Chem E 260
HW Solution #3
2-38 The work required to compress a spring is to be determined.
Analysis Since there is no preload, F = kx. Substituting this into the work expression
gives
2
2
2



1
1
1
k 2
( x 2  x12 )
2
W  Fds  kxdx  k xdx 

300 kN/m
(0.03 m) 2  0 2
2
 0.135 kN  m

F

x
 1 kJ 
 (0.135 kN  m) 
  0.135 kJ
 1 kN  m 
2-53 A room contains a light bulb, a TV set, a refrigerator, and an iron. The rate of
increase of the energy content of the room when all of these electric devices are on is to
be determined.
Assumptions 1 The room is well sealed, and heat loss from the room is negligible. 2 All the appliances are
kept on.
Analysis Taking the room as the system, the rate form of the energy balance can be
written as
 dEroom / dt  E in
E  E out

dEsystem / dt
in

Rate of net energy transfer
by heat, work, and mass

Rate of change in internal,kinetic,
p otential,etc. energies
since no energy is leaving the room in any form, and thus E out  0 . Also,
ROOM
E in  E lights  E TV  E refrig  E iron
 100  110  200  1000 W
 1410 W
Substituting, the rate of increase in the energy
content of the room becomes
Electricity
- Lights
- TV
- Refrig
- Iron
dEroom / dt  E in  1410 W
Discussion Note that some appliances such as refrigerators and irons operate
intermittently, switching on and off as controlled by a thermostat. Therefore, the rate of
energy transfer to the room, in general, will be less.
3-75 The pressure in a container that is filled with air is to be determined.
Assumptions At specified conditions, air behaves as an ideal gas.
Properties The gas constant of air is R = 0.287 kJ/kgK (Table A-1).
Analysis The definition of the specific volume gives
v
V
m

0.100 m 3
 0.100 m 3 /kg
1 kg
Using the ideal gas equation of state, the pressure is
P
RT
v

(0.287 kPa  m 3 /kg  K)(27  273 K)
0.100 m 3 /kg
 861 kPa
3-77 A balloon is filled with helium gas. The mole number and the mass of helium in the
balloon are to be determined.
Assumptions At specified conditions, helium behaves as an ideal gas.
Properties The universal gas constant is Ru = 8.314 kPa.m3/kmol.K. The molar mass of
helium is 4.0 kg/kmol (Table A-1).
Analysis The volume of the sphere is
V 
4 3 4
 r   (3 m) 3  113.1 m 3
3
3
Assuming ideal gas behavior, the mole numbers of He is determined from
He
D=6m
20C
200 kPa
PV
(200 kPa)(113.1 m3 )
N

 9.28 kmol
RuT (8.314 kPa  m3/kmol  K)(293 K)
Then the mass of He can be determined from
m  NM  (9.28 kmol)(4.0 kg/kmol)  37.15 kg
3-92 The specific volume of steam is to be determined using the ideal gas relation, the
compressibility chart, and the steam tables. The errors involved in the first two
approaches are also to be determined.
Properties The gas constant, the critical pressure, and the critical temperature of water
are, from Table A-1,
R = 0.4615 kPa·m3/kg·K,
Tcr = 647.1 K,
Pcr = 22.06 MPa
Analysis (a) From the ideal gas equation of state,
v
RT (0.4615 kPa  m3/kg  K)(723 K)

 0.09533 m3 /kg
P
3500 kPa
(3.7% error)
(b) From the compressibility chart (Fig. A-15),
P
3.5 MPa


 0.159 
Pcr 22.06 MPa

 Z  0.961
T
723 K

TR 

 1.12

Tcr 647.1 K
PR 
H2O
3.5 MPa
450C
Thus,
v  Zv ideal  (0.961)(0.09533 m 3 /kg)  0.09161m3 /kg
(c) From the superheated steam table (Table A-6),
(0.4% error)

P  3.5 MPa v  0.09196 m3/kg
T  450 C
4-8 A piston-cylinder device contains nitrogen gas at a specified state. The boundary
work is to be determined for the polytropic expansion of nitrogen.
Properties The gas constant for nitrogen is 0.2968 kJ/kg.K (Table A-2).
Analysis The mass and volume of nitrogen at the initial state are
m
V2 
P1V1
(130 kPa)(0.07 m3 )

 0.07802 kg
RT1 (0.2968 kJ/kg.K)(1 20  273 K)
N2
130 kPa
120C
mRT 2 (0.07802 kg)(0.2968 kPa.m 3 /kg.K)(100  273 K)

 0.08637 m 3
P2
100 kPa
The polytropic index is determined from
P1V1n  P2V 2n 
(130 kPa)(0.07 m 3 ) n  (100 kPa)(0.086 37 m 3 ) n 
 n  1.249
The boundary work is determined from
Wb 
P2V 2  P1V1 (100 kPa)(0.086 37 m 3 )  (130 kPa)(0.07 m 3 )

 1.86 kJ
1 n
1  1.249
4-19 Nitrogen gas in a cylinder is compressed polytropically until the temperature rises to
a specified value. The boundary work done during this process is to be determined.
Assumptions 1 The process is quasi-equilibrium. 2 Nitrogen is an ideal gas.
Properties The gas constant for nitrogen is R = 0.2968 kJ/kg.K (Table A-2a)
Analysis The boundary work for this polytropic process
can be determined from
P
2
P2V 2  P1V1 mR (T2  T1 )
Wb, out   P dV 

2
1
1 n
1 n
(2 kg)(0.2968 kJ/kg  K)(360  300)K

1  1.4
 89.0 kJ
PV n
=C
1
Discussion The negative sign indicates that work is done on
the system (work input).
V
4-25 A piston-cylinder device contains air gas at a specified state. The air undergoes a
cycle with three processes. The boundary work for each process and the net work of the
cycle are to be determined.
Properties The properties of air are R = 0.287 kJ/kg.K , k = 1.4 (Table A-2a).
Analysis For the isothermal expansion process:
V1 
mRT (0.15 kg)(0.287 kJ/kg.K)(3 50  273 K)

 0.01341 m 3
P1
(2000 kPa)
Air
2 MPa
350C
V2 
mRT (0.15 kg)(0.287 kJ/kg.K)(3 50  273 K)

 0.05364 m 3
P2
(500 kPa)
 0.05364 m3 
V 
  37.18 kJ
Wb,1 2  P1V1 ln  2   (2000 kPa)(0.013 41 m3 ) ln 
 0.01341 m3 
 V1 


For the polytropic compression process:
P2V 2n  P3V 3n 
(500 kPa)(0.053 64 m 3 )1.2  (2000 kPa)V 31.2 
V 3  0.01690 m 3
Wb, 23 
P3V 3  P2V 2 (2000 kPa)(0.016 90 m 3 )  (500 kPa)(0.053 64 m 3 )

 -34.86 kJ
1 n
1  1.2
For the constant pressure compression process:
Wb,31  P3 (V1 V 3 )  (2000 kPa)(0.013 41  0.01690)m
3
 -6.97 kJ
The net work for the cycle is the sum of the works for each process
Wnet  Wb,12  Wb, 23  Wb,31  37 .18  (34 .86 )  (6.97 )  -4.65 kJ