Thermodynamics Chapter 1 Solution Manual

SOLUTION MANUAL OF

THERMODYNAMICS

by Hipolito Sta. Maria

Answered by:

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

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

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

ii. (F-Fg) = mg/k

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

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 mps}2_.

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

= 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 =

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

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

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

(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,

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)

= 621.2 lb/min

γ = dg/k

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