# Thermodynamics Chapter 1 Solution Manual

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## SOLUTION MANUAL OF

### THERMODYNAMICS

by Hipolito Sta. Maria

### Engr. Naser A. Fernandez

Published by: ‘I Think, Therefore I’m An Atheist’ Enterprises and Priority Development Fund (PDF)

*This solution manual is an original work of Engr. Naser A. Fernandez. No part of this Manual may be reprinted, reproduced and distributed in any form or by any means

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Chapter 1

1. What is the mass in grams and the weight in dynes and in gram-force of 12 oz of salt? Local g is 9.65m/s2 1lb = 16oz

Given: m= 12 oz g= 9.65 m/s2 = 965 cm/s2 Solution: (a) 12 oz x x = 340.2 g (b) Fg = mg/k = Fg = 334.80 gf (c) Fg = 334.80 gf x = 328324.97 dynes

2. A mass of 0.10 slug in space is subjected to an external vertical force of 4 lb.

If the local gravity acceleration is g = 30.5 fps2 and if friction effects are

neglected, determine the acceleration of the mass if the external vertical force is acting (a) upward (b) downward.

Given:

m= 0.10 slug x

= 3.2174 lbm F= 4 lbf

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Solution: F (a) i. Fg = mg/k = Fg

### ii. (F-Fg) = mg/k

(4-3.05)lbf = a = 9.5 ft/s2 (b) i.(F+Fg) = mg/k (4+3.05)lbf = F Fg a = 70.5 ft/s2

3. The mass of a given airplane at sea level (g = 32.1 fps2) is 10 tons. Find its

mass in lb, slugs, and kg and its (gravitational) weight in lb when it is travelling at

a 50,000-ft elevation. The acceleration of gravity g decreases by 3.33 x 10-6 fps2

for each foot of elevation. Given:

g = 32.1 ft/s2 h = 50, 000 ft m = 10 tons

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Solution: (a) 10 tons x = 20,000 lbm (b) 20,000 lbm x = 621.62 slugs (c) h = 50,000 = (a-32.1)/(-3.33 x 10-6/ft) a = 31.9335 ft/s2 Fg = mg/k = Fg = 19850.50 lbf

4. A lunar excursion module (LEM) weighs 1500 kgf on earth where g = 9.75

mps2. What will be its weight on the surface of the moon where gm = 1.70 mps2.

On the surface of the moon, what will be the force in kgf anf in newtons required

to accelerate the module at 10 mps2?

Given: Fge = 1500 kgf gm = 1.70 m/s2 ge = 9.75 m/s2 Solution: (a) i. Fge = mge/k m = 1508.71 kgm ii. Fgm = mgm/k

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= Fgm = 261.5 kgf ( b) a = 10 m/s2 Fgm = Fgm = 1538.5 kgf (c) Fgm = 1538.5 kgf x = 15,087.45 N

5. The mass of a fluid system is 0.311 slug, its density is 30 lb/ft3 and g = 31.90

ft/s2. Find (a) the specific volume (b) the specific weight (c) and the total volume.

Given: m = 0.311 slug x = 10.006 g = 31.90 ft/s2 d= 30 lb/ft3 Solution: (a) v = 1/d (c) V = m/d = 1/(30 lb/ft3) = v = 0.0333 ft3/lb V = 0.3335 ft3 (b) γ = dg/k =

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γ = 29.7445 lb/ft3

6. A cylindrical drum (2-ft diameter, 3-ft in height) is filled with a fluid whose

density is 40 lb/ft3. Determine (a) the total volume of fluid, (b) its total mass in

pounds and slugs, (c) its specific volume, and (d) its specific weight where g =

31.90 fps2. Given: d = 40 lb/ft3 h = 3 ft diameter = 2 ft Solution: (a) V = πr2h (c) v = 1/d =1/40 = 0.025 ft3/lb = π(1)(3) V = 9.42 ft3 (b) i. m = dV (d)

### γ = dg/k

= (40 lb/ft3)( 9.42 ft3) = (40)(31.90)/(32.174) m = 377 lb γ = 39.66 lb/ft3 ii. m = 377 lbm x = 11.72 slugs

7. A weatherman carried an aneroid barometer from the ground floor to his office atop Sears Tower in Chicago. On the ground the barometer read 30.15 in.Hg absolute; topside it read 28.607 in. Hg absolute. Assume that the average

atmospheric air density was 0.075 lb/ft3 and estimate the height of the building.

Solution: ΔP = (30.15 – 28.607) in.Hg x x x x = 109.10 lb/ft2

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ΔP = dh

109.10 lb/ft2 = (0.075 lb/ft3)h

h = 1455 ft

8. A vacuum gauge mounted on a condenser reads 0.66 m.Hg. What is the absolute pressure in the condenser in kPa when the atmospheric pressure is 101.32 kPa. Given: Po = 101.3 kPa Pg = 0.66 m.Hg Solution: 1mmHg = 0.13332 kPa Pg = 0.66 m.Hg x x = 87.99 kPa P = Po - Pg = 101.32 – 87.99 P = 13.3 kPa

9. Convert the following readings of pressure to kPa absolute, assuming that the barometer reads 760 mmHg: (a) 90 cm.Hg gage; (b) 40 cm Hg vacuum; (c) 100 psig; (d) 8 in.Hg, and (e) 76 in. Hg gage.

Given: Po = 760 mm Hg x x = 101.32 kPa Solution: 1 mm.Hg = 0.13332 kPa

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(a) Pg = 90 cm.Hg x x = 119.99 kPa P = Po + Pg = 101.32 + 119.99 P = 221.31 kPa (b) Pg = 40 cm.Hg x x = 53.33 kPa P = Po - Pg = 101.32 – 53.33 P = 48 kPa (c) Pg = 100 psi = 100 lb/in2 x = 689.48 kPa P = Po + Pg = 101.32 + 689.48 P = 790.8 kPa (d) Pg = 8 in.Hg x x = 27.09 kPa P = Po - Pg = 101.32 – 27.09 P = 74.2 kPa (e) Pg = 76 in.Hg x x = 257.36 kPa P = Po + Pg = 101.32 + 257.36 P = 358. 68 kPa

10. A fluid moves in a steady flow manner between two sections in a flow line. At

section 1:A1=10ft2, Ʋ1=100ft/min, (specific volume)v1= 4ft3/lb. At section 2: A2=2ft2,

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Given: A1=10ft2 A2=2ft2 Ʋ 1=100ft/min d2= 0.20 lb/ft 3 v1= 4ft3/lb Solution:

(a) d1 = 1/v1 mass flow rate = (A1)( Ʋ 1)( d1)

= 1/(4ft3/lb) = (10ft2)( 100ft/min)(0.25 lb/ft3)( ) = 0.25 lb/ft3 = 15,000 lb/h (b) (A1)( Ʋ 1)(d1) = (A2)( Ʋ2)(d2) (10ft2)( 100ft/min)( 0.25 lb/ft3) = (2ft2)( 0.20 lb/ft3)( Ʋ2) d2 = (625 ft/min)(1min/60secs) Ʋ2 = 10.42 fps

11. if a pump discharges 75 gpm of water whose specific weight is 61.5

lb/ft3(g=31.95fps2). find a) the mass flow rate in lb/min, and b) and total time

required to fill a vertical cylinder tank 10ft in diameter and 12ft high. Given:

Volume flow rate = 75 gal/min g=31.95fps2 Specific weight, γ = 61.5 lb/ft3 Solution: (a) x = 10.03 ft 3 /min

mass flow rate = (10.03 ft3/min)( 61.93 lbm/ft3)

### γ = dg/k

61.5 lbf/ft3 = d = 61.93 lbm/ft3

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(b) diametercylinder = 10ft h = 12 ft T = (Volume)/(flowrate) = ( )(h)/(flowrate) = ( )(12)/10.03 T = 93.97 min

References

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