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Solutions Manual to accompany

Applied Fluid Mechanics

Eighth Edition Robert L. Mott Joseph A. Untener

Table of Contents

1. The Nature of Fluids and the Study of Fluid Mechanics

2. Viscosity of Fluids

3. Pressure Measurement

4. Forces Due to Static Fluids

5. Buoyancy and Stability

6. Flow of Fluids and Bernoulli’s Equation

7. General Energy Equation

8. Reynolds Number, Laminar Flow, Turbulent Flow, and Energy Losses Due to Friction

107

9. Velocity Profiles for Circular Sections and Flow in Noncircular Sections

129

10. Minor Losses

145

11. Series Pipe Line Systems

160

12. Parallel and Branching Pipeline Systems

212

13. Pump Selection and Application

239

14. Open-Channel Flow

245

15. Flow Measurement

262

16. Forces due to Fluid in Motion

267

17. Drag & Lift

278

18. Fans, Blowers, Compressors, & the Flow of Gases

287

19. Flow of Air in Ducts

296

CHAPTER ONE THE NATURE OF FLUIDS AND THE STUDY OF FLUID MECHANICS Conversion factors 1.1

1750 mm(1m/103mm)=1.75m

1.2

1800 mm 2 [1m 2 / (103 mm) 2 ]  1.8 ×103 m 2

1.3

3.65  103 mm3 [1m 3 / (103 mm)3 ]  3.65 ×106 m 3

1.4

2.05 m 2 [(103 mm)2 / m 2 ]  2.05 ×106 mm 2

1.5

0.391 m3[(103 mm)3 / m3 ]  391×106 mm 3

1.6

55.0 gal(0.00379 m3 /gal)= 0.208 m 3

1.7

80km 103 m 1h    22.2 m / s h km 3600s

1.8

25.3 ft(0.3048 m/ft) = 7.71 m

1.9

1.86 mi(1.609 km/mi)(103 m/km) = 2993 m

1.10

8.65 in(25.4 mm/in) = 220 mm

1.11

3570 ft(0.3048 m/ft) = 1088 m

1.12

560 ft 3 (0.0283 m3 / ft 3 )  15.85 m 3

1.13

6250 cm 3[1m 3 / (100 cm)3 ]  6.25 ×103 m 3

1.14

8.45 L(1 m3 /1000 L) = 8.45×103 m3

1.15

6.0 ft/s(0.3048 m/ft) = 1.83 m / s

1.16

2500 ft 3 0.0283 m3 1min    1.18 m 3 s 3 min ft 60 s

Consistent units in an equation 1.17

s 0.60 km 103 m   56.6 m s υ  t 10.6 s km

The Nature of Fluids

1.18

υ

s 1.50 km 3600 s    871 km /h t 6.2 s h

1.19

υ

s 1000 ft 1 mi 3600 s     45.5 mi /h t 15 s 5280 ft h

1.20

υ

s 1.0 mi 3600 s    632 mi /h t 5.7 s h

1.21

a

2 s (2)(3.2 km) 103 m 1min 2     8.05×102 m /s 2 2 2 2 t (4.7 min) km (60s)

1.22

t

2s (2)(13m)   1.63 s a 9.18 m/s 2

1.23

2s (2)(3.2 km) 103 m 1 ft 1min 2 ft     0.264 2 a 2  2 2 t (4.7 min) km 0.3048 m (60s) s

1.24

t

1.25

mυ2 (15 kg)(1.2 m s)2 kg . m 2 KE    10.8  10.8N  m 2 2 s2

1.26

mυ 2 (3600 kg)  16 km  (103 m) 2 1 h2 kg  m KE       35.6×103 2  2 2  h  2 2 km (3600 s) s

2s (2)(53in) 1 ft    0.524 s a 32.2 ft/s 2 12 in

KE = 35.6 kN  m 2

1.27

mυ2 75 kg  6.85 m  kg  m    1.76  103 2  1.76 kN  m KE    s  2 2 s

1.28

2( KE ) (2)(38.6 N m)  h  1 kg  m (3600s) 2 1 km 2      m  31.5 km  υ2 1 s2  N h2 (103 m) 2

m

(2)(38.6)(3600) 2 kg = 1.008 kg (31.5) 2 (103 )2

1.29

m

2( KE ) (2)(94.6 m N m) 103 N 1 kg  m 103 g    2   37.4 g υ2 (2.25 m/s) 2 mN s N kg

1.30

2( KE ) 2(15 N m) 1 kg  m/s 2 υ    1.58 m / s m 12 kg N

Chapter 1

The Nature of Fluids

The definition of pressure 1.43

p  F /A  2500 lb/[π(2.00 in) 2 /4]  796 lb /in 2  796 psi

1.44

p  F /A  6500 lb/[π(1.50in)2 /4]  3678 psi

1.45

p

1.46

F 38.8  103 N (103 mm) 2 N p    19.8  106 2  19.8 MPa 2 2 A  (50.0 mm) 4 m m

1.47

p

F 6000 lb   119 psi A  (8.0in) 2 / 4

1.48

p

F 1800 lb   3667 psi A  (2.50 in) 2 / 4

1.49

F  pA 

1.50

F  pA  (6000 lb/ in 2 )  [2.00 in]2 / 4  18850 lb

1.51

p

20.5  106 N  (50 mm)2 1 m2    40.25 kN m2 4 (103 mm) 2



D

1.52

F 14.0 kN 103 N (103 mm) 2 N     3.17  106 2  3.17 MPa 2 2 A  (75 mm) / 4 kN m m



F F 4F 4F   : Then D = 2 2 p A D / 4 D 4(20000 lb)  2.26 in  (5000 lb/ in 2 )

4F 4(30  103 N) D   50.5  103 m  50.5 mm 6 2 p  (15.0  10 N/ m )

Chapter 1

The Nature of Fluids

Bulk modulus 1.57

p   E (V / V )  130000 psi(0.01)  1300 psi p  896 MPa(  0.01) = 8.96 MPa

1.58

p   E (V / V )  3.59  106 psi(  0.01) = 35900 psi p  24750 MPa(  0.01) = 247.5 MPa

1.59

p   E (V / V )  189000 psi(  0.01) = 1890 psi p  1303 MPa(  0.01) = 13.03 MPa

1.60

V / V  0.01; V  0.01 V  0.01 AL Assume area of cylinder does not change. V  A(L)  0.01AL Then L  0.01 L  0.01(12.00 in)  0.120 in

1.61

V  p 3000 psi    0.0159  1.59% V E 189000 psi

1.62

V 20.0 MPa   0.0153  1.53% V 1303 MPa

1.63

Stiffness = Force/Change in Length = F/ΔL P  pV  V /V V But p = F /A;V  AL; V   A(L)

Bulk Modulus = E =

E

F AL FL   A  A(L) A(L)

F EA 189000 lb π (0.5 in) 2    884 lb /in (L) L in 2 (42 in)4 1.64

F EA 189000 lb π (0.5in) 2    3711 lb /in (L) L in 2 (10.0 in)(4)

4.2 times higher

1.65

F EA 189000 lb π (2.00 in) 2    14137 lb /in (L) L in 2 (42.0 in)(4)

16 times higher

1.66

Use large diameter cylinder and short storkes.

Force and mass 1.67

m

w 810 N 1 kg  m/s 2    82.6 kg g 9.18 m/ s 2 N

w 1.85  103 N 1 kg  m/s 2    189 kg 9.18 m/ s 2 N g

1.68

m

1.69

w  mg  825 kg  9.81 m/s 2  8093 kg  m/s 2  8093 N

1.70

w  mg  450 g 

1.71

w 7.8 lb lb  s 2 m   0.242  0.242 slugs g 32.2 ft/s 2 ft

1.72

m

1.73

1 lb  s 2 /ft w  mg  1.58 slugs  32.2 ft/ s   50.9 lb slug

1.74

w  mg  0.258 slugs  32.2 ft/ s 2 

1.75

1.76

1 kg  9.81 m/s 2  4.41 kg  m/s 2  4.41 N 103 g

w 42.0 lb   1.304 slugs g 32.2 ft/s 2 2

1lb  s 2 /ft  8.31 lb slug

w 160 lb   4.97 slugs g 32.2 ft/s 2 w  160 lb  4.448 N/lb = 712 N m = 4.97 slugs  14.59 kg/slug = 72.5 kg m

w 1.00 lb   0.0311 slugs g 32.2 ft/s 2 w  0.0311 slugs  14.59 kg/slug = 0.453 Kg

m

m = 1.00 lb  4.448 N/lb = 4.448 N 1.77

F  w  mg  1000 kg  9.81 m/s 2  9810 kg  m/s 2  9810 N

1.78

F  9810 N  1.0 lb/4.448 N = 2205 lb

1.79

(Variable Answer) See problem 1.75 for method.

Density, specific weight, and specific gravity

1.80

γ B  (sg) B γ w  (0.876)(9.81 kN/m3 )  8.59 kN /m 3 ρ B  (sg) B ρ w  (0.876)(1000 kg/m3 )  876 kg /m 3

1.81

ρ=

γ 12.02 N s2 1 kg  m/s 2     1.225 kg /m 3 3 g m 9.81 m N

The Nature of Fluids

1N  19.27 N /m 3 2 1 kg  m/s

1.82

γ =  g  1.964 kg/m3  9.81 m/s 2 

1.83

sg =

γo 8.860 kN/m3   0.903 at 5o C o 3 γ w @ 4 C 9.81 kN/m

sg =

γo 8.483 kN/m3   0.865 at 50o C o 3 γ w @ 4 C 9.81 kN/m

w w 3.50 kN  0.0268 m 3 ;V   3 V γ 130.4 kN/m

1.84

γ=

1.85

V  AL  πD 2 L / 4  π (0.150 m) 2 (0.100 m) / 4  1.767  103 m3

ρo 

m 1.56 kg   883 kg /m 3 3 3 V 1.767 10 m

1N 103 N kg  8.66  3  8.66 3 γ o  ρo g  883 kg/m  9.81 m/s  2 1 kg  m/s m m 3

sg = ρo / ρw @ 4o C = 883 kg/m3 /1000 kg/m3  0.883 1.86

γ = (sg)(γw @ 4o C)=1.258(9.81 kN/m3 ) = 12.34kN/m3  w / V

w  γV  (12.34 kN/ m3 )(0.50 m3 )  6.17 kN w 6.17 kN 103 N 1 kg  m/s 2 m     629 kg g 9.18 m/s 2 kN N 1.87

1.88

w  γV  (sg)(γ w )(V )  (0.68)(9.81 kN/ m3 )(0.095 m 3 )  0.634 kN  634 N

 1N  γ  ρg = (1200 kg/m 3 )(9.81m/s 2 )   11.77 kN /m 3 2  kg  m/s   3 ρ 1200 kg/m sg =   1.20 ρw @ 4 C 1000 kg/m 3

w 32.0 N 1 kN   3  3.95 × 103 m 3 3 γ (0.826)(9.81 kN/m ) 10 N

1.89

V

1.90

γ  ρg 

1080 kg 9.81 m 1N 1 kN    3  10.59 kN /m 3 3 2 2 m s 1kg  m/s 10 N

1080 kg/m3 sg = ρ /ρ w   1.08 1000 kg/m3 1.91

ρ  (sg)( ρw )  (0.789)(1000 kg/m 3 )  789 kg /m 3 γ  (sg)(γ w )  (0.789)(9.81 kN/m 3 )  7.74 kN /m 3

Chapter 1

wo  35.4 N  2.25 N = 33.15 N

1.92

Vo  Ad  (πD 2 /4)(d )  π (.150m) 2 (.20 m)/4  3.53 103 m3 w 33.15 N   9.38  103 N/m3  9.38 kN /m 3 3 3 V 3.53 10 m γ 9.38 kN/m3 sg = o   0.956 γ w 9.81 kN/m3 γo 

V  Ad  (πD 2 /4)(d )  π (10 m) 2 (6.75 m)/4  530.1 m 3

1.93

w  γV  (0.68)(9.81 kN/m 3 )(530.1 m 3 )  3.536 103 kN = 3.536 MN m  ρV  (0.68)(1000 kg/m 3 )(530.1 m 3 )  360.5 103 kg = 360.5 Mg

wcastor oil  γ co  Vco  (9.42 kN/m3 )(0.03 m3 )  0.283 kN

1.94

w 0.283 kN  Vm   2.13×103 m 3 γ m (13.54)(9.81 kN/m3 ) 1.95

w  γV  (2.32)(9.81 kN/m3 )(1.42  104 m3 )  3.23  103 kN = 3.23 N

1.96

γ  (sg)(γ w )  0.876(62.4 lb/ft 3 )  54.7 lb /ft 3 ρ  (sg)( ρw )  0.876(1.94 slugs/ft 3 )  1.70 slugs /ft 3

γ 0.0765 lb/ft 3 1slug    2.38×10-3 slugs /ft 3 2 2 g 32.2 ft/s 1 lb  s /ft

1.97

ρ

1.98

1 lb  s 2 /ft γ  ρg  0.00381 slug/ft (32.2 ft/s )  0.1227 lb /ft 3 slug

1.99

sg = γ o / (γ w @ 4 C)=56.4 lb/ft 3 / 62.4 lb/ft 3  0.904 at 40o F

sg = γ o / (γ w @ 4 C)  54.0 lb/ft 3 / 62.4 lb/ft 3  0.865 at 120o F 1.100

V  w / γ  500 lb/834 lb/ft 3 = 0.600 ft 3

1.101

γ

w 7.50 lb 7.48 gal    56.1 lb /ft 3 3 V 1 gal ft

γ 56.1 lb/ft 3 lb  s 2 ρ   1.74 4  1.74 slugs /ft 3 2 g 32.2 ft/s ft sg =

γo 5.61 lb/ft 3   0.899 γ w @ 4 C 62.4 lb/ft 3

1.102

w  γV  (1.258)

1.103

w  γV  ρgV 

The Nature of Fluids

(62.4 lb) (1 ft 3 ) (50 gal)  525 lb ft 3 7.84 gal

1.32 lb.s 2 32.2 ft 1ft 3    142 lb 25.0 gal ft 4 82 7.84 gal

Chapter 1

The Nature of Fluids

1.112

1.113 Required Volume  85 Gallons 

Tank Volume  19,636 in 3  Required Height 

1.114 Flow Rate 

1 ft 3 12 3 in 3  3 3  19,636 in 3 7.48 gal 1 ft

  (D) 2  (h) 4



  (38 in) 2  (h) 4

19,636 in  4  17.3 in   (38 in) 2 3

80 N 60 s N   960 5 s 1 min min

1.115 VREQ.  1.5 m  2.5 m  25 cm 

Time Required 

1m  0.938 m 3 100 cm

1 min 1L   0.938 m 3  15.6 min 60 L 0.001 m 3

 π  24 in 2 1.0 gal     18 in  4 231 in 3  Volume  gal   23.5 1.116 Flow Rate  1 min  Time min   90 s   60 s   1.117 $17,000  7500

X

$  X years year

$17,000  2.27 years $ 7500 year

1.118 Annual Cost  2 HP 

1.119 Displacement 

0.746 kW 365 days 24 hr $0.10  1 year     $1,307 Year 1 HP 1 year 1 day kW  HR

π  7.5 cm 2  10.0 cm 0.001 L   0.442 L 4 1 cm 3

Chapter 1

1.120 Flow Rate 

1.121 Volume 

2.2 L 80 rev 1 m 3 60 min m3     10.6 1 rev 1 min 1000 L 1 hr hr

π  1 in 2  2.5 in in 3  1.963 4 rev

gal 1.963 in 3 1 gal X rev    3 min 1 rev min 231 in

gal min  2,354 RPM X 3 1.963 in 1 gal 1 rev 231 in 3 20

The Nature of Fluids

CHAPTER TWO VISCOSITY OF FLUIDS 2.1

Shearing stress is the force required to slide one unit area layer of a substance over another.

2.2

Velocity gradient is a measure of the velocity change with position within a fluid.

2.3

Dynamic viscosity = shearing stress/velocity gradient.

2.4

Oil. It pours very slowly compared with water. It takes a greater force to stir the oil, indicating a higher shearing stress for a given velocity gradient.

2.5

N.s/m2 or Pa.s

2.6

lb.s/ft 2

2.7

1 poise = 1 dyne.s/cm2 = 1 g/(cm.s)

2.8

It does not conform to the standard SI system. It uses obsolete basic units of dynes and cm.

2.9

Kinematic viscosity = dynamic viscosity/density of the fluid.

2.10

m2/s

2.11

ft2/s

2.12

1 stoke = 1 cm2/s

2.13

It does not conform to the standard SI system. It uses obsolete basic unit of cm.

2.14

A newtonian fluid is one for which the dynamic viscosity is independent of the velocity gradient.

2.15

A nonnewtonian fluid is one for which the dynamic viscosity is dependent on the velocity gradient.

2.16

Water, oil, gasoline, alcohol, kerosene, benzene, and others.

2.17

Blood plasma, molten plastics, catsup, paint, and others.

2.18

6.5 10 4 Pas

2.19

1.5 10−3 Pas

2.20

2.0 10 5 Pas

−

Chapter 2

CHAPTER EIGHT REYNOLDS NUMBER, LAMINAR FLOW, TURBULENT FLOW, AND ENERGY LOSSES DUE TO FRICTION DUE TO FRICTION

REYNOLDS NUMBER, LAMINAR FLOW, TURBULENT FLOW, AND ENERGY LOSSES DUE TO FRICTION

107

108

Chapter 8

8.11

8.12

NR 

υ

Q 16.5 ft 3 /s   8.66 ft/s A 1.905 ft 2

υ

Q 0.40 gal ft 3 1 hr 1      0.732 ft/s A hr 7.48 gal 3600 s 2.029  105 ft 2

NR 

8.13

υD (8.66)(1.558)   9.64×105 v 1.40  105

NR 

υDρ

 υD





(0.732)(0.00508)(0.88)(1.94)  1.02 Laminar 6.2  103



(0.732)(0.00508)(0.88)(1.94)  33.4 Laminar 1.90  104

Note : sg of oil may be slightly lower at 160 F. 8.14

NR 



:υ  N R  D 

υDρ

(4000)(4.01105 )  0.424 ft / s (0.2423)(1.56)

Q  Aυ  4.609  102 ft 2  0.424 ft/s  1.96 × 10-2 ft 3 / s 8.15

υ

Q 45 L/ min 1 m3 /s    2.667 m s A 2.433  104 m 2 60000 L/ min υDρ

N    R

(2.667)(0.0176)(0.89)(1000)  5.22×103 Turbulent 8  103

Note :  from App.D. 8.16

NR 

8.17

υ





(2.667)(0.0176)(890)  13.9 very low  Laminar 3.0

Q 45 L/min 1 m3 / s    0.432 m s A 1.735  103 m 2 60000 L/min

NR 

8.18

υDρ





(0.432)(0.0470)(890)  2260 Critical Zone 8  103

Q 1.65 gal/min 1 ft 2 /s υ  14.65 ft/s  A 2.509  10 4 ft 2 449 gal/min NR 

υD (14.65)(0.01788)   1105 Laminar v 2.37  104

REYNOLDS NUMBER, LAMINAR FLOW, TURBULENT FLOW, AND ENERGY LOSSES DUE TO FRICTION

109

8.19

A

Q 500gal/min 1 ft 3 /s    0.1114 ft 2  5 - in Sch .40 pipe υ 10.0 ft/s 449 gal/min A  0.1390 ft 2 , D  0.4026 ft

Q (500/499)ft 3 /s   8.01 ft/s A 0.1390 ft 2

Actual υ  NR  8.20

υD





(8.01)(0.4026)(2.13)  2.12×104 3.38  104

N R v 2000(1.21  105 ft 2 / s)   0.3897 ft/ s 0.0621 ft D For N R  4000, υ2  2(0.3897 ft/ s )  0.7794 ft/ s

υ1 

Q1  Aυ1  (3.027  103 ft 2 )(0.3897 ft/ s) 449 gal/ min 1 ft 3 / s Q1 = 0.0530 gal / min Lower Limit Q2 = 2Q1 = 1.060 gal / min Upper Limit  1.180  103 ft 3 / s 

8.21

(See Prob. 8.20) N R v 2000(3.84  106 )   0.1237 ft/s; υ2  2υ1  0.2473ft/s 0.0621 D 449gal/min  0.1681 gal/min Q1  Aυ1  (3.027 103 ft 2 )(0.1237ft/s)  1ft 3 /s Q2  2Q1  0.3362 gal /min υ1 

8.22

υ = 1.30 cs 

1.076 105 ft 2 / s  1.40  10 5 ft 2 /s 1cs

Q  45 gal/min  υ

0.1002 ft 2 / s) Q   14.65ft/ s A 6.842  103 ft 2

NR  8.23

1ft 3 / s  0.1002ft 3 /s 449 gal/min

υD (14.65)(0.0933)   9.78×104 5 v 1.40  10

v  17.0 cs

106 m 2 /s  1.7  105 m 2 /s 1 cs

Q 215 L/min 1 m3 / s    7.142 m/s A 5.017  104 m 2 60000L/min υD (7.142)(0.0253) NR    1.06  104 v 1.70  105

υ

110

Chapter 8

8.24

v  1.20 cs 

υ

Q 200 L/ min 1m3 / s    8.69 m/ s A 3.835  104 m 2 60000 L/ min

NR 

8.25

106 m 2 / s  1.20  106 m 2 / s 1cs

υD (8.69)(0.0221)   1.60×105 6 v 1.20  10

p1 υ2 p υ2  z1  1  hL  2  z2  2 : υ1  υ2 γo 2g γo 2g p1  p2  γ o [ z 2  z1  hL ]

NR 

υDρ





(0.64)(0.0243)(0.86)(1000) 64  787(Laminar); f   0.0813 2 1.70  10 NR

L υ2 60 (0.64) 2 hL  f  0.0813    4.19 m D 2g 0.0243 2(9.81) p1  p2  (0.86)(9.82 kN/ m3 )[60 m  4.19 m]  471 kN/ m 2  471 kPa 8.26

p1 υ12 p2 υ2 2  z1   hL   z2  : υ1  υ2 ; z1  z2 ; p1  p2  γ w hL γw 2g γw 2g

Q 12.9 L/min 1m3 /s υ    1.724 m/s A 1.247  104 m 2 60000 L/min NR 

υD (1.724)(0.0126)   5.67 104 (turbulent) 7 v 3.83 10

D /ε  0.0126/1.50  106  8400; Then f  0.0207

hL  f

L υ2 45 (1.724)2 .  (0.0207).  11.20 m D 2g 0.0126 2(9.81)

p1  p2  γ w hL  9.56 kN / m 3  11.20 m  107.1 kN / m 2  107.1 kPa 8.27

Let N R  2000; f  64 / N R  0.032; N R 

υ

υDρ



N R (2000)(8.3  104 )   2.85 ft / s D (0.3355)(0.895)(1.94)

hL  f

L υ2 100 (2.85) 2 . ft = 1.20ft  1.20ft lb /lb  (0.032). D 2g 0.3355 2(32.2)

REYNOLDS NUMBER, LAMINAR FLOW, TURBULENT FLOW, AND ENERGY LOSSES DUE TO FRICTION

111

8.28

pA υ 2 p υ2  zA  A  hL  B  zB  B : υA  υB γo 2g γo 2g pB  pA    [ zA  zB  hL ] υ

(800)(4  104 ) N R   0.717 ft / s D  (0.2557)(0.90)(1.94)

64 5000 (0.717) 2 L υ2   12.5 ft hL  f . . D 2 g 800 0.2557 2(32.2) 1ft 2  37.3psig pB  50 psig + (0.90)(62.4 lb/ ft )[ 20 ft  12.5ft] 144 in 2 3

8.29

p1 υ2 p υ2 L υ2  z1  1  hL  2  z2  2 : z1  z2 : υ1  υ2 : p1  p2   b hL   b f D 2g γb 2g γb 2g υ

Q 20L / min 1m3 / s    0.719 m / s A 4.636 104 m 2 60000L / min



γ 8.62 kN s2 103 N 1kg  m / s 2     = 879 kg / m3 g m3 9.81 m kN N

NR 

υD 





(0.719)(0.0243)(879)  3.89  104 3.95  104

D /ε  0.0243 / 4.6 105  528; Then f  0.027 p1  p2 = 8.62 kN / m3 × 0.027× 8.30

100 (0.719) 2 × m = 25.2 kN/m 2  25.2kPa 0.0243 2(9.81)

From Prob.8.31, p1  p2 = γ w h L ; h L = p1  p2 /γ w

hL =

(1035 - 669)kN / m 2 L υ2 = 37.3 m = f 9.81kN / m3 D 2g

f 

hL D 2 g (37.3)(0.03388)(2)(9.81)   0.048 Lυ2 (30)(4.16)2

υ=

Q 225L / min 1 m3 / s = × = 4.16m / s A 9.017×10-4 m 2 60000L / min

NR =

υD (4.16)(0.03388) D = = 1.08×105 : Then = 55 for f = 0.048 -6 v 1.30×10 ε

  D / 55  0.036 / 55  6.16×10-4 m

112

Chapter 8

8.31

Pt.1 at tank surface. p1  0, υ1  0

p1 υ2 p υ2  z1  1  hL  2  z2  2 γw 2g w 2g 2 υ h  z1  z2  hL  2 2g

Pt. 2 in outlet stream. p2  0 D  0.5054 ft A  0.2006 ft 2

Q 2.50 ft 3 / s υ= = = 12.46 ft / s A 0.2006 ft 2

8.32

NR 

υD (12.46)(0.5054) D 0.5054   6.88  105 :   3369 : f  0.0165 6 v 9.15  10  1.5  10 4

h f

L υ2 υ2 550 (12.46) 2 (12.46) 2   0.0165     45.7 ft D 2g 2g 0.5054 2(32.2) 2(32.2)

From Prob 8.31, p1  p2 = γ w hL = γ w f υ=

L υ2 D 2g

Q 15.0 ft 3 / s = = 8.49 ft/s A π (1.50 ft) 2 / 4

NR 

υD (8.49)(1.50) D 1.50   9.09  105 ;   3750; f  0.0158 5 v 1.40  10 ε 4  104

L υ2 62.4 lb 5280 ft (8.49) 2 ft 2 /s 2 1 ft 2 p1  p2 = γ w f = × 0.0158× × × = 30.5 psi D 2g ft 3 1.50 ft 2(32.2 ft /s 2 ) 144 in 2 8.33

Q = 1500 gal/min × υA = a)

1 ft 3 /s = 3.34 ft 3 /s 449 gal/min

Q 3.34 ft 3 /s υ2 (6.097) 2 = = 6.097 ft / s; = = 0.577 ft AA 0.5479 ft 2 2g 2(32.2)

p1 υ2 p υ2 + z1 + 1  h Ls = A + z A + A Pt.1 at tank surface. p1 = 0, υ1 = 0 γw 2g γw 2g z1  zA  h 

NR 

υA D (6.097)(0.835) D 0.835   4.21  105 :   5567 : f  0.0155 5 v 1.21  10 ε 1.5  104

hL  f h

pA υ2 A   hL γw 2g

L υ2 45  (0.0155)   0.577 ft = 0.482 ft D 2g 0.835

5.0 lb  ft 3 144 in 2  0.577  0.482  12.60 ft in 2 62.4 lb ft 2

REYNOLDS NUMBER, LAMINAR FLOW, TURBULENT FLOW, AND ENERGY LOSSES DUE TO FRICTION

113

pA υ2 p υ2  z A  A  hLd  hA  B  z B  B γ 2g γ 2g

Q 3.34 ft 3 /s υB    9.62 ft/s AB 0.3472 ft 2 hA  

pB  p A γw

(85  5) lb ft 3 (144 in 2 ) υB 2  υA 2  ( zB  zA )   hLd   25 2g in 2 (62.4 lb) ft 2

(9.622  6.097 2 ) ft 2 /s 2  89.9  300.4 ft 2(32.2 ft/s 2 )

υ D (9.62)(0.6651) N RB  B B  v 1.21  105

D /ε 

0.6651  4434 : f  0.016 1.5  104

hLd  f

L υ2 2600 (9.62) 2  (0.016)    89.9 ft D 2g 0.6651 2(32.2)

PA  hA γ wQ  300.4 ft  8.34

 5.29  105

62.4 lb 3.34 ft 3 hp   113.8 hp 3 ft s 550 ft  lb/s

υ2 p2 υ 2 Pt.1 at well surface ( p 1 = 0 psig).  z1  1  hA  hL   z2  2 γ 2g γ 2 g Pt. 2 at tank surface. p2 υ1  υ2  0  ( z2  z1 )  hL hA  γw p1

Q

745 gal 1h 1 ft 3 /s = 0.0277 ft 3 /s   h 60 min 449 gal/min

υ

Q 0.0277 ft 3 /s = 4.61 ft/s in pipe  A 0.0060 ft 2

NR 

υD (4.61)(0.0874) D 0.0874   3.33 104 :   583 : f  0.0275 5 v 1.2110  1.5 104

hL  f

L υ2 140 (4.61) 2  (0.0275)  ft = 14.54 ft D 2g 0.0874 2(32.2)

(40 lb)ft 3 (144 in 2 ) +120 + 14.54 = 226.8 ft hA = in 2 (62.4 lb)ft 2 p A = hA γQ = (226.8 ft)(62.4 lb/ft 3 )(0.0277 ft 3 / s) / 550 ft  lb/s/hp = 0.713 hp

114

Chapter 8

8.35

p1 p υ2 υ2  z1  1  hL  `2  z2  2 γw 2g γw 2g

Pt .1 at tank surface. υ1  0 Pt .2 in outlet stream. p2  0

  υ2 2  hL  p1  γ w (z 2 z1)  2g  

υ2 

Q 75 gal/ min 1ft 2 / s υ 2 (11.8) 2 11.8 ft/ s :      2.167 ft A 0.01414 ft 2 449 gal/ min 2 g 2(32.2)

NR 

υD (11.8)(0.1342) D 0.1342   1.31 105 :   895 : f  0.0225 5 v 1.21 10  1.5  104

L υ2 300  (0.0225) (2.167 ft)  109.0 ft D 2g 0.1342

hL  f

62.4 lb 1ft 2 [ 3 ft  2.167 ft  109.0 ft] p1   46.9 psi ft 3 144 in 2 8.36

P1 p υ2 2  z1  1  hA  hL  2  z2  2 γ 2g γ 2g

p2 υ2  ( z2  z1 )  2  hL γ 2g

a) hA  υ

Pt .1 at tank surface. p1  0; υ1  0 Pt .2 in house at nozzle. Pt .3 in house at pump outlet . υ3  υ2

Q 95 L/ min 1m 3 / s υ 2 (3.23) 2 3.23 m/ s :      0.530 m A π (0.025 m) 2 / 4 60000 L/ min 2 g 2(9.81)

NR 

υDρ

hL  f hA 





(3.23)(0.025)(1100)  4.44  104 : f  0.021(smooth) 2.0  103

L υ2 85 (0.530) m  37.86 m  (0.021) D 2g 0.025

140 kN/ m 2  7.3m  0.530  37.86 m  58.67 m (1.10)(9.81kN/ m3 )

p A  hA γQ  (58.67 m)(1.10)(9.81kN/m3 )(95 / 60000)m3 /s  1.00 kN m/s  1.00 kW

p3 υ32 p2 υ2 2 b)  z3   hL   z2  : p3  p2  [( z2  z3 )  hL ]γ γ 2g γ 2g p3  140 kPa  (1.10)(9.81kN/ m3 )[8.5 m  37.86 m]  640 kPa

REYNOLDS NUMBER, LAMINAR FLOW, TURBULENT FLOW, AND ENERGY LOSSES DUE TO FRICTION

115

8.37

Q  1200 L/ min 1m3 / s/ 60000 L/ min  0.02 m3 / s

υ

Q 0.02 m3 / s   1.189 m/ s A 1.682  102 m 2

a) p2  p3  γ o hL  γ o f NR 

υD 





L υ2 D 2g

1.189(0.1463)(930)  1079 Laminar 0.15 2

 64   3200  (1.189) m  853 kPa p2  p3  γ o hL  (0.93)(9.81 kN/ m3 )   1079   0.1463 2(9.81) b)

p p1 υ 12 υ 32  z1   hA  hL  3  z3  : p1  p3 , υ1  υ3 , z1  z3 γ 2g γ 2g hA  hL 

853kN/ m 2  93.5 m (0.93)(9.81kN/ m 3 )

p A  hA γQ   93.5 m (0.93)(9.81 kN/ m3 )(0.02 m3 / s)  17.1 kN  m/ s  17.1kW 8.38

At 100o C,μ  7.9  10 3 Pas a) With pumping stations 3.2 km apart :

NR 

υD 





(1.189)(0.1463)(930)  2.05  104 turbulent 3 7.9 10

D /ε  0.1436 m/ 4.6  105 m  3180; f  0.026

L υ2 (3200) (1.189) 2 hA  hL  f m  40.98 m  (0.026) D 2g 0.1463 2(9.81) p A  hA γQ  (40.98)(0.93)(9.81)(0.02)  7.48kW

L υ2 b) Let h L  93.5 m (from Prob. 9.13) : hL  f D 2g L

116

hL D(2 g ) (93.5 m)(0.1463 m)(2)(9.81 m/ s 2 )   8682 m  8.68 km fυ 2 (0.026)(1.189 m/ s) 2

Chapter 8

CHAPTER TEN MINOR LOSSES

MINOR LOSSES

145

146

Chapter 10

MINOR LOSSES

147

148

Chapter 10

MINOR LOSSES

149

10.26 Sketches for the contractions for 15 and 40 gradual contractions:

10.27 Gradual contraction, θ = 120° D1 = 6.27 in, D2 = 4.17 in Ductile iron pipe D1/D2 = 1.50; K = 0.200

150

Chapter 10

MINOR LOSSES

151

152

Chapter 10