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/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
= 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 = πr h
2
3
(c) v = 1/d =1/40 = 0.025 ft /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