Applied Fluid Mechanics 8th edition By Joseph A. Untener, SOLUTIONS MANUAL

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

1

2. Viscosity of Fluids

14

3. Pressure Measurement

20

4. Forces Due to Static Fluids

28

5. Buoyancy and Stability

49

6. Flow of Fluids and Bernoulli’s Equation

68

7. General Energy Equation

91

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

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

2

2

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

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

2

Chapter 1

The Nature of Fluids

3

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

4

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

5

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

6

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

7

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

2

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

8

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

3

2

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

9

10

Chapter 1

The Nature of Fluids

11

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 

12

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 

20

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

13

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

14

−

−

Chapter 2

2.21

1.1 10 5 Pa  s

2.22

3.0 10−1 Pa  s

2.23

1.90 Pa  s

2.24

3.2 10 5 lb  s/ft2

2.25

8.9 10−6 lb  s/ft2

2.26

3.6 10−7 lb  s/ft2

2.27

1.9 10−7 lb  s/ft2

2.28

5.0 10−2 lb  s/ft2

2.29

4.1 10−3 lb  s/ft2

2.30

3.3 10 5 lb  s/ft2

2.31

2.8 10 5 lb  s/ft2

2.32

2.1 10 3 lb  s/ft2

2.33

9.5 10 5 lb  s/ft2

2.34

1.3 10−2 lb  s/ft2

2.35

2.2 10−4 lb  s/ft2

2.36

Viscosity index is a measure of how greatly the viscosity of a fluid changes with temperature.

2.37

High viscosity index (VI).

2.38

Rotating drum viscometer.

2.39

The fluid occupies the small radial space between the stationary cup and the rotating drum.

Therefore, the fluid in contact with the cup has a zero velocity while that in contact with the drum has a velocity equal to the surface speed of the drum.

2.40

A meter measures the torque required to drive the rotating drum. The torque is a function of the drag force on the surface of the drum which is a function of the shear stress in the fluid.

Knowing the shear stress and the velocity gradient, Equation 2-2 is used to compute the dynamic viscosity.

2.41

The inside diameter of the capillary tube; the velocity of fluid flow; the length between pressure taps; the pressure difference between the two points a distance L apart.

See Eq. (2-5).

−

−

−

−

−

−

Viscosity of Fluids

15

2.42

Terminal velocity is that velocity achieved by the sphere when falling through the fluid when the downward force due to gravity is exactly balanced by the buoyant force and the drag force on the sphere. The drag force is a function of the dynamic viscosity.

2.43

The diameter of the ball; the terminal velocity (usually by noting distance traveled in a given time); the specific weight of the fluid; the specific weight of the ball.

2.44

The Saybolt viscometer employs a container in which the fluid can be brought to a known, controlled temperature, a small standard orifice in the bottom of the container and a calibrated vessel for collecting a 60 mL sample of the fluid. A stopwatch or timer is required to measure the time required to collect the 60 mL sample.

2.45

No. The time is reported as Saybolt Universal Seconds and is a relative measure of viscosity.

2.46

Kinematic viscosity.

2.47

Standard calibrated glass capillary viscometer.

For questions 2.48 to 2.53, Refer to Section 2.8 and to Internet resource 19 for Tribology-abc. Tables for SAE viscosity grades for engine oils and automotive gear lubricants are listed on the Internet site, from standards SAE J300 and SAE J306 that can be used to determine appropriate values for viscosities and information on the testing procedures used. NOTE: It is essential that the latest version of the standards be used for critical applications. See References 14 and 15.

2.48

The kinematic viscosity of SAE 20 oil must be between 5.6 and 9.3 cSt at 100C using ASTM D 445. Its dynamic viscosity must be over 2.6 cP at 150C using ASTM D 4683, D 4741, or D 5481. The kinematic viscosity of SAE 20W oil must be over 5.6 cSt at 100C using ASTM D 445. Its dynamic viscosity for cranking must be below 9500 cP at −15C using ASTM D 5293. For pumping it must be below 60,000 cP at −20C using ASTM D 4684.

2.49

SAE 0W through SAE 60 for engine crankcase oils, depending on operating conditions.

2.50

SAE 70W through SAE 250 for gear-type transmissions, depending on operating conditions.

2.51

From Internet resource 19: 100C using ASTM D 445 testing method and at 150C using ASTM D 4683, D 4741, or D 5481.

2.52

From Internet resource 19: At −25C using ASTM D 5293; at −30C using ASTM D 4684; at 100C using ASTM D 445.

2.53

From Internet resource 19: The kinematic viscosity of SAE 5W-40 oil must be between 12.5 and 16.3 cSt at 100C using ASTM D 445. Its dynamic viscosity must be over 2.9 cP at 150C using ASTM D 4683, D 4741, or D 5481. The kinematic viscosity must be over 3.8 cSt at 100C using ASTM D 445. Its dynamic viscosity for cranking must be below 6600 cP at −30C using ASTM D 5293. For pumping it must be below 60 000 cP at−35C using ASTM D 4684.

2.54

v = SUS/4.632 = 500/4.632 = 107.9 mm2/s = 107.9 10 6 m2/s − v = 107.9 10 6 m2/s [(10.764 ft2/s)/(m2/s)] = 1.162 10−3 ft2/s

16

−

Chapter 2

2.55

From Table 2.5: Viscosities at 40C in mm2/s of cSt ISO Viscosity grade Minimum

Nominal

Maximum

10

9.0

10.0

11.0

68

61.2

68.0

74.8

220

198

220

242

1000

900

1000

1100

Viscosity of Fluids

17

18

Chapter 2

Problem 2.77 Convert kinematic viscosity for ISO grades from mm2/s to SUS ISO VG* Nominal 2 (mm /s) (SUS) 2.2 32.0 3.2 34.3 4.6 39.5 6.8 46.5 10 57.5 15 76.5 22 107 32 151 46 214 68 316 100 463 150 695 220 1019 320 1482 460 2131 680 3150 1000 4632 1500 6948 2200 10190 3200 14822 *

Minimum (mm /s) (SUS) 1.98 28.8 2.88 30.87 4.14 35.55 6.12 41.85 9.00 51.75 13.5 68.9 19.8 96.3 28.8 135.9 41.4 192.6 61.2 284.4 90 417 135 625 198 917 288 1334 414 1918 612 2835 900 4169 1350 6253 1980 9171 2880 13340 2

Maximum (mm /s) (SUS) 2.42 35.2 3.52 37.7 5.06 43.5 7.48 51.2 11.0 63.3 16.5 84.2 24.2 118 35.2 166 50.6 235 74.8 348 110 510 165 764 242 1121 352 1630 506 2344 748 3465 1100 5095 1650 7643 2420 11209 3520 16305 2

First four values are called VG 2, 3, 5, and 7 NOTES: For VG ≥ 100 values computed from SUS = 4.664*nmm2s) Other values read from graph in Figure 2.14 Minimum = Nom*0.9; Maximum = Nom*1.1

Viscosity of Fluids

19

CHAPTER THREE PRESSURE MEASUREMENT Absolute and gage pressure 3.1 3.2

Atmospheric pressure is the absolute pressure in the local area.

Gage pressure is measured relative to atmospheric pressure.

3.3 3.4

Perfect vacuum is complete absence of molecules. Absolutely NO pressure.

Absolute pressure is measured relative to a perfect vacuum.

3.5 3.6

pabs  pgage  patm True.

3.7

False. Atmospheric pressure varies with altitude and with weather conditions.

3.8

False. Absolute pressure cannot be negative because a perfect vacuum is the reference for absolute pressure and a perfect vacuum is the lowest possible pressure.

3.9

True.

3.10

False. A gage pressure can be no lower than one atmosphere below the prevailing atmospheric pressure. On earth, the atmospheric pressure would never be as high as 150 kPa.

3.11

At 4000 ft, patm =12.7 psia; from App. E by interpolation.

3.12

At 13,500 ft, patm =8.84 psia; from App. E by interpolation.

3.13

Zero gage pressure.

3.14

P  6 lbs/(pi  (0.3 in)2 ) / 4  84.9 psi

3.15

pgage  583  103  480 kPa(gage)

3.16

pgage  157 101  56 kPa(gage)

3.17

pgage  30 100   70 kPa(gage)

3.18

pgage  74  97   23 kPa(gage)

3.19

pgage  101  104   3kPa(gage)

3.20

pabs  284  100  384 kPa(abs)

3.21

pabs  128  98  226 kPa(abs)

20

Chapter 3

3.22

pabs  4.1  101.3  105.4 kPa(abs)

3.23

pabs  29.6  101.3  71.7 kPa(abs)

3.24

pabs  86  99  13 kPa(abs)

3.25

pgage  84.5  14.9  69.6 psig

3.26

pgage  22.8  14.7  8.1 psig

3.27

pgage  6.3  14.6  8.3 psig

3.28

pgage  12.8  14.0  1.2 psig

3.29

pgage  14.7  15.1  0.4 psig

3.30

pabs  41.2  14.5  55.7 psia

3.31

pabs  18.5  14.2  32.7 psia

3.32

pabs  0.6  14.7  15.3 psia

3.33

pabs  4.3  14.7  10.4 psia

3.34

pabs  12.5  14.4  1.9 psia

Pressure – Elevation Relationship 3.35

p  γh  1.08(9.81 kN/m3 )(0.550 m) = 5.83 kN/m 2  5.83 kPa(gage)

3.36

p  γh  (sg)γ w h : sg  p / γ w h) 1.820 lb ft 3 144 in 2 sg = 2   1.05 in (62.4 lb)(4.0 ft) ft 2

p 52.75 kN m3   6.70 m γ m 2 7.87 kN

3.37

h

3.38

p  γh 

64.00 lb 1 ft 2    5.56 psig 12.50 ft ft 3 144 in 2

3.39

p  γh 

62.4 lb 1 ft 2 50.0 ft    21.67 psig ft 3 144 in 2

3.40

p  γh  (10.79 kN/m3 )(3.0 m) = 32.37 kN/m2  32.37 kPa(gage)

3.41

p  γh  (10.79 kN/m3 )(12.0 m) = 129.5 kPa(gage)

Pressure Measurement

21

3.42

pair  pA  γ o  64 in   180 psig   0.9   62.4 lb/ft 3   64 in  1 ft 3 /1728 in 3  pair  180 psig  2.08 psi  177.9 psig

3.43 3.44

pi  γh  1.15   9.81 kN/m3   0.375 m   4.23 kPa  gage 

patm = 24.77 kPa  abs By interpolation  App. E : γ m  (13.54)9.81 kN/m 3 γ m  132.8 kN/m3 pB  patm  γh  24.77 kPa  (132.8 kN/m3 )(0.325 m)  67.93 kPa(abs)

3.45

p  γh  (0.95)(62.4 lb/ft 3 )(28.5 ft)(1 ft 2 /144 in 2 )  11.73 psig

3.46

p  50.0 psig  γh = 50.0 psig  11.73 psi  61.73 psig (Sec 3.44 for γh  11.73 psig)

3.47

p  10.8 psig  γh  10.8 psig  (0.95)

(62.4 lb) 1 ft 2 6.25ft   ft 3 144 in 2

p  10.8 psig  2.57 psi  8.23 psig 3.48

pbot  ptop

ptop  γ o h  pbot : h 

γo

(35.5  30.0)lb ft ) 144 in 2 h 2   13.36 ft in (0.95)(62.4 lb) ft 2 3

3.49

0  γ o ho  γ w hw  pbot pbot  γ w hw 52.3 kN/m 2  (9.81 kN/m3 )(2.80 m) ho    2.94 m γo (0.86)(9.81 kN/m3 )

3.50

0  γ o ho  γ w hw  pbot pbot  γ o ho 125.3 kN/m 2  (0.86)(9.81 kN/m3 )(6.90 m) hw    6.84 m γw 9.81 kN/m3

3.51

0  γ o h1  γ w h2  pbot ; but h1  18.0  h2 γ o (18  h2 )  γw h2  pbot 18γ o  γ o h2  γ w h2  pbot  h2 (γ w  γ o )  18γ o pbot  18 o 158 kN/m 2  (18 m)(.86)(9.81 kN/m3 )   4.47 m h2  [9.81  (0.86)(9.81)]kN/m3  w o

3.52

p  γh  (1.80)(9.81 kN/m3 )(4.0 m)=70.6 kN/m 2  70.6 kPa(gage)

3.53

p  γh  (0.89)(62.4 lb/ft 3 )(32.0 ft)(1 ft 2 /144 in 2 )  12.34 psig

3.54

p  γh  (10.0 kN/m3 )(11.0  103 m) = 110  103 kN/m 2  110 MPa

3.55

patm  γ m (.457 m)  γ w (1.381 m)  γG (0.50 m)= pair pair  (13.54)(9.81 kN/m3 )(.457 m)  (9.81 kN/m3 )(1.381 m)  (.68)(9.81)(.50) pair  (60.7  13.55  3.34) kN/m 2  43.81 kPa(gage)

22

Chapter 3

3.56

pbot  34.0 kPa + γ o ho  γ w hw  34 kPa + (0.85)(9.81 kN/m3 )(0.50 m) + (9.81 kN/m3 )(0.75 m) pbot  34.0 kPa + 4.17 + 7.36 =  22.47 kPa(gage)

3.57

pbot  pair  γ o ho  γ w hw  200 kPa + [(0.80)(9.81)(1.5) + (9.81)(2.6)]kN/m 2 pbot  200  11.77  25.51  237.3 kPa(gage)

Manometers (See text for answers to 3.57 to 3.61.)

3.62

patm  γ m (.075 m)  γ w (0.10 m) = pA pA   (13.54)(9.81 kN/m3 )(0.075 m)  (9.81)(0.10)  10.94 kPa(gage)

3.63

pA  γ o (13 in) + γ w (9 in)  γ o (32 in) = pB 62.4 lb 1 ft 3 (0.85)(62.4)(19) pB  pA  γ w (9 in)  γ o (19 in) =  9 in   3 3 ft 1728 in 1728 pB + pA  0.325 psi  0.583 psi   0.258 psi

3.64

pB  γ w (33 in) + γ o (8 in)  γ w (13 in) = pA pA  pB  γ o (8 in)  γ w (20 in) =

(.85)(62.4) lb 1 ft 3 (62.4)(20)  8 in   3 3 ft 1728 in 1728

pA  pB  0.246 psi  0.722 psi =  0.477 psi 3.65

pB  γ o (.15 m)  γ m (0.75 m)  γ w (0.50 m) = pA pA  pB  (.90)(9.81 kN/ m3 )(.15 m) + (13.54)(9.81)(.75)  (9.81)(0.50) pA  pB  (1.32  99.62  4.91) kPa  96.03 kPa

3.66

pB  γ w (.15 m) + γ m (0.75 m)  γ o (0.60 m)  pA pA  pB  (9.81 kN/m3 )(0.15 m) + (13.54)(9.81)(0.75)  (0.86)(9.81)(0.60) pA  pB  (1.47  99.62  5.06)kPa = 96.03 kPa

3.67

patm  γ m (.475 m)  γ w (.30 m)  γ w (.25 m)  γ o (.375 m) = pA pA  γ m (.725 m)  γ w (.30 m)  γ o (.375 m) pA  (13.54)(9.81 kN/m3 )(.725 m)  (9.81)(.30)  (.90)(9.81)(.375) pA  (96.30  2.94  3.31)kPa = 90.05 kPa(gage)

3.68

pB  γ w (6 in) + γ m (6 in)  γ w(10 in)  γ m (8 in)  γ o (6 in) = pA pA  pB  γ m (14 in)  γ w (4 in)  γ o (6 in) pA  pB  (13.54) 

62.4 lb 1 ft 3 (62.4)(4) (.9)(62.4)(6) (14 in)     3 3 ft 1728 in 1728 1728

pA  pB  (6.85  0.14  0.195) psi = 6.51 psi

Pressure Measurement

23

3.69

pB  γ w (2 ft)  γ o (3 ft) + γ w (11 ft) = pA p A  pB  γ w (9 ft)  γ o (3 ft) =

62.4 lb 1 ft 2 (.90)(62.4)(3)  9 ft   3 2 ft 144 in 144

pA  pB  3.90 psi  1.17 psi = 2.73 psi

62.4 lb 1 ft 3  6.8 in   0.246 Psi ft 3 1728 in 3

3.70

patm  γ w (6.8 in) = pA  0 

3.71

patm  γGF h  pA : h  L sin 15  0.115 m sin 15  0.0298 m pA  (0.87)(9.81 kN/m3 )(0.0298 m)  0.254 kPa(gage)

3.72

a)

patm  γ m (.815 m)  γ w (.60m)  pA pA  (13.54)(9.81kN/m3 )(0.815 m)  (9.81)(.60)  102.4kPa(gage)

b)

patm  γ m h  (13.54)(9.81)(.737)  97.89 kpa pA  102.4  97.89  200.3kP(abs)

Barometers 3.73

A barometer measures atmospheric pressure.

3.74

The height of the mercury column is convenient.

3.75

h 

3.76

h = 29.29 in

See Example Problem 3.13

3.77

h = 760 mm

See Example Problem 3.11

3.78

The vapour pressure above the mercury column and the specific weight of the mercury change.

3.79

h 

3.80

101.3 kPa → 760 mm of Mercury (See Ex. Prob. 3.11) h 

patm 14.7lb ft 3 144 in 2  2   33.92 ft very large (10.34m) γw in 62.4lb ft 2

1.0 in of Mercury  1250 ft =  1.25 in 1000 ft

 85 mm .3048 m  5200 ft   134.7mm 1000 m 1ft

h  760 134.7  625.3 mm patm  γ m h  133.3

24

kN  0.6253 m  83.35 kPa m3

Chapter 3

3.81

Patm  γ m h 

848.7 lb 1 ft 3 28.6 in    14.05 psia ft 3 1728 in 3

3.82

Patm  γ m h 

848.7 lb 1 ft 3 30.65in    15.05 psia ft 3 1728 in 3

3.83

h 

3.84

patm  γ m h  133.3

patm 14.2 lb ft 3 1728 in 3     28.91 in γm in 2 848.7 lb ft 3 kN  0.745 m= 99.3kPa(abs) m3

Expressing Pressures as the Height of a Column of Liquid

3.85

p  4.37 inH 2 O(1.0 psi/27.68inH 2 O)  0.158 psi p  4.37inH 2 O(249.1 Pa/1.0 inH 2 O)  1089 Pa

3.86

p  3.68inH 2 O(1.0 psi/27.68inH 2 O)  0.133 psi p  3.68inH 2 O(249.1 Pa/1.0 inH 2 O) =  917pa

3.87

p  3.24 mmHg(133.3 Pa/1.0 mmHg) = 431.9 pa p  3.24 mmHg(1.0 psi/57.71 mmHg) = 0.0627 psi

3.88

p  21.6 mmHg(133.3 Pa/1.0 mmHg) = 2879 Pa = 2.88 kPa p  21.6 mmHg(1.0 psi/57.71 mmHg) = 0.418 psi

3.89

p  68.2 kpa (1000 Pa/kpa)(1.0 mmHg/133.3 Pa)=  512mmHg

3.90

p  12.6 Psig(2.036 inHg/psi)=  25.7 inHg

3.91

p  12.4 inWc = 12.4 inH 2 O(1.0 psi/27.68 inHL 2 O) = 0.448 psi p  12.4 inH 2 O(249.1 Pa/1.0 inH 2 O) = 3089 Pa = 3.09 kPa

3.92

p  115 inWc = 115 inH 2 O(1.0 psi/27.68 inH 2 O) = 4.15 psi p  115 inH 2 O(249.1 Pa/1.0 inH 2 O) = 28.646 Pa = 28.6 kPa

Pressure Measurement

25

3.93

P = γ w  h P= 9.810

3.94

kN kN  16m=157 2  157kPa 3 m m

P = γ w  h  h =

P γw

kN m 2  16.3 m h= kN 9.810 3 m 160

3.95

P = γ m  h  h =

P γw

lb 2 122 in 2 in h=  2 2  1.874 ft lb 844.9 3 1 ft ft 12 in 1.874 ft   22.5 in 1 ft 11

3.96

P = sg c  γ w  h P = 2.6  9.810

3.97

kN  3 m = 76.5 kPa m3

P = sg sw  γ w  h P = 1.030  9.810

P = γ w  h  h =

3.98

14.7 h=

26

kN 1000 m  11 km   111 MPa 3 m 1 km P γw

lb 122 in 2  1.5 x in 2 12 ft 2  50.9 ft =15.4 m lb 62.4 3 ft

Chapter 3

3.99

P  γ w  h = 62.4

lb 1 ft 12 ft 2 lb  1 in    0.036 3 2 2 ft 12 in 12 in in 2

P = sg gf  γ w  h  h = Sin 25  X OIL 

P sg gf  γ w

1 in ; X WATER  2.37 in X WATER

X WATER ; X OIL  2.86 in sg

kN  0.790 m = 105 kPa m3 kN P738mm  sg m  γ w  h = 13.54  9.810 3  0.738 m = 98.0 kPa m P = P.790  P738  105 kPa  98.0 kPa = 7.0 kPa

3.100

P790mm  sg m  γ w  h = 13.54  9.810

3.101

P10  γ w  h = 62.4

3.102

lb 12 ft 2 lb    4.33 2 (g) 10 ft 3 2 2 ft 12 in in

lb lb  8400 ft = 630 2 3 ft ft 2 2 lb 1 ft lb 630 2  2 2  4.38 2 ft 12 in in P = γ a  h = 0.075

Pressure Measurement

27

CHAPTER FOUR FORCES DUE TO STATIC FLUIDS Forces due to gas pressure

4.1

F  pA; where p  Patm  Pinside ; A  Patm  γ m h 

 (12 in) 2  113 .1 in 2 4

844.9 lb 1 ft 3  30.5 in   14.91 psi ft 3 1728 in 3

F  (14.91  0.32 )lb/in 2  113.1 in 2  1650 lb 4.2

F  pA  (23.6 lb/in 2 )(π(30 in 2 ) / 4)  16 682 lb

4.3

F  pA; A  36  80 in 2  2880 in 2 p  γ w h 

62.4 lb 1 ft 3  1.80 in   0.065 lb/in 2 ft 3 1728 in 3

F  (0.065 lb/in 2 )(2880 in 2 )  187 lb 4.4

F  p A; A  0.9396 ft 2 (App.F) F  (280 lb/in 2 )(0.9396 ft 2 )(144 lb 2 /ft 2 )  37 885 lb

π (0.030 m) 2  7.07  104 m 2 4 F  (3.50  106 N/m 2 )(7.07  104 m 2 )  2.47  103 N = 2.47 kN

4.5

F  pA; A 

4.6

F  pA; A  π(0.050 m) 2 / 4  1.963  10 3 m 2 F  (20.5  10 6 N/m 2 )(1.963  10 3 m 2 )  40.25  10 3 N = 40.25 kN

4.7

F  pA; A  (0.800 m) 2  0.640 m 2 F  (34.4  103 N/m 2 )(0.64 m) 2  22.0  10 3 N = 22.0 kN

Forces on horizontal flat surfaces under liquids 4.8

F  pA; A  24  18 in 2  432 in 2 56.78 lb 1 ft 2 p  γA h =  12.3 ft   4.85 lb/in 2 ft 3 144 in 2 F  (4.85 lb/in 2 )(432 in 2 ) = 2095 lb

4.9

F  pA; A  π(0.75 in) 2 / 4  0.442 in 2 844.9 lb 1 ft 3 p  γmh   28.0 in   13.69 lb/in 2 ft 3 1728 in 3 F  (13.76 lb/in 2 )(0.442 in 2 ) = 6.05 lb

28

Chapter 4

Forces Due to Static Fluids

29

30

Chapter 4

Forces Due to Static Fluids

31

32

Chapter 4

Forces Due to Static Fluids

33

34

Chapter 4

Forces Due to Static Fluids

35

36

Chapter 4

Forces Due to Static Fluids

37

38

Chapter 4

Forces Due to Static Fluids

39

40

Chapter 4

Forces Due to Static Fluids

41

42

Chapter 4

Forces Due to Static Fluids

43

44

Chapter 4

4.58

4.59

#  # #  P  γ  h   2.4  62.4 3   14 ft  2097 2  14.6 2 ft  ft in 

hc1 = 1.75 m hc2 = 3 m Lc1 = 1.75 m Lc2 = 3 m A1 = 3.5m x 4m = 14 m2 A2 = 6m x 4m = 14 m2 N FR1  γ  h C1  A  9810 3  1.75 m  (3.5 m  4 m)  240 kN m N FR2  γ  h C 2  A  9810 3  3 m  (6 m  4 m)  706 kN m

Forces Due to Static Fluids

45

B  H 3 4 m  (3.5 m) 3 B  H 3 4 m  (6 m) 3   14.3 m 4 IC 2    72 m 4 12 12 12 12 4 I C1 14.3 m L P1  L C   1.75 m   2.3 m L C1  A1 1.75 m  14 m 2

I C1 

IC 2 72 m 4 LP 2  LC   3m   4m L C2  A 2 3 m  24 m 2 4.60 h C  L C  36 in A  10 in  28 in  280 in 2

FR  γ  h C  A  62.4

#  1 ft   12 ft 2  2    364 #  36 in   280 in    12 in   ft 3  12 2 in 2 

FR 364 # #   91 # of bolts 4 bolts bolt Note: If accounting for slight differences between top and bottom bolts, Ic =0.1125 ft4, Lp = 3.019 ft, top bolt = 91.4 lb and bottom = 90.7 lb.

FBOLT 

4.61

hc = Lc = 1.5 m A WALL  L  W  3 m  7 m  21 m 2 h kN 3 m FR  γ SW   A WALL  10.1 3   21 m 2  318 kN 2 2 m

B  H 3 7 m  (3 m) 3   15.74 m 4 12 12 IC 15.75 m 4 LP  LC   1.5 m   2m LC  A 1.5 m  21 m 2 F P   F  PA A IC 

4.62

46

Chapter 4

π  (3 in) 2  7.07 in 2 4 π  (0.875 in) 2 A ROD   0.601 in 2 4 # FEXT .  P  A CYL  60 2  7.07 in 2  424 # in # FRET  P  (A CYL  A ROD )  60 2  (7.07 in 2  0.601 in 2 )  388 # in 2 π  (3 m) A  3.53 m 2 8 A CYL 

4.63

Y  0.212  D  0.212  3 m  0.636 m I C  6.86  10 3  (D) 4  6.86  10 3  (3m) 4  0.556 m 4

h c  L C  1.2 m  (1.5 m - 0.636 m)  2.064 m kN FR  γ  h c  A  9.81 3  2.064 m  3.53 m 2  71.5 kN m IC 0.556 m 4 LP  LC   2.064 m   2.14 m LC  A 2.064 m  3.53 m 2

4.64   180  110  70

20 m 20 m X  21.284 m X cos (20) hp 16 m L C  21.284 m   21.284 m   13.284 m 2 2 h c  L C  sin ( )  13.284 m  sin (70)  12.48 m cos (20) 

A G  L  W  16 m  8 m  128 m 2 kN FR  γ  h c  A  10.1 3  12.48 m  128 m 2  16.1  10 6 N m

Forces Due to Static Fluids

47

B  H 3 8 m  (16 m) 3   2731 m 4 12 12 IC 2731 m 4 LP  LC   13.284 m   14.9 m LC  A 13.284 m  128 m 2 IC 

M  0 A

0  (16.1 10 6 N  9.616 m)  FS  16 m  FS  9.7  10 6 N

4.65

F  PA   hA 2

1 ft   π  8   # 12 in   F  1.03  62.4 3  30 ft   673 # 4 ft

48

Chapter 4

CHAPTER FIVE BUOYANCY AND STABILITY 5.1

 FV  0  w  T  Fb Fb  γ f V  (10.5 kN/ m3 )(0.45)(0.60)(0.30)m 3  0.814 kN = T  Fb  w  814  258  556 N

5.2

814 N

If pipe is submerged Fb  γ f V  (1.26)(9.81 kN/ m 3 )(π(0.168 m 2 )/4)(1.00 m) Fb = 0.2740 kN = 274 N; because w = 277 N > Fb  It will sink. wc  Fb  0; wc  Fb ; γ cVc  γ f Vd  γ f 0.9 Vc

5.3

Then, γ c  0.9γ f  (0.90)(1.10)(62.4 lb/ft 3 )  61.78 lb / ft 3 wc  Fb  0; γ cVc  γ f Vd Vd  Vc

γc γf

 (0.30 m)2 7.90  D2 1.2 m  0.0683 m3  .X 4 9.81 4 4(0.0683)m3 X  0.9664 m = 966 mm  (0.30 m) 2 Y  1200  966  234 mm

5.4



5.5

w  Fb  0  γ cVc  γ f Vd γc  γ f

5.6

Vd  (0.90)(9.81 kN/m 3 )(75 /100)  6.62 kN / m 3 Vc

w  Fb  T  0  γ CVC  γ f VC  T  Vc (γ C  γ f )  T VC 

T 2.67 kN   0.217 m 3 (γC  γ f ) [23.6  (1.15)(9.81)] kN/m3

Buoyancy and Stability

49

5.7

w  Fb  FSP  0  W  γ 0Vd  FSP FSP  w  γ Vd  14.6 lb  (0.90)(62.4 lb/ft 3 )(40 in 3 )

1 ft 3 1728 in 3

FSP  13.3 lb 5.8

 FV  0  wF  wS  FbF  FbS FbS  γ wVS  9.81 kN/m3  (0.100 m)3  9.81 N 0  γ FVF  80 N  γ wVF  9.81 N 0 = VF (γ F  γ w )  70.19 N VF 

70.19 N 70.19 N  γF  γw (470  9810) N/m 3

 7.515×10-3 m 3

5.9

 (2) 2 .3 ft 3  588.1 lb 4 wA  Fb  wc  588.1  30  558.1 lb = γ AVA wc  wA  Fb  γ wVc  62.4 lb/ft 3.

VA  5.10

wA 558.1 lb in 3 1 ft 3  .  3.23 ft 3 γA 0.100 lb 1728 in 3

wc  wA  Fbc  FbA  0 wA  Fbc  Fbc  wc  588.1  30  558.1 lb(See Prob 5.10) γ AVA  γ wVA  558.1 lb γ A  (0.100 lb/in 3 )(1728 in 3 / ft 3 )  172.8 lb/ft 3 558.1 lb 558.1 lb   5.055 ft 3 γ A  γ w (172.8  62.4)lb/ft 3 If both concrete block and sphere are submerged:

VA 

Upward forces = FU  Fbs  FbC  γ wVs  γ wVc  γ w (Vs  VC ) Vs   D 3 / 6   (1.0 m)3 / 6  0.5326 m3 5.11

 Vtot  0.6973 m3 wC 4.10 kN 3 VC    0.1737 m γC 23.6 kN/m3   FU  FD It will float.

Downward forces = FD  wC  wS  4.1  0.20  4.30 kN  FU  γ wVtot  (9.81 kN/m3 )(0.6973 m3 )  6.84 kN

5.12

wc  Fb  0  γ cVc  γ f Vd  γ c S 3  γ f S 2 X X

γc S 3 γc S  γf S2 γf

wH  wS  Fb  0 5.13

50

  (1.0) 2  in 3 ft 3  (.25) 2 Fb  γ w (V1  V2 )  62.4 lb/ ft 3  .1.50   1.30  3 4  4  1728 in  0.04485 lb wS  Fb  wH  0.04485  0.020  0.02485 lb Chapter 5

Buoyancy and Stability

51

52

Chapter 5

FbS  γ wVB  9.81 kN/m 3 

 (.45) 2  .03 m3 4

 0.0468 kN FbC  γ wVd  wC  wB  FbS  0.771  0.4008  0.0468  1.125 kN FbC  γ wVd  γ w AX Fbc

X

γw A



1.125 kN  0.721 m submerged = 721 mm (9.81 kN/m3 )( (.45) 2 m 2 / 4 29 mm above surface

5.26

γ CT  15.60 kN/m3 wc  Fbc  wB  FbB  0

 (.45) 2  .75 m3  0.7753 kN 4  (.45) 2 Fbc  γ CTVd  15.60 kN/m3   .70 m3  1.737 kN 4 w B  FbB  γ BVB  γ CTVB  VB (γ B  γ CT )  Fbc  wc  1.737  0.7753 wc  γ cVc  6.50 kN/m3 

 0.9614 kN VB 

0.9614 kN 0.9614 kN   0.01406 m3  At 3 γ B  γ CT (84.0  15.60)kN/m

VB 0.01406 m3 t   0.0884 m = 88.4 mm A  (0.45) 2 / 4 m 2

w  Fb  γ f Vd  (1.16)(9.81 kN/m 3 )(0.8836 m 3 )  10.05 kN 5.27

Vd   D 3 /12   (1.50 m)3 /12  0.8836 m 3 Entire hemisphere is submerged.

5.28

w  Fb  γ f Vd  γ w  A  X X

w 0.05 N 4 kN w   3  0.965 103 m = 0.965 mm 3 2 γ A (9.81 kN/m )( (0.082 m ) 10 N

 D2 L 4 2 103 N 2  (.038 m) wS  76.8 kN/ m . .0.08 m .  6.97 N 4 kN wS  wC  6.97  0.05  7.02 N  Fb  γ w AX w 7.02 N 4 1 kN X T  . . 3  0.135 m = 135 mm 3 2 γ w A (9.81 kN/m )  (0.0802 m) 10 N

5.29A Wt. of steel bar = wS  γ SVS  γ S AL  γ S

Buoyancy and Stability

53

5.29 B From Prob. 5.29, wT  7.02 N

D2  (.038 m) 2 .L  .0.080 m = 9.073  10 5 m 3 4 4 FbS = γ wVS  9.81 kN/m3  9.073 10 5 m 3 103 N/kN = 0.890 N

VS 

wT  FbS  FbC  0 FbC  wT  FbS  7.02 N  0.890 N = 6.13 N = γ w AX X

5.30

6.13 N 4 1 kN   3  0.118 m = 118 mm 3 2 9.81 kN/m  (0.082 m) 10 N

wT  Fb  4γ wVdrum  4(62.4 lb/ft 3 )

( (21 in) 2 ) (36 in) 3 ft  1801 lb 4 1728 in 3

Drums Weight 4(30 lb) =120 lb Wt. of platform and load = 1801  120  1681 lb 5.31

Vol. of wood: 2(6.0 ft)(1.50 in)(5.50 in)(1 ft 2 / 144 in 2 )

 0.6875 ft 3 ends

4(96  3)in(1.5 in)(5.50 in)(1 ft 3 /1728 in 3 )  1.776 ft 3 main boards (0.50 in)(6 ft)(8 ft) (1 ft/12 in)

= 2000 ft 3 plywood 4.464 ft 3 total

ww  γ wV  (40. 0 lb/ft 3 )(4.464 ft 3 )  178.5 lb 5.32

wD  wP  Fb  γ wVd Vd 

wD  wP 120 lb +178.5 lb   4.78 ft 3 total 3 γw 62.4 lb/ft

VD  4.78 / 4  1.196 ft 3 sub. each drum VD  AS  L AS 

VD 1.196 ft 3 144 in 2   0.399 ft 2   57.4 in 2 L 3.0 ft ft 2

By trial: X = 4.67 in when As  57.4 in 2

wdrums  wwood  wload  FbD  FFw  0 wdrums  4(30 lb)  120 lb (Prob. 5.31) 5.33

wwood  178.5 lb(Prob. 5.32) FbD  1801 lb (Prob. 5.31) Fbw  γ wV w  62.4 lb/ft 3  4.464 ft 3  278.6 lb (Prob. 5.32) D

w

wload  FbD  Fbw  wD  ww  1801  278.6  120  178.5  1781 lb

54

Chapter 5

5.34

Given: γ F  12.00 lb/ft 3 , γ C  150 lb/ft 3 , wC  600 lb Find: Tension in cable Float only :  FV  0  wF  T  FbF T  FbF  wF But w F  γ FVF  12.0 lb/ft 3  9.0 ft 3  108 lb (18.0 in) 2 (48 in) VF   9.00 ft 3 3 3 1728 in / ft FbX  γ wVd  (64.0 lb/ft 3 )(6.375 ft 3 )  408 lb (18.0 in) 2 (34 in)  6.375 ft 3 3 3 1728 in / ft T  408  108  300 lb Check concrete block: Fnet  wC  FbC  T w 600 lb  4.00 ft 3 wC  600 lb; VC  C  3 γ C 150 lb/ft FbC  γ wVC  (64.0 lb/ft 3 )(4.00 ft 3 )  256 lb

Vd 

5.35

Fnet  600  256  300  44 lb down  OK  block sits on bottom.

Rise of water level by 18.00 in would tend to submerge entire float. But additional buoyant force on float is sufficient to lift concrete block off sea floor.

With block suspended: T = wC  FbC T  600  256  344 lb (see Problem 5.35) Float : wF  T  FbF  0 FbF  wF  T  108 lb + 344 lb = 452 lb = γ wVd FbF

3 452 lb 3 1728 in Vd    7.063 ft  γ w 64.0 lb/ft 3 ft 3

5.36

 12204 in 3 Vd  (18.0 in 2 )( X ) Vd 12204 in 3 X   37.67 in submerged (18.0 in) 2 324 in 2 Y  48  X  10.33 in above surface With concrete block suspended, float is unrestrained and it would dirft with the currents.  FV  0  wA  Fbw  Fbo  T T  wA  Fbw  Fbo 0.10 lb  (6.0 in)3  21.6 lb 3 in (6 in)2 (3.0in) Fbw  γ wVd w  62.4 lb/ft 3   3.90 lb 1728 in 3 / ft 3 Fbo  γ oVdo  0.85 Fbw  3.315 lb

wA  γ AVT 

Then: T  21.6  3.90  3.315  14.39 lb

Buoyancy and Stability

55

5.37

FS  wc  Fbc  γ cVc  γ f Vc  Vc (γ c  γ f ) Vc 

 D2  (6.0 in) 2 ft 3 .L   10.0in  4 4 1728 in 3

 0.1636 ft 3  0.284 lb 1728 in 3 62.4 lb  FS  0.1636 ft    ft 3 ft 3   in 3 3

 70.1 lb FS acts up on cylinder; down on tank bottom Stability 5.39

w  Fb  0 γ cVc  γ f Vd  γ f AX

 cVc γ c A(1.0 m) 8.00 kN/m3 / (1.0 m) X   γf A γf A 9.81 kN/m 3  0.8155 m ycg  1.00 m/2 = 0.500 m ycb  X /2  0.8155 m/ 2  0.4077 m MB =

I  D 2 / 64  (1.0 m) 4 / 64   Vd ( D 2 / 4)( X ) [ (1.0 m) 2 / 4](0.8155 m)

0.04909 m 4   0.0766 m 0.6405 m3 ymc  ycb  MB = 0.4077 + 0.0766 = 0.4844 m < ycg  unstable 5.40

w  Fb  γ wVd  γ w AX w 250 lb(144 in 2 / ft 2 ) X   0.4808 ft γ w A (62.4 lb/ft 3 )(30in)(40in) X  5.77 in; γ cb  X /2 = 2.88 in I  (40 in)(30 in) 3 /12  9000 in 4 Vd  AX  (30 in)(40 in)(5.77 in) = 6923 in 3 MB = I /Vd  90000 in 4 / 6923 in 3  13.0 in ymc  ycb  MB =2.88 in + 13.0 in  15.88 in > ycg  stable

56

Chapter 5

Buoyancy and Stability

57

58

Chapter 5

5.51

From Prob. 5.30, X = 118 mm VS = Vol. of steel bar = 9.073 ´ 104 mm3 p DC2 VCS = Sub. vol. of cup = ´X 4 p (82)2 VCS = (118) = 6.232 ´ 105 mm3 4 Vd = VS + VCS = 7.139 ´ 105 mm3 D yS = cb of steel bar = S = 19 mm 2 yCS = cb of sub. vol. of cup X 118 yCS = DS + = 38 + = 97 mm 2 2 y V + yCSVCS ycb = S S Vd (19)(9.073 ´ 104 ) + (97)(6.232 ´ 105 ) = 87.1 mm 7.139 ´ 105 I p (82) 4/64 MB = mm = 3.11 mm = Vd 7.139 ´ 105 ymc = ycb + MB = 90.2 mm ycg is very low because wt. of bar is >> wt. of cup—stable

=

5.52

From Prob. 5.22, Fig. 5.23: X = 600 mm ycb = X/2 = 300 mm ycg = H/2 = 750/2 = 375 mm 2 2 I p D 4 /64 D = (450) = 21.1 mm MB = = = Vd (p D 2 /4)( X ) 16 X 16(600) ymc = ycb + MB = 300 + 21.1 = 321.1 mm < ycg—unstable

Buoyancy and Stability

59

60

Chapter 5

Buoyancy and Stability

61

62

Chapter 5

Buoyancy and Stability

63

p DX4

I = 64

I

=

p (11.51/ 2) 4 64

= 53.77 in4

53.77 = 0.539 in V d 99.69 ycb = 0.75X = 0.75(11.51) = 8.633 in ymc = ycb + MB = 8.633 + 0.539 = 9.172 in ycg = 9.00 in < ymc—stable

MB =

5.63

(a)

=

å F = 0 = Fb - Wc - Wv v

Wc = Weight of contents; Wv = Weight of vessel; Find Wc + Wv Wc + Wv = Fb; But Fb = gfVd Vd = Vhs + Vcyl-d; Where Vhs = Vol. of hemisphere; Vcyl-d = Vol. of cyl. below surface Vhs = pD3/12 = p(1.50 m)3/12 = 0.8836 m3 Vcyl-d = pD2hd/4 = p(1.50 m)2(0.35 m)/4 = 0.6185 m3 Then Vd = Vhs + Vcyl-d = 0.8836 + 0.6185 = 1.502 m3 Fb = gfVd = (1.16)(9.81 kN/m3)(1.502 m3) = 17.09 kN = Wc + Wv (Answer) (b)

Find gv = Specific weight of vessel material = Wv/VvT; Given Wc = 5.0 kN From part (a), 17.09 kN = Wc + Wv; Then Wv = 17.09 - Wv = 17.09 - 5.0 = 12.09 kN VvT = Total volume of vessel = Vhs + Vcyl-T Vcyl-T = ( Do2 - Di2 ) (0.60 m)/4 = p[(1.50 m)2 - (1.40 m)2](0.60 m)/4 = 0.1367 m3 VvT = Vhs + Vcyl- = 0.8836 m3 + 0.1367 m3 = 1.020 m3 gv = Specific weight of vessel material = Wv/VvT; = (12.09 kN)/(1.020 m3) = 11.85 kN/m3 = gv

64

Chapter 5

(c)

Evaluate stability. Find metacenter, ymc. See figure for key dimensions.

Given: ycg = 0.75 + 0.60 - 0.40 = 0.950 m from bottom of vessel.

We must find: ymc = ycb + MB = ycb + I/Vd I = pD4/64 = p(1.50 m)4/64 = 0.2485 m 4 For circular cross section at fluid surface.

Vd = 1.502 m3 From part (a).

MB = (0.2548 m4)/(1.502 m3) = 0.1654 m The center of buoyancy is at the centroid of the displaced volume. The displaced volume is a composite of a cylinder and a hemisphere. The position of its centroid must be computed from the principle of composite volumes. Measure all y values from bottom of vessel. (ycb)(Vd) = (yhs)(Vhs) + (ycyl-d)(Vcyl-d) ycb = [(yhs)(Vhs) + (ycyl-d)(Vcyl-d)]/Vd We know from part (a): Vd = 1.502 m3; Vhs = 0.8836 m3; Vcyl-d = 0.6185 m3 yhs = D/2 - y = D/2 - 3D/16 = (1.50 m)/2 - 3(1.50 m)/16 = 0.4688 m ycyl-d = D/2 + (0.35 m)/2 = (1.50 m)/2 + 0.175 m = 0.925 m Then ycb = [(0.4688 m)(0.8836 m3) + (0.925 m)(0.6185 m3)]/(1.502 m3) = 0.657 m Now, ymc = ycb + MB = 0.657 m + 0.1654 m = 0.822 m From bottom of vessel.

Because ymc < ycg, vessel is unstable.

5.64

Let Fs be the supporting force acting vertically upward when the club head is suspended in the water. å Fv = 0 = Fs + Fb - W; Then Fs = W - Fb Fb = gwVal; W = galVal; Where Val = Volume of aluminum club head Val = W/gal = (0.500 lb)/(0.100 lb/in3) = 5.00 in3 Fs = W - Fb = 5.00 lb - (62.4 lb/ft3)(5.00 in3)(1 ft3)/(1728 in3) Fs = 0.319 lb

5.65

FB  γ F  VD 32 ft 2 1 1 ft V  1.8 yd2  2 2  in   0.338 ft 3 12 in 1 yd 4 # # FB W  ((1.03  62.4 3 ) - 38 3 )  0.338 ft 3  8.9 # ft ft

5.66

π  (0.5m) 2 VCYL.   2 m  0.393m 3 4 N WCYL.  0.393m 3  535 3  210 N m kN FB  γ F  VD  10.1 3  0.393 m 3  3.97 kN m F  0  F  F  W  FT  FB  W Y B T FT  3,760 N

Buoyancy and Stability

65

5.67

A. When hanging above the water, there is no buoyant force, so the tension in the cable is equal to the weight of the diving bell; 72 kN kN B. FB  γ F  VD  10.1 3  6.5 m 3  65.7 kN m F  0  T  W  F Y B  T  W  FB

T  72 kN  65.7 kN  6.3 kN C. When released from the cable, the bell will sink since when in the water, there is still tension in the cable holding the diving bell up.

5.68

m Weight Lifted  125 kg  9.81 2  1226 N s 1226 N  VD   441 m 3 N (11.81 - 9.03) 3 m 3 V 3 4 π r VSPHERE   r  3 SPHERE 3 4 π 441m 3  6  9.45 m diameter π The load must be carried below the balloon because the center of mass must be below the center of volume to be stable.

D3

5.69

5.70

5.71

66

When neutrally buoyant, 0 = FBD + FBL – WD - WL 0  γ SW  VD   78  9.81N  VLEAD  γ SW  γ L  N 1 m3 1.03  9810 3  82.5 L   78  9.81 N 1000 L m VLEAD  N   N   1.03  9810 3   11.35  9810 3  m   m   4 3 VLEAD  6.759  10 m  7.67 kg

N π  (0.12 m) 2 FB  γ F  VD  2.6  9810 3  4 m N FB  288  FT m Steel will float in any fluid that has a specific gravity higher than its own. Steel has a specific gravity of 7.85. From the table of common fluids listed in appendix, steel will float in Mercury, which has a specific gravity of 13.54.

Chapter 5

5.72

Naturally it will float since it has a specific weight less than that of water.

N 13 m 3 W  9125 3  125 cm 3   1.141 N m 100 3 cm 3 N 13 m 3 FB  9810 3  125 cm 3   1.226 N m 100 3 cm 3

Weights  FB  W  1.226 N  1.141 N  0.085 N

5.73

WCAM  WF  FB CAM  FB F WCAM  (γ F  VF )  (γ W  VCAM )  (γ W  VF ) WCAM  (γ W  VCAM )  VF  (γ W  γ F )

VF 

WCAM  (γ W  VCAM )  0.401 ft 3 γW  γF

π  (D) 2 1 ft  6 in  4 12 in VF  4 0.401 ft 3  4 D   1.01 ft 1 ft 1 ft      π  π  6in   6 in  12 in 12 in     VF 

5.74

WCAM  WF  FB F WCAM  (γ F  VF )  (γ W  VF )

WCAM  VF  (γ W  γ F )  VF 

WCAM 40 #   0.754 ft 3 γW  γF  #  #   62.4 3   0.15  62.4 3   ft  ft   

π  (D) 2 1 ft VF   6 in  4 12 in VF  4 0.754 ft 3  4 D   1.39 ft 1 ft 1 ft 6in  π 6 in  π 12 in 12 in

Buoyancy and Stability

67

CHAPTER SIX FLOW OF FLUIDS and BERNOULLI’S EQUATION Conversion factors 6.1

Q  8.0 gal/min  6.309  105 m3 /s/1.0 gal/min = 5.05  104 m 3s/

6.2

Q  759 gal/min  6.309  105 m3 /s/1.0 gal/min = 4.79  102 m 3 /s

6.3

Q  8720 gal/min  6.309  105 m3 /s/1.0 gal/min = 0.550 m 3 /s

6.4

Q  84.3 gal/min  6.309  105 m3 /s/1.0 gal/min = 5.32  103 m 3 /s

6.5

Q  125 L/min  1.0 m3 /s/60000 L/min = 2.08  103 m 3 /s

6.6

Q  4500 L/min  1.0 m3 /s/ 60000 L/min = 7.50  102 m 3 /s

6.7

Q  15000 L/min 1.0 m 3 /s/ 60000 L/min = 0.250 m 3 /s

6.8

Q  259 gal/min  3.785 L/min/ 1.0 gal/min = 980 L /min

6.9

Q  3720 gal/min  3.785 L/min/1.0 gal/min = 1.41  104 L /min

6.10

Q  23.5 cm3 /s  m3 / 100 cm  = 2.35  10 5 m 3 /s

6.11

Q  0.296 cm3 /s  1 m/ 100 cm  = 2.96  107 m 3 /s

6.12

Q  0.105 m3 /s  60000 L/min/1.0 m3 /s = 6300 L /min

6.13

Q  3.58  103 m3 /s  60000 L/min/1.0 m3 /s = 215 L /min

6.14

Q  5.26  106 m3 /s  60000 L/min/1.0 m3 /s = 0.316 L /min

6.15

Q  459 gal/min  1.0 ft 3 /s/ 449 gal/min = 1.02 ft /s

6.16

Q  15 gal/min1.0  ft 3 /s/449 gal/min = 3.34102 ft 3 /s

6.17

Q  6500 gal/min  1.0 ft 3 /s/449 gal/min = 14.5 ft 3 /s

6.18

Q  2.50 gal/min  1.0 ft 2 /s/449 gal/min = 5.57  103 ft 3 /s

6.19

Q  1.25 ft 3 /s  449 gal/min/1.0 ft 3 /s = 561 gal /min

6.20

Q  0.06 ft 3 /s  449 gal/min/1.0 ft 3 /s = 26.9 gal /min

68

3

3

Chapter 6

6.21

Q  7.50 ft 3  449 gal/min /1.0 ft 3 /s  3368 gal /mins

6.22

Q  0.008 ft 3 /s  449 gal/ min/1.0 ft 3 /s  3.59 gal /min

6.23

Q  500 gal/min  1.0 ft 3 /s / 449 gal/min  1.11 ft 3 /s Q  2500/449  5.57 ft 3 /s Q  500 gal/min  6.309  105 m3 /s/1.0 gal/min  3.15 ×102 m 3 /s Q  2500(6.309  105 )  0.158 m 3 /s

6.24

Q  3.0 gal/min  1.0 ft 3 /s/449 gal/min  6.68×103 ft 3 /s Q  30.0/449  6.68×102 ft 3 /s Q  3.0 gal/min  6.309  105 m3 /s/1.0 gal/min  1.89×104 m 3 /s Q  30(6.309  105 )  1.89×103 m 3 /s

6.25

745 gal 1h 1.0 ft 3 /s Q    2.77 ×102 ft 3 /s h 60 min 449 gal/min

6.26

Q

0.85 gal 1h 1.0 ft 3 /s    3.16×105 ft 3 /s h 60 min 449 gal/min

6.27

Q

11.4 gal 1h 1.0 ft 3 /s    1.76×105 ft 3 /s 24 h 60 min 449 gal/min

6.28

Q

19.5 mL 1.0 L 1.0 m3 /s  3   3.25×107 m 3 /s min 10 mL 60000 L/min

Fluid flow rates

6.29

W  γQ  (9.81 kN/m3 )(0.075 m3 /s)  0.736 kN/s(103 N/kN)  736 N /s M  ρQ  (1000kg/m3 )(0.075 m3 /s)  75.0 kg /s

6.30

W  γQ  (0.90)(9.81 kN/m3 )(2.35  103m3 /s)  2.7  102 kN/s  20.7 N /s M  ρQ  (0.90)(1000 kg/m3 )(2.35  103 m3 /s)  2.115 kgs

6.31

Q

W 28.5 N m3 1.0 kN 1.0 h    3   7.47 × 107 m 3 /s γ h 1.08(9.81 kN) 10 N 3600s

M  ρQ  (1.08)(1000 kg/m 3 )(7.47×10 7 m 3 /s)  8.07 × 104 kg /s

6.32

W 28.5 N m3 1h Q     6.33 × 104 m 3 /s γ h 12.50 N 3600 s

6.33

M  ρQ 

1.20 kg 840 ft 3 1.0 slug 1 min 0.0283m3     m3 min 14.59 kg 60 s ft 3

 3.26×102 slug /s M  ρQ  γQ /g  W /g 32.2 ft 3.26 102 slug 1 lb  s2 /ft 3600 s W  gM      3779 lb /hr s3 s slug hr Flow of Fluids

69

6.34

W  γQ  (0.075 lb/ft 3 )(61000 ft 3 /min)  4575 lb /min M  ρQ 

6.35 6.36

Q

γQ W 4575 lb/min 142 lb  s 2 /ft     142 slugs /min g g 32.2ft/s 2 min

W 1200 lb ft 3 1 hr     5.38 ft 3 /s γ hr 0.062 lb 3600 s

62.4 lb 1.65 gal 1.0 ft 3 w    13.76 lb/min  3 ft min 7.48 gal t w 7425 lb min 1 hr t     8.99 hr W 13.76 lb 60 min W  γQ 

Continuity equation

6.37

Q  Aυ : A =

Q 75.0 ft 3 /s πD 2   7.50 ft 2  υ 10.0 ft/s 4

D  4A/π  4(7.50)/π  3.09ft 2

2

6.38

D A 1.65 ft  12  A1υ1 = A2υ2 ; υ2  υ1 1  υ1  1      26.4ft /s A2 s  3  D2 

6.39

Q  2000 L/ min × υ1 

1.0 m 3 /s  0.0333 m 3 /s 60000 L/min

Q 0.0333m 3 / s   0.472 m /s A1 π (.30 m) 2 / 4

Q 0.0333m 3 /s υ2    1.89 m /s A2 π (.15 m) 2 /4 2

2

6.40

D A  150  A1υ1  A2υ2 ; υ2  υ1 1  υ1  1   1.20 m/s   0.300m /s  300  A2  D2 

6.41

Q1  A1υ1  A2υ2  A3υ3  Q2  Q3 Q2  A2υ2  1.735  103 m 2  12.0 m/ s  0.02082 m3 / s Q3  Q1  Q2  0.072  0.02082  0.05118 m3 / s υ3 

6.42

70

Q3 0.05118 m3 / s   7.882 m / s A3 6.793  103 m2

Amin 

Q 1ft 3 / s 1  10 gal/ min    0.02227 ft 2 : 2 - in Sch. 40 pipe υ 449 gal/ min 1.0 ft/ s

Chapter 6

Flow of Fluids

71

Flow of Fluids

72

Flow of Fluids

73

6.62.

Pt A at gage; Pt B outside nozzle:

pA υ2 p υ2  Z A  A  B  zB  B ; p B  0 γ γ 2g 2g

υA2  υB 2 p 620 kN   z B  z A   A  3.65 m  2   59.55 m 2g γ m  9.81kN / m3  A A  4.347 103 m 2 A B  0.962 103 m 2 υB  υA ( AA / AB )  υA 

υB 2  20.419 υA2

4.347  103 m 2  4.5187υA 0.962  103 m 2

υA 2  υB 2  υA 2  20.419υA 2  19.419  19.419 υA 2  2 g ( 53.94 m) 19.419 υA 2  2 g (53.94 m) 2 g (59.55 m) 2(9.81m/ s 2 )(59.55 m)   7.76 m/ s 19.419 19.419 Q  A A υA  4.387  103 m 2  7.76 m/ s  0.0340 m3 / s  3.40×10-2 m 3 / s UA 

6.63

6.64

pA υ2 p υ2  z A  A  B  z B  B ; p B  0, z A  z B γ 2g γ 2g 2 2 2 2 2 2 2  υ  υA  60.6 lb  (75  42.19 ) ft / s  1ft pA  γ  B   ft 3   144 in 2  25.1 psig 2(32.2 ft/ s 2 )    2g  2 2 D  A  .75  υB  75 ft/ s; υA  υB B  75  B   75    42.19 ft/ s  1.0  AA  DA  Pt . A before nozzle; Pt . B outside nozzle :

Q  10 gal/ min 

1ft 3 / s 0.0223ft 3 Q 0.0223 ft 3 / s 3.71ft ; υA     449 gal/ min s AA 0.0060 ft 2 s

Q 0.0223 0.955ft p A υ A 2 pB υB 2 υB  ; ; z A  zB    zA    zB  AB 0.02333 s γK 2g γK 2g  2   A 2  50 lb  (0.9552  3.712 ) ft  1ft 2 p A  pB  γ K  B  0.0694 psi  3  ft  2(32.2 ft/ s 2 )  144 in 2  2g  6.65

Pt. 1 at water surface; Pt. 2 outside nozzle.

p1 υ2 p υ2  z1  1  2  z2  2 ; p1  0, υ1  0, p2  0 γ 2g γ 2g υ 2  2 g ( z1  z2 )  2(9.81m/ s 2 )(6.0 m)  10.85 m/ s

Q  A 2 2 

 (0.050 m) 2

 10.85 m/ s  2.13×10-2 m 3 / s

4 Q 0.0213 m3 / s  A2 1.2222 m / s 2 A  1.222 m / s;     0.0761 m AA 1.744  102 2g 2(9.81 m / s 2 )

p1 2 p 2  z1  1  A  z A  A ; p1  0, 1  0 γ 2g γ 2g   2  9.81 KN p A  γ w (z1  z A )  A    6.0 m  0.0761 m   58.1kPa 2g  m3 

74

Chapter 6

Flow of Fluids

75

76

Chapter 6

Flow of Fluids

77

78

Chapter 6

Flow of Fluids

79

80

Chapter 6

Flow of Fluids

81

82

Chapter 6

Flow of Fluids

83

84

Chapter 6

Flow of Fluids

85

86

Chapter 6

2

6.107

2

P1 V1 P V   Z1  2  2  Z 2 γ 2g γ 2g 2

V2 m  V2  Z1  2g  3 m  2  9.81 2 2g s m V2  7.67 s π  (0.020 m) 2 m m3 Q  AV   7.67  2.41  10 3  2.41 L S 4 s s

 Z1 

Flow of Fluids

87

6.108 Q 6 

 A V  A V ....  A V 1

1

2

2

Q 6  A  V  6 Jets 

Q  AV  A 

6

6

  (0.012m) 2 4

 12

m m3  6 Jets  8.14  10 3 s s

Q π  D2 Q   D V 4 V

4Q πV

 m3   4   8.14  10 -3 s   D  64 mm m π  2.5 s 3 16 oz 1 gal 1 ft 3 3 ft Q     2.79  10 6.109 6 s 128 oz 7.48 gal s 3 ft 2.79  10 3 Q ft s Q  AV  V    0.908 1 ft 2  A  s )   π  (0.75 in  12 in   4       2 2 2 V  P1 V1 P V   Z1  2  2  Z 2  P1   2  Z 2    γ 2g γ 2g  2g  2   ft    0.908   1ft  # # # s P1     52 in   62.4 3  1.08  293 2  2.03 2   ft  12in ft ft in   2  32.2 2   s     2 2 2 P1 V1 P V V 6.110   Z1  2  2  Z 2  Z 2  1  V1  Z 2  2g γ 2g γ 2g 2g

m m V1  7m  2  9.81 2  11.7 s s 2 π  (0.005m) m 1000 L L Q  AV   11.7   0.23 3 4 s s 1m V 3L Time    13 s L Q 0.23 s

88

Chapter 6

6.111

 L 13 m 3 1 min  12    Q  min 1000 L 60 s  m Q  AV  V    2.55 2 A s  π  0.010 m       4  

 π  0.010 m 2 m    2.55 4 s  A1 V1  m A1 V1  A 2 V2  V2    20.82 2 A2 s  π  0.0035 m       4   2 2  V1 2  V2 2 P1  P1 V1 P2 V2   Z1    Z 2  P2      γ γ 2g γ 2g 2g γ  2 2     2.55 m    20.82 m   N  s  s  180,000 Pa  P2     9810 3   33.5 kPa  m N m  2  9.81 2 9810 3    s m  

2

2

2

P1 V1 P V V   Z1  2  2  Z 2  Z 2  1  50ft.

6.112 γ 2g γ 2g 2g ft ft ft V1  Z1  2  32.2 2  50ft  2  32.2 2  56.7 s s s 2

D  ft  2 in  ft V1  V2   2   56.7     9.07 s  5 in  s  D1  2 2  V 2  V1 2  P1 V1 P V  γ   Z1  2  2  Z 2  P1   2  γ 2g γ 2g 2g   2 2     56.7 ft    9.07 ft   # # #  s  s  P1     62.4 3  3,035 2  21.1 2  ft ft in in   2  32.2 2   s   3 gal 1 ft ft 3   1.34 6.113 Q CM  10 min 7.48 gal min 2

1 ft 3 Q S   Q CM  0.079 17 min 16 ft 3 Q M   Q CM  1.261 17 min

Flow of Fluids

89

D

ft min  0.059 ft  12 in  0.71in ft 60 s 1 ft π8  s 1 min 4  1.34

4Q  D CM  πV

DS  0.174in

D M  0.694in  gal 1 ft 3  1000   min 7.48 gal  Q  ft 6.114 Q  A  V  V    1532 2 2 2 A  π  4 in  min 1 ft     2 2  4 12 in   2

2

2

P1 V1 P V V   Z1  2  2  Z 2  Z 2  1 γ 2g γ 2g 2g 2

ft 1 min    1532  min 60 s   Z2   10.1 ft ft 2  32.2 2 s 2 2 2 P1 V1 P2 V2 V2   Z1    Z 2  Z1  6.115 γ 2g γ 2g 2g 2

Z1 

V2 ft ft  V2  Z1  2g  12 ft  2  32.2 2  27.8 2g s s 2

1 ft   π   0.5in   ft ft 3 12 in   Q  AV   27.8  0.038 4 s s As the water level in the pool decreases, so will the flow rate of the water through the hose due to a loss of elevation head.  gal 1 min 231 in 3  2    1 gal  1 ft Q  min 60 s ft 6.116 Q  A  V  V     212 2 A 12 in s  π  0.062in       4  

90

Chapter 6

CHAPTER SEVEN GENERAL ENERGY EQUATION

General Energy Equation

91

92

Chapter 7

General Energy Equation

93

94

Chapter 7

General Energy Equation

95

96

Chapter 7

General Energy Equation

97

98

Chapter 7

General Energy Equation

99

100

Chapter 7

General Energy Equation

101

102

Chapter 7

7.46

7.47

2

2

2

2

P1 V1 P V   Z1  h A  h L  2  2  Z 2  h A  Z 2  h L γ 2g γ 2g h A  2.5 m  1.8 m  4.3 m Nm   1   s  0.65  125 W  1W     P m3   P  hA  γ Q  Q    0.0019 N  hA  γ s   4.3 m  9810 3  m   m 3 1000 L 60 s L 0.0019    114 3 s 1 min min 1m

2

P1 V1 P V V   Z1  h A  h L  2  2  Z 2  h A  2 γ 2g γ 2g 2g ft # 550 s  31,570 ft # PA  70 HP  0.82  1 HP s

General Energy Equation

103

V P  2g PA  h A  γ  Q  2  γ  A  V  V 3  A 2g γA 2

ft # ft  2  32.2 2 P  2g s s  69.2 ft V3 A 3 2 γA s # π  0.35417ft  62.4 3  4 ft ft 3600 s 1 mile 69.2    47.2 mph s 1 hr 5,280 ft 31,570

2

7.48

A.

2

2

P1 V1 P V V   Z1  h A  h L  2  2  Z 2  Z 2  1 γ 2g γ 2g 2g m m VY  Z 2  2  g  15m  2  9.81 2  17.2 s s VY m V  19.8 Sin 60 s

B.

m π  0.045 m  m3 Q  A  V  19.8   0.0315 s 4 s

C.

P1 V1 P V V   Z1  h A  h L  2  2  Z 2  h A  2 γ 2g γ 2g 2g

2

2

2

2

2

m  19.8  s hA    20.0 m m 2  9.81 2 s

PA  h A  γ  Q  20.0 m  9810 D. P 

7.49

104

N m3  0.0315  6.2 KW s m3

PA 6.2 KW   7.5 KW ε 0.82

 gal 1 hr 1 ft 3  1600    hr 3600 s 7.48 gal  Q  ft Q  AV  V    9.90 A 0.006 ft 2 s 2 2 2 P1 V1 P V V   Z1  h A  h L  2  2  Z 2  h A  2  Z 2  h L γ 2g γ 2g 2g

Chapter 7

2

ft    9.90  s hA    9 ft  3.8 ft  14.3 ft ft 2  32.2 2 s # gal 1 hr 1 ft 3 ft # PA  h A  γ  Q  14.3 ft  62.4 3  1600    53.0 hr 3600 s 7.48 gal s ft P 800 W  h P  ET  T    11.1 hrs E    ft # 1.356 W   53   ft #  s  1   s   2

7.50

2

V2  Z1  h L  2g  Q  AV 

3 m  2.8 m   2  9.81 m2   1.98 m 

s 

π  0.020 m m m  2.00  6.22  10 4 4 s s

2

7.51

2

P1 V1 P V V   Z1  h A  h L  2  2  Z 2  2  Z1  h L γ 2g γ 2g 2g

2

s

3

2

P1 V1 P V   Z1  h A  h L  2  2  Z 2  Z1  h R  2.5 m γ 2g γ 2g N L 1 min 1 m 3 PA  h R  γ  Q  2.5 m  9810 3  150    61.3 W min 60 s 1000 L m P 61.3 W Pε    36.8 W ε 0.6 2

7.52

 m 7  2 2 2 P1 V1 P2 V2 V2 s   Z1  h R  h L    Z2  h R     2.50 m m γ 2g γ 2g 2g 2  9.81 2 s 2 N m π  0.008 m N-m P  h R  γ  Q  2.50 m  9810 3  7   40  344 @ 100 % Eff . s 4 s m 745.7W N-m 0.5HP   373 2 s 344W  100  92% , Which gives 8% allowable inefficiency 373W

General Energy Equation

105

2

7.53

106

2

P1 V1 P V   Z1  h A  h R  h L  2  2  Z 2 γ 2g γ 2g # 1200 2 P1  P2 in hR    37,129 in γ # 13 ft 3 0.895  62.4 3  3 3 ft 12 in 80 HP 1 ft # PR  h R  γ  Q   37,129 in   62.4 3  Q 0.87 12 in ft ft-#   550   s   80 HP  1 HP     0.87     ft 3   Q  0.293  131 gal min 1 ft # s 37,129 in   62.4 3  0.895 12 in ft

Chapter 7

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

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

b)

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

8.39

p1 υ2 p  2  z1  1  hL  B  z B  B γw 2g γw 2g   υ 2 pB  γ w ( z1  z B )  B  hL  2g  

Pt .1 at tank surface p1  0, υ1  0 Q

900 L/ min 1m3 / s  0.015 m3 / s 60000 L/ min

υ

Q 0.015 m3 / s   2.208 m/ s A 6.793  103 m 2

υD (2.208)(0.093) D 0.093   1.58 105 :   62000 : f  0.0162 6 v 1.30 10  1.5 106 L υ2 80.5 (2.208) 2 hL  f  (0.0162)   3.485 m D 2g 0.093 2(9.81) NR 

8.40

  (2.208) 2 pB  9.81 kN/ m 3 12   3.485  m  81.1kPa 2(9.81)   1 ft 2 / s Q 0.1114 ft 3 / s Q  50 gal/ min  0.1114 ft 3 / s : υ    18.56 ft/ s 449 gal/ min A 0.0060 ft 2 p p1 υ 12 υ 22  z1   hA  hL  2.  z2  : p1  0, υ1  υ2  0 γw 2g γw 2g p hA  2  ( z2  z1 )  hL γw υD (18.56)(0.0874) D 0.0874 NR    1.34  105 :   583 : f  0.0243 5 v 1.21  10 ε 1.5  104 L υ2 225 (18.56)2 hL  f ft  335ft  (0.0243)  D 2g 0.0874 2(32.2) hA 

(40 lb) ft 3 144in 2   220 ft  335ft  647 ft in 2 62.4 lb ft 2

p A  hA  wQ  (647 ft)(62.4 lb/ft 3 )(0.1114ft 3 /s ) / 550  8.18 hp (b) Increase the pipe size to 1 1/2 - in Schedule 40. Results : υ  7.88ft/ s; N R  8.74 104 ; D /   895; f  0.0232;Then, hL  37.5 ft; hA  349.8ft; Power = PA = 4.42 hp.

8.41

p1 p υ 12 υ 22  z1   hL  2  z2  : p1  p2   0 hL γo 2g γo 2g υ=

Q 60 gal/ min 1ft 3 / s    9.45 ft/s A 0.01414 ft 2 449 gal/ min

NR 

υD 





(9.45)(0.1342)(0.94)(1.94)  272 Laminar 8.5 103

L υ2 64 40 (9.45) 2 hL  f    ft  97.23ft D 2 g 272 0.1342 2(32.2) p1  p2  γ o hL  (0.94)

(62.4 lb) 1ft 2 (97.23 ft)  39.6 psi ft 3 144in 2

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

117

8.42

pA υ2 A pB. υ2 B  zA   hA  hL   zB  γ 2g γ 2g 2 2 p  pA υ  υA hA  B  zB  zA  B  hL γ 2g Q 0.50 ft 3 / s Q 0.50 υA    7.28ft/ s : υB    15.03 ft/s 2 AA 0.06868 ft AB 0.03326 N RB 

υB D



(15.03)(0.2058)(1.026)(1.94)  1.54  105 5 4.0  10

 D 0.2058   1372 : f  0.020  1.5  104 L υ2 80 (15.03) 2 hL  f ft  27.28ft  (0.020)  D 2g 0.2058 2(32.2) hA 

 25.0  (3.50) lb ft 3 144in 2  80 ft  (15.03)2  (7.28)2 ft 2 / s2  27.28 ft  174.1ft in 2 (1.026)(62.4 lb) ft 2

2(32.2 ft/ s 2 )

p A  hA γQ  (174.1 ft)(1.026)(62.4 lb/ ft 3 )(0.50 ft 3 / s ) / 550  10.13hp 8.43

υDρ QDρ QDρ 4Qρ 4Qρ : Dmin     2 πD η Aη πN R η  Dη  4 4(0.90 ft 3 / s)(1.24)(1.94 lb  s 2 / ft 4 )  0.184 ft Dmin   (300)(5.0  102 lb  s/ ft 2 ) NR 

2 1/2  in Type K Copper Tube : D  0.2029 ft : A  0.03234 ft 2

υ=

Q 0.90 ft 3 /s   27.8 ft/s A 0.03234 ft 2

NR 

υDρ (27.8)(0.2029)(1.24)(1.94)   272 η 5.0  102

p1  p2  γ g hL  γ g f

8.44

L υ2 64 55 (27.8) 2 lb 1ft 2  (1.24)(62.4)   D 2g 272 0.2029 2(32.2) ft 2 144in 2

 411 psi p p1 υ 12 υ 22  z1   hL  2.  z2  γw 2g γw 2g   υ 12  hL  p1  γ w  ( z2  z1 )  2g  

Pt .1 at pump outlet in pipe.

Pt . 2 at reservoir surface. p2  0, υ2  0 Q 4.00ft 3 /s υ   11.52ft/s A 0.3472 ft 2

υD (11.52)(0.6651) D 0.6651   6.33  105 :   4434 : f  0.0155 5 v 1.21 10  1.5 104 L υ2 2500 (11.52) 2 hL  f . ft  120.1ft  (0.0155) D 2g 0.6651 2(32.2)

NR 

p1 

118

 ft 1ft 2 62.4 lb  (11.52) 2 210 120.1    144 in 2  142.1psi ft 3  2(32.2)  Chapter 8

8.45

Pt . 0 at pump outlet inlet .

p0 p1. υ 02 υ 12  z0   hA   z1  γw 2g γw 2g

hA 

Pt .1 at pump outlet .

p1  p0 [142.1  (2.36)]lb  γw (62.4 lb/ft 3 )(in 2 )(1 ft 2 /144in 2 )

Assume z0  z1 , υ0  υ1

hA  335.5ft  lb/lb

333.5ft  lb 62.4 lb 4.00 ft 3 1 hp . . .  151 hp p A  hA γ wQ  3 lb ft s 550 ft  lb/ s 8.46

pA υ 2 p υ B2  zA  A  hL  B  zB  γg 2g γw 2g

pA  pB  γ g [( zB  zA )  hL ]

Q 4.00 ft 3 /s   7.76ft/s A 0.5479 ft 2 Assume sg = 0.68  From App.D.

υ

υDρ (7.76)(0.8350)(1.32)   1.19  106 6 η 7.2.  10 D 0.8350   5567 : f  0.0145 s 1.5  104

NR 

8.47

L υ2 3200 (7.76) 2 hL  f  0.0145.   51.9 ft D 2g 0.8350 2(32.2) 42.4 lb ft 1 ft 2 [85  51.9] pA  40.0 Psig   80.3psig ft 3 144 in 2 p1 p υ 12 υ 22 Pt .1 at tank surface, p1  0, υ1  0  z1   hA  hL  2  z2  γo 2g γo 2g Pt . 2 in outlet stream from 3  in pipe. p2  0 2 υ2 300 gal/min 1ft 3 /s hA  ( z2  z1 )   hL Q  0.668ft 3 /s 2g 449 gal/min Q 0.668ft 3 / s oil-App.C υ4    7.56 ft/ s A4 0.08840 ft 2 Q 0.668 ft 3 /s   13.02 ft/s  υ2 A3 0.05132 ft 2 L υ2 L υ2 hL  hL3  hL4  f 3 3 3  f 4 4 4 D3 2 g D4 2 g υ3 D3 (13.02)(0.2557) 64 N R3    1548(Laminar) : f 3   0.0413 3 v NR 2.15  10

υ3 

υ4 D4 (7.56)(0.3355) 64   1180(Laminar) : f 4   0.0543 3 2.15  10 v NR 75 (13.02) 2 25 (7.56) 2 hL  0.0413    0.0543    35.5ft 0.2557 2(32.2) 0.3355 2(32.2) (13.02) 2 hA  1.0 ft  ft  35.5ft  39.1ft 2(32.2) (62.4 lb) (0.668ft 3 ) 1hp PA  hA γ o Q  (39.1 ft)(0.890)  2.64 hp 3 ft s 550 ft lb/ s N R4 



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

119

8.48

p p1 υ 12 υ 22  z1   hL  2.  z2  : υ1  υ2 γo 2g γo 2g p1  p2  γ o [( z2  z1 )  hL ] υDρ (3.65)(0.0176)(930) 64   1805 (Laminar) : f   0.0355 2 v 3.31  10 NR L υ2 17.5 (3.65)2 hL  f  (0.0355)   23.97 m D 2g 0.0176 2(9.81) p1  p2  9.12 kN/m3 [1.88 m  23.97 m]  201 kPa NR 

8.49

p1  p2  γ g [( z2  z1 )  hL ](From 9.24)

Q

180 L/min 1m3 /s  0.003m3 /s 60000 L/min

υDρ (0.690)(0.0744)(1258)  0.003 Q  0.690 m/s υ  η 0.960 A 4.347  103 υDρ (0.690)(0.0744)(1258)  NR  η 0.960 64 N R  67.3(Laminar) : f   0.951 NR L υ2 25.8 (0.690) 2 hL  f  (0.0951)    8.00 m D 2g 0.0744 2(9.81) NR 

p1  p2  12.34 kN/m3 [0.68 m  8.00 m]  107 kPa Note : For problems 8.52 through 8.62 the objective is to compute the value of the friction factor , f , fromtheSwamee - Jain equation(8  7) from section 8.8,shown below : 0.25 f  2   1 5.74    0.9    log    3.7( D /ε) N R   For each problem, the calculation of the Reynolds number and relative roughness

are shown followed by the result of the calculations for f .

8.50

Water at 75°C; v  3.83  107 m 2 /s

Q 12.9 L/min 1 m3 /s υ  .  1.724 m/s A 60000 L/min 1.247  104 m 2 υD (1.724)(0.0126)   5.67  104 7 v 3.83  10 D /ε  0.0126 /1.5  106  8400; f  0.0207 NR 

8.51

Benzene at 60 C :   0.88(1000)  880 kg/ m3 ;  3.95  104 Pas Q 20 L/ min 1 m3 / s  0.719 m/ s υ  .

A 60000 L/ min 4.636  104 m 2

(0.719)(0.0243)(880)  3.89  104 4 3.95  10  D /   0.0243 / 4.6  105  528; f  0.0273 NR 

120

υDρ



Chapter 8

8.52

Water at 80 F : v  9.15  106 ft 2 / s D  0.512 ft; A  0.2056 ft 2

Q 2.50 ft 3 / s   12.16 ft/ s A 0.2056 ft 2 υD (12.16)(0.512) 0.512 NR    6.80  105 : D /    1280 6 v 9.15  10 4  104

υ

f  0.0191 8.53

Water at 50 F : v  1.40  105 ft 2 / s D  18in(1ft/ 12 in)  1.50 ft; A 

πD 2  1.767 ft 2 4

Q 15.0 ft 3 / s   8.46 ft/ s A 1.767 ft 2 1.50 υD (8.49)(1.50)   9.09  105 : D /    3750 NR  5 1.40  10 4  104 v f  0.0155

υ=

8.54

Water at 60 F : v  1.21  105 ft 2 / s



1ft 3 / s Q 1500 gal/ min    6.097 ft/ s 0.5479 ft 2 449 gal/ min A

NR 

D v



(6.097)(0.835) 0.835  4.21  105 : D/    5567 5 1.21  10 1.5  104

f  0.0156 8.55

A  D 2 / 4  (0.025) 2 / 4  4.909 104 m 2

Q 95 L/ min 1m3 / s υ    3.23m/ s A 4.909 ft 2 104 m 2 60000 L/ min NR 

υD 





(3.23)(0.025)(1.10)(1000)  4.44  104 : D /   Smooth [LargeD/  ] 3 2.0  10

f  0.0213 8.56

Crude oil(sg  0.93) at100 C   (0.93)(1000 kg/ m3 )  930 kg/ m3 ;  7.8 103 Pas

υ

Q 1200 L/ min 1m3 / s    1.19 m/ s A 1.682 102 m 2 60000 L/ min

NR 

 D  (1.19)(0.1463)(9.30)   2.07 104 3 7.8 10 

D/  

0.1463  3180; f  0.0264 4.6  105

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

121

8.57

Water at65C; v  4.39  107 m 2 / s D  0.0409m υD (10)(0.0409) NR    9.32  105 7 4.39  10  0.0409 D/   889; f = 0.0206 4.6  105

8.58

Propyl alcohol at 25 C;    kg / m3

  1.90  103 Pas Q 0.026m3 / s   5.98 m / s A 4.347  103 m 2 υD (5.98)(0.0744)(802) NR    1.86  105  1.92  103 (0.0744) D/   49600; f = 0.0159 1.5  106

υ=

8.59

Water at 70F; v = 1.05  105 ft 2 / s Q 3.0 ft 3 / s   3.82 ft/ s A 0.7854 ft 2 υD (3.82)(1.00) NR    3.64  105  1.05  105 1.00 D/   2500; f = 0.0175 4.0  104

υ=

8.60

Heavy fuel oil at 77F;ρ = 1.76 slugs / ft 3

  2.24 103 lb.s / ft 2 υ  12ft / s; D  0.05054 ft υD 

(12.0)(0.5054)(1.76)  4.77  103 3 2.24 10  0.5054  3369; f  0.0388 D/  1.5  104 NR 



Hazen - Williams Formula 8.61

Q  1.5 ft 3 /s, L  550 ft, D  0.512 ft, A  0.2056 ft 2 R  D / 4  0.128 ft, Ch  140 1.852

  Q hL  L  0.63  1.32 ACh R 

1.852

  1.50 hL  550  0.63   (1.32)(0.2056)(140)(0.128) 

122

 15.22ft

Chapter 8

8.62

1 m3 /s Q  1000 L/ min   0.0167m3 /s; L  45 m 60000 L/min 120mm OD 3.5 mm wall copper tube; D  113mm  0.113 m, A  1.003  102 m 2 R  D /4  0.0283 m, Ch  130 1.852

  Q hL  L  0.63   0.85 ACh R 

1.852

  0.0167  45  0.63  2  0.85(1.003  10 )(130)(0.0283)  hL  1.220m 8.63

Q  7.50ft 3 /s; L  5280 ft, D  18 in  1.50ft, A  1.767 ft 2 R  D /4  0.375 ft; Ch  100 1.852

8.64

  7.50 hL  5280   28.51 ft 0.63   (1.32)(1.767)(100)(0.375)  1 ft 3 /s Q  1500 gal/min   3.341ft 3 /s; L  1500 ft 449gal/min D  10.02 in  0.835ft; A  0.5479 ft 2 Ch  100; R  D /4  0.2088 1.852

  3.341 hL  1500  0.63   (1.32)(0.5479)(100)(0.2088)  hL  31.38 ft 8.65

Q  900 L/min 

1 m3 /s  0.015m3 /s; L  80 m 60000 L/min

D  112 mm  0.112 m; A  1.003  102 m2 R  D /4  0.0283 m, Ch  130 1.852

  0.015 hL  80  2 0.63   0.85(1.003  10 )(130)(0.0283)  8.66

 1.78 m

Q  0.20 ft 3 /s; D  2.469 in  0.2058 ft; A  0.03326 ft 2 Ch  100; R  D /4  0.0515 ft υ = Q /A = 6.01 ft/s (OK) 1.852

  0.20 hL  80  0.63   (1.32)(0.03326)(100)(0.0515) 

 8.35 ft

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

123

8.67

Q  2.0 ft 3 /s; L  2500 ft a)

8 - in Schedule 40 steel pipe; D  0.6651 ft; A  0.3472 ft 2 R  D /4  0.1663 ft; Ch  100 1.852

  2.0 hL  2500  0.63   (1.32)(0.3472)(100)(0.1663) 

 61.4 ft

b) Cement lined 8-in ductile iron pipe D  8.42 in  0.702 ft; A  0.3867 ft 2 ; Ch  140 R  D /4  0.1755 ft 1.852

  2.0 hL  2500  0.63   (1.32)(0.3867)(140)(0.1755)  8.68

 25.3 ft

Specify a new Schedule 40 steel pipe size. Use Ch  130 1 ft 3 /s Q  300 gal/min   0.668 ft 3 /s 449 gal/min s  10 ft/1200 ft  0.00833 ft/ft  2.31(0.668)  D 0.54   (130)(0.00833) 

0.380

 0.495 ft

6-in Schedule 40 steel pipe; D = 0.5054 ft Actual hL for 6-in pipe D  0.5054 ft; R  D /4  0.1264 ft A  0.2006 ft 2 1.852

  Q hL  L  0.63  1.32 ACh R 

1.852

  0.668 hL  1200  0.63   (1.32)(0.2006)(130)(0.1264)  8.69

 9.05 ft

From 8.70 Q  0.668 ft 3 /s D  0.5054 ft  6.065 in; A  0.2006 ft 2 R  D /4  0.1264 ft; Ch  100 1.852

  0.668 hL  1200  0.63  1.32(0.2006)(100)(0.1264)  hL  14.72 ft

124

Chapter 8

8.70

Q  100 gal/min  1 ft 3 /s / 449 gal/min = 0.2227 ft 3 /s L = 1000 ft; Ch  130 (New Steel)

a)

2-in pipe : D  2.067 in  0.1723 ft; A  0.02333 ft 2 R  D /4  0.0431 ft 1.852

  0.2227 hL  1000ft  0.63   (1.32)(0.02333)(130)(0.0431)  hL  186 ft

b) 3-in pipe; D = 3.068 in = 0.2557 ft; A = 0.05132 ft 2 R = D /4  0.0639 ft 1.852

8.71

8.72

  0.2227 hL =1000  0.63   (1.32)(0.05132)(130)(0.0639)  hL = 27.27 ft m3 6.22  104 Q s  1.98 m Q= AVV =  A   (0.020 m) 2 s 4 m 1.98  0.020m VD s NR    4.4  104 ;    Smooth); Then f = 0.021 2 m v 8.94  107 s 2 m  1.98  L V2 1200 m  s hL  f    0.021    252 m D 2g 0.020 m 2  9.81 m s2 gal 1 ft 3 1 min 750   Q ft min 7.48gal 60 s Q = AV V =   9.2 2 A 0.181ft s ft 9.2  0.4801ft VD s NR    1.17  105 ;     4 m ; Then f = 0.0195 2 ft v 3.79  105 s 2 P1 V1 P V2 P P   Z1  h A  h L  2  2  Z2  h L  1 2  P1  P2  h L  γ γ 2g γ 2g γ hL  f 

ft    9.2  s

2

L V2 (2  5, 280 ft)   563 ft   0.0195  ft D 2g 0.4801 ft 2  32.2 2 s

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

125

# #  563ft  0.852  29,970 2 3 ft ft # gal 1 ft 3 1 min ft  #    50, 020 PA  h A  γ  Q  563 ft  0.852  62.4 3  750 ft min 7.48 gal 60 s s P = γ  h L  62.4

8.73

P1 V12 P V2   Z1  h A  h L  2  2  Z 2  P1  P2  h L  γ γ 2g γ 2g m3 1 hr  Q m hr 3600s  4.30 Q = AVV =  2 2 A 1.291  10 m s 200

NR 

VD  v

m  0.1282m s  157, 054;     5 m ; Then f = 0.0186 2 m 3.51  10 6 s

4.30

2

m  4.30   2  L V 15,000 m s  0.0186   hL  f    2, 051 m D 2g 0.1282 m 2  9.81 m s2 N N P1  P2  2, 051  8,360 3  17.15  106 2 m m ΔP 17,150 kPa # of Stations    3.57  4 Stations PMAX 4800 kPa

mi 5, 280 ft 1 hr 1 ft    8.5 in  VD hr 1 mi 3600 s 12in A.) N R    3.61  105 2 ft v 1.58  104 s km 1000 m 1 hr 135    0.070 m VD hr 1 km 3600 s B.) N R    1.74  105 2 m v 1.51  105 s mi 5, 280 ft 1 hr 1 ft 95    2.88 in  VD hr 1 mi 3600 s 12in C.) N R    2.12  105 2 ft v 1.58  104 s m 440  0.013 m VD s D.) N R    3.79  105 2 m v 1.51  10 5 s 55

8.74

126

Chapter 8

8.75

Q1 = A1  V1  V =

Q 2 = A 2  V2  V =

Q1  A1

150

Q2  A2

gal 1 ft 3 1 min   ft min 7.48 gal 60 s  55.7 2 0.006 ft s

150

gal 1 ft 3 1 min   ft min 7.48 gal 60 s  14.3 2 0.02333ft s

ft  0.0874 ft VD s N R1    402,329 2 v 5 ft 1.21  10 s ft 14.3  0.1723 ft VD s N R2    203, 627 2 v 5 ft 1.21  10 s 55.7

1&2  1.5  104 ;Then f1 = 0.0231& f 2 = 0.0206 2

ft    55.7  L V2 100 ft s hL  f    0.0231   1, 273 ft D 2g 0.0874 ft 2  32.2 ft s2 2

8.76

ft   14.3   2 L V 100 ft s hL  f    0.0206    38.0 ft D 2g 0.1723 ft 2  32.2 ft s2 3  m 1 hr  8   Q  hr 3600 s  m Q = A  V  V1 =   4.53 2 A  π  (0.025m)  s   4  m 4.53  0.025 m V1  D s N R1    111, 029 2 v 6 m 1.02  10 s  m3 1 hr  8   Q  hr 3600 s  m Q = A  V  V2 =   0.503 2 A  π  (0.075m)  s   4

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

127

m  0.003 m V2  D s N R2    1, 479 2 v 6 m 1.02  10 s 4.53

Yes, a Reynolds Number of 1, 479 is in the laminar flow region while 111, 029 is in the turbulent flow region on a Moody's diagram.

8.77

From PIPE-FLO ® ; V = 7.425 ft/s Pressure Drop = 0.153 psi h L  24.65 ft N R  66, 055 f  0.0252

128

Chapter 8

CHAPTER NINE VELOCITY PROFILES FOR CIRCULAR SECTIONS AND FLOW IN NONCIRCULAR SECTIONS

VELOCITY PROFILES FOR CIRCULAR SECTIONS AND FLOW IN NONCIRCULAR SECTIONS

129

130

Chapter 9

VELOCITY PROFILES FOR CIRCULAR SECTIONS AND FLOW IN NONCIRCULAR SECTIONS

131

132

Chapter 9

VELOCITY PROFILES FOR CIRCULAR SECTIONS AND FLOW IN NONCIRCULAR SECTIONS

133

134

Chapter 9

9.21

QT  AT υ = (0.01414 ft 2 )(25 ft/s)  0.3535 ft 3 /s Qs  Asυ = (0.0799 ft 2 )(25 ft/s) = 1.998ft 3 /s

3π (1.900in) 2 ft 2 As  0.1390 ft    0.0799 ft 2 2 4 144 in 2

9.22

D  4R  4((0.01*0.04)/(0.02  0.08)

N R  (3m/s)(0.016)/(4.67  10 7 )  1.03  105

Noncircular cross sections 9.23

A  (0.05 m) 2  (0.5)(0.05 m) 2   (0.025 m) 2 /4  3.259  103 m 2

WP  0.05  0.05  0.10  .052  .052   (0.025)  0.3493 m A 3.259  103 m 2 R   9.33  103 m WP 0.3493m Q 150 m3 1hr 1    12.78 m/s υ  A hr 3600 s 3.259  103 m 2

9.24

γ 12.5 N s 2 1kg  m/s 2 ρ  3   1.274 kg/m3 g m 9.81m in υ(4 R) ρ (12.78)(4)(9.33  103 )(1.274) NR    3.04×104  2.0  10 5 A  (12 in)(6 in)  2 (4.0 in) 2 /4  46.87 in 2 (1 ft 2 /144 in 2 )  0.3255 ft 2 WP  2(6 in)  2(12in)  2 (4.0in)  61.13 in(1 ft/12in)  5.094 ft A 0.3255ft 2   0.0639 ft R WP 5.094 ft Q 200 ft 3 1 min 1    10.24 ft/s υ  A min 60s 0.3255 ft 2 γ 0.114 lb  s 2 1 slug ρ  3   0.00354 slugs/ft 3 g ft (32.2 ft) 1 lb  s 2 /ft 2 (10.24)(4)(0.0639)(0.00354)  2.77 ×104 7 3.34  10  A  (10in) 2   (6.625in) 2 /4  65.53 in 2 (1 ft 3 /144in 2 )  0.455 ft 2 NR 

9.25

υ(4 R) ρ



WP  4(10in)   (6.625 in)  60.81 in(1 ft/12 in)  5.067 ft R  A/WP  0.455 ft 2 /5.067 ft  0.0898ft Q 4.00ft 3 /s  8.79 ft/s υ  A 0.455ft 2

υ4 R (8.79)(4)(0.0898)   3.808 105 6 v 8.29 10 Tube : D  0.402 in  0.03350 ft; A  8.814  104 ft 2 ft 3 /s Q  4.75 gal/min   0.1058 ft 3 /s 449 gal/min Q 0.1058 ft 3 /s  12.00 ft/s υ  A 8.814  104 m 2 υD (12.00)(0.03350) NR    1.20×105 6 v 3.35  10

NR  9.26

VELOCITY PROFILES FOR CIRCULAR SECTIONS AND FLOW IN NONCIRCULAR SECTIONS

135

A  3.293  103 ft 2   (0.0417 ft) 2 /4  1.929  103 ft 2

Shell :

WP   (0.06475ft)   (0.0417 ft)  0.3393 ft R = A/WP = 0.00176 m = 5.770 ×103 ft

1 ft 3 /s  0.06682 ft 3 /s 449 Q 0.06682 ft 3 /s = = 34.63 ft 3 /s A 1.929 ×103 ft 2 Q  30.0 gal/min 

NR  9.27

υ(4 R ) 34.63 ft/s (4)(5.77  103 )   5027 Turbulet υ 1.59  104

Q

Pipe :

υ

450 L/min  1 m3 /s  0.00750 m3 /s 60000 L/min

Q 0.00750 m3 /s   0.402 m/s A 1.864  102 m 2

NR 

υD (0.402)(0.1541)   6.08×104 6 v 1.02  10 Water at 20C

Duct : Benzene at 70C; ρ  sg(ρ w )  0.862(1000)  862 kg/m3   3.5  104 Pa s(App.D) A  (0.40 m)(0.20 m)  2( )(0.1683 m) 2 /4  0.0355 m 2

WP  2(0.40)  2(0.20)  2( )(0.1683)  2.257 m R=

A 0.0355 m 2 = = 0.0157 m WP 2.257 m

NR 

υ(4 R)( ρ)



;υ 

N R  (6.08  104 )(3.5  104 )   0.392 m/s 4 Rρ 4(0.0157)(862)

Q  Aυ  (0.0355 m 2 )(0.392 m/s)  1.39 ×10-2 m 3 /s 9.28

υD (25)(0.1342)   1.00×106 6 v 3.35  10 2 A  0.1390 ft  3( )(0.1583ft) 2 /4  0.0799 ft 2

Inside pipes : N R  Shell :

1 Do  1.900 in(1 ft/12in)  0.1583 ft  Outside dia . of 1 in pipe.

2 WP   (0.4206 ft)  3 (0.1583ft)  2.813ft

A 0.0799 ft 2   0.0284 ft WP 2.813 ft υ(4 R ) (25)(4)(0.0284)   2.35  105 NR   1.21  105 ) R

136

Chapter 9

9.29

Q  850 L/min 

1 m3 /s  0.01417 m3 /s 60000 L/min

A  Ashell  3 Atube  D  1.735  103 m 2  3 (0.0150 m) 2 /4  1.205  103 m 2 WP   (0.04970 m )  3 (0.0150 m)  0.2890 m

A 1.205  103 m 2   4.169  103 m WP 0.289 m Q 0.01417 m3 /s  11.76 m/s υ  A 1.205  103 m 2 υ4 R (11.76)(4)(4.169  103 ) NR    1.51  105 v 1.30  10 6 R

9.30

A  (0.25in)(1.75in)  4(0.25 in)(0.50 in)  0.9375 in 2 (1 ft 2 /144 in 2 )  0.00651ft 2

WP  (1.75in )  7(0.25in)  2(0.75 in)  6(0.50in) WP  8.00in(1 ft/12in)  0.667 ft R  A/WP  0.00651 ft 2 /0.667 ft  9.77  10 3 ft

NR 

N R  (1500)(3.38  104 ) :υ   6.09 ft/s  4 Rρ 4(9.77  10 3 )(2.13)

υ4 Rρ

Q  Aυ  (0.00651ft 2 )(6.09 ft/s)  3.97 ×10-2 ft 3 /s 9.31

Air flows in cross-hatched area .

A  [4(3.00)  2(4.50)](50)  1050 mm 2

WP  12(50)  4(4.50)  8(3.00)  642 mm A 1050 mm 2 R   1.636 mm WP 642 mm

Q A 50 m3 1 106 mm 2 1h = . . .

2 2 h 1050 mm m 3600 s υ  13.23 m/s

υ=

NR  9.32

υ(4 R)( ρ)





(13.23)(4)(0.001636)(1.15)  6.11×103 5 1.63  10

A  (0.45)(0.30)   (0.30) 2 /4  2(0.15) 2  0.1607 m 2

WP  2(0.45)   (0.30)  8(0.15)  3.042 m R  A/WP  0.1607 m 2 /3.042 m  0.0528 m υ  Q /A  0.10 m3 / s/ 0.1607 m 2  0.622 m/s

NR 

υ(4 R) ρ





(0.622)(4)(0.0528)(1260)  552 0.30 (App.D)

VELOCITY PROFILES FOR CIRCULAR SECTIONS AND FLOW IN NONCIRCULAR SECTIONS

137

9.33

A  (0.14446) 2  0.02087 m 2

WP  4(0.14446)  0.5778 m

A 0.02087 m 2   0.0361m R WP 0.5778 m 2 υ(4 R) (35.9)(4)(0.0361)   1.61×107 7 v 3.22  10 3 Q 0.75m /s  35.9 m/s υ  A 0.02087 m 2 NR 

9.34

A  (0.75)(0.75) 

 (1.50) 2 8

 1.446 in 2 (1 ft 2 /144 in 2 )  0.01004 ft 2

WP  0.75  2.25   (1.50)/2  5.356 in(1 ft/12 in)  0.4463ft R  A/WP  0.01004 ft 2 /0.4463ft  0.02250 ft

υ

Q 78.0 gal/min 1 ft 3 /s    17.30 ft/s A 0.01004 ft 2 449 gal/min

NR 

9.35

A

υ(4 R) (17.30)(4)(0.02250)   1.112×105 5 1.40  10 v

 π (0.75) 2 π (0.25) 2  (2(π )(0.50) 2  2(0.50)(0.50)     8 4 4  

A  1.089in 2 (1 ft 2 /144 in 2 )  0.00756 ft 2

WP  π (0.50)  4(0.50) 

(π )(.25)(2) (π )(0.75)(2)  4 4

WP  5.142 in(1ft/12in)  0.428ft R  A/WP  0.00756 ft 2 /0.428 ft  0.0177 ft

υ

N R v (1.5  105 )(1.40  105 )   29.66 ft/s (High) 4R 4(0.0177)

Q  Aυ  (0.00756 ft 2 )(29.66 ft/s)  0.224ft 3 /s

138

Chapter 9

9.36

sin 45   1.098 in  sin 105   By Law of Sines sin 30 b  1.50  0.7765in   sin 105 a  1.50

PART

A

WP

1 2 3 4 5 6 7 8 9

0.2209 in 2 0.0552 0.0736 0.0920 0.5625 2.250 0.4118 0.4118 0.2912

1.178 in 0.2945 0.3927 0.4909 1.500 2.000 0 1.0981  a 0.7765  b

A

4.3690 in 2

7.7307  WP

R  A/WP  0.5651 in(1 ft/12 in)  0.04710 ft N R (2.6  104 )(6.60  106 ) υ   0.595 ft/s 4 Rρ (4)(0.0471)(1.53) (4.369in 2 )(0.595 ft/s) Q  Aυ   0.0181ft 3 /s 2 2 (144in /ft ) 9.37

A  (2)(8)  16 mm 2 (1 m 2 /106 mm 2 )  1.60  105 m 2

WP  2(8)  2(2)  20 mm  0.020 m

R

A 1.60  105 m 2   8.00  104 m WP 0.020 m

Q  Aυ  (1.60  105 m 2 )(25.0 m/s)  4.00  104 m 3 /s each Qtot = (4.00  10-4 m 3 /s)6 = 2.40  103 m 3 /s (25)(4)(8.00  104 )(1.20)  6.40×103 1.50  105  From Prob. 9.24, R  0.0898 ft; N R  3.808  105 ; υ  8.79 ft/s; Water at 90 F

NR  9.38

4R





υ(4 R)( ρ)



4(0.0898)  4226; then f  0.0165 8.5  105

L υ2 (0.0165)(30)(8.79) 2    1.65 ft hL  f 4 R 2 g (4)(0.0898)(2)(32.2) p  γ w hL  (62.1 lb/ft 3 )(1.65 ft)(1 ft 2 /144 in 2 )  0.713 Psi

VELOCITY PROFILES FOR CIRCULAR SECTIONS AND FLOW IN NONCIRCULAR SECTIONS

139

9.39

Some date from Prob. 9.25 and Fig 9.13.

L  5.35 m  3.281 ft/m  17.23ft

Tube : Water at 200 F; γ = 60.1 u /ft 3 : N R  1.20  105 D

(0.0335)  223; then f  0.030  1.5  10 4 L υ2 (0.030)(17.23)(12.0) 2   34.50 ft hL  f D 2g (0.0335)(2)(32.2) 

Δp  γ w hL  (60.1 u / ft 3 )(34.50 ft)(1 ft 2 /144in 2 )  14.40 psi Shell : Ethlylene glycol at 77 C;   68.47 u / ft 3 :

v  4.63 ft/s : R  5.77  103 ft; N R  5027  5.027  103 4R/ε = (4)(5.77 ×103 ft)/(1.5 ×104 ft) = 154; Then f = 0.044 L v2 525 10.562  0.044  611.7 ft hL  f 4R 2 g 4(0.00577) 2(32.2) Δp  γhL  (68.47 u )(611.7 ft)  (1 ft 2 /144 in 2 )  290.9 Psi [Very high] 9.40

Some data from Prob. 9.26 and Fig . 9.11.

Each Pipe : Water at 20C; γ  9.79 kN/m3 ; N R  6.08  104 D (0.1541)   3350; then f  0.021  4.6  105

L υ 2 (0.021)(3.80)(0.402) 2   0.00427 m hL  f D 2g (0.1541)(2)(9.81) Δp  γ w hL  (9.79 kN/ m3 )(0.00427 m)  0.00418 kPa  41.8 Pa Duct : Benzene at 70 C;  B  (0.862)(9.81 kN/m3 )  8.46 kN/m3 4 R (4)(0.0157) N R  6.08  104 ; R  0.0157 m;   1365 4.6  105  L  2 (0.023)(3.80)(0.392) 2 .   0.0109 m f  0.023  hL  f 4 R 2 g (4)(0.0157)(2)(9.81) p   B hL  (8.46 kN/m3 )(0.0109 m)  0.0920 kPa  92.0 Pa 9.41

Some data from Prob. 9.27 and Fig . 9.12.

Each Pipe : Water at 200F; γ w  60.1 lb/ft 3 ; N R  1.0  106 D

0.1342  895 : then f  0.021  1.5  104 L υ2 (0.021)(50)(25) 2 hL  f   75.9 ft D 2 g (0.1342)(2)(32.2) 

p  γ w hL  (60.1 lb/ft 3 )(75.9 ft)(1 ft 2 /144in 2 )  31.7 psi Shell : Water at 60F; γ  62.4 lb/ ft 3 ; N R  2.35  105 ; R  0.0284 ft 4 R (4)(0.0284)   758; then f  0.0225 1.5  104 

hL  f

L 2 (0.0225)(50)(25) 2    96.1ft 4 R 2 g (4)(0.0284)(2)(32.2)

Δp  γ w hL  (62.4 lb/ft 3 )(96.1 ft)(1 ft 2 /144in 2 )  41.6 psi 140

Chapter 9

9.42

Data from Prob. 9.28 and Fig . 9.18. Copper tubes.

Water at 10C; γ w  9.81 kN/m3 ; N R  1.51  105 : R  4.169  103 m (4)(4.169  103 )  11117; f  0.017  1.5  106 L υ2 (0.017)(3.60)(9.05) 2   25.86 m hL  f .

4 R 2 g (4)(4.169  103 )(2)(9.81)

4R



Δp  γ w hL  (9.81 kN/m3 )(25.86 m)  254 kPa 9.43

Data from Prob. 9.29 and Fig. 9.19. N R  1500 Laminar f 

64 64 57 in   0.0427 : R  9.77  103 ft : L   4.75ft 12 in/ft N R 1500

hL  f

L υ2 (0.0427)(4.75)(6.09) 2   2.99 ft .

4 R 2 g (4)(9.77  103 )(2)(32.2)

Ethlylene glycol at 77 F -  68.47 lb/ft 3

p  γhL  (68.47 lb/ft 3 )(2.99 ft)(1 ft 2 /144in 2 )  1.42 psi 9.44

Data from Prob. 9.31 and Fig . 9.21. N R  552 Laminar

64 64   0.116 : R  0.0528 m N R 552 L υ 2 (0.116)(22.6)(0.622) 2    0.244 m hL  f 4 R 2 g (4)(0.0528)(2)(9.81) p  γG hL  (1.26)(9.81 kN/ m3 )(0.244 m)  3.02 kPa f 

9.45

Data from Prob. 9.32 : R  0.0361m : N R  1.61  107 4R





(4)(0.0361)  96267; f  0.008 approx .

1.5  106

L υ 2 (0.008)(22.6)(35.9) 2    82.2 m hL  f 4 R 2 g (4)(0.0361)(2)(9.81) p  γhL  (9.47 kN/ m3 )(82.2 m)  779 kPa

9.46

Water at 90C Data from Prob. 9.23 and Fig. 9.22. N R  1.112  105 4 R (4)(0.0225) R  0.0225ft;   3600; f  0.019 2.5  105  L  (105 in)(1 ft/12in)  8.75 ft

hL  f

L υ 2 (0.019)(8.75)(17.30) 2    8.58ft 4 R 2 g (4)(0.0225)(2)(32.2)

Δp  γhL  (62.4 lb/ft 3 )(8.58ft)(1 ft 2 /144in 2 )  3.72 psi

VELOCITY PROFILES FOR CIRCULAR SECTIONS AND FLOW IN NONCIRCULAR SECTIONS

141

9.47

Data from Prob. 9.34 and Fig . 9.23. N R  1.5  105 : R  0.0177 ft; υ = 29.66 ft/s 4R

(4)(0.0177)  472 : f  0.025  1.5  104 L  (45 in)(1 ft/12 in)  3.75 ft 

[Steel]

L υ 2 (0.025)(3.75)(29.66)2 hL  f .   18.1ft 4 R 2 g (4)(0.0177)(2)(32.2) p  γhL  (62.4 lb/ft 3 )(18.1ft)(1 ft 2 /144 in 2 )  7.84 psi 9.48

A  (2.25)(1.50)  7( 2 /4)  2.602 in 2 (1 ft 2 /144in 2 )  0.01807 ft 2 WP  2(2.25)  2(1.50)  7( )  15.75 in(1 ft/12in)  1.312 ft R

A 0.01807 ft 2   0.0138ft WP 1.312 ft

NR 

υ4 Rρ



N R η (8000)(3.38  104 ) :υ    23.05ft/s 4R ρ 4(0.0138)(2.13) 4R

f  0.0325

4(0.0138)  11000  5 106 L υ2 10.67 (23.05) 2 hL  f .  (0.0325)  51.9 ft 4 R 2g 4(0.0138) 2(32.2) L  128 in(1 ft/12in)  10.67 ft :



Q  Aυ = (0.01807 ft 2 )(23.05ft/s)  0.417 ft 3 /s  9.49

449 gal/min  187 gal /min 1 ft 3 /s

A  (0.100 m) 2  4(0.02)2  0.0076 m 2  A  0.0119 m R WP WP  4(0.10)  8(0.03)  0.64 m  Q 3000 L/min 1 m3 /s υ=    6.58 m/s A 0.0076 m 2 60000 L/min υ(4 R )  (6.58)(4)(0.0119)(789) NR    4.40  105 ; f  0.0135 5.60  104  L υ 2 (0.0135)(2.25)(6.58) 2 .   1.41 m 4 R 2g 4(0.0119)(2)(9.81) A  (28)(14) - 3[(8)(2)  π2 /4]  334.6 in 2 (1 ft 2 /144in 2 )  2.32 ft 2

hL  f

9.50

WP  2(28)  2(14)  3[2(8)  π)  150.8 in(1 ft/12in)  12.57 ft R  A/WP  0.1848 ft (20)(4)(0.1848)(2.06  103 ) NR    7.36  104 7  4.14  10 Each Tube : Ethy1 Alcohal at 0F;assume ρ  1.53 slugs/ft 3 υDρ   5  105 lb-s/ft 2 (App.D); N R  υ(4 R) ρ

9.51



N R (3.5  10 )(5.0  10 )   26.02 ft/s Dρ (0.044)(1.53) D  13.4 mm(1.0 in/25.4 mm)  0.5276 in  1 ft/12in  0.044 ft

υ

142

4

5

Chapter 9

π (0.044 ft)2  26.02 ft  0.0395 ft 3 /s 4 Total Flow for 3 Tubes : QT  3(0.0395)  0.118 ft 3 /s L υ 2 (0.0232)(10.5)(26.02) 2 hL  f   58.2 ft D 2g (0.044)(2)(32.2) D 0.044   8793; f  0.0232  5  10 6 p  γhL  (49.01 lb/ft 3 )(58.2 ft)(1 ft 2 /144 in 2 )  19.8 psi Shell : Methy1 Alcohol at 77F (App.B) Q  Aυ 

A  (2.00)(1.00)  π (1.00) 2 /4  3π (0.625) 2 /4  1.865 in 2 (1 ft 2 /144 in 2 ) A  0.01295 ft 2 WP  2(2.0)   (1.0)  3   13.03 in(1 ft/12in)  1.086 ft R  A/WP  0.0119 ft υ

N R (3.5  104 )(1.17  105 )  5.61 ft/s  4(0.0119)(1.53) 4 Rρ

Q  Aυ  (0.01295ft 2 )(5.61 ft/s)  0.0727 ft 3 /s 4R

4(0.0119)  9540; f  0.0232  5  106 L υ 2 (0.0232)(10.5)(5.61) 2 hL  f   2.50 ft 4 R 2g 4(0.0119)(2)(32.2)

9.52



p  γhL  (49.1 lb/ft 3 )  2.50 ft  (1 ft 2 /144 in 2 )  0.851 psi Given : Water at 40F; N R  3.5  104 ; Figure 9.30; Sections is semicircular .

Find : Volume fiow rate of water .

N R  v(4 R)/v; Then, v  N R v /(4R)  Average velocity of flow v  1.67  10 5 ft 2 /s  Kinematic viscosity A  (1.431  102 ft 2 )/2  7.155  103 ft 2 ID  0.1350 ft; WP  ID  π ( ID )/2  ID (1  π /2)  0.347 ft R  A/WP  (7.155  103 ft 2 )/(0.347 ft)  0.0206 ft Then, v  N R v /(4 R )  (3.5  104 )(1.67  105 )/(4)(0.0206)  7.09 ft/s Q  Aυ  (7.155  10 3 ft 2 )(7.09 ft/s)  0.0507 ft 3 /s = Q Energy Loss for 92 in(7.667 ft) of drawn steel : h L  f ( L / 4R )(v /2 g ) 4 R /ε  4  0.0206 ft  / 5.0  10 –6 ft  16 480 Then ƒ  0.023 2 hL  (0.023)[7.667/(4)(0.0206)][(7.09) /(2)(32.2)]  1.67 ft Given : Figure 9.31. Three semcirculur sections. Velocity  v  15 ft/s in each .

Ethylene glycol at 77F; ρ  2.13 slugs/ft 3 ;   3.38  104 lb s/ft 2 (App.B) Find : Reynolds number in each passage. N R  v(4 R) ρ / Top channel : 2-in Type K.copper tube (Half).ID  0.1632 ft; Atot  2.093  102 ft 2 A  Atot /2  (2.093  102 ft 2 )/2  1.0465  102 ft 2 WP  ID  π ( ID)/2  ID(1  π /2)  0.4196 ft R  A/WP  (1.0465  10 2 ft 2 )(0.4916 ft)  0.0249 ft N R  v(4 R) ρ /  (15)(4)(0.0249)(2.13)(3.38  104 )  9.43  103  9430  N R Turbulent



9.53



VELOCITY PROFILES FOR CIRCULAR SECTIONS AND FLOW IN NONCIRCULAR SECTIONS

143

Both side Channels : 1 1 2  in Type K copper tube (Half).ID = 0.1234 ft ; Atot  1.196  102 ft 2 A  Atot / 2  (1.196  102 ft 2 ) / 2  5.98  103 ft 2 WP  ID  π ( ID) / 2  ID(1  π /2)  0.3172 ft R  A / WP  (5.98  103 ft 2 ) / (0.3172 ft)  0.01885 ft N R  v(4 R) ρ /  (15)(4)(0.01885)(2.13)(3.38  104 )  7.127  103  N R Turbulent Energy Loss for 54 in (4.50 ft) of drawn copper : hL  f ( L /4 R)(v 2 /2 g ) Top : 4 R /  4(0.0249 ft)/(5.0  106 ft)  19 920 Then f  0.0315 hL  (0.0315)[4.50/(4)(0.0249)][(15.0) 2 /(2)(32.2)]  4.97 ft Each side : R /  4(0.01885 ft)/(5.0  106 ft)  15 080 Then f  0.0340 hL  (0.0340)[4.50/(4)(1.01S85)][(15.0) 2 /(2)(32.2)]  7.09 ft (each of two sides) Total loss for all three channels: hLT  4.97  2(7.09)  19.15 ft 9.54

Given : Figure 9.32 Rectangulur channel with three fins. Q  225 L/min Brine (20 NaCl), sg =1.10 at 0C;ρ  1.10(1000 kg/m3 )  1100 kg/m3

  2.5 103 N s/m2 (App.D) Find : Reynolds number for the flow . N R  v(4 R) ρ / A  (20)(50)  3(5)(10)  (850 mm 2 [(1 m 2 )/(103 mm)2 ]  8.5  104 m 2 WP  2(20)  2(50)  6(10)  200 mm R  A/WP  (850 mm 2 ) / (200 mm) = (4.25 mm)(1 m2 /1000 mm) = 0.00425 m v = Q /A  (0.00375 m3 /s)(8.5  104 m2 )  4.41 m/s N R  v(4 R)  /   (4.41)(4)(0.00425)(1100)/(2.5  103 )  3.30 × 104 = N R Turbulent Energy Loss for 1.80 m of commercial steel: hL  f ( L /4 R)(v 2 /2 g ) 4 R /  4(0.00425 m)/(4.6  105 )  370 Then f  0.030 hL  (0.030)[1.80/(4)(0.00425)][(4.41) 2 /(2)(9.81)]  3.149 m

144

Chapter 9

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

MINOR LOSSES

153

154

Chapter 10

MINOR LOSSES

155

156

Chapter 10

10.60

Dp = sg(Q/Cv)2 = 0.989(600/640)2 = 0.869 psi sg = 61.7 lb/ft3/62.4 lb/ft3 = 0.989

10.61

Dp = sg(Q/Cv)2 = 0.997(15/25)2 = 0.359 psi sg = 62.2 lb/ft3/62.4 lb/ft3 = 0.997

For Problems 10.62 to 10.70, values of sg are found from Appendix B.

MINOR LOSSES

10.62

Dp = sg(Q/Cv)2 = 1.590(60/90)2 = 0.707 psi

10.63

Dp = sg(Q/Cv)2 = 0.68(300/330)2 = 0.562 psi

10.64

Dp = sg(Q/Cv)2 = 0.495(5000/4230)2 = 0.692 psi

10.65

Dp = sg(Q/Cv)2 = 1.590(60/34)2 = 4.952 psi

10.66

Dp = sg(Q/Cv)2 = 0.68(300/160)2 = 2.391 psi

10.67

Dp = sg(Q/Cv)2 = 0.495(1500/700)2 = 2.273 psi

10.68

Dp = sg(Q/Cv)2 = 1.030(18/25)2 = 0.534 psi

10.69

Dp = sg(Q/Cv)2 = 0.823(300/330)2 = 0.680 psi

10.70

Dp = sg(Q/Cv)2 = 1.258(3500/2300)2 = 2.913 psi

157

158

Chapter 10

MINOR LOSSES

159

CHAPTER ELEVEN SERIES PIPE LINE SYSTEMS 11.1

Class I; Pt. A at tank surface :

PA υ2 p υ2  zA  A  hL  B  zB  B : p A  0, υA  0 γ 2g γ 2g

  υ2 L   pB  γ w  zA  zB  B  hL   γ w 14 m  hυB  0.5hυB  3 fr  30  hυB  f hυB  2g D     Entr. 3Elbows Friction 2

hυB  Velocity head in pipe =

υ2B 1 Q (1.99 m/s) 2    0.202 m   2g 2g  A  2(9.81 m/s)2

N R  υD /v=(1.99)(0.098)/(1.30  10 6 )  1.50  105 D /  0.098/1.5  106  65300  f  0.0165; f T  0.010 in zone of complete turbulence 9.81 kN  80.5  14 m  0.202  0.101  (3)(0.010)(30)(0.202)  (0.0165) (0.202)  3  m 0.098  

pB 

11.2

pB  105.7 kN/m 2  105.7 kPa P υ2 P υ2 Class I; A  zA  A  hL  B  zB  B: υA  υB; pB 0 γ 2g γ 2g pA  γ[ zB  zA  hL ]  γ[3.5 m + h L ]; f T  0.019

38 hυ  hυ [7.10  724 f ] 0.0525 Ent . Chk. value Ang. V. Elbow Exit Friction

hL  0.78hυ  100 fT hυ  150 fT hυ  30 fT hυ  1.0hυ  f

hυ 

υ2 Q 435 L/min 1 m3 /s (in pipe); υ     3.444 m/s 2g A 2.168  103 m 2 60000 L/min

hυ  (3.344) 2 /(2)(9.81)  0570 m NR 

υDρ





(3.344)(0.0525)(820) 0.0525 D  8.47  104 ;   1141; f  0.0222 3 1.70  10  4.6 105

pA  γ m + hL ]  (0.82)(9.81 kN/m3 )[3.5 m +0.570 m(7.10 +724(0.0222)] =134.4 kPa

160

Chapter 11

11.3 Class I;

PA υ2 p υ2  zA  A  hL  B  zB  B : pA  pB  γ[( zB  zA )  hL ] γ 2g γ 2g

Q 60 gal/min 1 ft 3 /s υ 2 (5.73)2    5.73 ft/s: h    0.509 ft υ A 0.02333ft 2 449 gal/min 2 g 2(32.2) 50 hL  6.5hυ  2(30) f T hυ  f hυ  (6.5  60 f T  290 f )hυ : fT  0.019 0.1723 υDρ (5.73)(0.1723)(0.90)(1.94) D 0.1723 NR    2.87  104 :   1150 5 6.0  10   1.5  104 f  0.0260 : Then h L  [6.5  60(0.019)  290(0.0260)](0.509 ft) =7.73 ft

υ

(0.90)(62.4 lb) 1 ft 2 pA  200 psig + [25 ft + 7.73 ft]  212.8 psig ft 3 144 in 2 11.4 Class I

PA υ2 p υ2  zA  A  hL  B  zB  B γ 2g γ 2g

1 ft 3 /s Q  750 gal/min   1.67ft 3 /s 449 gal/min

1.67ft 3 / s Q   20.9 ft/s   υA  υ2B  υ2A AA 0.07986 ft 2 pA  pB  γ  ( zB  zA )   hL  2g   Q 1.67ft 3 / s υB    9.23 ft/s 0.181 ft 2 AB υ 2 A (20.9) 2 υ2 ft=6.79 ft; B  1.32 ft  2 g 2(32.2) 2g

fT  0.0162 is zone of complete turbulance

υ2A υ2 υ2 40 ft υ 2 A f  0.28 A  (1.30  25 f ) A 2g 0.3188 ft 2 g 2g 2g Elbows Friction Enlarge Where D2 /D1  0.4801/0.3188  1.51  K = 0.28 (Table 10.1)

hL  2(30) fT

υA DA (20.9)(0.3188)   9.25  103 v 7.21  104 D /   0.3188 /1.5  104  2125  f  0.032

At 100°F, v = 7.21  104 ft 2 /s; N RA 

hL  [1.25  125(0.032)](6.79 ft)=35.6 ft pA  500 psig +

(0.895)(62.4 lb) ft(1 ft 2 ) [3.5 (1.32 6.79) 35.6]     513.0 psig ft 3 144 in 2

Similarly: At 210F,v = 7.85  10 5 ; N R  8.49  104  f  0.0205 hL  25.9 ft; pA  509.3 psig

SERIES PIPE LINE SYSTEMS

161

162

Chapter 11

SERIES PIPE LINE SYSTEMS

163

164

Chapter 11

Class II systems 11.8

Class II hL = 30.0 ft = f Then

Method IIC Iteration

2 ghL D 2(32.2)(30)(0.3188) 24.64 = = fL f (25) f

=

(0.3188) D 0.3188 4 = 2120 = −6 = 4.33 ´ 10 ( ): 7.37 ´ 10 1.5 ´ 10 -4 Try f = 0.02; then = 24.64 / f = 35.1 ft/s

NR =

D

L 2 D 2g

=

NR = 4.33 ´ 104(35.1) = 1.52 ´ 106; New f = 0.0165 = 24.64 / 0.0165 = 38.6 ft/s; NR = 1.67 ´ 106; f = 0.0165 OK Q = A = (0.07986 ft2)(38.6 ft/s) = 3.08 ft3/s Class II [Repeated using computational approach - Sec. 11.4] hL = 30.0 ft; D/ε = 2120; Use Eq. 11-3. é 1 gDhL 1.784n ù 2 log + ê ú Q =−2.22D L êë 3.7 D/ε D gDh L/L úû Q =−2.22(0.3188)

2

Method IIA

é (32.2)(0.3188)(30) 1 (1.784)(7.37 ´ 10-6 ) ù log ê + ú 25 (0.3188) 12.32 û ë 3.7(2120)

­ Q = 3.05 ft3/s 11.9

Class II

p1

+ z1 +

2 1

2g

−hL =

p2

+ z2 −

2 2

2g

:

Method IIC Iteration p p 68 kN/ m 2 L 2 2 ghL D hL = 1− 2 = = 7.70 m = f ; = ((0.90)(9.81 kN/ m3 ) D 2g fL =

2(9.81)(7.70)(0.0470) 0.2367 D 0.0470 = 31333 : = = ε 1.5 ´ 10-6 f (30) f

Try f = 0.03; = 0.236 7 / 0.03 = 2.81 m/s Dρ (0.0470)(900) NR = = 1.41 ´ 104( ) = 3.96 ´ 104 =  3.0 ´ 10−3 New f = 0.022; υ = 3.28 m/s; NR = 4.62 ´ 104; f = 0.0210 = 3.357 m/s; NR = 4.73 ´ 104; f= 0.0210 No change Spreadsheet solutions to Problems 11.8, 11.9, and 11.10 are on next page.

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166

Chapter 11

SERIES PIPE LINE SYSTEMS

167

168

Chapter 11

SERIES PIPE LINE SYSTEMS

169

170

Chapter 11

SERIES PIPE LINE SYSTEMS

171

11.14

Class II Pt. 1 at surface of tank A; Pt. 2 in stream outside pipe. u 1 = 0, p2 = 0 2 2 2 p1 p p 2 + z1 + 1 −h L = 2 + z2 : 1 + ( z1 - z 2 ) − 2 −hL Copper Tube: 2g 2g 2g OD = 50 mm ; 1.5 mm wall 2

150 kN/m 2 −5 m = 13.59 m = 2 + hL 2g 8.07 kN/m 3 2 2

+ 160 fT

+ 2(19) fT

2 2

Method IIC

D = 47 mm = 0.047 m A = 1.735 ´ 10-3 m2

2 30 mm 22 = (2.36 + 638)f 2 2g 2g 2g 0.0470 m 2 g 2g Entrance Valve Bends Friction ë r/D = 300 mm = 6.38 ® Le = 19 (Fig. 10.28) 47.0 mm D 0.0470 m D/ε = = 31333 1.5 ´ 10−6 m fT = 0.0094

hL = 0.50

2 2

I

+ f

In Eq. I: 13.59 m =

2 2

2g

+ (2.36 + 638f )

2 2

2g

= (3.36 + 638f )

2g (13.59 m) 2(9.81)(13.59) = = 3.36 + 638f 3.36 + 638f

=

2 2

2g

266.6 3.36 + 638f

Try f = 0.02 Dρ (4.07)(0.0470)(823) 266.6 = = 4.07 m/s; NR = = = 9.59 ´ 104  1.64 ´ 10−3 3.36 + 638(0.02) New f = 0.018; = 4.24 m/s; NR = 1.00 ´ 105; New f = 0.018 No change Q = A = (1.735 ´ 10-3 m2)(4.24 m/s) = 7.36 ´ 10-3 m3/s 11.15

Class II with 2 pipes: Pts. A and B at tank surfaces.

Method IIC 2 2 p = pB = 0 pA u p u + zA + A - hL = B + zB + B : zA - zB = hL = 10 m: A uA =uB = 0 g 2g g 2g hL = 0.78

u 42 2g

+ 2(30) f4 T

Entrance Elbows + f6

u 42

u2 ü 55 u 42 + 0.27 4 ý Based on 4-in pipe; f4T = 0.0203 2g 0.1059 2g 2g þ D 0.1593 Friction Enlarge. 2 = = 1.504 D1 0.1059 + f4

u2 u2 u2 ü 30 u 62 + 30 f 6T 6 + 45 f 6T 6 + 1.0 6 ý Based on 6-in pipe; f6T = 0.0184 0.1593 2g 2g 2g 2g þ Friction Elbow Valve Exit

h L = (2.268 + 519.4 f3)

u 32

+ (2.38 + 188.3) f6 u 6 2g 2g 2

2

æD ö A But u 4 = u 6 6 = u 6 ç 6 ÷ = u 6 æç 0.1593 ö÷ = 2.263u 6; u 32 = 5.120u 62 A4 è 0.1059 ø è D4 ø

172

2

Chapter 11

SERIES PIPE LINE SYSTEMS

173

[11.15 solution continued]

hL = (2.268 + 519.4 f4)

5.120u 62 2g

(2.38 + 188.3 f 6 )

u 62 2g

(13.99 + 2659 f 4 + 188.3 f 4 )

u 62 2g

Solve for υ6

u6 =

2 gh L = 13.99 + 2659 f4 + 188.3 f 6

2(9.81)(10) 13.99 + 2659 f4 + 188.3 f 6

196.2 13.99 + 2659 f4 + 188.3 f 6

=

Iterate for both f4 and f6: D4 u D u (0.1059) 0.1059 m = = 883: N R4 = 4 4 = 3 =1.614 ´ 105 (u 4 ) -4 6.56 ´ 10-7 e 1.2 ´ 10 m n D6

e

=

u D u (0.1593) 0.1593 = 1328 : NR6 = 6 6 = 6 = 2.865 ´ 105 (u 6 ) -4 1.2 ´ 10 n 6.56 ´ 10-7

Try f4 = f6 = 0.02

u6 =

196.2 = 1.663 m/s 13.99 + 2659(0.02) + 188.3(0.02)

u 4 = 2.263 u 6 = 3.76 m/s

NR4 = 1.614 ´ 10 (3.76) = 6.07 ´ 10 ® New f3 = 0.0206 5

5

N R6 = 2.805 ´ 10 5 (1.663) = 4.76 ´ 10 5® New f6 = 0.0192

u6 =

196.2 = 1.368 m/s 13.99 + 2659(0.0195)+ 188.3(0.02)

u 4 = 2.263u 6 = 3.085 m/s

N R3 = 1.614 ´ 105(3.085) = 4.99 ´ 105 ® f4 = 0.0206 No change N R6 = 2.865 ´ 105(1.368) = 3.92 ´ 10 5® f6 = 0.0192 No change

Q = A 6u 6 =

174

p (0.1593 m)2 4

´ 1.368 m/s = 0.02726 m3/s

Chapter 11

11.16

Class II with two pipes Pt. B in stream outside pipe. pB = 0 pA p u2 u2 p u 2 - u A2 + zA + A - hL = B + zB + B : A + zA - zB = B + hL 2g 2g g o 2g go go pA

go

u u 175 kN/m 2 - 4.5 m = 14.68 m = B - A + hL I 3 2g 2g 0.93(9.81 kN/m ) 2

+ z A - zB =

Method IIC

2

30 u A2 + 0.52 u A2 + 100 u B2 + 2(30) u2 f8 fT 8 B : fT8 = 0.0088 0.0470 2g 2g 0.093 2g 2g Friction Enlarge. Friction Elbows Copper Tube: A = 50 mm OD ´ 1.5 mm wall; D A = 47mm 0.093 ë D2/D1 = = 1.98 B = 100 mm OD ´ 3.5 mm wall; D B = 0.3 mm 0.0470 AB = 6.793 ´ 10-3 m2

hL = f A

u A DA r u A (0.0470)(930) = = 4.60 ´ 103(u A) h 9.50 ´ 10 -3 u D r u (0.093)(930) DB/ε = 0.093/1.5 ´ 10-6 = 62000; N R B B = B = 9.10 ´ 103(u B) h 9.5 ´ 10 -3 DA/ε = 0.0470/1.5 ´ 10-6 = 31333; N RA = B

hL = (0.52 + 638f )

u A2 + (0.53 + 1075 f ) u B2 2g

8

2g

2

But u A = u B

æD ö AB = u B ç B ÷ = u B (1.979)2= 3.92u B: u A2 = 15.3u B2 AA è DA ø

Then: hL = (0.52 + 638fA)

2 2 15.3u B2 + (0.53 + 1075 f B ) u B = u B [8.49 + 9761fA + 1075fB] 2g 2g 2g

In Eq. I 14.68 =

uB =

u B2 2g

-

15.3u B2 u B2 u2 + [8.49 + 9761 fA + 1075 fB ] = B [- 5.81 + 9761f A + 1075f B ] 2g 2g 2g

2 g (14.68) 288.1 = - 5.81 + 9761 fA + 1075fB - 5.81 + 903 fA + 1020 fB

Iterate for both fA and fB .

Try f2 = f4 = 0.02

uB =

288.1 = 1.169 m/s; u A = 3.92u B = 4.58 m/s -5.81 + 9761(0.02)+ 1075(0.02)

N RA = 4.60 ´ 103(4.58) = 2.11 ´ 104; N RB = 9.10 ´ 103(1.169) = 1.06 ´ 104

New fA = 0.0255, fB = 0.0305 288.1 uB = = 1.02 m/s; u A = 3.92u B = 4.01 m/s -5.81 + 9761(0.0255) + 1075(0.0305) N RA = 1.84 ´ 10 4 ® fA = 0.0255 No change; N RB = 9.23´ 10 3 ® fB = 0.0305 No change

Q = ABu B = (6.793 ´ 10-3m2)(1.02 m/s) = 6.93 ´ 10-3 m3/s SERIES PIPE LINE SYSTEMS

175

176

Chapter 11

SERIES PIPE LINE SYSTEMS

177

178

Chapter 11

SERIES PIPE LINE SYSTEMS

179

180

Chapter 11

Practice problems for any class 11.23 Class I Pt. 1 at tank surface, Pt. 2 in stream outside pipe: p1 = p2 = 0; u 1 = 0 2 2 2 p1 p + z1 + 1 −hL = 2 + z2 + 2 : z1− z 2 = 2 + hL I 2g 2g 2g 2 Q 1500 L/min 1 m 3 /s (3.92) 2 2 = = 3.92 m/s: = = 0.782 m A2 6.38 ´ 10-3 m 2 60000 L/min 2g 2(9.81) fT = 0.017

2=

hL = 0.5

2 2

2

2

+ fT (160) 2 + fT (30) 2 2g 2g 2g

2 2

2g

[0.5 + 190 fT ] + 0.782[0.5 +190(0.017)]

= 2.917 m In Eq. I z1 − z2 =

2 2

2g

+h L + 0.782 + 2.917 = 3.70 m

But h = z1− z2 −0.5 m = 3.70 m − 0.5 m = 3.20 m 11.24

Class I Pt. 1 at collector tank surface. Pt. 2 at pump inlet. p1 = 0, u 1 = 0 2 2 é u2 ù p1 p + z1 + 1 + hL = 2 + z2 + 2 : p2 = g ê ( z1 - z2 ) - hL - 2 ú I 2g û 2g 2g ë Q = 30 gal/min ´

u2 =

1 ft 3 /s 1 = 0.0668 ft3/s 449 gal/min

fT = 0.019 for 2-in pipe

Q 0.0668 ft 3 /s u 22 (2.86) 2 = = 2.86 ft/s; = = 0.127 ft A2 0.02333 ft 2 2g 2(32.2)

10.0 ft u 22 u2 u2 + fT (8) 2 = 2 (2.50 + 58.0 f ) 2g 2g 0.1723 ft 2 g 2g 2g Entrance Filter Friction Valve u Dr (2.86)(0.1723)(0.92)(1.94) D 0.1723 NR = = = 2.45 ´ 104 : = = 1149 -5 h 3.6 ´ 10 e 1.5 ´ 10 -4 f = 0.0265 hL = (0.127 ft)[2.50 + 58.0(0.0265)] = 0.513 ft In Eq. I : 62.4 lb ft ft 2 p2 = (0.92) [3.0 0.513 0.127] = 0.94 psig ft 3 144 in 2

hL = 0.5

u 22

+ 1.85

u 22

+ f

SERIES PIPE LINE SYSTEMS

181

5

182

Chapter 11

SERIES PIPE LINE SYSTEMS

183

11.27

184

Chapter 11

SERIES PIPE LINE SYSTEMS

185

11.29

Class I Q = 475 L/min (1 m3/s/60000 L/min) = 7.917 ´ 10-3m3/s Pt. 1 at reservoir surface; Pt. 2 at pump inlet. p1 = 0, u 1 = 0 é ù u 22 p1 p1 u 12 u 22 p = g ( z z ) - hL ú I + z1 + - hL = + z2 + : 2 ê 1 2 2g 2g 2g g g ë û

u2 =

Q 7.917 ´ 10-3 m3 /s u 22 (2.56) 2 = = 2.56m/s: = = 0.335 m A 3.090 10-´3 m 2 2 g 2(9.81)

12.90 m u 22 u2 u2 + (0.018)(2)(30) 2 + (0.018)(340) 2 = 4.016 m 2g 0.0627 m 2 g 2g 2g Entrance Friction 2 Elbows Valve u D (2.56)(0.0627) D 0.0627 NR = = 4.46 ´ 105 : = = 1363 ® f = 0.0195 -7 n 3.60 ´ 10 e 4.6 ´ 10-5 fT = 0.018 L = 11.5 m + 1.40 m = 12.90 m z1 - z2 = 0.75 m - 1.40 m = -0.65 m In Eq. I 9.53 kN p2 = [-0.65 m - 0.335 m - 4.016 m] = -47.66 kPa m3

hL = 0.78

u 22

+ (0.0195)

See also spreadsheet solution.

11.30

Design problem with variable solutions: Pressure at pump inlet can be increased by: lowering the pump, raising the reservoir, reducing the flow velocity in the pipe by using a larger pipe, reducing the hL in the valve by using a less restrictive valve (gate, butterfly), eliminating elbows or using long-radius elbows, using a well-rounded entrance, and shortening the suction line.

See spreadsheet for an example of a modified system.

11.31 Class I Pt. B in stream outside nozzle. pB = 0 pA

g

+ zA +

u A2 2g

- hL + hA =

hA = ( zB - zA ) +

u B2 - u A2 2g

pB

g

-

+ zB +

pA

g

In 2 1/2-in discharge line:

ud =

Q 0.50 = = 15.03 ft/s Ad 0.03326

u d2 2g

186

=

(15.03) 2 = 3.509 ft 2(32.2)

+ hL

u B2 2g

uA = u A2

Q 0.50 ft 3 /s = = 9.743 ft/s AA 0.05132 ft 2

(9.743) 2 ft = 1.474 ft 2g 2(32.2) Q 0.50 uB = = = 54.2 ft/s AB p (1.3/12) 2 4 =

u B2 = 45.7 ft 2g r=

g g

=

64.0 lb/ft 3 slugs = 1.988 2 32.2 ft/s ft 3

Chapter 11

SERIES PIPE LINE SYSTEMS

187

188

Chapter 11

32

31

33

SERIES PIPE LINE SYSTEMS

189

190

Chapter 11

Problems 11.34 and 11.35 are solved using the spreadsheet on the following two pages. These problems are of the Class II type and are solved using the procedure described in Section 11.4.

Method II-B is used because the system contains significant minor losses. In fact, one objective of these two problems is to compare the performance of two design approaches for the same system, using a different, more efficient valve in Problem 11.33 as compared with Problem 11.32.

Recall that Method II-A is set up and solved first in the spreadsheet. This ignores the minor losses and gives an upper limit for the volume flow rate that can be delivered through the system with a given pressure drop or head loss. Then, Method II-B is used iteratively to hone in on the maximum volume flow rate that can be carried with the minor losses considered. The designer enters a series of estimates for the value of Q and observes the resulting pressure at the outlet from the piping system.

Obviously this pressure for this problem should be exactly zero because the pipe discharges into the free atmosphere. Each trial requires only a few seconds to complete and the designer should continue making estimates until the pressure at the outlet is at or close to zero.

Problem 11.32 shows that Q = 0.00285 m 3/s (171 L/min) can be delivered through the given system with a fully open globe valve installed and with a pressure of 300 kPa at point A at the water main.

In Problem 11.33, the globe valve is replaced with a fully open gate valve having much less energy loss. The result is that Q = 0.003398 m3/s (204 L/min) can be delivered with the same pressure at point A. This is approximately a 20% increase in flow. Also, it is only about 3% less than the flow that could be delivered through the pipe with no minor losses at all, as shown in the results of Method II-A at the top of the spreadsheet.

SERIES PIPE LINE SYSTEMS

191

192

Chapter 11

SERIES PIPE LINE SYSTEMS

193

11.36

Class I Pt. 1 at surface of tank A; Pt. 2 outside pipe in tank B. u 1 = 0 2 2 é ù u2 p p1 + z1 + 1 − hL= 1 + z2 + 2 : p1 = p2 + g ê ( z2 - z1 ) + 2 + hL ú I 2g 2g 2g ë û Q 250 gal/min 1 ft 3 /s u 22 (23.87)2 = × = 23.87 ft/s: = = 8.844 ft A 0.02333 ft 2 449 gal/min 2g 2(32.2) u Dr (23.87)(0.1723)(1.53) NR = = = 3.00 ´ 105 : -5 h 2.10 ´ 10 D 0.1723 = = 1149 ® f = 0.0205 e 1.5 ´ 10-4 2 u2 u2 u2 æ 110 ft ö u 2 hL = 0.5 2 + fT (160) 2 + 2 fT (30) 2 + f ç : fT = 0.019 ÷ 2g 2g 2g è 0.1723 ft ø 2 g Entr. Valve 2 Elbows Friction

u2 =

hL = I

194

u 22 2g

[4.68 + 638f] = 8.844 ft[4.68 + 638(0.0205)] = 157.1 ft

p1 = 40.0 psig +

49.01 lb 1 ft 2 [20 ft + 8.844 ft + 157.1 ft] = 103.3 psig 3 ft 144 i n 2

Chapter 11

SERIES PIPE LINE SYSTEMS

195

196

Chapter 11

SERIES PIPE LINE SYSTEMS

197

198

Chapter 11

SERIES PIPE LINE SYSTEMS

199

200

Chapter 11

SERIES PIPE LINE SYSTEMS

201

44

202

Chapter 11

SERIES PIPE LINE SYSTEMS

203

204

Chapter 11

SERIES PIPE LINE SYSTEMS

205

206

Chapter 11

SERIES PIPE LINE SYSTEMS

207

208

Chapter 11

SERIES PIPE LINE SYSTEMS

209

11.51

210

Chapter 11

11.52

SERIES PIPE LINE SYSTEMS

211

CHAPTER TWELVE PARALLEL PIPE LINE SYSTEMS

212

Chapter 12

PARALLEL PIPE LINE SYSTEMS

213

214

Chapter 12

PARALLEL PIPE LINE SYSTEMS

215

216

Chapter 12

PARALLEL PIPE LINE SYSTEMS

217

218

Chapter 12

PARALLEL PIPE LINE SYSTEMS

219

220

Chapter 12

Compute NR values for Trial 1: N Ra = (5.114 ´ 105)(Qa) = (5.114 ´ 105)(0.50) = 2.557 ´ 105

Similarly, N Rb = 3.580 ´ 105; N Rc = 5.114 ´ 104; N Rd = 2.045 ´ 105 N Ra = 2.557 ´ 105; N Rf = 5.114 ´ 104

Compute f values for Eq. 9.9: 0.25

fa =

5.74 log (1.970 ´ 10-4 ) + (2.557 ´ 105 ) 0.9

2

= 0.0197

ka = (3410)(0.0197) = 67.18 Similarly: PIPE

f

k

Eq. for k

a

0.0197

67.18

ka = 3410fa

b

0.0194

66.00

kb = 3410fb

c

0.0233

47.65

kc = 2046fc

d

0.0200

68.26

kd = 3410fd

e

0.0197

67.18

ke = 3410fe

f

0.0233

47.65

kf = 2046ff

PARALLEL PIPE LINE SYSTEMS

221

222

Chapter 12

PARALLEL PIPE LINE SYSTEMS

223

224

Chapter 12

PARALLEL PIPE LINE SYSTEMS

225

226

Chapter 12

PARALLEL PIPE LINE SYSTEMS

227

228

Chapter 12

PARALLEL PIPE LINE SYSTEMS

229

230

Chapter 12

PARALLEL PIPE LINE SYSTEMS

231

232

Chapter 12

PARALLEL PIPE LINE SYSTEMS

233

234

Chapter 12

PARALLEL PIPE LINE SYSTEMS

235

236

Chapter 12

PARALLEL PIPE LINE SYSTEMS

237

12.13

238

Chapter 12

CHAPTER THIRTEEN PUMP SELECTION AND APPLICATION 13.1 to 13.14: 13.15

Answers to questions in text.

Affinity laws relate the manner in which capacity, head and power vary with either speed or

impeller diameter.

N2 0.5 N 1  Q1  0.5Q1 : Capacity cut in half.

N1 N1

13.16

Q 2  Q1

13.17

 N2   0.5 N 1  ha2  ha 2    ha1    0.25ha1 : ha divided by 4.  N1   N1 

13.18

 N2   0.5 N 1  P 2  P1    P1    0.125P1 : P divided by 8.  N1   N1 

2

3

13.19 Q 2  Q 1

2

3

D2 0.75 D 1  Q1  0.75Q 1 : 25% reduction.

D1 D1 2

2

13.20

 D2   0.75 D1  ha 2  ha1    ha1    0.5625ha1 : 44% reduction.  D1   D1 

13.21

 D2   0.75 D 1  P 2  P1    P1    0.422 P1 : 58% reduction.  D1   D1 

3

13.22 1

3

1 3 6 2 6 in casing-size of largest impeller 3 in nominal suction connection size 1

1 in nominal discharge connection size 2

13.23 1

1  3  10 2

13.24

1

1 36 2

13.25

Q  230 gal/min; P  22hp; e  53%; NPSHR  10.1 ft

PUMP SELECTION AND APPLICATION

239

13.26 At ha = 248 ft, emax = 57%: Q = 165 gal/min; P = 17.0 hp; NPSHR = 7.5 ft 13.27 From Problem 13.26, ha1 = 248 ft: Let ha2 = 1.15 ha1 = 285 ft Then Q 2 = 65 gal/min; P 2 = 8.5 hp; e 2 = 44%; NPSHR = 5.5 ft (approximate values)

13.28

6 in

7 in

8 in

9 in

10 in

ha

120 ft

195 ft

248 ft

320 ft

390 ft

Q

90 gal/min

130 gal/min

165 gal/min

210 gal/min

245 gal/min

emax

52%

55%

57%

58%

58.7%(Est)

13.29 NPSHR increases.

13.30 Throttling valves dissipate energy from fluid that was delivered by pump. When a lower speed is used to obtain a lower capacity, power required to drive pump decreases as the cube of the speed. Variable speed control is often more precise and it can be automatically controlled.

13.31 As fluid viscosity increases, capacity and efficiency decrease, power required increases.

13.32 Total capacity doubles.

13.33 The same capacity is delivered but the head capability increases to the sum of the ratings of the two pumps.

13.34 a.

Rotary or 3500 rpm centrifugal

b.

Rotary

c.

Rotary

d.

Reciprocating

e.

Rotary or high speed centrifugal

f.

1750 rpm centrifugal

g.

1750 rpm centrifugal or mixed flow

h.

Axial flow

1 3.35 Q  350 gal / min; H  550 ft; D  12in; N  3560 rpm

Ns =

N Q H

3/4



3560 350 0.75

(550)

1/4

= 586; Ds = DH Q

0.25

(12)(550)  350

= 3.1

Point in Fig. 13.53 lies in radial flow centrifugal region.

13.36 Q  2525 gal / min; H  200 ft; D  15in; N  1780 rpm

Ns =

N Q H

3/4



1780 2525 0.75

(200)

1/4

= 1682; Ds = DH Q

0.25

15(200)  2525

= 1.12

Point in Fig. 13.34 lies in radial flow centrifugal region.

N Q

3/4

13.37 Ns = 3/4 : N = NsH Q H 240

0.75

(5000)(40) = 1000

= 795 rpm Chapter 13

13.38

Ns 

N Q (1750) 5000   3913 H 3/ 4 (100)0.75

13.39

Ns 

N Q (1750) 12000   2659 H 3/4 (300) 0.75

13.40

Ns 

N Q (1750) 500   1237 H 3/ 4 (100)0.75

13.41

Ns 

N Q (3500) 500   2475; Twice N s from 13.40 H 3/4 (100)0.75

13.42

Same method as Problems 13.38 to 13.41

a.

N s = 1463 radial

b.

N s = 260 too low

c.

N s = 3870 radial or mixed

d.

N s = 18.5 too low

e.

N s = 104 too low

f.

N s = 2943 radial

g.

Ns = 7260 mixed

h.

Ns = 24277 axial

13.43 to 13.46 See Text.

13.47

At inlet to pump. Pressure at this point and fluid properties affect pump operation, particularly the onset of cavitation. Pump manufacturer's NPSHR rating related to pump inlet.

13.48

Elevating reservoir raises pressure at pump inlet and increases NPSHa.

13.49

Large pipe sizes reduce flow velocity and reduce energy losses, thus increasing NPSHa.

13.50

Air pockets will not form in an eccentric reducer as they will in a concentric reducer.

13.51

N   2850   4.97 ft ( NPSH R ) 2  ( NPSH R )1 2   7.50 ft   3500  N 

2

2

1

13.52

NSPH a  hsp + hs − h p : Some data from Prob.7.14.

14.4 lb ft 3 144in 2 = 33.34 ft in 2 62.2lb ft 2 hs  10.0 ft; h f  6.0 ft;

a. hsp =

patm

=

b. Water at 180°F, hvp = 17.55 ft

h p  1.17 (Table 13.2) NPSH a  33.34 □10 □6 □1.17  16.17 ft

13.53

= 60.0 lb / ft 3 hsp = (14.4)(144) / (60.0) = 34.22 ft NSPH a 34.22□10□ 6□17.55 = 0.67 ft Cavitation will likely occur!

NPSH a  hsp − hs −h f −h p  34.48 −4.8 − 2.2 −6.78  20.70 ft 14.7 lb ft 3 144in 2 patm hsp      34.48 ft in 2 61.4 lb ft 2 □ hs  4.8 ft; h f  2.2 ft; h p  6.78 ft (Table 13.2) Water at 140°F

PUMP SELECTION AND APPLICATION

241

13.54

NSPH a  hsp  hs□h f □h p  11.47  2.6 □0.80 □1.55  11.72 m patm 98.5 KN m3   11.47 m; hs  2.6 m; h f  0.80 m □ m 2 8.59 KN

hsp  h p  13.55

p□p □



13.3 KN m3  1.55 m m 2 8.59 KN

NPSH a  hsp□h s□h f □h p

patm 101.8 KN m 2 hsp    10.68 m; hs = 2.0 m; hup = 4.97 m □ m2 9.53 KN 2 2 □2 □2  L  □3  L  □2 h f  f3    K1 3  f3T (20) 3  f 2    D  3 2g  D  3 2g 2g 2g

Friction 3 in Foot valve Elbow Friction 2 in

NOTE : f3T  0.018 K1  75 f3T  75(0.017)  1.28

Q 300 L/min 1 m3 /s 1.05 m □22 (1.05)2    0.0560 m ; □3   A3 4.76810 □3 m 2 60000 L/min s 2 g 2(9.81) □ D (1.05)(0.0779) 5 D3 0.0779 N R3  3 3   2.3 10 ;  11694; f3  0.019 □ □ 4.610 □5 3.60 10□7 □2 

Q 300 / 60000 2.31m □22 (2.31) 2     0.271 m ; A2 2.168  10 □3 s 2 g 2(9.81)

N R2 

D □2 D2 (2.31)(0.0525)   3.4  105 ; 2  1141; f 2  0.0202 □7 □ □ 3.60  10

hsp  psp /□ patm /□ (100.2 KN / m 2 ) / 1.48(9.81 KN / m3 )   6.90 m

For all problems 13.56 □ 13.65: NPSHa = hsp ± hs □ hf □ hvp See Section 13.11, Equation 13-14. See Figure 13.37 for vapor pressure head hvp.

13.56

Find NSPH a : Carbon tetrachloride at 150°F; sg = 1.48; patm = 14.55 psia ; hs = □3.6 ft; hf = 1.84 ft hvp = 16.3 ft Open tank: hsp  psp / / □ patm  (14.55 lb / in 2 )(144 in 2 / ft) 1.48(62.4 lb / ft 3 )   22.69 ft NSPH a : = 22.69 □ 3.6 □ 1.84 □16.3 = 0.95 ft (Low)

13.57

Find NPSHa: Carbon tetrachloride at 65C; sg = 1.48; patm = 100.2 kPa; hs = □1.2 m; hf = 0.72 m hvp = 4.8 m Open tank: hsp  psp / □ patm / □ (100.2 KN / m 2 ) / 1.48(9.81 KN / m3 )   6.90 m NPSHa = 6.90 □1.2 □0.72 □4.8 = 0.18 m (Very low)

242

Chapter 13

13.58

Find NPSHa: Gasoline at 40°C; sg = 0.65; patm = 99.2 kPa; hs = -2.7 m; hf = 1.18 m hvp = 14.0 m Open tank: hsp = psp/g = patm/g = (99.2 kN/m2)/[0.65(9.81 kN/m3)] = 15.55 m NPSHa = 15.55 - 2.7 - 1.18 - 14.0 = -2.33 m (Cavitation expected)

13.59

Find NPSHa: Gasoline at 110°F; sg = 0.65; patm = 14.28 psia; hs = +4.8 ft; hf = 0.87 ft hvp = 51.0 ft Open tank: hsp = psp/g = patm/g = (14.28 lb/in2)(144 in2/ft2)/[0.65(62.4 lb/ft3)] = 50.70 ft NPSHa = 50.70 + 4.8 - 0.87 - 51.0 = 3.63 ft

13.60

Find NPSHa: Carbon tetrachloride at 150°F; sg = 1.48; patm = 14.55 psia; hs = +3.66 ft; hf = 1.84 ft hvp = 16.3 ft Open tank: hsp = psp/g = patm/g = (14.55 lb/in2)(144 in2/ft2)/[1.48(62.4 lb/ft2)] = 22.69 ft NPSHa = 22.69 + 3.67 - 1.84 - 16.3 = 8.22 ft

13.61

Find NPSHa: Gasoline at 110°F; sg = 0.65; patm = 14.28 psia; hs = -2.25 ft; hf = 0.87 ft hvp = 51.0 ft Open tank: hsp = psp/g = patm/g = (14.28 lb/in2)(144 in2/ft2)/[0.65(62.4 lb/ft3)] = 50.70 ft NPSHa = 50.70 - 2.25 - 0.87 - 51.0 = -3.42 ft (Cavitation expected)

13.62

Find NPSHa: Carbon tetrachloride at 65°C; sg = 1.48; patm = 100.2 kPa; hs = +1.2 m; hf = 0.72 m hvp = 4.8 m Open tank: hsp = psp/g = patm/g = (100.2 kN/m2)/[1.48(9.81 kN/m3)] = 6.90 m NPSHa = 6.90 + 1.2 - 0.72 - 4.8 = 2.58 m

13.63

Find NPSHa: Gasoline at 40°C; sg = 0.65; patm = 99.2 kPa; hs = +0.65 m; hf = 1.18 m hvp = 14.0 m Open tank: hsp = psp/g = patm/g = (99.2 kN/m2)/[0.65(9.81 kN/m3)] = 15.55 m NPSHa = 15.55 + 0.65 - 1.18 - 14.0 = 1.02 m

13.64

Find required pressure above the fluid in a closed, pressurized tank so that NPSHa ³ 4.0 ft.

Propane at 110°F; sg = 0.48; patm = 14.32 psia; hs = +2.50 ft; hf = 0.73 ft hvp = 1080 ft Solve Eq. 13-14 for required hsp = NPSHa - hs + hf + hvp = 4.0 - 2.5 + 0.73 + 1080 = 1082.2 ft Psp = g hsp = (0.48)(62.4 lb/ft3)(1082.2 ft)(1 ft2/144 in2) = 225.1 lb/in2 = 225.1 psia Gage pressure: ptank = psp - patm = 225.1 psia - 14.32 psia = 210.8 psig

13.65

Find required pressure above the fluid in a closed, pressurized tank so that NPSHa ³ 150.0 m.

Propane at 45°C; sg = 0.48; patm = 9.47 kPa absolute; hs = -1.84 m; hf = 0.92 m hvp = 340 m Solve Eq. 13-14 for required hsp = NPSHa - hs + hf + hvp = 1.50 + 1.84 + 0.92 + 340 = 344.3 m Psp = g hsp = (0.48)(9.81 kN/m3)(344.3 m) = 1621 kN/m2 = 1621 kPa absolute Gage pressure: ptank = psp - patm = 1621 kPa - 98.4 kPa = 1523 kPa gage

PUMP SELECTION AND APPLICATION

243

13.66

244

Chapter 13

CHAPTER FOURTEEN OPEN - CHANNEL FLOW

OPEN- CHANNEL FLOW

245

246

Chapter 14

14.14

See Prob. 14.8 for d = 0.50 m; Prob. 14.9 for d = 2.50 m S = 0.50% = 0.005, n = 0.017 a. d = 0.50 m; A = 0.50 m2; R = 0.25 m 1.00 1.00 Q= AR 2 / 3 S 1/ 2 = (0.50)(0.25)2/3(0.005)1/2 = 0.825 m3/s n 0.017 b. d = 2.50 m; A = 9.72 m2; R = 0.909 m 1.00 Q= (9.72)(0.909) 2 / 3 (0.005)1/ 2 = 37.9 m3/s 0.017

14.15

a.

Depth = 3.0 ft: é1 ù A = (3)(12) + 2 ê (3)(3) ú = 45 ft 2 ë2 û

WP = 12 + 2(4.243) = 20.485 ft R = A/WP = 45/20.485 = 2.197 ft 1.49 Q= (45)(2.197) 2 / 3 (0.00015)1/ 2 = 34.7 ft3/s 0.04 Depth = 6.0 ft:

b.

é1 ù A = (4)(12) + 2 ê (4)(4) ú ë2 û é1 ù + (2)(40) + 2 ê (2)(2) ú ë2 û A = 148 ft2 WP = 2(2.828) + 2(10) + 2(5.657) + 12 = 48.97 ft R = A/WP = 148/48.97 = 3.022 ft 1.49 Q= (148)(3.022)2 / 3 (0.00015)1/ 2 = 141.1 ft3/s 0.04

14.16

nQ (0.015)(150) = 47.75 = 1/ 2 1.49 S (1.49)(0.001)1 / 2 A 10 y A = 10y; WP = 10 + 2y; R = = WP 10 + 2 y

AR2/3 =

AR

2/3

é 10 y ù = 10y ê ú ë10 + 2 y û

2/3

Find y such that AR2/3 = 47.75

By trial and error, y = 3.10 ft; AR

2/3

é 10(3.1) ù = 10(3.1) ê ú ë10 + 2(3.1) û

OPEN- CHANNEL FLOW

2/3

= 47.78

247

248

Chapter 14

OPEN- CHANNEL FLOW

249

250

Chapter 14

OPEN- CHANNEL FLOW

251

252

Chapter 14

Trapezoid: y =

A 0.4545 = = 0.5126 ft; R = y/2 = 0.2563 ft 1.73 1.73 2

é 0.02768 ù S= ê = 0.00471; 2/ 3 ú ë (0.2563) û A A 0.4545 yh = = = 0.3841 ft = T 2.309 y 2.309(0.5126)

NF =

2.75 (32.2)(0.3841)

Semicircle: A =

p y2 2

;y=

= 0.782 < 1.0 Subcritical

2A

p

=

2(0.4545)

p

= 0.5379 ft

2

é 0.02768 ù R = y/2 = 0.269 ft: S = ê = 0.00441 2/3 ú ë (0.269) û A p y2 p y yh = = = = 0.4225 ft T 2(2 y ) 4

NF =

14.39

a.

2.75 (32.2)(0.4225)

When y = yc, NF = 1.0 = é Q ù Then y = ê ú êë b g N F úû

b.

= 0.746 < 1.0 Subcritical

2/3

u gy h

=

Q Q Q = = A gy (by ) gy b g y 3 / 2

é ù 5.5 =ê ú ë 2.0 9.81(1.0) û

2/3

= 0.917 m = yc

Minimum E occurs when y = yc: From Eq. 14.18: Q2 Q2 Q2 Emin = yc + = + = y + y c c 2 gA2 2 g (byc )2 2 gb 2 yc2 = 0.917 +

5.52 2(9.81)(2.0) 2 (.917) 2

Emin = 1.375 m c. d.

e.

See spreadsheet and graph for values of y versus E. 5.52 For y = 0.50 m; E = 0.50 + 2(9.81)(2.0) 2 (0.5) 2 = 2.042 m From spreadsheet, alternate depth = 1.934 m Q Q 5.5 u = = ; NF = A by 2.0 y gy For y = 0.5 m, u = 5.50 m/s; NF = 2.48 For y = 1.934 m, u = 1.418 m/s; NF = 1.325

u=

OPEN- CHANNEL FLOW

253

254

Chapter 14

14.40

Circular Channel Q= 1.45 m3/s D= 1.20 m

y(m) q (rad) y less than D 0.10 1.171 0.20 1.682 0.25 1.896 0.30 2.094 0.40 2.462 0.50 2.807 0.60 3.142 y greater than D 0.658 3.335 0.70 3.476 0.80 3.821 0.90 4.189 0.913 4.238 1.00 4.601 1.199 6.168 1.20 6.283

n=

0.015 Finished concrete

A(m2)

T(m)

yh(m)

NF

E(m)

0.0450 0.1239 0.1707 0.2211 0.3300 0.4460 0.5655

0.6633 0.8944 0.9747 1.0392 1.1314 1.1832 1.2000

0.068 0.139 0.175 0.213 0.292 0.377 0.471

39.478 10.039 6.481 4.539 2.597 1.690 1.193

52.982 7.181 3.928 2.492 1.384 1.039 0.935

Velocity (m/s) 32.211 11.703 8.494 6.558 4.394 3.251 2.564

0.6349 0.6849 0.8010 0.9099 0.9231 1.0071 1.1309 1.1310

1.1944 1.1832 1.1314 1.0392 1.0240 0.8944 0.0693 0.0000

0.532 1.000 0.579 0.888 0.708 0.687 0.876 0.544 0.901 0.528 1.126 0.433 16.330 0.101 #DIV/0! #DIV/0!

0.924 0.928 0.967 1.029 1.039 1.106 1.283 1.284

2.284 2.117 1.810 1.594 1.571 1.440 1.282 1.282

Given y Critical depth

Alt depth for given y Nearly full pipe depth Full pipe (yh and NF undefined)

Part f of problem: Slopes for given y and alternate depth y(m) 0.50 0.913

R(m) 0.2649 0.3630

S 0.0140 0.0021

S for given y S for alternate depth

Problem 14.40 Procedure: Refer to Table 14.2 for geometry of a partially full circular pipe. a)

For given Q, D, and y: Compute q , A, T using equations in Table 14.2.

(

)

Compute NF = v / gyh = Q A gyh .

Iterate values of y until NF = 1.000. See spreadsheet: yc = 0.658 m. b)

Minimum specific energy: E= y + v2/2g = y + Q2/(2gA2) From spreadsheet, with y = yc = 0.658 m: Emin = 0.924 m.

c)

Specific energy versus y: See spreadsheet using equation in b).

d)

Specific energy for y = 0.50 m: E = 1.039 m from spreadsheet.

Iterate on y to find alternate depth for which E = 0.1039 m.

See spreadsheet: yalt = 0.913 m.

e)

Velocity = v = Q/A, NF = v / gyh = Q / A gyh . See spreadsheet

(

)

For y = 0.50 m: v = 3.251 m/s, NF = 1.690. Supercritical For y = 0.913 m: v = 1.571 m/s, NF = 0.528. Subcritical.

OPEN- CHANNEL FLOW

255

f)

Compute WP = q D/2 (See Table 14.2). Compute R = A/WP. 2

é nQ ù Compute S: ê 2 / 3 ú ë AR û See spreadsheet: For y = 0.50 m, S = 0.0140. For y = 0.913 m, S = 0.0021.

14.41

Triangular channel z= 1.5 Q= 0.68 ft3/s y(ft) 0.20 0.25 0.30 0.40 0.418 0.50 0.60 0.70 0.80 0.90 1.00 1.065 1.10 1.20 1.30 1.40 1.50

A(ft2) 0.060 0.094 0.135 0.240 0.262 0.375 0.540 0.735 0.960 1.215 1.500 1.701 1.815 2.160 2.535 2.940 3.375

n = 0.022 V(ft/s) 7.253 2.594

0.400

T(ft) 0.60 0.75 0.90 1.20 1.254 1.50 1.80 2.10 2.40 2.70 3.00 3.20 3.30 3.60 3.90 4.20 4.50

yh(ft) 0.100 0.125 0.150 0.200 0.209 0.250 0.300 0.350 0.400 0.450 0.500 0.533 0.550 0.600 0.650 0.700 0.750

NF 6.316 3.615 2.292 1.116 1.000 0.639 0.405 0.276 0.197 0.147 0.113 0.097 0.089 0.072 0.059 0.049 0.041

E(ft) 2.194 1.067 0.694 0.525 0.523 0.551 0.625 0.713 0.808 0.905 1.003 1.067 1.102 1.202 1.301 1.401 1.501

Given y Critical depth

Alternate depth

Slopes at given depth and alternate depth y(ft) R(ft) S 0.25 0.10401 0.521 Slope for given depth 1.065 0.44307 0.000229 Slope for alternate depth Problem 14.41 Procedure: Refer to Table 14.2 for geometry of a triangular channel. a)

For given Q, z, and y: Compute A, T using equations in Table 14.2.

(

)

Compute NF = v / gyh = Q / A gyh .

Iterate values of y until NF = 1.000.

See spreadsheet: yc = 0.418 ft.

256

b)

Minimum specific energy: E = y + v2/2g = y + Q2/(2gA2) From spreadsheet, with y = yc = 0.418 ft: Emin = 0.523 ft.

c)

Specific energy versus y: See spreadsheet using equation in b).

d)

Specific energy for y = 0.25 ft: E = 1.067 ft from spreadsheet.

Iterate on y to find alternate depth for which E = 0.1067 ft.

See spreadsheet: yalt = 1.065 ft.

Chapter 14

e)

(

)

Velocity = v = Q/A, NF = v / gyh = Q / A gyh . See spreadsheet For y = 0.25 ft: v = 7.253 ft/s, NF = 3.615. Supercritical For y = 1.065 ft: v = 0.400 ft/s, NF = 0.097. Subcritical.

f)

Compute WP = 2 y 1 + z 2 (See Table 14.2). Compute R = A/WP. 2

é nQ ù Compute S: ê 2 / 3 ú ë AR û See spreadsheet: For y = 0.25 ft, S = 0.0521. For y = 1.065 ft, S = 0.000229.

14.42

Trapezoidal channel z= 0.75 Q= 0.80 ft3/s y(ft) 0.05 0.1 0.1288 0.20 0.25 0.30 0.40 0.4770 0.50 0.60 0.70 0.80 0.90 1.00 1.065 1.10 1.20 1.30 1.40 1.50

A(ft2) 0.152 0.308 0.399 0.630 0.797 0.968 1.320 1.602 1.688 2.070 2.468 2.880 3.308 3.750 4.046 4.208 4.680 5.168 5.670 6.188

n = 0.013 b = 3.000 ft V(ft/s) 5.267 2.602 2.006 1.270 1.004 0.827 0.606 0.499 0.474 0.386 0.324 0.278 0.242 0.213 0.198 0.190 0.171 0.155 0.141 0.129

T(ft) 3.08 3.15 3.19 3.30 3.38 3.45 3.60 3.72 3.75 3.90 4.05 4.20 4.35 4.50 4.60 4.65 4.80 4.95 5.10 5.25

Yh(ft) 0.049 0.098 0.125 0.191 0.236 0.280 0.367 0.431 0.450 0.531 0.609 0.686 0.760 0.833 0.880 0.905 0.975 1.044 1.112 1.179

NF 4.177 1.467 1.000 0.512 0.364 0.275 0.176 0.134 0.125 0.093 0.073 0.059 0.049 0.041 0.037 0.035 0.031 0.027 0.024 0.021

E(ft) 0.481 0.205 0.191 0.225 0.266 0.311 0.406 0.481 0.503 0.602 0.702 0.801 0.901 1.001 1.066 1.101 1.200 1.300 1.400 1.500

Given y Critical depth

Alternate depth

Slopes at given depth and alternate depth y(ft) R(ft) S 0.05 0.0486 0.264 Slope for given depth 0.477 0.3820 0.000152 Slope for alternate depth Problem 14.42 Procedure: Refer to Table 14.2 for geometry of a trapezoidal channel. a)

For given Q, b, z, and y: Compute A, T using equations in Table 14.2.

(

)

Compute NF = v / gyh = Q / A gyh .

Iterate values of y until NF = 1.000. See spreadsheet: yc = 0.1288 ft. b)

Minimum specific energy: E = y + v2/2g = y + Q2/(2gA2) From spreadsheet, with y = yc = 0.1288 ft: Emin = 0.191 ft.

c)

Specific energy versus y: See spreadsheet using equation in b).

OPEN- CHANNEL FLOW

257

258

Chapter 14

14.48

Q = 2.48H5/2 H(in) H(ft) 0 0 2 .167 4 .333 6 .500 8 .667 10 .833 12 1.000

Q(ft3/sec) 0 .0283 .159 .439 .900 1.57 2.48

14.49

Q = 3.07H1.53 H1.53 = Q/3.07 \ H = (Q/3.07)1/1.53 Min Q = 0.09 ft3/sec H = (0.09/3.07)1/1.53 = (0.0293)0.654 = 0.10 ft Max Q = 8.9 ft3/sec H = (8.9/3.07)1/1.53 = (2.90)0.654 = 2.01 ft

14.50

L = 8.0 ft; Qmin = 3.5 ft3/s; Qmax = 139.5 ft3/s Q = 4.00 LHn; H = [Q/4.00)L)1/n = [Q/(4.00)(8.0)]1/1.61 = [Q/32]0.621 Qmin = 3.5 ft3/s; H = [3.5/32]0.621 = 0.253 ft Qmax = 139.5 ft3/s; H = [139.5/32]0.621 = 2.496 ft H(ft) 0.25 1.00 1.50 2.00 2.25 2.50

14.51

Q(ft3/sec) 3.434 32.000 61.469 97.681 118.077 139.905

a)

Q = 50 ft3/s; L = 4.0 ft; Q = 4.00 LHn; n = 1.58 H = [Q/(4.00)L]1/n = [50/(4.00)(4.0)]1/1.58 = [3.125]0.633 = 2.06 ft

b)

L = 10.0 ft Q = (3.6875L + 2.5)H1.6 = 39.375H1.6 1/1.6

æ Q ö H= ç ÷ è 39.375 ø

14.52

æ 50 ö =ç ÷ è 39.375 ø

0.625

= 1.155 ft

Trapezoidal channel—Long-throated flume - Design C: H = 0.84 ft; Q = K1(H + K2)n K1 = 16.180; K2 = 0.035; n = 1.784 Q = 16.180[0.84 + 0.035]1.784 = 12.75 ft3/s = Q

OPEN- CHANNEL FLOW

259

14.53

Trapezoidal channel—Long-throated flume  Design B: H = 0.65 ft; Q = K1(H + K2)n K1 = 14.510; K2 = 0.053; n = 1.855 Q = 14.510[0.65 + 0.053]1.855 = 7.547 ft3/s = Q

14.54

Rectangular channel—Long-throated flume  Design A: H = 0.35 ft; Q = bcK1(H + K2)n bc = 0.500 ft; K1 = 3.996; K2 = 0.000; n = 1.612 Q = (0.500)(3.996)[0.35 + 0.000] 1.612 = 0.368 ft3/s = Q

14.55

Rectangular channel—Long-throated flume  Design C: H = 0.40 ft; Q = bcK1(H + K2)n bc = 1.500 ft; K1 = 3.375; K2 = 0.011; n = 1.625 Q = (1.500)(3.375)[0.40 + 0.011] 1.625 = 1.194 ft3/s = Q

14.56

Circular channel—Long-throated flume  Design B: H = 0.25 ft; Q = D2.5 K1 (H/D + K2)n D = 2.00 ft; K1 = 3.780; K2 = 0.000; n = 1.625 Q = (2.00)2.5(3.780)[0.25/2.00 + 0.000]1.625 = 0.729 ft3/s = Q

14.57

Circular channel—Long-throated flume  Design A: H = 0.09 ft; Q = D2.5 K1 (H/D + K2)n bc = 1.00 ft; K1 = 3.970; K2 = 0.004; n = 1.689 Q = (1.00)2.5(3.970)[0.09/1.00 + 0.004]1.689 = 0.0732 ft3/s = Q

14.58

Rectangular channel—Long-throated flume  Design B: Q = 1.25 ft3/s; Find H. Q = bc K1(H + K2)n; bc = 1.00 ft; K1 = 3.696; K2 = 0.004; n = 1.617 Solving for H: H = [Q/(bcK1)]1/n  K2 = [1.25/(1.0)(3.696)]1/1.617  0.004 = 0.507 ft = H

14.59

Circular channel—Long-throated flume  Design C: Q = 6.80 ft3/s; Find H. Q = D2.5 K1(H/D + K2)n; D = 3.000 ft; K1 = 3.507; K2 = 0.000; n = 1.573 Solving for H: H = D{[Q/D2.5)(K1)]1/n  K2} = 3.0{[6.80/(3.02.5)(3.507)]1/1.573  0.00} = 0.797 ft = H

14.60

Select a long-throated flume for 30 gpm < Q < 500 gpm. Using 449 gpm = 1.0 ft 3/s, 0.0668 ft3/s < Q < 1.114 ft3/s; Select Circular channel; Design A; Q  D2.5 K1(H/D + K2)n H = D{[Q/D2.5)(K1)]1/n  K2}; D = 1.000 ft; K1 = 3.970; K2 = 0.004; n = 1.689 For Q = 0.0668 ft3/s: H = 1.0{[0.0668/(1.02.5)(3.970)]1/1.689  0.004} = 0.0851 ft = H For Q = 1.114 ft3/s: H = 1.0{[1.114/(1.02.5)(3.970)]1/1.689  0.004} = 0.467 ft = H H 0.10 0.20 0.30 0.40

260

Q(ft3/s) 0.087 0.271 0.531 0.859

Q(gpm) 39.06 121.7 238.4 385.7

Chapter 14

14.61

Given 50 m3/h < Q , 180 m3/h; Convert to ft3/s; 0.4907 ft3/h < Q < 1.766 ft3/h Specify Rectangular channel long-throated flume, Design B. Find H for each limiting flow rate.

Q = bc K1(H + K2)n; bc = 1.00 ft; K1 = 3.696; K2 = 0.004; n = 1.617 Solving for H: Hmin = [Q/(bcK1)]1/n  K2 = [0.4907/(1.0)(3.696)]1/1.617  0.004 = 0.2829 ft = Hmin Converting to m: Hmin = 0.0863 m for Q = 50 m3/h Hmax = [Q/bcK1)]1/n  K2 = [1.766/(1.0)(3.696)1/1.617  0.004 = 0.629 ft = Hmax Converting to m: Hmax = 0.1917 m for Q = 180 m3/h H(m) 0.100 0.125 0.150 0.175

OPEN- CHANNEL FLOW

H(ft) 0.328 0.410 0.492 0.524

Q(ft3/s) 0.622 0.888 1.190 1.524

Q(m3/h) 63.38 90.49 121.3 155.3

261

CHAPTER FIFTEEN FLOW MEASUREMENT

262

Chapter 15

FLOW MEASUREMENT

263

15.6

Find Dp across Venturi, Q = 600 gal/min(1 ft 3/s)(449 gal/min) = 1.336 ft3/s u 1 = Q/A1 = (1.336 ft3/s)/(0.0844 ft2) = 15.12 ft/s Kerosene at 77°F; g = 51.2 lb/ft3; n = 2.14 ´ 10-5 ft2/s NR = u 1D1/v = (15.12)/(0.3355)/(2.14 ´ 10-5) = 2.37 ´ 105; Then C = 0.984 From Eq. (15-4), solving for p1 - p2 = Dp Dp = [u 1/C]2 [(A1/A2)2 - 1] [g /2g] = [15.12/0.984]2 [(0.0884/0.01227)2 - 1][51.2/(2(32.2))] Dp = 9551 lb/ft2 (1 ft2/144 in2) = 66.3 psi

15.7

Find Q through an orifice meter. D1 = 97.2 mm; A1 = 7.419 ´ 10-3 m2; d = 50 mm = 0.050 m A2 = 1.963 ´ 10-3 m2; A1A2 = 3.778; d/D = 0.05/0.0972 = 0.514; Trial 1: C = 0.608 for NR = 1 ´ 105 Ethylene glycol at 25°C: g = 10.79 kN/m3; n = 1.47 ´ 10-5 m2/s é 2 gh[g m / g eg - 1] ù ú 2 ë ( A1 / A2 ) - 1 û

1/ 2

u1 = C ê

é 2(9.81)(0.095) /[132.8 /10.79 - 1] ù = (0.608) ê ú [3.778]2 - 1 ë û

1/ 2

u = 0.766 m/s; Iteration: New NR = 5.07 ´ 103; New C = 0.623 New u 1 = 0.785 m/s; New NR = 5.19 ´ 103; New C = 0.623 - Unchanged.

Final value of Q = A1u 1 = (7.419 ´ 10-3 m2)(0.785 m/s) = 5.824 ´ 10-3 m3/s = Q 15.8

Orifice meter. Propyl alcohol at 25°C; g = 7.87 kN/m3; n = 2.39 ´ 10-6 m2/s 40 mm OD ´ 3.0 mm wall steel tube: D1 = 34.0 mm = 0.0340 m; A1 = 9.079 ´ 10-4 m2 Let b = 0.40 = d/D; Then d = 0.40 D = 0.40(34.0 mm) = 11.8 mm = 0.01176 m A2 = p D2/4 = p (0.01176 m)2/4 = 1.09 ´ 10-4 m2; A1/A2 = 8.36 From Eq. (15-6), solve for h in mercury manometer; g m = 132.8 kN/m3 h=

[( A1 / A2 ) 2 - 1]u 12 2 gC 2 [(g m / g a ) - 1)]

u 1min = Qmin/A1 = (1.0 m3/h)/(9.079 ´ 10-4 m2)(1 h)/(3600 s) = 0.306 m/s NR1 = u 1D1/n = (0.306)(0.0340)/(2.39 ´ 10-6 ) = 5.1 ´ 103; C = 0.619 hmin =

2 2 2 2 [( A1 / A2 ) - 1]u 1min [(8.36) - 1](0.306 m/s) = = 0.0273 m = 27.3 mm 2 gC 2 [(g m / g a ) - 1] 2(9.81 m/s 2 )(0.619) 2 [(132.8 / 7.87) - 1]

Similarly, u 1max = 0.730 m/s; NR = 1.06 ´ 103; C = 0.610 hmax = 0.1754 m = 175.4 mm 15.9

Flow nozzle — Design: Install in 5 1/2 inch Type K copper tube; D1 = 4.805 in; A1 = 0.1259 ft2 Specify the throat diameter d. NOTE: Multiple solutions possible.

Fluid: Linseed oil at 77°F; g = 58.0 lb/ft3; n = 3.84 ´ 10-4 ft2/s Use mercury manometer with scale range 0 - 8 inHg Range of flow rate: Qmin = 700 gpm = 1.559 ft 3/s; Qmax = 1000 gpm = 2.227 ft 3/s Velocity in pipe: u 1-min = Qmin/A1 = 12.38 ft/s; u 1-max = Qmax/A1 = 17.69 ft/s Reynolds No.: NR-min = u 1-min D1/n = 1.29 ´ 104; NR-max = u 1-max D1/n = 1.84 ´ 104 From Figure 15-5: Cmin = 0.955; Cmax = 0.961

264

Chapter 15

a)

Use Eq. (15-6) and solve for A2 from which we can obtain the throat diameter d A2 =

A1 é BhC 2 ù ê u 2 + 1ú ë 1 û

1/ 2

Where B = 2g[(g m/g Lo) - 2(32.2)[844.9/58.0) - 1] = 873.7 Let h = 8.0 in = 0.667 ft when u = u 1-max = 17.69 ft/s and C = 0.961 A2 =

0.1259 1/ 2

é (873.7)(0.667)(0.961) ù + 1ú ê 2 (17.69) ë û 2

= 0.07635 ft2 = p d2/4

Then d = (4A2/p )1/2 = [(4)(0.07635)/p ]1/2 = 0.3118 ft = 3.741 in = Throat diameter b)

Use Eq. (15-6) and solve for h to determine the manometer reading when Q = Qmin. (u 1 / C )2 [( A1 / A2 ) 2 - 1] B For u 1 = u 1-min = 12.38 ft/s and C = 0.955,

h=

h=

(u 1 / C ) 2 [( A1 / A2 ) 2 - 1 [(12.38 / 0.955) 2 [(0.1259 / 0.07635) 2 - 1] = = 0.3306 ft = 3.97 in B 873.73

Summary:

15.10

Nozzle throat diameter = d = 3.741 in When Q = 1000 gpm, h = 8.00 in manometer deflection When Q = 700 gpm, h = 3.97 in manometer deflection

Orifice Meter — Design: Install in 12 inch ductile iron pipe; D1 = 12.51 in = 1.043 ft; A1 = 0.8536 ft2 Specify the orifice diameter d. NOTE: Multiple solutions possible.

Fluid: Water at 60°F; g = 62.4 lb/ft3; n = 1.21 ´ 10-5 ft2/s Use mercury manometer with scale range 0 - 12 inHg Range of flow rate: Qmin = 1500 gpm = 3.341 ft 3/s; Qmax = 4000 gpm = 8.909 ft 3/s Velocity in pipe: u 1-min = Qmin/A1 = 3.914 ft/s; u 1-max = Qmax/A1 = 10.44 ft/s Reynolds No.: NR-min = u 1-min D1/n = 3.23 ´ 105; NR-max = u 1-max D1/n = 8.63 ´ 105 From Figure 15-7: Cmin = 0.610; Cmax = 0.612 [Assumed b = d/D = 0.70] a)

Use Eq.(15-6) and solve for A2 from which we can obtain the throat diameter d.

A2 =

A1 é BhC 2 ù ê u 2 + 1ú ë 1 û

1/ 2

Where B = 2g[(g m/g w) -1] = 2(32.2)[(844.9/62.4) - 1] = 807.6 Let h = 10.0 in = 0.833 ft when u = u 1-max = 10.44 ft/s and C = 0.610 FLOW MEASUREMENT

265

266

Chapter 15

CHAPTER SIXTEEN FORCES DUE TO FLUIDS IN MOTION

DRAG AND LIFT

267

268

Chapter 17

DRAG AND LIFT

269

270

Chapter 17

DRAG AND LIFT

271

272

Chapter 17

DRAG AND LIFT

273

274

Chapter 17

DRAG AND LIFT

275

276

Chapter 17

DRAG AND LIFT

277

CHAPTER SEVENTEEN DRAG AND LIFT 17.1

17.2

FD = CD æç 1 r u 2 ö÷ A: A = D ´ L = (0.025 m)(1 m) = 0.025 m 2 è2 ø a.

Water at 15°C: ρ = 1000 kg/m3; n = 1.15 ´ 10-6 m2/s u D (0.15)(0.025) NR = = = 3.26 ´ 103; Then CD = 0.90 (Fig. 17.3) n 1.15 ´ 10-6 FD = (0.90)(0.50)(1000 kg/m3)(0.15 m/s)2(0.025 m2) = 0.253 kg×m/s2 = 0.253 N

b.

Air at 10°C: ρ = 1.247 kg/m3; n = 1.42 ´ 10-5 m2/s u D (0.15)(0.025) NR = = = 2.64 ´ 102; Then CD = 1.30 n 1.42 ´ 10-5 FD = (1.30)(0.5)(1.247 kg/m2)(0.15 m/s)2(0.025 m2) = 4.56 ´ 10-4 N

Assume a smooth sphere a.

15 km 103 m 1h p D 2 p (2.0) 2 ´ ´ = 4.17 m/s A= = = 3.142 m 2 h km 3600 s 4 4 u D (4.17)(2.0) = = 6.27 ´ 105; Then CD = 0.20 (Fig. 17.3) NR= -5 1.33 ´ 10 n FD = CD æç 1 r u 2 ö÷ A = (0.20)(0.5)(1.292 kg/m3)(4.17 m/s)2(3.142 m2) = 7.06 N è2 ø

u=

Similarly for (b), (c), and (d):

278

u (km/h)

u (m/s)

NR

CD

FD(N)

(a)

15

4.17

6.27 ´ 105

0.20

7.06

(b)

30

8.33

1.25 ´ 106

0.20

28.16

(c)

60

16.67

2.50 ´ 106

0.20

112.7

(d)

120

33.33

5.01 ´ 106

0.20

450.7

(e)

160

44.44

6.68 ´ 106

0.20

801.9

Chapter 17

DRAG AND LIFT

279

280

Chapter 17

DRAG AND LIFT

281

282

Chapter 17

a.

Square cylinder: L = 9 in ´ 1 ft/12 in = 0.75 ft u L (147)(0.75) NR = = 9.42 ´ 105 ® Use CD » 2.10 Fig. 17.3 Extrapolated = n (1.17 ´ 10-4 ) ë [See Prob. 17.14 for n ] NOTE: Extrapolated values FD = (30.12)(2.10)(3.75) = 237 lb should be verified.

b.

Assume CD = 1.60 — Square cylinder — Point first orientation FD = (30.12)(1.60)(5.303) = 256 lb Highest

Circular cylinder: D = 9.0 in = 0.75 ft u D (147)(0.75) NR = = 9.42 ´ 105 ® CD = 0.30 (Fig. 17.3) = n (1.17 ´ 10-4 ) FD = (30.12)(0.30)(3.75) = 33.9 lb Note that CD would rise to approximately 1.30 at lower speed. But, because FD is proportional to u 2, drag force would likely be lower. c.

d.

17.17

Elliptical cylinder: L = 18 in = 1.50 ft; h = 9.00 in = 0.75 ft; L/h = 2.0 u L (147)(1.50) NR = = = 1.88 ´ 106 ® Use CD » 0.25 Fig. 17.5 Extrapolated -4 n 1.16 ´ 10 FD = (30.12)(0.25)(3.75) = 28.2 lb Lowest

65 mi 5280 ft 1h FD = CD æç 1 r u 2 ö÷ A : u = ´ ´ = 95.3 ft/s h mi 3600 s è2 ø

A = DL = (3.50 in)(92 in)1 ft2/144 in2 = 2.236 ft2 u D (95.3)(3.50 /12) NR = = = 2.38 ´ 105 ® CD = 1.10 Fig. 17.3 -4 n 1.17 ´ 10 ë [See Problem 17.14 for v] FD = (1.10)(0.5)(2.80 ´ 10-3)(95.3)2(2.236) = 31.3 lb 17.18

FDtot = FDdisks + FDtubes : u = 100 mi/hr = 147 ft/s (See Prob. 17.16): CDdisks = 1.11 (Table 17.1) N Rtubes =

u D (147)(4.5 /12) = = 4.71 ´ 105 ® Use CD = 0.33 Fig. 17.3 -4 n 1.17 ´ 10 T

ë [See Prob. 17.14]

p D (3)(p )(56 /12)2

æ 4.5 öæ 90 ö 2 = 51.31 ft2; Atubes = DL = ç ÷ç ÷ = 2.813 ft 4 4 è 12 øè 12 ø æ1 ö æ1 ö FDtot = CDdisks ç r u 2 ÷ Ad + CD ç r u 2 ÷ At = (1.11)(0.5)(2.80 ´ 10-3)(147)2(51.31) è2 ø è2 ø 2

Adisks = 3

+ (0.33)(0.5)(2.80 ´ 10-3)(147)2(2.813) = 1723 lb + 28.1 lb = 1751 lb Disks Tubes Total

DRAG AND LIFT

283

284

Chapter 17

DRAG AND LIFT

285

286

Chapter 17

CHAPTER EIGHTEEN FANS, BLOWERS, COMPRESSORS, AND THE FLOW OF GASES Units and conversion factors 18.1

Q = 2650 cfm ´

1 ft 3 /s = 44.17 ft3/s 60 cfm

18.2

Q = 8320 cfm ´

1 ft 3 /s = 138.7 ft3/s 60 cfm

18.3

Q = 2650 cfm ´

1 m 3 /s = 1.25 m3/s 2120 cfm

18.4

Q = 8320 cfm ´

1 m 3 /s = 3.92 m3/s 2120 cfm

18.5

u = 1140 ft/min ´

18.6

u = 5.62 m/s ´

18.7

p = 4.38 in H2O ´

18.8

Q = 4760 cfm ´

18.9

p = 925 Pa ´

1.0 in H 2 O = 3.72 in H2O 248.8 Pa

18.10

p = 925 Pa ´

1.0 psi = 0.134 psi 6895 Pa

1.0 m/s = 5.79 m/s 197 ft/min

3.28 ft/s = 18.43 ft/s m/s 1.0 psi = 0.158 psi 27.7 in H 2 O

1.0 m 3 /s = 2.25 m3/s 2120 cfm 248.8 Pa p = 0.75 in H2O ´ = 186.6 Pa 1.0 in H 2 O

FANS, BLOWERS, COMPRESSORS, AND THE FLOW OF GASES

287

288

Chapter 18

FANS, BLOWERS, COMPRESSORS, AND THE FLOW OF GASES

289

290

Chapter 18

FANS, BLOWERS, COMPRESSORS, AND THE FLOW OF GASES

291

292

Chapter 18

FANS, BLOWERS, COMPRESSORS, AND THE FLOW OF GASES

293

294

Chapter 18

FANS, BLOWERS, COMPRESSORS, AND THE FLOW OF GASES

295

CHAPTER NINETEEN FLOW OF AIR IN DUCTS

296

Chapter 19

19.8

3 ´ 10 in duct: 1.3( ab) 0.625 1.3[ (3)(10)] De = = ( a + b)0.250 (3 + 10) 0.250

0.625

= 5.74 in: Qmax = 95 cfm For hL = 0.10 in H2O

19.9

42 ´ 60 in duct: 0.625 1.30[ (42)(60) ] De = = 54.7 in: Qmax = 37000 cfm (42 + 60)0.250 for hL = 0.10 in H2O

19.10

250 ´ 500 mm duct: De =

1.30[ (250)(500) ] (250 + 500)

0.625

= 381 mm: Qmax = 0.60 m3/s

0.250

for hL = 0.80 Pa/m 19.11

75 ´ 250 mm duct: De =

19.12

1.30[ (75)(250) ]

0.625

= 125 mm: Qmax = 0.0295 m3/s

(75 + 500) 0.25

Q = 1500 cfm; hLmax = 0.10 in H2O per 100 ft ® De = 16.2 in Duct: 10 ´ 24, 12 ´ 20

19.13

Possible sizes from Table 19.2

Q = 300 cfm; hLmax = 0.10 in H2O per 100 ft ® De = 8.8 in Duct: 6 ´ 12 (De = 9.1 in)

Energy losses in ducts with fittings 19.14

Q = 650 cfm; D = 12 in round; u = 830 ft/min (Fig. 19.2) æ u ö æ 830 ö Hu = ç ÷ =ç ÷ = 0.0429 in H2O; C = 0.42 3-pc elbow è 4005 ø è 4005 ø 2

2

HL = CHu = 0.42(0.0429) = 0.0180 in H2O 19.15

C = 0.33 5-pc elbow; HL = CHu = 0.33(0.0429) = 0.0142 in H2O

19.16

Q = 1500 cfm; D = 16 in; u = 1080 ft/min (Fig. 19.2) æ u ö æ 1080 ö Hu = ç ÷ =ç ÷ = 0.0727 in H2O; C = 0.20 è 4005 ø è 4005 ø 2

2

HL = CHu = 0.20(0.0727) = 0.0145 in H2O

FLOW OF AIR DUCTS

297

298

Chapter 19

Damper: HL = CHu = 0.20(0.0505) = 0.0101 in H2O 2, 3-pc elbows: HL = 2(0.42)(0.0505) = 0.0424 in H2O Outlet grille: HL = 0.06 in H2O (Table 19.3) H Ltotal = 0.1629 in H2O

19.25

12 ´ 20 rect ® 16.8 in Circ. Eq.; use Q = 1500 cfm; u = 980 ft/min 2

æ 980 ö hL = 0.08 in H2O per 100 ft; Hu = ç ÷ = 0.060 in H2O è 4005 ø æ 38 ö Duct: HL = 0.08 in H2O ç ÷ = 0.0304 in H2O è 100 ø

Damper: HL = CHu = 0.20(0.060) = 0.0120 in H 2O 3 elbows: HL = 3(0.22)(0.060) = 0.0396 in H2O Outlet grille: HL = 0.060 in H2O (Table 19.3) H Ltotal = 0.1420 in H2O

19.26

Q = 0.80 m3/s For square duct: De1 =

u1 =

1.3[ (800)(800) ]

5/8

(800 + 800)1/ 4

= 875 mm Circ. Eq.

Q 0.80 m 3 /s = = 1.33 m/s A p (0.875 m) 2 / 4

æ u ö æ 1.33 ö Hu = ç ÷ =ç ÷ = 1.065 Pa è 1.289 ø è 1.289 ø 2

2

De1 / D 2 = 875 / 400 = 2.19 ® K = 0.40 (Fig. 10.7) Sudden contraction H L1 = 0.40(1.065 Pa) = 0.426 Pa H L 2 = 17 Pa (Table 19.3) louvers

Duct 1, HL too low for chart in Fig. 19.3 — Neglect Duct 2, u 2 = 6.30 m/s; HL = (1.10 Pa/m)(9.25 m) = 10.2 Pa H Ltot = 0.426 + 17 + 0 + 11.1 = 27.6 Pa Pressure drop from patm

Then pfan = -27.6 P

FLOW OF AIR DUCTS

299