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 without prior permission from him. j/k :-D

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 F= 4 lbf g= 30.5 ft/s2

= 3.2174 lbm

Solution: F (a)

i. Fg = mg/k

=

Fg

ii. (F-Fg) = mg/k (4-3.05)lbf =

a = 9.5 ft/s2

i.(F+Fg) = mg/k

(b)

(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 m = 10 tons

h = 50, 000 ft

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 ge = 9.75 m/s2

gm = 1.70 m/s2

Solution: (a) i. Fge = mge/k

m = 1508.71 kgm ii. Fgm = mgm/k

=

Fgm = 261.5 kgf ( b) a = 10 m/s

2

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 = 1/(30 lb/ft3) v = 0.0333 ft3/lb (b) γ = dg/k =

(c) V = m/d = V = 0.3335 ft3

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 (d) γ = dg/k

(b) i. m = dV 3

3

= (40 lb/ft )( 9.42 ft )

= (40)(31.90)/(32.174) γ = 39.66 lb/ft3

m = 377 lb 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 = 109.10 lb/ft2

x

x

x

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 Solution: 1 mm.Hg = 0.13332 kPa

x

= 101.32 kPa

(a) Pg = 90 cm.Hg x

x

= 119.99 kPa

x

= 53.33 kPa

x

= 27.09 kPa

x

= 257.36 kPa

P = Po + Pg = 101.32 + 119.99 P = 221.31 kPa (b) Pg = 40 cm.Hg x 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 P = Po - Pg = 101.32 – 27.09 P = 74.2 kPa (e) Pg = 76 in.Hg x 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, (density)d2= 0.20 lb/ft3.Calculate the (a) mass flow rate and (b) speed at section 2.

Given:

A1=10ft2 Ʋ 1=100ft/min v1= 4ft3/lb

A2=2ft2 d2= 0.20 lb/ft3

Solution:

mass flow rate = (A1)( Ʋ 1)( d1)

(a) d1 = 1/v1 = 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 Specific weight, γ = 61.5 lb/ft3

g=31.95fps2

Solution:

(a)

x

= 10.03 ft3/min

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

= 621.2 lb/min γ = dg/k 61.5 lbf/ft3 = d = 61.93 lbm/ft3

)

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

)(h)/(flowrate)

=(

)(12)/10.03

T = 93.97 min