Fundamentals of Hydraulic Engineering Systems Solutions
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 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 E2 = (1200 g)(597 cal/g)
E1 = 5.97x107 cal The energy remaining (E2) is: E2 = E – E1 E2 = 6.80x107 cal – 5.97x107 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 = 7.16x105 cal
ΔT = 83°C, making the temperature
Etotal = E1 + E2 = 7.79x105 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 10-2 m3 _________________________________________
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 in Equation 1.1
E1 = (5 slugs)(32.2 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:
γ = 133 kN/m3 S.G. = γliguid/γwater at 4°C
2.41 x 104 BTU = (10 slugs)(32.2 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) _________________________________________
The energy lost by the water (to lower its temp. to 45.2°F) is that required to 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 . (Vol)] (g)
= [(5 slugs)(32.2 lbm/slug)(T2 - 32°F)(1 BTU/lbm·°F)] T2 = 40.8°F
F = [1.94 slugs/ft3 (π · (5 ft)2 · 3 ft)] (32.2 ft/sec2) F = 1.47 x 104 lbs (Note: 1 slug = 1 lb·sec2/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 as mass on 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)
[μ(air)/μ(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 units of (lb·s2)/(ft)
[μ(air)/μ(H2O)]80°C = 5.90x10-2
Volume of 1 gal = 0.134 ft3
[ν(air)/ν(H2O)]20°C = (1.509x10-5)/(1.003x10-6)
S.G. = γliguid/γwater at 4°C 3
[ν(air)/ν(H2O)]20°C = 15.04 3
γ = (8.31 lb)/(0.134 ft ) = 62.0 lb/ft S.G. = (62.0 lb/ft3)/(62.4 lb/ft3)
S.G. = 0.994 _________________________________________
[ν(air)/ν(H2O)]80°C = (2.087x10-5)/(0.364x10-6) [ν(air)/ν(H2O)]80°C = 57.3 Note: The ratio of absolute and kinematic 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 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. Why?
and even though volume changes with temperature, mass does not. Thus,
_________________________________________
(ρ1)(Vol1) = (ρ2)(Vol2) = constant; or Vol2 = (ρ1)(Vol1)/(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] = 2.088x10-3 lb⋅sec/ft2
y = 0 ft, τ = - 9.00 N/m2
alternatively,
y = 1/12 ft, τ = 0 N/m2 y = 1/6 ft, τ = 9.00 N/m2
2
1 lb⋅sec/ft = 478.9 poise
y = 1/4 ft, τ = 18.0 N/m2
2
(b) 1 stoke = 1 cm /sec 2
2
2
2
2
(1 cm /s)[(0.3937 in) /(1 cm) ][(1 ft) /(12 in) ] = 1.076x10-3 ft2/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 N)(sin15°)]
τ = μ(dv/dy) = μ(Δv/Δy) τ = (2.09x10-5 lb⋅sec/ft2)[(5 ft/sec)/(0.25 ft)]
Δy = 2.12 x 10-4 m = 2.12 x 10-2 cm _________________________________________
τ = (4.18x10-4 lb/ft2) F = τ·A = (4.18x10-4 lb/ft2)(10 ft)(30 ft) F = 0.125 lbs _________________________________________ 1.4.5 v = y2 – 2y, where y is in inches and v is in ft/s 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: dv/dy = 288y – 24 sec-1 τ = μ(dv/dy) τ = (0.375 N⋅sec/m2)( 288y – 24 sec-1)
1.4.7 ∑Fy = 0 (constant velocity motion) W = Tshear force = τ⋅A; where A is the surface area (of the cylinder) in contact with the oil film: A = (π)[(5.48/12)ft][(9.5/12)ft] = 1.14 ft2 Now, τ = W/A = (0.5 lb)/(1.14 ft2) = 0.439 lb/ft2 τ = μ(dv/dy) = μ(Δv/Δy), where Δv = v (the velocity of the cylinder). Thus, v = (τ)(Δy)/μ v = [(0.439 lb/ft2){(0.002/12)ft}] / (0.016 lb·s/ft2) v = 4.57 x 10-3 ft/sec
4 © 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.
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/ft2)(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/m3)(0.03m)] D = 9.71 x 10-4 m = 9.71 x 10-2 cm; thus,
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 _________________________________________
Thus; μ = (Torque/R)/[(A)(Δv/Δy)] μ=
μ=
Torque / R Torque ⋅ Δy = (2π )( R)(h)(ω ⋅ R / Δy ) (2π )( R 3 )(h)(ω ) (1.50 N ⋅ m)(0.0002m) ⎛ 2πrad / sec ⎞ ⎟⎟ (2π )(0.025m) 3 (0.04m)(2000rpm)⎜⎜ ⎝ 60rpm ⎠
1.5.2 The concept of a line force is logical for two reasons: 1) The surface tension acts along the perimeter of the tube pulling the column of water upwards due to adhesion between the water and the
μ = 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) -2
2
μ = 1.603x10 N⋅sec/m R
R
0
0
equation for capillary rise. _________________________________________
Torque = (r )dF = r ⋅ τ ⋅ dA ∫ ∫
1.5.3
Torque = (r )( μ )( Δv )dA ∫
σ = [(h)(γ)(D)] / [(4)(sinθ)]
Torque = (r )(μ )( (ω )(r ) − 0 )(2πr )dr ∫0 Δy
σ = [(0.6/12)ft(1.94slug/ft3)(32.2 lb/ft3)(0.02/12)ft]/ [(4)(sin 54°)]
R
Δy
0
R
R Torque = (2π )( μ )(ω ) (r 3 )dr
Δy
Torque =
∫
σ = 1.61 x 10-3 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 109 N/m2 _________________________________________
h = [(4)(σ)(sinθ)] / [(γ)(D)] 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 104 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) 4 (1.00 ⋅10 N / m 3 )(0.0025m)
ΔVol/Vol = -ΔP/Eb
h = 6.62x10-3 m = 0.662 cm _________________________________________
ΔVol/Vol = -(4.5 x 105 N/m2 – 2.5 x 106 N/m2)/ (2.2x109 N/m2) 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)(0.9σ1)(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+(ΔVol/Vol)]; see example 1.3
alternatively, h2 = 1.204(h1), about a 20% increase! _________________________________________
ρ = 1.94 slug/ft3/[1+(-0.545/120)] = 1.95 slug/ft3 _________________________________________ 1.6.4
1.5.6 R ΔP
Pi = 30 N/cm2 = 300,000 N/m2 = 3 bar ΔP = 30 bar – 3 bar = 27 bar = 27x105 N/m2 Amount of water that enters pipe = ΔVol Volpipe= [(π)(1.50 m)2/(4)]·(2000 m) = 3530 m3
∑Fx = 0; 2π(R)(σ) – ΔP(π)(R2) = 0
ΔVol = (-ΔP/Eb)(Vol) = [(-27x105 N/m2)/ (2.2x109 N/m2)]*(3530 m3) 3 ΔVol = -4.33 m
ΔP = 2σ/R
Water in the pipe is compressed by this amount.
ΔP = Pi – Pe (internal pressure minus external pressure)
∴ The volume of H2O that enters the pipe is 4.33 m3
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Fundamentals of Hydraulic Engineering Systems Solutions
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