Solution Manual For Thermodynamics: An Interactive Approach, 1st Edition

Name: Solution Manual For Thermodynamics: An Interactive Approach, 1st Edition
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Author: Sofia Garcia

Solution Manual For Thermodynamics: An Interactive Approach, 1st Edition offers a comprehensive guide to solving every question in your textbook, helping you master the material.

Sofia Garcia

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0-1-1 [US] Two thermodynamics books, each with a mass of 1 kg, are stacked one on top
of another. Neglecting the presence of atmosphere, draw the free body diagram of the
book at the bottom to determine the vertical force on its (a) top and (b) bottom faces in
kN.
SOLUTION
(a) From the free body diagram of the book at the top, a
vertical force balance produces:  
1 2
1
m kN
kg k
0.00981 kN
N
1000 s N
1 9.81
1000
mg
F
F
 
   
  
(b) Using the free body diagram of the book at the bottom, 
  
2 1
2
kN
1000
1 9.81
0.00981 1000 0.01962 kN
mg
F F
F
 
   1Fmg2Fmg1F

0-1-2 [UA] Determine (a) the pressure felt on your palm to hold a textbook of mass 1 kg
in equilibrium. Assume the distribution of pressure over the palm to be uniform and the
area of contact to be 25 cm2. (b) What-if Scenario: How would a change in atmospheric
pressure affect your answer (0: No change; 1: increase; -1 decrease)?
SOLUTION
(a)2
kN
; =kPa
m
F
p A
 
     
  
1 9.81
; ;
1000 1000 0.0025
mg
p p
A
   3.924 kPap 
(b) A change in atmospheric pressure would not affect the answer because the net force
created by a uniform pressure around any object is zero. Answer: 0

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0-1-6 [UM] According to Newton's law of gravity, the value of g at a given location is
inversely proportional to the square of the distance of the location from the center of the
earth. Determine the weight of a textbook of mass 1 kg at (a) sea level and (b) in an
airplane cruising at an altitude of 45,000 ft. Assume earth to be a sphere of diameter
12,756 km.
SOLUTION
(a) At sea level0 02 ;
GmM
w g m
R
   0 9.81 1 ;w 0 9.81 Nw 
(b)  3 km
At 45,000 ft 0.3048 10 13.716 km
ft
h 
       
2 2
02 2 22
; ; ; ;
GmM GmM R R
w w w g m w gm
RR h R h R h
      
    
2
2
2 2
m
9.81 (6,378 km) ms ; 9.76 ;
(6,378 km 13.716 km) s
9.767 1 ;
g g
w
 
 
    

 9.76 Nw 

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0-1-7 [UJ] The frictional force on a block of mass mA resting on a table (see
accompanying figure) is given as F = μN, where N is the normal reaction force from the
table. Determine the maximum value for mB that can be supported by friction. Assume
the pulley to be frictionless.
SOLUTION
The maximum value ofBm that can be supported by the friction: 
 
; kN
1000
; kN
1000
;
A
A
B
B
A B
m g
F
m g
F
F F



;A Bm g m g
 B Amm
 

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0-1-9 [XR] A block with a mass of 10 kg is at rest on a plane inclined at 25o to the
horizontal. If μs = 0.6, determine the range of the horizontal push force F if the block is
(a) about to slide down, and (b) about to slide up.
SOLUTION
(a) As the block is about to slide down, friction acts upward and a force
balance along the slope yields:  sin
cos cos sin ;
1000
s
mg
F mg F

 
    
 
  
  
(sin cos )
cos sin ;
1000
(sin cos ) ;
1000 cos sin
10 9.81 (sin 25 0.6 cos 25 ) ;
1000 cos 25 0.6sin 25
s
s
s
s
mg
F
mg
F
F
  
  
  
  

  

  
  
    0.010248 kNF  
The negative sign indicates that a slight pull force is necessary to overcome friction
for the block to slide down.
(b) As the block is about to slide up, friction acts downward and a force balance along
the slope yields: cos sin sin
cos ;
1000
s mg F mg
F

 

  
   
  
  
( cos sin ) ( cos sin )
cos sin ;
1000 1000 cos sin
10 9.81 (0.6cos 25 sin 25 ) ;
1000 cos 25 0.6sin 25
s s
s
s
mg mg
F F
F
  
  
  
  
 
     
  
    0.1452 kNF 

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0-1-10 [XO] A vertical piston cylinder device contains a gas at an unknown pressure. If
the outside pressure is 100 kPa, determine (a) the pressure of the gas if the piston has an
area of 0.2 m2 and a mass of 20 kg. Assume g = 9.81 m/s2. (b) What-if Scenario: What
would the pressure be if the orientation of the device were changed and it were now
upside down?
SOLUTION
(a)piston 2
piston 0 piston 2
m kN
; kPa m kg =kN
1000 s N
i
m g
p A p A  
      
  
  
piston
0 2 2 2
piston
kN m kN 1
; kPa= kg
1000 m s N m
20 9.81
100 ;
1000 0.2
i
i
m g
p p A
p
 
     
  100.981 kPaip 
(b)piston 2
piston 0 piston 2
m kN
; kPa m kg =kN
1000 s N
i
m g
p A p A  
      
  
  
piston
0
piston
;
1000
20 9.81
100 ;
1000 0.2
i
i
m g
p p A
p
  
  99.019 kPaip 
po
pi
mg

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0-1-11 [XB] Determine the mass of the weight necessary to increase the pressure of the
liquid trapped inside a piston-cylinder device to 120 kPa. Assume the piston to be
weightless with an area of 0.1 m2, the outside pressure to be 100 kPa and g = 9.81 m/s2.
SOLUTION2
piston 0 piston 2
m kN
; kPa m kg =kN
1000 s N
w
i
m g
p A p A  
        
     
   
0
piston
0 piston
; kPa
1000
1000 ; kg
120 100 0.1 1000 ;
9.81
w
i
i
w
w
m g
p p A
p p A
m g
m
  

 

 203.873 kgwm 

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0-1-12 [XS] A mass of 100 kg is placed on the piston of a vertical piston-cylinder device
containing nitrogen. The piston is weightless and has an area of 1 m2. The outside
pressure is 100 kPa. Determine (a) the pressure inside the cylinder. The mass placed on
the piston is now doubled to 200 kg. Also, additional nitrogen is injected into the cylinder
to double the mass of nitrogen. (b) Determine the pressure inside under these changed
conditions. Assume temperature to remain unchanged at 300 K and R for nitrogen to be
0.296 kJ/kg-K.
SOLUTION
(a) A vertical force balance on the piston yields:2
piston 0 piston 2
m kN
; kPa m kg =kN
1000 s N
w
i
m g
p A p A  
        
  
  
0
piston
; kPa
1000
100 9.81
10
100.98 k
0 ;
1000
a
1
P
w
i
i
i
m g
p p A
p
p
  
  
 
(b) As the weight is doubled, the force balance on the piston yields:  
  
200 9.81
= 100;
1 1000
ip 101.96 kPa=ip
Note the amount of nitrogen (or the identity of the gas) is not relevant.

Loading page 13...

0-1-13 [XN] A piston with a diameter of 50 cm and a thickness of 5 cm is made of a
composite material with a density (ρ) of 4000 kg/m3. (a) If the outside pressure is 101
kPa, determine the pressure inside the piston-cylinder assembly if the cylinder contains
air. (b) What-if Scenario: What would the inside pressure be if the piston diameter were
100 cm instead? (c) Would the answers change if the cylinder contains liquid water
instead? (1: yes; 2: no)
SOLUTION
(a)   22
20.5
; ; 0.1963 m ;
4 4
d
A A A

     piston ; kgm At
    2
piston piston
0.5
4000 0.05; 39.25 kg;
4
m m
      piston
0
piston
; kPa
1000
i
m g
p p A
   
  
39.25 9.81
101 ;
1000 0.1 102.96
6 kPa
9 3
i ip p    
(b)      
2
piston piston
1
4000 0.05 ; 157 kg;
4
m m
   
piston
0
piston
;
1000
i
m g
p p A
   
  
157 9.81
101 ;
1000 0.785 102.96 kPa
4
i ip p    
(c) No, the force balance on the piston does not depend on the identity of the fluids
exerting pressures. Answer: 2
po
pi
mg

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0-1-14 [XA] A piston-cylinder device contains 0.02 m3 of hydrogen at 300 K. It has a
diameter of 10 cm. The piston (assumed weightless) is pulled by a connecting rod
perpendicular to the piston surface. If the outside conditions are 100 kPa and 300 K, (a)
determine the pull force necessary in kN to create a pressure of 50 kPa inside. (b) The
piston is now released; as it oscillates back and forth and finally comes to equilibrium,
the temperature inside is measured as 600 K (reasons unknown). What is the pressure of
hydrogen at equilibrium? (c) What-if Scenario: What would be the answer in part (a) if
the gas were oxygen instead?
SOLUTION
(a) A vertical force balance on the piston yields: 0 ; kNiF p A p A  
   
0
2
;
100 50 0.393 kN0.1 ;
4
iF p p A
F F

  
 
     
 
(b) With the external force absent, a new force balance on the piston after it comes to
equilibrium yields: 0 ; kNip A p A0 ; 100 kPai ip p p   
(c) The identity of the gas has nothing to do with the force balance. Therefore, the
answer does not change.
F = 0.393 kN.

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0-1-15 [XH] Air in the accompanying piston-cylinder device is in equilibrium at 200oC.
If the mass of the hanging weight is 10 kg, atmospheric pressure is 100 kPa, and the
piston diameter is 10 cm, (a) determine the pressure of air inside. Assume g = 9.81 m/s2.
(b)What-if Scenario: What would the pressure be if the gas were hydrogen instead?
Molar mass of air is 29 kg/kmol and that of hydrogen is 2 kg/kmol. Neglect piston mass
and friction.
SOLUTION
(a) A vertical force balance on the piston yields: 0 ; kN
1000
i p p
mg
p A p A    
   
  
0
2
; kPa
1000
10 9.81 4
100 ; (100 12.49);
1000 (0.1)
i
p
i i
mg
p p A
p p

  
     87.51 kPainp 
(b) The force balance does not depend on the identity of the gas. Hence,87.51 kPaip 

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Solution Manual For Thermodynamics: An Interactive Approach, 1st Edition

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