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Solutions Manual For Applied Fluid Mechanics 8th Edition By Joseph Untener, Robert Mott (All Chapters, 100% Original Verified, A+ Grade) $28.49   Add to cart

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Solutions Manual For Applied Fluid Mechanics 8th Edition By Joseph Untener, Robert Mott (All Chapters, 100% Original Verified, A+ Grade)

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  • Applied Fluid Mechanics 8e Joseph Untener, Robert
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This Is The Original 8th Edition Of The Solution Manual From The Original Author All Other Files In The Market Are Fake/Old Editions. Other Sellers Have Changed The Old Edition Number To The New But The Solution Manual Is An Old Edition. Solutions Manual For Applied Fluid Mechanics 8th Edition ...

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  • Applied Fluid Mechanics 8e Joseph Untener, Robert
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Solutions Manual
for


Applied Fluid Mechanics


Eighth Edition


By


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

80km 103 m 1h
1.7    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
2500 ft 3 0.0283 m3 1min
1.16  3
  1.18 m 3 s
min ft 60 s

Consistent units in an equation

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




The Nature of Fluids 1

, s 1.50 km 3600 s
1.18 υ    871 km /h
t 6.2 s h

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

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

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

2s (2)(13m)
1.22 t   1.63 s
a 9.18 m/s 2

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

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

mυ2 (15 kg)(1.2 m s)2 kg . m 2
1.25 KE    10.8  10.8N  m
2 2 s2
2 2
mυ 2 (3600 kg)  16 km  (103 m) 2 1 h2 kg  m
1.26 KE        35.6×103 2
2 2  h  km 2
(3600 s) 2
s
KE = 35.6 kN  m

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

2
2( KE ) (2)(38.6 N m)  h  1 kg  m (3600s) 2 1 km 2
1.28 m    31.5 km    
υ2 1 s2  N h2 (103 m) 2
(2)(38.6)(3600) 2
m kg = 1.008 kg
(31.5) 2 (103 )2

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

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


2 Chapter 1

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