Solution Manual for Fundamentals of Hydraulic Engineering Systems 4th Edition by Houghtalen

Chapter 1 – Problem Solutions 1.2.1

1.2.3

E1 = energy req’d to bring ice temperature to 0°C

E1 = energy required to change water to ice

E1 = (250 L)(1000 g/L)(20°C)(0.465 cal/g·°C)

E1 = (100 g)(79.7 cal/g)

E1 = 2.33x106 cal

E1 = 7.97x103 cal

E2 = energy required to melt ice

E2 = energy required to change vapor to ice

E2 = (250 L)(1000 g/L)(79.7 cal/g·°C)

E2 = (100 g)(597 cal/g) + (100 g)(79.7 cal/g)

E2 = 1.99x107 cal

E2 = 6.77x104 cal

E3 = energy required to raise the water temperature to 20°C

Total energy removed to freeze water and vapor.

E3 = (250 L)(1000 g/L)(20°C)(1 cal/g·°C) E3 = 5.00x106 cal

Etotal = E1 + E2 = 7.57x104 cal  _________________________________________  ____________________________________ _____

1.2.4 7

Etotal = E1 + E2 + E3 = 2.72x10 cal  _________________________________________  _________________________________________

E1 = energy needed to vaporize the water E1 = (100 L)(1000 g/L)(597 cal/g)

1.2.2

E1 = 5.97x107 cal

At 0.9 bar (ambient pressure), the boiling temperature of water is 97°C (see Table 1.1). E1 = energy required to bring the water temperature to 97°C E1 = (1200 g)(97°C - 45°C)(1 cal/g·°C) E1 = 6.24x104 cal E2 = energy required to vaporize the water

The energy remaining (E2) is: E2 = E – E 1 E2 = 6.80x107 cal – 5.97x10 7 cal E2 = 8.30x106 cal The temperature change possible with the remaining energy is: 8.30x106 cal = (100 L)(1000 g/L)(1 cal/g·°C)(ΔT)

E2 = (1200 g)(597 cal/g)

ΔT = 83°C, making the temperature

E2 = 7.16x105 cal 5

Etotal = E1 + E2 = 7.79x10 cal

T = 93°C when it evaporates. Therefore, based on Table 1.1, P = 0.777 atm

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1.2.5

1.3.1

E1 = energy required to raise the temperature to 100°C

The weight of water in the container is 814 N.

E1 = (5000 g)(100°C – 25°C)(1 cal/g·°C)

m = W/g = (814 N)/(9.81 m/sec2) = 83.0 kg

E2 = 3.75x105 cal

At 20 C, 998 kg = 1 m3

E2 = energy required to vaporize 2.5 kg of water

Therefore, the volume can be determined by

E2 = (2500 g)(597 cal/g)

Vol = (83.0 kg)(1 m3/998 kg)

E2 = 1.49x106 cal

Vol = 8.32 x 10  m  _________________________________________  ____________________________________ _____

˚

-2

3

Etotal = E1 + E2 = 1.87x106 cal Time required = (1.87x106 cal)/(500 cal/s) = 3740 sec = 62.3 min  _________________________________________  ______________________________________ ___

1.3.2

1.2.6

W = m·g, dividing both sides of the equation by volume yields γ = ρ·g  _________________________________________  ____________________________________ _____

E1 = energy required to melt ice

F = m·a

Letting a = g results results in Equation 1.1

E1 = (5 slugs)(32.2 lbm/slug)(32 lbm/slug)(32°F - 20 °F)(0.46 BTU/lbm· °F) + (5 slugs)(32.2 lbm/slug)(144 BTU/lbm)

1.3.3

E1 = 2.41 x 104 BTU

γ = ρ·g = (13,600 kg/m3)(9.81 m/s2)

To melt the ice, the temperature of the water will decrease to:

 = 133 kN/m3

S.G. = γliguid/γwater at 4°C

2.41 x 10  BTU = (10 slugs)(32.2 lbm/slug)(120 lbm/slug)(120°F – T1)(1 BTU/lbm· °F)

S.G. = (133,000 N/m3)/(9810 N/m3)

T1 = 45.2°F

S.G. = 13.6 (mercury)  _________________________________________  ____________________________________ _____

4

The energy lost by the water (to lower its temp. to 45.2°F) is that required to melt melt the ice. Now you have 5 slugs of water at 32 °F and 10 slugs at 45.2 °F.

1.3.4

Therefore, the final temperature of the water is:

The force exerted on the tank bottom is equal to the weight of the water body.

[(10 slugs)(32.2 lbm/slug)(45.2°F – T2)(1 BTU/lbm· °F)]

F = W = m . g = [ρ [ρ . (Vol)] (g)

= [(5 slugs)(32.2 lbm/slug)(T2 - 32°F)(1 BTU/lbm· °F)] T2 = 40.8 F

F = [1.94 slugs/ft 3 (π  (π · (5 ft) 2 · 3 ft)] (32.2 ft/sec 2) F = 1.47 x 104 lbs

(Note: 1 slug = 1 lb·sec 2/ft)

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1.3.5

1.3.8

Weight of water on earth = 7.85 kN

(1 N)[(1 lb)/(4.448 N)] = 0.2248 lb  _________________________________________  ____________________________________ _____

m = W/g = (7,850 N)/(9.81 m/s2) m = 800 kg

1.3.9

 Note: mass on moon is the same same as mass on earth earth

(1 N⋅m)[(3.281 ft)/(1 m)][(0.2248 lb)/(1 N))]

W (moon) = mg = (800 kg)[(9.81 m/s2)/(6)]

= 7.376 x 10-1 ft lb  _________________________________________  ____________________________________ _____

W(moon) = 1310 N  _________________________________________  _________________________________________ 1.4.1 1.3.6

[μ(air)/μ(H2O)]20°C = (1.817x10-5)/(1.002x10-3)

W = mg = (0.258 slug)(32.2 ft/s2)

[ (air)/ (H2O)]20 C = 1.813x10-2

W = 8.31 lb

[μ(air)/μ(H2O)]80°C = (2.088x10-5)/(0.354x10-3)

 Note: a slug has has units of (lb·s (lb·s2)/(ft) Volume of 1 gal = 0.134 ft 3 S.G. = γliguid/γwater at 4°C

γ = (8.31 lb)/(0.134 ft3) = 62.0 lb/ft3 S.G. = (62.0 lb/ft 3)/(62.4 lb/ft3) S.G. = 0.994  _________________________________________  _________________________________________

[ (air)/ (H2O)]80 C = 5.90x10-2

[ν(air)/ν(H2O)]20°C = (1.509x10-5)/(1.003x10-6) [ (air)/ (H2O)]20 C = 15.04

[ν(air)/ν(H2O)]80°C = (2.087x10-5)/(0.364x10-6) [ (air)/ (H2O)]80 C = 57.3

 Note: The ratio ratio of absolute and kinematic viscosities viscosities of air and water increases with temperature because the

1.3.7

viscosity of air increases with temperature, but that of water decreases with with temperature. Also, the values of

Density can be expressed as:

kinematic viscosity (ν) for air and water are much

ρ = m/Vol

closer than those of absolute viscosity. viscosity. Why?

and even though volume changes with temperature, mass does does not. Thus,

 _________________________________________  ____________________________________ _____

(ρ1)(Vol1) = (ρ (ρ2)(Vol2) = constant; or Vol2 = (ρ (ρ1)(Vol1)/(ρ )/(ρ2) Vol2 = (1000 kg/m3)(100 m3)/(958 kg/m3) Vol2 = 104.4 m3 (or a 4.4% change in volume)

1.4.2

μ(water)20°C = 1.002x10-3 N⋅sec/m2 ν(water)20°C = 1.003x10-6 m2/s (1.002x10-3 N⋅sec/m2)·[(0.2248 lb)/(1 N)]· [(1 m)2/(3.281 ft)2] = 2.092x10-5 lb sec/ft2 (1.003x10-6 m2/s)[(3.281 ft)2/(1 m)2] = 1.080x10-5 ft2/s

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1.4.3

1.4.5 (cont.)

(a) 1 poise = 0.1 N ⋅sec/m2

Solutions:

(0.1 N⋅sec/m2)[(0.2248 lb)/(1 N)][(1 m) 2/(3.281 ft)2] = -3 2 2.088x10  lb sec/ft

y = 0 ft, τ = - 9.00 N/m 2

alternatively,

y = 1/12 ft, τ = 0 N/m2 y = 1/6 ft, τ = 9.00 N/m2

1 lb⋅sec/ft2 = 478.9 poise

y = 1/4 ft, τ = 18.0 N/m2

(b) 1 stoke = 1 cm 2/sec (1 cm2/s)[(0.3937 in)2/(1 cm)2][(1 ft)2/(12 in)2] = -3 2 1.076x10  ft /sec

y = 1/3 ft, τ = 27.0 N/m2  _________________________________________  ____________________________________ _____

alternatively,

1.4.6

1 ft2/sec = 929.4 stoke  _________________________________________  ______________________________________ ___

Based on the geometry of the incline Tshear force = W(sin15°) = τ⋅A = μ(dv/dy)A

1.4.4

Δy = [(μ)(Δv)(A)] / [(W)(sin15°)]

Assuming a Newtonian relationship:

Δy = [(1.29 N⋅sec/m2)(0.025 m/sec) (0.50m)(0.75m)]/[(220 (0.50m)(0.75m)]/[(220 N)(sin15°)]

v/Δy) τ = μ(dv/dy) = μ(Δv/Δ

-4

τ = (2.09x10

 lb⋅sec/ft )[(5 ft/sec)/(0.25 ft)]

-5

2

-2

y = 2.12 x 10  m = 2.12 x 10  cm  _________________________________________  ____________________________________ _____

τ = (4.18x10-4 lb/ft2) F = τ·A = (4.18x10-4 lb/ft2)(10 ft)(30 ft)

1.4.7

∑Fy = 0 (constant velocity motion) F = 0.125 lbs  _________________________________________  ______________________________________ ___

W = Tshear force = τ⋅A; where A is the surface surface area area (of the cylinder) in contact with the oil film:

1.4.5 2

v = y  – 2y, where y is in inches and v is in ft/s

A = (π)[(5.48/12)ft][(9.5/12)ft] )[(5.48/12)ft][(9.5/12)ft] = 1.14 ft2  Now, τ = W/A = (0.5 lb)/(1.14 ft 2) = 0.439 lb/ft2

Making units consistent yields v = 144y2 – 24y, where y is in ft and ν is in ft/s Taking the first derivative w/respect to y:

v/Δy), where τ = μ(dv/dy) = μ(Δv/Δ Δv = v (the velocity velocity of the cylinder). cylinder). Thus, v = (τ)(Δ )(Δy)/μ

dv/dy = 288y – 24 sec -1

v = [(0.439 lb/ft2){(0.002/12)ft}] ){(0.002/12)ft}] / (0.016 lb·s/ft2)

τ = μ(dv/dy) τ = (0.375 N⋅sec/m )( 288y – 24 sec 2

-1

)

v = 4.57 x 10-3  ft/sec

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1.4.8

1.5.1

τ = μ(Δv/Δy)

h = [(4)(σ)(sin θ)] / [(γ)(D)]

τ = (0.0065 lb⋅sec/ft2)[(1 ft/s)/(0.5/12 ft)

But sin 90  = 0, σ = 7.132x10-2 N/m

τ = 0.156 lb/ft2

and γ = 9790 N/m3 (at 20 C)

F = (τ)(A) = (2 sides)(0.156 lb/ft 2)(2 ft2)

thus, D = [(4)(σ)] / [(γ)(h)]; for h = 3.0 cm

F = 0.624 lb  _________________________________________  ______________________________________ ___

D = [(4)( 7.132x10-2 N/m)] / [(9790 N/m 3)(0.03m)]

1.4.9

for h = 3.0 cm, D = 0.0971 cm

μ = τ/(dv/dy) = (F/A)/( Δv/Δy);

for h = 2.0 cm, D = 0.146 cm

Torque = Force·distance = F·R; R = radius

for h = 1.0 cm, D = 0.291 cm  _________________________________________  _________________________________________

˚

˚

D = 9.71 x 10 -4 m = 9.71 x 10-2 cm; thus,

Thus; μ = (Torque/R)/[(A)( (Torque/R)/[(A)(Δv/Δy)]

μ =

Torque / R

=

(2π )( R)( h)(ω  ⋅ R / Δ y )

μ =

Torque ⋅ Δ y

1.5.2

( 2π )( R 3 )( h)(ω )

The concept of a line force is logical for two reasons: 1)

(1.50 N  ⋅ m)(0.0002m)

⎛ 2π rad / sec ⎞ ⎟⎟ ⎝  60rpm  ⎠

the tube pulling the column of water upwards

( 2π )(0.025m) 3 (0.04m)(2000rpm)⎜⎜

due to adhesion between the water and the

 = 3.65x10-1 N sec/m2  _________________________________________  ______________________________________ ___

tube. 2)

The surface tension is be multiplied by the tube perimeter, a length, to obtain the upward

1.4.10

force used in force balance development of the

μ = (16)(1.002x10-3 N⋅sec/m2)

equation for capillary rise.

μ = 1.603x10-2 N⋅sec/m2  R

 _________________________________________  ____________________________________ _____

R

Torque = (r )dF  = r ⋅ τ  ⋅ dA

∫

∫

0

1.5.3

0

Torque = (r )(μ )( Δv )dA  R

∫

σ = [(h)(γ)(D)] / [(4)(sin θ)]

Δ y

0

 R

Torque = (r )( μ )(

∫

(ω )(r ) − 0

0

Δ y

[(0.6/12)ft(1.94slug/ft3)(32.2 lb/ft3)(0.02/12)ft]/ σ = [(0.6/12)ft(1.94slug/ft [(4)(sin 54°)]

)(2π r ) dr 

 R

Torque = (2π )( μ )(ω ) (r 3 )dr 

Δ y

Torque =

The surface tension acts along the perimeter of

-3

∫

 = 1.61 x 10  lb/ft

0

(2π )(1.603 ⋅ 10−2 N  ⋅ sec/ m2 )(0.65rad / sec) ⎡ (1m)4 ⎤ 0.0005m

⎢ ⎢⎣

4

⎥ ⎥⎦

Torque = 32.7 N m

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1.5.4

1.6.1

Capillary rise in the 0.25 cm. tube is found using:

Eb = -ΔP/(ΔVol/Vol) = 9.09 x 10  N/m  _________________________________________  ____________________________________ _____

h = [(4)(σ)(sinθ)] / [( γ)(D)]

9

2

where σ = (6.90 x 10 -2)(1.2) = 8.28 x 10 -2 N/m

1.6.2

and γ = (9752)(1.03) = 1.00 x 10 4 N/m3

P1 = 25 bar = 25 x 105 N/m2 = 2.50 x 106 N/m2

−2 h = 4(8.28 ⋅ 10  N  / m)(sin 30) (1.00 ⋅10 4 N  / m 3 )(0.0025m)

ΔVol/Vol = - ΔP/E b ΔVol/Vol = -(4.5 x 10 5 N/m2 – 2.5 x 106 N/m2)/ (2.2x109 N/m2)

-3

h = 6.62x10  m = 0.662 cm  _________________________________________  _________________________________________

ΔVol/Vol = 9.3 x 10 -4 = 0.093% (volume increase) Δρ/ρ = -ΔVol/Vol = -0.093% (density decreases)

1.5.5

 _________________________________________  ____________________________________ _____

Condition 1: h1 = [(4)(σ1)(sinθ1)] / [( γ)(D)] h1 = [(4)( σ1)(sin30°)] / [( γ)(0.7 mm)] Condition 2: h2 = [(4)(σ2)(sinθ2)] / [( γ)(D)] h2 = [(4)(0.9 σ1)(sin42°)] / [( γ)(0.7 mm)] h2/h1 = [(0.9)(sin42 °)] / (sin30°) = 1.204

1.6.3

ρo = 1.94 slugs/ft3 (based on temp. & pressure) m = ρo·Volo = (1.94 slug/ft3)(120 ft3) = 233 slugs W = mg = (233 slugs)(32.2 ft/sec2) = 7,500 lb

ρ = ρo/[1+(Δ /[1+(ΔVol/Vol)]; see example 1.3

alternatively, h2 = 1.204(h1), about a 20% increase!  _________________________________________  ____________________________________ _____

= 1.94 slug/ft3/[1+(-0.545/120)] = 1.95 slug/ft3  _________________________________________  ____________________________________ _____ ρ

1.6.4

1.5.6

Pi = 30 N/cm2 = 300,000 N/m2 = 3 bar R P

ΔP = 30 bar – 3 bar = 27 bar = 27x10 5 N/m2 Amount of water that enters pipe =

ΔVol

Vol pipe= [(π)(1.50 m)2/(4)]·(2000 m) = 3530 m3

ΔP = Pi – Pe (internal pressure minus external pressure)

ΔVol = (-ΔP/E b)(Vol) = [(-27x105 N/m2)/

∑Fx = 0;

ΔVol = -4.33 m

2π(R)(σ) – ΔP(π)(R 2) = 0

P = 2 /R

(2.2x109 N/m2)]*(3530 m3)

3

Water in the pipe is compressed by this amount. The volume of H2O that enters the pipe is 4.33 m3

6 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.