5-2
Moving Boundary Work
5-2 Nitrogen gas in a cylinder is compressed at constant temperature until its pressure 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.
Analysis The boundary work is determined from its definition to be
1
kJ 1
kPa 150
lnln
3
2
1 2
1
11
1
2
11out,
P
P
PPdPW
b
V
V
V
VV
5-3 Helium is compressed in a piston-cylinder device. The initial and final temperatures of helium and the work required to
compress it are to be determined.
Assumptions The process is quasi-equilibrium.
Properties The gas constant of helium is R = 2.0769 kJ/kgK (Table A-1).
Analysis The initial specific volume is
m 5 3
3
1
V
K 173.3 )K 3.433(
m 5
m 2
3
3
1
1
2
2TT
V
V
P
V
(m3)
3
7
P
2
T = 300 K
5-5
5-8 Water is expanded isothermally in a closed system. The work produced is to be determined.
Assumptions The process is quasi-equilibrium.
/kgm 10200.0
)001157.012721.0(80.0001157.0
3
The definition of specific volume gives
3
3
3
2
/kgm 0.10200
v
1
5-9 A gas in a cylinder is compressed to a specified volume in a process during which the pressure changes linearly with
volume. The boundary work done during this process is to be determined by plotting the process on a P–
V
diagram and also
by integration.
Assumptions The process is quasi-equilibrium.
Analysis (a) The pressure of the gas changes linearly with volume, and thus the process curve on a P–
V
diagram will be a
straight line. The boundary work during this process is simply the area under the process curve, which is a trapezoidal.
Thus,
kPa 456kPa) (600)m )(0.12kPa/m 1200(
kPa 96kPa) (600)m )(0.42kPa/m 1200(
33
22
33
11
baP
baP
V
V
and
kJ82.8
mkPa 1
kJ 1
0.42)m(0.12
2
456)kPa(96
)(
2
Area
3
3
12
21
out,
VV
PP
Wb
(b) The boundary work can also be determined by integration to be
2
)m0.42(0.12
)(
2
)(
3
622
3
12
2
1
2
2
2
1
2
1
out,
kJ 82.8
VV
VV
VVV
badbadPWb
Discussion The negative sign indicates that work is done on the system (work input).
V
(m3)
1555
1
88.16
V
(m3)
P
(kPa)
P1
1
2
P2
0.12
0.42
P = a
V
+ b
P = a
V
+ b
5-7
5-12 Problem 5-11 is reconsidered. The process described in the problem is to be plotted on a P–
V
diagram, and the
effect of the polytropic exponent n on the boundary work as the polytropic exponent varies from 1.1 to 1.6 is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.
Function BoundWork(P[1],V[1],P[2],V[2],n)
“This function returns the Boundary Work for the polytropic process. This function is required
V[2] = 0.2 [m^3]
Gas$=’AIR’}
“System: The gas enclosed in the piston-cylinder device.”
“Process: Polytropic expansion or compression, P*V^n = C”
P[2]*V[2]^n=P[1]*V[1]^n
5-8
n
Wb [kJ]
1.1
1.156
1.211
1.267
1.322
1.378
1.433
1.489
1.544
1.6
18.14
17.25
16.41
15.63
14.9
14.22
13.58
12.98
12.42
11.89
0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2
0
20
40
60
80
100
120
140
160
V [m3]
P [kPa]
13
14
15
16
17
18
19
n
Wb [kJ]
5-9
5-13 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)
kJ408
1.41
0)K25K)(360kJ/kg kg)(0.2968 (5
1
)(
1
12
2
1
1122
out,
n
TTmR
n
PP
dPWb
VV
V
Discussion The negative sign indicates that work is done on
the system (work input).
5-14 A gas whose equation of state is
TRP u
)/10(2
vv
expands in a cylinder isothermally to a specified volume. The
unit of the quantity 10 and the boundary work done during this process are to be determined.
Assumptions The process is quasi-equilibrium.
Analysis (a) The term
2
/10
v
must have pressure units since it
is added to P.
Thus the quantity 10 must have the unit kPa·m6/kmol2.
(b) The boundary work for this process can be determined from
2
2
22
10
)/(
10
/
10
V
V
V
V
v
v
N
TNR
N
N
TRTR
and
3
12
2
1
2
2
1 2
2
2
1
out,
m 4
11
10ln
10
VVV
V
V
V
V
V
NTNRd
N
TNR
dPW u
u
b
V
1
2
P
V
n =C
P
V
T = 350 K
2
4
5-12
5-18E Hydrogen gas in a cylinder equipped with a spring is heated. The gas expands and compresses the spring until its
volume doubles. The final pressure, the boundary work done by the gas, and the work done against the spring are to be
determined, and a P–
V
diagram is to be drawn.
Assumptions 1 The process is quasi-equilibrium. 2 Hydrogen is an ideal gas.
Analysis (a) When the volume doubles, the spring force and the final pressure of H2 becomes
lbf 75,000
ft 3
ft 15
lbf/ft) (15,000
2
3
2
A
kkxF
s
V
Btu 281.7
3
3
12
21
out,
ftpsia 5.40395
Btu 1
15)ft(30
2
14.7)psia(188.3
)(
2
Area
VV
PP
Wb
Btu240.9 8.407.281
spring no,spring bb WWW
(ft3)
15
30
5-13
5-19 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:
3
K) 27350kJ/kg.K)(3 287.0(kg) 15.0(
mRT
5-17
5-24 Problem 5-23 is reconsidered. The effect of the spring constant on the final pressure in the cylinder and the
boundary work done as the spring constant varies from 50 kN/m to 500 kN/m is to be investigated. The final pressure and
the boundary work are to be plotted against the spring constant.
A=0.1[m^2]
k=100 [kN/m]
DELTAx=20 [cm]
Spring_const=k/A^2 “[kN/m^5]”
V[1]=m*spvol[1]
5-18
10-4 10-3 10-2 10-1 100101102
100
101
102
103
104
105
v [m3/kg]
P [kPa]
25°C
111.4°C
138.9°C
Steam
1
2
3
50 100 150 200 250 300 350 400 450 500
300
500
700
900
1100
1300
P[3] [kPa]
5-19
Closed System Energy Analysis
5-25E The table is to be completed using conservation of energy principle for a closed system.
Analysis The energy balance for a closed system can be expressed as
)( 1212outin
energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
outin
eemEEWQ
EEE
Application of this equation gives the following completed table:
Qin
(Btu)
Wout
(Btu)
E1
(Btu)
E2
(Btu)
m
(lbm)
e2 e1
(Btu/lbm)
350
510
1020
860
3
–53.3
350
130
550
770
5
44.0
560
260
600
900
2
150
–500
0
1400
900
7
–71.4
–650
–50
1000
400
3
–200
5-26E The heat transfer during a process that a closed system undergoes without any internal energy change is to be
determined.
Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero. 2 The compression or
expansion process is quasi-equilibrium.
