storm sewers b) culverts c) street gutters d) street inlets e) all of these structures Ans. (e) All of these structures.

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23. (T or F) The Soil Conservation Service (SCS) presented five dimensionless rainfall distributions for different regions of the United States.

Ans. False – there are four different regions.

24. (T or F) The SCS dimensionless rainfall distributions are for 24-hour storms. However, shorter and more intense rainfalls are nested within the 24-hr duration, and therefore these distributions are appropriate for both small and large watersheds.

Ans. True 25. Define the term interception.

Ans. Rain caught by the leaves, other vegetation, and manmade structures like rooftops.

26. (T or F) Rainfall excess or effective rainfall is the part of rainfall that becomes runoff and is calculated as the total rainfall minus abstractions.

Ans. True 27. Identify three types of abstractions or losses from rainfall that do not produce runoff.

Ans. Interception by trees and vegetation, surface depression storage, evaporation, transpiration through plants, and infiltration into the underlying soil.

28. Define the term surface depression storage.

Ans. Rainfall that is retained on the ground as puddles or small isolated wetlands.

29. Which of the following abstractions is the dominant mechanism through which losses from rainfall occurs. a) infiltration b) evaporation c) interception d) surface depression e) transpiration Ans. (a) Infiltration.

30. (T or F) The actual rate of infiltration is always equal to the infiltration capacity of the soil.

Ans. False. The rate of infiltration equals the rainfall rate if it falls below infiltration capacity.

31. (T or F) For the Green and Ampt model, it is assumed that the soil has a uniform degree of saturation (i.e., the fraction of the pore volume containing water) when rainfall begins.

Ans. True 32. Which two watershed characteristics are required to determine the SCS curve number (CN)? a) hydraulic length b) slope c) hydrologic soil group d) land use e) drainage area Ans. (c) and (d).

33. Define the term “rainfall excess.”

Ans. That part of the rainfall that becomes runoff, sometimes called direct runoff or just runoff.

34. (T or F) The theoretical range of the SCS curve number (CN) is 40 to 100.

Ans. False; it is 0 to 100.

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35. Sketch a typical total runoff hydrograph (TRH) and label the base flow and volume of direct surface runoff.

Ans.

36. A plot of stream discharge vs. time is called a: a) hydrograph b) hyetograph c) continuous simulation d) single event simulation Ans. (a) hydrograph.

37. (T or F) The base flow component of a total runoff hydrograph corresponding to a design storm is not generated by that particular storm.

Ans. True 38. Define the term “time of concentration.”

Ans. The time it takes runoff to reach the design point from the hydrologically most remote point in the watershed.

39. Sheet flow occurs when surface runoff moves across the ground toward defined channels.

Identify two surface flow features that produce resistance to sheet flow (Manning’s n values).

Ans. Sheet flow resistance incorporates raindrop impact, surface drag, resistance from obstacles (e.g., grass, stones, litter), erosion, and sediment transport.

40. According to SCS, the time of concentration is composed of which of the following times? a) sheet flow b) shallow concentrated c) channel d) a and c only e) a, b, and c Ans. (e).

41. One popular procedure for determining the time-of-concentration promoted by the NRCS (formerly SCS) breaks down the flow time into three components. Name these components.

Ans. Sheet (overland) flow, shallow concentrated flow, and open-channel flow.

42. Define the term “hydraulic length.”

Ans. The length of the longest flow path in the watershed (from the design point to the hydrologically most remote point in the watershed located on the drainage divide).

43. What is the difference between a one-hour and a two-hour unit hydrograph (i.e., UH1 and UH2)? (Hint: the answer is not one hour!)

104

Ans. A 1-hour unit hydrograph is produced by one inch (or one centimeter) of runoff that occurs over a one hour time period, and a 2-hour unit hydrograph is produced by one inch (or one centimeter) of runoff that occurs over a two hour time period.

44. (T or F) A unit hydrograph is a plot of flow over time at a stream location that results from one inch (or one centimeter) of rainfall over a specified duration.

Ans. False. One inch (or one centimeter) of runoff, not rainfall.

45. The “unit” in the term unit hydrograph (UH) refers to a) one hour storm b) one inch of rainfall c) one inch of runoff d) one hour time increment Ans. (c); one inch of runoff.

46. (T or F) A watershed has an infinite number of unit hydrographs.

Ans. True (i.e., 30-min, unit hydrograph, 1-hr unit hydrograph, 2-hr unit hydrograph, etc.).

47. (T or F) Synthetic unit hydrographs are useful design tools since most small watersheds do not have stream gages.

Ans. True.

48. Define “synthetic” in the term “synthetic unit hydrograph.”

Ans. A unit hydrograph that is derived based on watershed characteristics.

49. (T or F) The SCS time-to-peak is the elapsed time from the beginning of the effective (runoffproducing) rainfall to the time when the peak discharge occurs.

Ans. True.

50. The effects of rainfall on a watershed can be separately accounted for and accumulated. This concept is know as: a) superposition b) time-of-concentration c) linearity d) none of these Ans. (a); superposition 51. What two principles are assumed in using a unit hydrograph to predict a watershed’s response to a future (design) rainfall event?

Ans. Linearity and superposition.

52. (T or F) The total runoff hydrograph (TRH) or streamflow hydrograph produced by a design storm is called the direct runoff hydrograph.

Ans. False – it is the design runoff hydrograph.

53. As a flood wave moves through a reservoir or pond, the a) flood peak is reduced b) flood wave broadens c) flow volume is unchanged d) a, b, & c Ans. (d); all of these things occur 54. Define the term “storage routing.”

Ans. The process of evaluating the changes in a storm hydrograph as it passes through a pond or reservoir. In other words, an outflow hydrograph is developed using the inflow hydrograph, storage characteristics of the site, and the hydraulic properties of the outlet device.

105

55. Show all steps in the proof of the mass balance equation in going from dS/dt = I – O to the following: Ii + Ij + (2S/Δt – O)i = (2S/Δt + O)j Ans. Starting with dS/dt = I – O and going to discrete time steps, we have: ΔS/Δt = Iavg – Oavg which can be restated as: (Sj – Si)/Δt = (Ii + Ij)/2 - (Oi + Oj)/2 Finally, these terms can be rearranged as: Ii + Ij + (2S/Δt – O)i = (2S/Δt + O)j 56. (T or F) The orifice and mass balance equations are often used in elevation-outflow determinations for stormwater management pond outlet devices.

Ans. False – the orifice and weir equations.

57. To perform reservoir (storage) routing computations, all of the following are required except a) inflow hydrograph b) outflow hydrograph c) elevation-storage relationship d) elevation-outflow relationship e) routing time step Ans. (b) The outflow hydrograph is obtained by performing the routing computations.

58. (T or F) The area in between the inflow and outflow hydrographs represents the storage of water in a reservoir or pond.

Ans. True.

59. Peak flows resulting from rainfall events are the only hydrologic requirement to design all of the following hydraulic structures except: a) stormwater inlets b) stormwater pipes c) drainage channels d) culverts e) drainage ponds Ans. (e). The design of stormwater ponds require a full hydrograph, not just the peak flow.

60. Define the C term in the rational equation (Qp = CIA) and its units.

Ans. C is a dimensionless runoff coefficient. Basically it is the ratio of runoff to rainfall, or more precisely the rate of runoff to the rate of rainfall.

61. (T or F) Time of concentration does not equal the storm duration in the rational method.

Ans. False.

62. What information is needed before you can determine the storm intensity from an IDF curve?

Ans. You need a storm duration and a return interval.

63. Which parameter is not needed in solving for peak flow using the rational equation? a) hydrologic soil group b) return interval c) time of concentration d) drainage area Ans. (a). The hydrologic soil group is needed in the SCS procedure.

64. (T or F) The runoff coefficient (C) increases with increasing slope and imperviousness.

Ans. True 65. State the upper and lower limits of the runoff coefficient (C) value in the rational equation and describe the meaning of each.

Ans. The lower limit is zero (no runoff) and the upper limit is 1 (complete runoff).

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66. The runoff coefficient (C) in the rational equation decreases with all of the following except: a) smaller slopes b) larger soil permeability c) less vegetative cover d) less imperviousness Ans. (c). The runoff coefficient increases when the vegetative cover is reduced.

67. Define the “I” term in the rational equation (Qp = CIA) and its units.

Ans. I is the rainfall intensity (in./hr or cm/hr) and it is found from a local IDF curve.

68. Define the term “pavement encroachment criteria.”

Ans. The limiting spread of water in the street before a drainage inlet is needed. .

69. Locations that require the placement of a drainage inlet include all of the following except: a) after an intersection b) in sumps where water collects without an outlet c) upgrade of all bridges d) where the spread criteria is exceeded Ans. (a). Inlets need to be placed before an intersection, not after.

70. (T or F) Inlet location design to limit street spread combines Manning’s equation for gutter capacity and SCS pocedures for the rate of surface runoff.

Ans. False. The rational equation is used to determine the rate of surface runoff.

71. Give three locations that require the placement of a drainage inlet.

Ans. In all sumps, when the curb height is exceeded, along the curb when the pavement encroachment criteria is exceeded, upgrade of all bridges, and prior to an intersection.

72. Give three locations that require the placement of a manhole.

Ans. Manholes are needed at pipe junctions, changes of grade, and changes of alignment.

73. (T or F) Pipe slopes match overlying ground slopes when possible to minimize excavation.

Ans. True.

74. Locations that require the placement of manholes include all of the following except: a) pipe junctions b) changes of grade c) upgrade of intersections d) changes of alignment Ans. (c). Inlets need to be placed upgrade of intersections, but not manholes.

75. Why are pipe sizes not reduced downstream when increased slopes provide adequate capacity?

Ans. Clogging is a concern.

76. (T or F) Peak flow calculations for stormwater inlets are used to size the subsequent pipes.

Ans. False. Inlet location design only accounts for local surface water contributions.

Independent analyses must be performed for pipes accounting for all upstream stormwater.

77. Maximum manhole spacing in straight reaches is usually specified (e.g., 400 ft ). Why? Ans. Cleaning equipment in local areas may limit the distance between manholes.

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Problems 1. Delineate the watershed contributing flow from the tributary into Sugar Creek at the design point (circle) on the topographic map shown below.

Ans.

108

2. Delineate the two watersheds contributing flow to the tributary streams at the design points (marked with X’s) on the topographic map shown below.

Ans.

109

3. Delineate the two watersheds contributing flow to the tributary streams at the design points (marked with X’s) on the topographic map shown below.

Ans.

110

4. Fill in the cells with question marks for the storm hyetograph table below. t (hr) 0

ΔP (in.)

i (in./hr) ?

P (in.)

0.0

t/td

P/PT

0.00

0.00

0.4

0.25

0.4

0.25 2.8

?

0.7

?

?

1.2

0.50

0.69

?

0.75

?

0.8

?

0.88

0.2

1

1.6

1.00

?

P (in.)

0.0

t/td

P/PT

0.00

0.00

0.4

0.25

0.25

1.1

0.50

0.69

1.4

0.75

0.88

1.6

1.00

1.00

Ans. t (hr) 0

ΔP (in.)

i (in./hr) 1.6

0.4

0.25 2.8

0.7

1.2

0.3

0.8

0.2

0.5 0.75 1

5. Fill in the cells with question marks for the storm (synthetic block) hyetograph table below. td (min) 0

iavg (cm/hr)

P (cm) 0.00

10

15.32

2.55

20

?

3.83

30

9.20

?

40

7.67

?

ΔP (cm)

ΔP/Δt (cm/hr)

?

?

t (min) 0

i (cm/hr) 3.08

10 ?

7.68

? ?

0.77

4.61

15.32 30

0.51

3.08

4.61 40

Note: Table continues through the 60 minute storm.

111

Ans. td (min) 0

iavg (cm/hr)

P (cm) 0.00

10

15.32

2.55

20

11.50

3.83

ΔP (cm)

ΔP/Δt (cm/hr)

2.55

15.32

40

9.20 7.67

i (cm/hr) 3.08

10 1.28

30

t (min) 0

7.68

7.68 20

0.77

4.61

0.51

3.08

4.60

15.32 30

5.11

4.61 40

Note: Table continues through the 60 minute storm.

6. The 10-year, 24-hr precipitation for Miami, Florida is 9.0 inches. The table below represents a solution procedure for determining the storm hyetograph. Fill out the rest of the table. t1 (hr) 8.0 9.0 10.0 11.0 11.5

t2 (hr) 9.0 10.0 11.0 11.5 12.0

P1/PT

P2/PT

0.115 0.148 0.189 0.250 0.298

0.148 0.189 0.250 0.298 0.600

P1 (in.)

P2 (in.)

∆P (in.)

i (in./hr)

P2 (in.)

1.332 1.701 2.250 2.682 5.400

∆P (in.)

0.297 0.369 0.549 0.432 2.718

i (in./hr) 0.297 0.369 0.549 0.864 5.436

Ans. The following table provides the solutions. t1 (hr) 8.0 9.0 10.0 11.0 11.5

t2 (hr) 9.0 10.0 11.0 11.5 12.0

P1/PT

P2/PT

0.115 0.148 0.189 0.250 0.298

0.148 0.189 0.250 0.298 0.600

P1 (in.)

1.035 1.332 1.701 2.250 2.682

7. A rural watershed contains a soil with the following properties:  = 0.40, K = 0.12 in./hr, and Pf = 7.87 in. The initial degree of saturation is Si = 0.40 (i.e., 40% saturated). A storm occurs over the watershed with rainfall rates tabulated in column 4 of the table below. Determine the losses due to infiltration using the Green and Ampt method by filling in the table. Also determine the rates of effective rainfall assuming that the losses occur due to infiltration only.

112

1

2

3

4

5

6

7

8

9

10

Time

t1

t2

i.

Z1

fp

f

ΔZ

Z2

ie

Step

(min)

(min)

(in./hr)

(in.)

(in./hr)

(in./hr)

(in.)

(in.)

(in./hr)

1 2

0 20

20 40

1.0 2.0

1

2

3

4

5

6

7

8

9

10

Ans.

Time

t1

t2

i.

Z1

fp

f

ΔZ

Z2

ie

Step

(min)

(min)

(in./hr)

(in.)

(in./hr)

(in./hr)

(in.)

(in.)

(in./hr)

1 2

0 20

20 40

1.0 2.0

0.00 1.39

∞ 0.80

1.00 0.80

1.39 1.11

1.39 2.50

0.00 1.20

8. An 80 hectare agricultural watershed (clay soil) contains 20 hectares of meadows and the remainder is in row crops. Determine the SCS depth of runoff (in cm) from a 5-year storm that produces a total of 10.6 cm of rainfall.

Ans. Clay soil is hydrologic soil group D. For meadows, CN1 = 78 (Table 11.7) For row crops, CN2 = 86 (Table 11.7). The area-weighted average CN is: CN = [(A1)(CN1) + (A2)(CN2)]/(A1+A2) = [(20)(78) + (60)(86)]/(80) = 84 From Eq’n 11.7: S = [25,400 – 254(CN)]/CN = [25,400 – 254(84)]/84 = 48.4 mm From Eq;n 11.5: R = (P – 0.2S)2/(P + 0.8S) = [106 – 0.2(48.4)]2/[106 + 0.8(48.4)] Depth of runoff = R = 64.1 mm = 6.41 cm 9. A 200-acre watershed near Chicago, Illinois includes 169 acres of open space with 90% grass cover and a 31-acre industrial development park. The existing soil has course to fine texture based on sieve tests. Determine the runoff depth (inches) and volume (acre-feet) for the 10year, 24-hour rainfall event.

Ans. B soil based on soil description. Based on Table 11.7, CN = 61 (open space, B soil) and CN = 88 (industrial pard, B soil). The area-weighted composite curve number is: CN = [(169 ac)(61) + (31 ac)(88)]/200 ac = 65 Then based on Figure 11.5 and the location of Chicago, the 10-year, 24-hour rainfall depth is 4.0 inches. From Eq’n 11.6: S = {1000 – 10(65)}/65 = 5.4 in.; and from Eq’n 11.5, R = [P – 0.2(S)]2/[P + 0.8(S)] = [4.0 – 0.2(5.4)]2/[4.0 + 0.8(5.4)] = 1.0 in.

Alternatively, Figure 11.11 yields the same answer. The volume of runoff is: Vol. of R/O = (1.0 in.)(1 ft/12 in.)(200 ac) = 16.7 ac. ft.

113

10. A watershed in a industrial district possesses a CN of 91. Determine the rainfall excess for the first 30 minutes of a 1-hour storm (i.e., fill in the blanks to the table below). t1 (min) 0 15

t2 (min) 15 30

i (in./hr) 2.8 3.6

ΔP (in.)

P1 (in.)

P2 (in.)

R1 (in.)

R2 (in.)

ΔR (in.)

ΔP (in.)

0.70 0.90

P1 (in.)

0.00 0.70

P2 (in.)

0.70 1.60

R1 (in.)

0.00 0.17

R2 (in.)

0.17 0.82

ΔR (in.)

0.17 0.65

Cont.

Ans. t1 (hr) 0.00 0.25

t2 (hr) 0.25 0.50

i (in./hr) 2.8 3.6

Cont.

11. A 40.5-hectare watershed (B soils) has been commercially developed. The longest flow path is 915 meters and the average watershed slope is 2.5%. Determine the time of concentration.

Ans. Applying Equation (11.17): Tc = [2.586∙L0.8(S+1)0.7]/(1140Y0.5) requires the soil moisture storage deficit (S). Thus, from Equation (11.7) S = (25,400/CN) - 254 = (25,400/92) - 254 = 22.1 mm, where the curve number is found in Table 11.7 (commercial land use with B soils). Substituting into Equation (11.17) yields Tc = {2.586(915)0.8[(22.1/25.4)+1]0.7}/(1140(2.5)0.5) = 0.52 hours 12. A new development is planned for a 250-acre watershed with a hydraulic length of 4500 feet.

The predominant land use is woodlands (C soils) with an average land slope of 8%. However, the land use in the watershed is scheduled to change over the next two years to single family residential (1/2-acre lots). Determine the SCS peak discharge after development.

Ans. The time-to-peak requires the time of concentration. Applying Equation (11.16) Tc = [L0.8(S+1)0.7]/(1140Y0.5) Also; S = (1000/CN) - 10 = (1000/80) - 10 = 2.50 in. where the curve number is found in Table 11.7 (residential, ½-acre lots, C soils).

Substituting the hydraulic length and watershed slope into Equation (11.16) yields Tc = [(4500)0.8(2.50+1)0.7]/(1140(8)0.5) = 0.624 hours (37.4 minutes) By applying Equation (11.21): Tp = 0.67Tc = 0.67(0.624) = 0.418 hours (25.1 minutes) The peak discharge for the watershed is calculated from Equation (11.22) as qp = (KpA)/Tp = 484[250 acres (1 sq.mi./640 acres)]/0.418 hrs = 452 cfs/in.

114

13. The 20-minute (1/3 hr) unit hydrograph (UH1/3) and base flow (BF) for an urban watershed are provided in the table below. A 1 hour design storm produces 2 cm of runoff in the first 20 minutes, 3 cm of runoff in the next 20 minutes, and 1 cm of runoff in the final 20 minutes.

Determine the direct runoff hydrograph (DRF) and the total runoff hydrograph (TRH) for the design storm by filling out the table.

Time UH1/3 (min.) (m3/s) 0 0 20 6 40 14 60 30 80 24 100 18 120 12 140 6 160 0 180

DRH (m3/s)

BF (m3/s) 2 2 1 2 3 4 5 5 5 5

TRH (m3/s)

Time UH1/3 (min.) (m3/s) 0 0 20 6 40 14 60 30 80 24 100 18 120 12 140 6 160 0 180

2×UH1/3 3×UH1/3 1×UH1/3 DRH (m3/s) (m3/s) (m3/s) (m3/s) 0 0 12 0 12 28 18 0 46 60 42 6 108 48 90 14 152 36 72 30 138 24 54 24 102 12 36 18 66 0 18 12 30 0 0 6 6

BF (m3/s) 2 2 1 2 3 4 5 5 5 5

TRH (m3/s) 2 14 47 110 155 142 107 71 35 11

Ans.

14. A 20-minute unit hydrograph (UH1/3) for an urban watershed (310 acres) is given below.

Determine the stream flow (TRH) that would result from a rainfall event on the watershed if the runoff in a 40-minute storm is 1.5 inches in the first 20 minutes and 0.5 inches in the next 20 minutes. Assume a base flow of 20 cfs. Also determine the runoff volume in acre feet.

Time (min) Q (cfs)

0 0

10 140

20 320

30 500

40 420

50 280

60 160

70 60

80 0

115

Ans. Time UH1/3

1.5×UH1/3 0.5×UH1/3

DRH

BF

TRH

(cfs) 0 210 480 820 790 670 450 230 80 30

(cfs) 20 20 20 20 20 20 20 20 20 20

(cfs) 20 230 500 840 810 690 470 250 100 50

20 min lag

(min) 0 10 20 30 40 50 60 70 80 90

(cfs) 0 140 320 500 420 280 160 60 0

(cfs) 0 210 480 750 630 420 240 90 0 0

(cfs)

0 70 160 250 210 140 80 30

15. Build a 2S/Δt + O vs. O relationship for a stormwater management pond by filling in the blank spaces of the table below. The pond drains through a 2 ft by 2 ft box culvert that behaves like an orifice (discharge coefficient of 0.6) with the invert elevation on the bottom of the pond (stage = 0.0 feet). Assume a routing interval of 5 minutes and note that 1 ac-ft = 43,560 ft3.

Stage (ft) 0.0 2.0 4.0 6.0 8.0

Outflow (O) (ft3/sec) 0

Storage (S) (ac-ft) 0.0 0.4 1.0 1.6 2.4

2S/Δt (ft3/sec) 0

(2S/Δt)+O (ft3/sec) 0

Ans. The outflow column is found using the orifice equation: Q = CdA(2gh)1/2 where Cd = 0.6, A = 4 ft2, g = 32.2 ft/sec2, and h = (stage - 1 ft) (i.e., the difference between the stage and the middle of the culvert opening).

The remaining columns are just algebraic expressions with Δt = 300 seconds.

Stage (ft) 0.0 2.0 4.0 6.0 8.0

Outflow (O) (ft3/sec) 0 19 33 43 51

Storage (S) (ac-ft) 0.0 0.4 1.0 1.6 2.4

2S/Δt (ft3/sec) 0 116 290 465 697

(2S/Δt)+O (ft3/sec) 0 135 324 508 748

116

16. A portion of a storage routing table is given below (90 minutes into the storm) along with a graph of 2S/t + O vs. Outflow (O) for the reservoir. Fill in the blank spaces (?) in the table.

Time (min) 90 100 110 120

Inflow (Ii)

Inflow (Ij)

(cfs) 50 45

(cfs) 45

(2S/Δt)-O (cfs) 468 ?

?

? ?

30 26

30

554

(2S/Δt)+O Outflow, O (cfs) (cfs) 514 23 ? ?

599 ?

614 30

2S/Δt + O vs. Outflow (O) 35 30

Outflow (O), (cfs)

25 20 15 10 5 0 0

100

200

300

400

500

600

700

(2S/Δt)+O, (cfs)

Ans. The storage routing table is filled in using normal calculation procedures or using back calculations to fill in prior cells based on the information available in the table.

Time (min) 90 100 110 120

Inflow (Ii) (cfs) 50 45

Inflow (Ij) (cfs) 45

(2S/Δt)-O (cfs) 468

(2S/Δt)+O (cfs) 514

Outflow, O (cfs) 23

43

563

43

30 26

511 541

26 29

30

554

599 614

30

117

17. The 2S/t + O vs. Outflow (O) relationship for a reservoir is graphed below. Given an inflow flood hydrograph (in the table below), determine the peak outflow from the reservoir by filling in the missing values in the routing table. Note that the flood occurs before the storage pool of the reservoir is empty since 2S/t + O = 70 m3/sec when the flood begins to enter the reservoir.

2S/Δt + O vs. Outflow (O)

60.0

Outflow (O), (m3/sec)

50.0 40.0 30.0 20.0 10.0 0.0 0

Time (hr) 0:00 2:00 4:00 6:00 8:00 10:00 12:00 14:00

50

100

150 200 250 (2S/Δt)+O, (m3/sec)

Inflow (Ii)

Inflow (Ij)

3

3

(m /sec) 4 20 42 80 66 52 38 16

(m /sec)

(2S/Δt)-O (m3/sec) 58

300

(2S/Δt)+O (m3/sec) 70

350

400

Outflow, O (m3/sec) 6

Ans. The missing values are in bold type in the table below. The peak outflow is 50 m3/sec.

Time (hr) 0:00 2:00 4:00 6:00 8:00 10:00 12:00 14:00

Inflow (Ii)

Inflow (Ij)

3

3

(m /sec) 4 20 42 80 66 52 38 16

(m /sec) 20 42 80 66 52 38 16

(2S/Δt)-O (m3/sec) 58 68 106 174 232 250 244 218

(2S/Δt)+O (m3/sec) 70 82 130 228 320 350 340 298

Outflow, O (m3/sec) 6 7 12 27 44 50 48 40 118

18. Given the following 2S/t + O vs. Outflow (O) graph, compute the outflow hydrograph from the reservoir using the design storm (inflow hydrograph) provided. Identify the peak inflow, peak outflow, and the storage volume required to reduce the peak by this amount.

2S/Δt + O vs. Outflow (O) 30

Outflow (O), (cfs)

25 20 15 10 5 0 0

50

100

150

200

(2S/Δt)+O, (cfs)

The reservoir inflow hydrograph is given below (in the first two columns).

Time (min) 0 20 40 60 80 100 120 140 160

Inflow (Ii)

Inflow (Ij)

(cfs) 2 10 34 56 46 30 22 16 10

(cfs) 10 34 56 46 30 22 16 10

(2S/Δt)-O (cfs) 0 6 28 76 122 138 132 116 94

(2S/Δt)+O (cfs) 12 50 118 178 198 190 170 142

Outflow, O (cfs) 2 3 11 21 28 30 29 27 24

Ans. The storage routing table is shown above. The solution values are in bold type. The peak inflow is 56 cfs and the outflow peak is 30 m3/sec. To determine the storage, note that 2S/t + O = 198 cfs at the time that the peak discharge occurs in the routing table.

Subracting; (2S/t + O) – O = 2S/t; 198 – 30 = 168 cfs = 2S/t. Since t = 20 min.

2S/t = 168 ft3/sec; 2S/(1200 sec) = 168 ft3/sec; Speak = 100,800 ft3.

119

19. Determine the 10-year peak discharge using the rational equation for a concrete parking lot that is 300 feet wide by 600 feet long. The longest flow path is 300 feet of sheet flow across the lot on a slope of 0.5% and 600 feet of shallow concentrated flow along the parking lot’s length in a paved channel on a slope of 1.5%. Use SCS procedures to determine the time of concentration using a 2-year, 24-hr rainfall of 2.4 inches. Use Figure 11.4 for the rainfall intensity in the rational equation.

Ans. Using the SCS sheet flow and shallow concentrated flow equations (11.8 & 11.10) yields: Tt1 = [0.007(0.011300)0.8]/{(2.4)0.5 ∙(0.005)0.4} = 0.0978 hrs (5.87 min) V = 20.3282(0.015)0.5 = 2.49 fps, and Tt2 = 600/[(2.49 ft/sec)(60 sec/min)] = 4.02 min Applying the rational equation (11.27) yields Q10 = CIA = (0.90)(6.0 in./hr)(300 ft)(600 ft)(1 acre/43,560 ft2) = 22.3 cfs where C is found using Table 11.20 (mid-range) and I is obtained from Figure 11.4 (with a storm duration equal to the roughly 10 minute time of concentration; i.e., sheet flow time plus shallow concentrated flow).

20. The paved parking lot depicted in the figure below (left) drains as sheet flow into a concrete side channel that discharges into a stormwater management pond. The travel time across the parking lot as sheet flow is five minutes, and the travel time through the entire drainage channel is two minutes. Determine the 5-year peak discharge for design purposes using the rational equation. Also determine the peak discharge resulting from a 5-year, 5-minute storm. Assume that Figure 11.4 contains the appropriate IDF curves. (Hint: Recall that in the rational method, the time-of-concentration storm yields the highest peak flow because the entire drainage area is contributing flow to the design point (i.e., the stormwater management pond in this case).

Ans. Contributing area for 5-min. storm.

120

Ans. Applying the rational equation (11.27) to determine the peak flow for design yields: Qp = CIA = (0.90)(5.8 in./hr)[(270 ft)(600 ft)](1 acre/43,560 ft2) = 19.4 cfs where C is found using Table 11.20 (mid-range), I is obtained from Figure 11.4 (with a storm duration equal to 7 minutes, the time of concentration), and the full drainage area (A) is contributing flow to the design point. For the 5-year, 5-minute storm: Qp = CIA = (0.90)(6.1 in./hr)[(162 ft + 270 ft)/2](600 ft)(1 acre/43,560 ft2) = 16.3 cfs where C remains the same and I is obtained from Figure 11.4 (with a storm duration equal to 5 minutes). However, the drainage area (A) contributing flow five minutes after the storm began (based on flow travel times) is depicted above. Note that the peak discharge of 16.3 cfs is lower than the 19.4 cfs peak flow for the design storm. Even though the rainfall intensity is less for the 7-minute duration storm, the full parking area is contributing runoff which yields a larger peak flow.

21. A 100-ft wide parking lot drains as sheet flow across its width to a curb and gutter section on one side. Curb opening inlets are to be placed along the gutter (asphalt; n = 0.015) which has a 3/8-inch per foot cross slope and a longitudinal slope of 1.5%. If the sheet flow time is 5.0 minutes and the gutter flow time is negligible, determine how far along the length of the parking lot you should go before placing the first inlet based on a 10-year design storm. The allowable pavement encroachment is 8 feet and the IDF curve in Figure 11.4 applies.

Ans. Based on the cross slope and allowable spread, a 3.0-inch curb depth results. Thus Q = (1.49/n)ARh2/3Se1/2 = (1.49/0.015)(1.00 ft2)(1.00/8.25)2/3(0.015)1/2 = 2.98 cfs Applying the rational equation (with Tc = 5 min., Figure 11.4, & Table 11.20) yields A = Q/(CI) = 2.98 cfs/[(0.90)(7.1 in/hr)] = 0.466 acres (approximately 20,300 ft2) The first inlet is placed approximately 200 feet (i.e., 20,300 ft2/100 ft = 203 ft) from the high end of the parking lot along its length.

102

22. Stormwater collection pipes (concrete; 12-in. min. diameter) are needed for the subdivision shown below. The data for each of the drainage areas (basins) contributing flow to the manholes is provided in the table below, including the inlet time (Ti) and the runoff coefficient (C). In addition, stormwater pipe information is given, including the ground elevation at each manhole. The rainfall intensity-duration relationship for the design storm can be described by i = 18/(td0.50), where i = intensity in in./hr, and td = storm duration = time of concentration in minutes. Fill in the blanks in the stormwater pipe design table.

Basin 1

A Inlet/manh Basin 3

Basin 2

C Basin 1 2 3 4

Area (acres) 2.2 1.8 2.2 1.2

C 0.3 0.4 0.3 0.5

River

D

B Ti (min.)

12 10 13 10

Storm pipe Surface

Basin 4

Stormwater Pipe AB CB BD DR

Length (ft) 200 300 300 200

Upstream Elev. (ft) 24.9 24.4 22.9 22.0

Downstream Elev. (ft) 22.9 22.9 22.0 21.6

Ans. The blanks in the design table are filled in below (in bold and red).

Stormwater Pipe Length (ft) Inlet Time, Ti (min) Time of Concentration, Tc (min) Runoff Coefficient, C R/F Intensity, I (in/hr) Drainage Area, A (acres) Peak Discharge, Qp (cfs) Slope (ft/ft) Required Pipe Diameter, Dr (in.)

Design Pipe Diameter, D (in.)

Full Pipe Area, Af (ft2) Full Pipe Velocity, Vf (ft/sec) Full Pipe Flow, Qf (cfs) Qp/Qf (or Q/Qf) y/D V/Vf Flow Depth, y (in.)

Pipe Velocity, V (ft/sec) Pipe Flow Time (min)

AB 200 12 12 0.3 5.2 2.2 3.4 0.01 11.8 12 0.79 4.55 3.57 0.96 0.78 1.16 9.4 5.28 0.6

CB 300 10 10 0.4 5.7 1.8 4.1 0.005 14.4 15 1.23 3.73 4.58 0.89 0.73 1.14 11.0 4.25 1.2

BD 300 13 13 0.33 5.0 6.2 10.2 0.003 22.3 24 3.14 3.95 12.4 0.82 0.68 1.13 16.3 4.5 1.1

DR 200 10 14.1 0.36 4.8 7.4 12.8 0.002 26.2 30 4.91 3.75 18.4 0.69 0.60 1.09 18.0 4.1 0.8

103

23. Stormwater collection pipes (concrete; 12-in. min. diameter, min. full pipe velocity of 2 ft/sec) are needed for the subdivision shown below. The data for each of the drainage areas (basins) contributing flow to the manholes is provided in the table below, including the inlet time (Ti) and the runoff coefficient (C). In addition, stormwater pipe information is given, including the ground elevation at each manhole. The rainfall intensity-duration relationship for the design storm can be described by i = 18/(td0.50), where i = intensity in in./hr, and td = storm duration = time of concentration in minutes. Fill in the last column in the stormwater pipe design table.

Basin 1

A Inlet/manh Basin 3

Basin 2

C Basin 1 2 3 4

Area (acres) 2.2 1.8 2.2 1.2

C 0.3 0.4 0.3 0.5

River

D

B Ti (min.)

12 14 13 10

Storm pipe Surface

Basin 4

Stormwater Pipe AB CB BD DR

Length (ft) 200 300 300 200

Upstream Elev. (ft) 24.9 23.5 22.9 22.0

Downstream Elev. (ft) 22.9 22.9 22.0 22.0

Ans. The last column of the design table is filled in below (in bold and red).

Stormwater Pipe AB CB BD DR Length (ft) 200 300 300 200 Inlet Time, Ti (min) 12 14 13 10 Time of Concentration, Tc (min) 12 14 15.7 16.8 Runoff Coefficient, C 0.3 0.4 0.33 0.36 R/F Intensity, I (in/hr) 5.2 4.8 4.5 4.4 Drainage Area, A (acres) 2.2 1.8 6.2 7.4 Peak Discharge, Qp (cfs) 3.4 3.5 9.3 11.7 Slope (ft/ft) 0.01 0.002 0.003 0.0005* Required Pipe Diameter, Dr (in.)

11.8 16.1 21.5 32.9 Design Pipe Diameter, D (in.)

12 18 24 36 Full Pipe Area, Af (ft2) 0.79 1.77 3.14 7.07 Full Pipe Velocity, Vf (ft/sec) 4.55 2.67 3.95 2.12 Full Pipe Flow, Qf (cfs) 3.57 4.71 12.4 15.0 Qp/Qf (or Q/Qf) 0.96 0.74 0.75 0.78 y/D 0.78 0.63 0.64 0.67 V/Vf 1.16 1.11 1.12 1.12 Flow Depth, y (in.)

9.4 11.3 15.4 24.1 Pipe Velocity, V (ft/sec) 5.28 2.96 4.4 2.4 Pipe Flow Time (min) 0.6 1.7 1.1 1.4 Note: *Slope was adjusted up from 0% (ground slope) to 0.05% to get a 2 ft/sec full pipe velocity.

104

TEST QUESTIONS & PROBLEMS - CHAPTER #12 Short Answer Questions 1. Identify two changes in a watershed that would affect the hydrologic homogeneity of a data set containing annual maximum stream flows.

Ans. Hydrologic homogeneity would be affected by urbanization, relocated gages, streamflow diversions, and construction of dams and reservoirs during the data collection period.

2. (T or F) An annual maximum series is formed using the highest N values on a data record which is N years long.

Ans. False. This is annual exceedence series.

3. Random hydrologic variables include all of the following except: a) mean sea level b) precipitation c) flood flows d) annual evaporation e) wind speed Ans. (a) Mean sea level is found from random variables (sea level elevation).

4. (T or F) The 50 largest stream flows in a 50-year gage record represents a population.

Ans. False. It represents a sample of all of the largest flows that could occur on that stream.

5. Identify three statistical parameters that are useful in analyzing probability distributions.

Ans. Mean, standard deviation, and skewness.

6. Which of the following probability distributions is rarely used in hydrologic applications? a) Normal b) Log-Normal c) Log-Pearson Type III d) Gumbel Ans. (a) Normal. Most hydrologic random variables are not normally distributed.

7. (T or F) The probability distribution recommended to model annual maximum streamflow series by the Interagency Advisory Committee on Water Data, U.S. Geological Survey is the Log-Pearson Type III distribution.

Ans. True.

8. Define the term return period, also called the recurrence interval.

Ans. Return period is the average number of years between occurrences of a hydrologic event with a certain magnitude or greater.

9. A 3,000 cfs flood has an annual probability of exceedence of 20%. What is its return period?

Ans. Tr = 1/p = 1/0.20 = 5 years.

10. (T or F) Hydrologic risk is the probability that the design discharge will be exceeded once during the service life of the project.

Ans. False. It could be exceeded one time or more during the service life of the project.

11. In the closed form frequency function: xT = m + KT(s), what does xT represent??

Ans. xT is the magnitude of a hydrologic variable corresponding to specified return period, T. 122

12. Frequency analysis may be performed by all of the following distributions except: a) Normal b) Rational c) Log-Pearson Type III d) Gumbel Ans. (b) Rational. The Rational Method is a deterministic hydrologic equation.

13. (T or F) In the Gumbel method, KT is a function of the skew coefficient and the return period.

Ans. False. It is only a function of the return period.

14. (T or F) In the Log Pearson Type III frequency analysis method, KT is a function of the skew coefficient and the return period.

Ans. True.

15. What purpose is fulfilled by the chi-square test as a statistical procedure in hydrology?

Ans. It allows the user to determine the goodness of fit of data to a probability distribution. k

2

( - ) 16. The chi-square test statistic is  =  Oi E i . Define the variable Oi and Ei. Ei i=1 Ans. Oi and Ei are the observed and expected number of data values in the ith class interval.

2

17. (T or F) The width of a confidence interval depends upon the size of the sample and the confidence level trying to be achieved (i.e., 90%).

Ans. True.

18. Commercial graph paper is available for all of the following probability distributions except: b) Normal b) Log-Normal c) Log-Pearson Type III d) Gumbel Ans. (c) Log-Pearson Type III. For this distribution, there would have to be a different graph paper for each different value of the skew coefficient, which makes it impractical.

19. Hydrologic data is often plotted on specially-designed probability paper. What is plotted on the ordinate and what is plotted on the abscissa?

Ans. The ordinate usually represents the value of the hydrologic variable, and the abscissa represents the return period T, or the exceedence probability, p.

20. (T or F) Normally, the Log-Pearson III distribution is used for rainfall frequency analysis (i.e., the development of intensity-duration-frequency, or IDF, curves).

Ans. False. The Gumbel distribution is often used.

123

Problems 1. The table below is used to evaluate the statistical parameters of an annual peak discharge (m3/sec) data record. Fill in the blanks in the table and evaluate the mean, standard deviation, and skewness of the original data set.

Ans. The blanks in the table are filled in and given below (in bold and red). The statistical parameters are evaluated and shown below the table.

m=

Year

Qi

Qi - m

(Qi - m)2

(Qi - m)3

2006 2007 2008 2009 2010 2011 2012 2013 2014 2015

114 294 113 101 222 276 215 250 147 178

-7.70E+01 1.03E+02 -7.80E+01 -9.00E+01 3.10E+01 8.50E+01 2.40E+01 5.90E+01 -4.40E+01 -1.30E+01

5.93E+03 1.06E+04 6.08E+03 8.10E+03 9.61E+02 7.23E+03 5.76E+02 3.48E+03 1.94E+03 1.69E+02

-4.57E+05 1.09E+06 -4.75E+05 -7.29E+05 2.98E+04 6.14E+05 1.38E+04 2.05E+05 -8.52E+04 -2.20E+03

Sum

1910

0.00E+00

4.51E+04

2.08E+05

1 N  Q = (1,910)/10 = 191 m3/sec N i=1 i 1/2

 1 N  s=  (Qi - m) 2  = [45,100/9]1/2 = 70.8 m3/sec   N - 1 i=1  N

N  (Qi - m) G=

3

i=1

( N - 1) ( N - 2) s3

= (10)(2.08E+05)/[(9)(8)(70.8)3] = 0.0814

2. The table below is used to evaluate the statistical parameters of an annual flood flow record.

Fill in the blanks in the table and evaluate the mean, standard deviation, and skewness of the log transformed data set.

124

Ans. The blanks in the table are filled in and given below (in bold and red). The statistical parameters are solved for below the table.

ml =

Year

Qi

log Qi

(logQi-ml)

(logQi-ml)2

(logQi-ml)3

2006 2007 2008 2009 2010 2011 2012 2013 2014 2015

114 294 113 101 222 276 215 250 147 178

2.06E+00 2.47E+00 2.05E+00 2.00E+00 2.35E+00 2.44E+00 2.33E+00 2.40E+00 2.17E+00 2.25E+00

-1.95E-01 2.17E-01 -1.99E-01 -2.47E-01 9.45E-02 1.89E-01 8.06E-02 1.46E-01 -8.45E-02 -1.38E-03

3.80E-02 4.69E-02 3.95E-02 6.12E-02 8.94E-03 3.58E-02 6.50E-03 2.14E-02 7.14E-03 1.91E-06

-7.40E-03 1.02E-02 -7.85E-03 -1.52E-02 8.45E-04 6.76E-03 5.24E-04 3.12E-03 -6.03E-04 -2.65E-09

Sum

1910

2.25E+01

-2.22E-15

2.65E-01

-9.60E-03

1 N  log Qi = (22.5)/10 = 2.25 N i=1 1/2

 1 N  sl =  (log Qi - ml ) 2  = [0.265/9]1/2 = 0.172   N - 1 i=1  N

N  (log Pi - ml ) Gl =

3

i=1

( N - 1) ( N - 2) sl

3

= (10)(-9.60E-03)/[(9)(8)(0.172)3] = -0.262

3. A riparian property owner has been informed that the first floor of her garage is at the 5-year flood elevation. She would like to determine the flood damage risk of leaving old furniture (which she hopes to sell within a few years) in the garage. What is the probability that the garage will be flooded next year? What is the probability that it is flooded at least once in the next three years? What is probability that it will not be flooded in the next three years?

Ans. The probability of being flooded next year (Eq’n 12.22), p = 1/T = 1/5 = 0.20 = 20.0 % The probability of being flooded at least once in the next three years (Equation 12.23) is, R = 1 – (1 – 0.20)3 = 0.488 = 48.8 % The probability of not being flooded in 3 years: 1 – R = 1 – 0.488 = 0.512 = 51.2%

125

4. A 10-year record of annual peak discharges (cfs) are provided below for the Wabash River.

The statistical parameters for this annual series are: m = 191 cfs, s = 70.8 cfs, and G = 0.0814.

Determine the 10-year flood assuming the data fits a normal distribution. Also determine the return interval for the 2005 flood (276 cfs) assuming the data fits a Gumbel distribution.

Year Qi

2000 114

2001 294

2002 113

2003 101

2004 222

2005 276

2006 215

2007 250

2008 147

2009 178

Ans. Normal: K10 = 1.282 from Table 12.2 for p = 0.10 (T = 10 years). Then from Eq’n (12.24) Q10 = m + K10(s) = 191 + 1.282(70.8) = 282 cfs Gumbel: Using Eq’n (12.24) with QT = 276 cfs QT = m + KT(s); 276 = 191 + KT (70.8); KT = 1.20; from Table 12.2, we can see that this is a little less than a 10-year flood, T ≈ 9 years.

5. A 10-year record of annual peak discharges (cfs) are provided below for the Wabash River.

The statistical parameters for this annual series are: ml = 2.25, sl = 0.172, and Gl = -0.262.

Determine the 10-year flood assuming the data fits a Log-Pearson III distribution. Also determine the return interval for the 2005 flood (276 cfs) assuming the data fits a Log-Normal distribution.

Year Qi

2000 114

2001 294

2002 113

2003 101

2004 222

2005 276

2006 215

2007 250

2008 147

2009 178

Ans. Log-Pearson Type III: Use Equations (12.31a), (12.29a), (12.28), and (12.30). k = Gl/6 = -0.262/6 = -0.0437; w = [ln T2]1/2 = [ln 102]1/2 = 2.15

z = 2.15 -

2.515517+ 0.802853 2.15 + 0.010328 (2.15) 2 = 1.29 1+ 1.432788 2.15 + 0.189269 (2.15) 2 + 0.001308 (2.15)3

KT = z + (z2 – 1)k + (z3 – 6z)(k2/3) – (z2 – 1)k3 + zk4 + k5/3 KT = 1.29 + (1.292 – 1)(-0.0437) + (1.293 – 6∙1.29){(-0.0437)2/3) – (1.292 – 1)(-0.0437)3 + 1.29∙(-0.0437)4 + (-0.0437)5/3 = 1.29; Now from Equation (12.25) log Q10 = ml + KT(sl) = 2.25 + 1.29(0.172) = 2.47; therefore Q10 = 295 cfs Log-Normal: Use Eq’n (12.24) with QT = 276 cfs log QT = ml + KT(sl); log (276) = 2.25 + KT (0.172); KT = 1.11; From Table 12.2, we can this is a little less than a 10-year flood, T ≈ 8 years.

126

6. Annual maximum discharges (39 years) on a river of interest are examined to determine the goodness of fit to the Gumbel distribution. The sample statistics are m = 9,820 cfs and s = 4,660 cfs. Data to determine the goodness of fit test are shown in the table below. Fill in the blanks of the table if the significance level of α = 0.50 is used. Use five class intervals (k = 5).

Ans. The blanks in the table are filled in and given below (in bold and red). The exceedence probabilities of p = 0.8, 0.6, 0.4, and 0.2 divide the class intervals with frequency factors obtained from Table 12.2. The discharges are determined, by using Equation (12.24) p= KT = Class Interval

0.8 -0.821

0.6 -0.382

0.4 0.074

Exceedence Probability Limits

Discharge Limits (cfs)

i

Higher

Lower

Lower

Upper

1

1

0.8

0

2

0.8

0.6

3

0.6

4 5

0.2 0.719 Ei

Oi

(OiEi)2/Ei

5,994

7.8

9

0.185

5,994

8,040

7.8

9

0.185

0.4

8,040

10,165

7.8

8

0.005

0.4

0.2

10,165

13,171

7.8

5

1.005

0.2

0

13,171

Infinity

7.8

8

0.005

Totals

39

39

1.385

7. A forensic engineer is studying the past history of flooding in Wisconsin over 20 year periods.

The annual maximum flood discharges (Qi in m3/sec) for the Wolf River (tributary to Lake Michigan) at New London, Wisconsin, from 1950 to 1969 yield the following statistical parameters: m = 214 m3/sec and s = 94.6 m3/sec. Determine the 90-percent confidence limits for the 25-year flood discharge assuming the data fits the Gumbel distribution.

Ans. Using Table 12.2, K25 = 2.044 (Gumbel distribution). Also, when a 90-percent confidence level is used, z = 1.645 for β = 0.90. From Eq’ns (12.40), (12.41), (12.38), and (12.39) a =1 -

1.6452 z2 =1= 0.9288; 2 ( N - 1) 2 (20 - 1)

b = K T2 -

1.6452 z2 = 2.044 2 = 4.043; N 20

2 K - K T2 - ab K T + K T2 - ab 2.044 + 2.044 - (0.9288)(4.043) = 2.90; K 25L = T = 1.50 = K 25U = a a 0.9288

Then using Eq’ns (12.34) and (12.35), the Gumbel distribution confidence limits are U25 = m + K25U(s) = 214 + 2.90(94.6) = 488 m3/sec L25 = m + K25L(s) = 214 + 1.50(94.6) = 356 m3/sec

127

8. Annual maximum discharges (39 years) on a river of interest are examined to determine the 90% confidence limits for the 1.25-, 2-, 10-, 25-, 50-, 100- and 200-year peak discharges. The sample statistics are ml = 3.95 and sl = 0.197. If the Log-Normal distribution is shown to fit the data, fill in the blanks of the computational table below.

Ans. The blanks in the table are filled in and given below (in bold and red). This is actually the solution to Example 12.8 in the textbook.

1

2

3

4

5

6

7

8

9

10

11

12

T

KT

log QT

QT

a

b

KTU

KTL

log UT

log LT

UT

LT

1.25

-0.841

3.78

6.09E+03

0.964

0.638

-0.557

-1.187

3.840

3.716

6.92E+03

5.20E+03

2

0

3.95

8.91E+03

0.964

-0.069

0.268

-0.268

4.003

3.897

1.01E+04

7.89E+03

10

1.282

4.20

1.59E+04

0.964

1.574

1.697

0.962

4.284

4.140

1.92E+04

1.38E+04

25

1.751

4.29

1.97E+04

0.964

2.997

2.251

1.381

4.393

4.222

2.47E+04

1.67E+04

50

2.054

4.35

2.26E+04

0.964

4.150

2.613

1.647

4.465

4.274

2.92E+04

1.88E+04

100

2.327

4.41

2.56E+04

0.964

5.346

2.941

1.884

4.529

4.321

3.38E+04

2.10E+04

200

2.576

4.46

2.87E+04

0.964

6.566

3.242

2.100

4.589

4.364

3.88E+04

2.31E+04

9. Annual precipitation measurements (Pi in inches) at location X for a 20-year period are listed below. The distribution is Normal based on a chi-square test. If the data were to be plotted on Normal distribution (probability) paper, determine plotting positions (exceedence probability and return period) for two of the rainfall depths: 47.6 in. and 35.8 in. In addition, the mean, standard deviation, and the skew coefficient for this annual series were computed to be: m = 40.0 in., s = 3.50 in., and G = 0.296. Determine the 2-, 5-, and 20-year rainfall depths in order to draw the theoretical (straight line) probability distribution on Normal distribution paper.

Year Pi

1989 44.2

1990 47.6

1991 38.5

1992 35.8

1993 40.2

1994 41.2

1995 38.8

1996 39.7

1997 40.5

1998 42.5

Year Pi

1999 39.2

2000 38.3

2001 46.1

2002 33.1

2003 35.0

2004 39.3

2005 42.0

2006 41.7

2007 37.7

2008 38.6

Ans. The rank, r, of the two precipitation depths is determined and the exceedence probability, p, and the return period, T, are found using Equations (12.43) and (12.44): For P = 47.6 in., r = 1; p = r/(N + 1) = 1/(20 + 1) = 0.0476; T = 1/p = 21.0 years For P = 35.8 in., r = 18; p = r/(N + 1) = 18/(20 + 1) = 0.857; T = 1/p = 1.17 years The theoretical probability distribution is found using Equation (12.24) and Table 12.2: P2 = m + K2(s) = 40.0 + 0.00(3.50) = 40.0 in.

P5 = m + K10(s) = 40.0 + 0.841(3.50) = 42.9 in.

P20 = m + K25(s) = 40.0 + 1.645(3.50) = 45.8 in.

128