1
Chapter 4
Energy Analysis of Closed Systems
4.1
Boundary Work
Consider the gas enclosed in a piston-cylinder device below. The initial pressure is P, the total
volume is V and the cross-sectional area of the piston is A. When the piston moves a distance
ds in a quasi-equilibrium manner, the differential work done during this process is
Wb  Fds  PAds  PdV
The total boundary work done during the entire process is obtained by adding all the
differential works from the initial to the final state
2
Wb   PdV
………… ( kJ )
1
The integral can only be evaluated when P is stated in terms of V , i.e. P = f (V).
F
A gas does Wb work as it forces
the piston to move by ds
ds
P
gas
Consider a quasi-equilibrium expansion process as shown below. On the P-V diagram, the
differential area dA is equal to PdV, which is Wb . The total area under the process curve 1–2
is obtained by adding the differential areas:
2
2
1
1
Area = A =  dA   PdV
That is, “the area under the process curve on a P-V diagram is equal, in magnitude, to the work
done during a quasi-equilibrium expansion or compression process of a closed system”.
P
1
process path
dA = PdV
2
V
V
V1
dV
P
V2
2
For different paths followed by the gas as it expands from state 1 to state 2, the area underneath
the curve (and the work done) will be different. It is expected since work is a path function – it
depends on path and end states.
P is the pressure at the inner surface of the piston. It is only equal to the gas pressure in the
cylinder if the process is quasi-equilibrium.


During an expansion process, Wb is transferred from the system.
During a compression process, Wb is transferred to the system
4.1.1 Boundary Work for Constant Volume Process

For a constant volume process, Wb = 0.
Example.
A rigid tank contains air at 500 kPa and 150 oC. As a result of heat transfer to the
surroundings, the temperature and pressure inside the tank drop to 65 oC and 400 kPa,
respectively. Determine the boundary work done during this process.
0
2
For a rigid tank, its volume is constant. Thus dV = 0, for which Wb   PdV = 0.
1
500 kPa
150 oC
.
1
400 kPa
65 oC
cooling
V = constant
P, kPa
500
1
2
400
2
V
* Note : Wb = area under process curve
For constant volume, area = 0 !
4.1.2 Boundary Work for a Constant-Pressure Process
2
2
1
1
For a constant-pressure process, Wb   PdV  P  dV  PV2  V1  = mP v2  v1  ….… (kJ)
where P = P1 = P2.
Example
A frictionless piston-cylinder device contains 5 kg steam at 400 kPa and 200 oC. Heat is
transferred to the steam until the temperature reaches 250 oC. Determine the work done by the
steam during this process.
P, kPa
1
2
5 kg
400 kPa
200 oC
1
.
heating
P=C
400
250 oC
2
v
v1 = 0.53434
v2 = 0.59520
3
Although not explicitly stated, this is a constant-pressure process since the weight of the piston
and the atmospheric pressure are constant.
Wb  P1 V2  V1   mP1 v2  v1 
State 1 is at (400 kPa, 200 oC)
superheated
v1 = 0.53434 m3/kg
o
State 2 is at (400 kPa, 250 C)
superheated
v2 = 0.59520 m3/kg
Thus,
Wb  54000.59520  0.53434 =121.7 kJ
The +ve sign indicates work is done by the system (steam) on the surroundings.
4.1.3 Boundary Work for Isothermal Compression of an Ideal Gas
Example
A piston-cylinder device contains 0.4 m3 of air at 100 kPa and 80 oC. The air is compressed to
0.1 m3 in such a way that the temperature inside the cylinder remains constant. Determine the
work done during this process.
P, kPa 2
.
0.4 m3
100 kPa
80 oC
air
T = 80 oC
compression
PV = C
3
T=C
1
0.1 m
80 oC
1
100
2
V
.
0.1
Assume air is an ideal gas and the process is quasi-equilibrium.
C
P
PV  mRT = C
V
mRT is constant since m, R and T are constants.
2
Wb   PdV =
1
2
C
1
0.4
 V2 

1 
 V dV  C  V dV  C ln  V
1
where C = mRT1 = mRT2 = P1V1 = P2V2
can use here !
V 
 0.1 
Wb  P1V1 ln  2   1000.4 ln 
 = – 55.45 kJ
 0.4 
 V1 
V2 P1
Note : Since P1V1 = P2V2

V1 P2
The –ve sign indicates that work is done on the system by the surroundings, which is always
true for compression processes.
m3
4
4.1.3 Boundary Work for a Polytropic Process
In practice, expansion and compression processes of gases often obey the equation : PVn = C,
where n and C are constants. Such a process is called polytropic process. For such a process
between state 1 and state 2,
P  CV  n
PV n  C
V2
V2 n1  V1 n1 
 V  n 1 
Wb   PdV   CV dV  C 
Thus,
=
C



1 n
  n  1 V 1
1
1


n
n
since C = P1V1  P2V2 then,
2
2
n
P2V2nV2 n 1  P1V1nV1 n 1
P V  P1V1
= 2 2
1 n
1 n
----------------------------------For an ideal gas, PV = mRT , then the equation can be written as:
Wb 
Wb =
P2V2  P1V1
mRT2  mRT1 mRT2  T1 
=

1 n
1 n
1 n
general equation !
……………… ( kJ )
for n  1 .
-----------------------------------For a special case, when n = 1, it follows that PV = C or P = CV 1 for which
2
Wb   PdV =
1
2
C
 V2 

1 
1
 V dV  C  V dV  C ln  V
1
…………. general, n = 1
where C = P1V1 = P2V2 . This is also the equation for an ideal gas undergoing an isothermal
process !
Example
A piston-cylinder device contains 0.05 m3 of a gas initially at 200 kPa. At this state, a linear spring (k =
150 kN/m) is touching the piston but exerts no force on it. Heat is transferred to the gas causing the
piston to rise and to compress the spring until the volume inside the cylinder doubles. The crosssectional area of the piston is 0.25 m2. Determine (a) the final pressure inside the cylinder, (b) the total
work done by the gas, and (c) the fraction of this work done against the spring to compress it.
P, kPa
.
.
.
.
.
x
2
P2
Heating
II
x=0
1
200 kPa
0.05 m3
1
V2 = 2V1
= 0.1 m3
2
200
I
0.05
0.10
V, m3
5
(a)
The final volume (state 2) is, V2 = 0.2 V1 = 2(0.05) = 0.10 m3
From the undisturbed position, the spring is compressed a distance of,
x 
V  Ax
V 0.10  0.05
=
= 0.2 m
A
0.25
F = k x = 150 (0.2) = 30 kN
At state 2, the force exerted by the spring is,
At state 1, force balance on the piston (Fspring = 0), gives
mg  Patm A  P1 A
At state 2, force balance on the piston gives,
Fspring  mg  Patm A  P2 A
P2  P1 
Fspring  P1 A  P2 A
Fspring
A
 200 
30
 200 + 120 = 320 kPa
0.25
(b) The work done is the area under the process curve (a trapezoid) from state 1 to state 2,
W  area 
200  320 0.1  0.05
2
= 13 kJ
The work done is positive because integral of PdV is positive (P and dV are positive).
(c)
The rectangular area (region I) is work done against the piston and the atmosphere, and the
work done against the spring is represented by the triangular area (region II). Thus,
1
320  2000.10  0.05 = 3 KJ
2
Note: A similar result could also be obtained from :
1
1
Wspring  k x 22  x12  = 150 0.2 2  0 2  = 3 kJ
2
2
Wspring  area II 
6
4.2
Energy Balance for Closed Systems
Energy balance for any system undergoing any kind of process is
Ein  Eout  E system
Net energy transfer
by heat, work, mass
=
……………………….... ( kJ )
Change in internal, kinetic,
potential, etc energies
or in rate form, it is written as
Ein  E out 
dE system
………………………. ( kW )
dt
Rate of et energy transfer
by heat, work, mass
=
Rate of change in internal, kinetic,
potential, etc energies
The energy balance on a per unit mass basis is,
ein  eout  e system
………. ( kJ/kg )
and in differential form as,
Ein  E out  dE system
ein  eout  desystem
or
For a closed system undergoing a cycle, the initial and final states are identical, thus
E system  E final  Einitial  0
Ein  Eout  0 or Ein  Eout
Since a closed system has no mass flow across its boundaries, the only energy interactions
possible are work and heat transfer only. Therefore, for a cycle:
Wnet ,out  Qnet ,in
or
Wnet ,out  Qnet ,in
For a closed system, the general energy balance equation is :
Qnet ,in  Wnet ,out  E system
or
Q  W  E
where,
Q  Qnet ,in  Qin  Qout
is net heat input
W  Wnet ,out  Wout  Win
is net work output
Other forms of first-law relation for closed systems are :
q  w  e per unit mass ( kJ/kg )

q  w  de per unit mass; differential form ( kJ/kg )

Note : The 1st law cannot be proven mathematically !
7
Example
A piston-cylinder device contains 25 g saturated water vapor that is maintained at a constant
pressure of 300 kPa. A resistance heater within the cylinder is turned on and passes a current of
0.2 A for 5 min from a 120 V source. At the same time, a heat loss of 3.7 kJ occurs. Determine
the final temperature of the steam.
P, kPa
.
.
.
.
.
Heating
1
300
300 kPa
25 g
P=C
.
heater
1
2
Let Wother = work done on the system other than boundary work.
For the process,
0
0
Qin  Wother  Wb  U  KE  PE
Qin  Wother  Wb  U
Qin  Wother  PV2  V1   U 2  U 1 
= P2V2  P1V1  U 2  U1
= U 2  P2V2  U1  P1V1 
= H 2  H1 = mh2  h1 
where enthalpy is defined as
H = U + PV
Conclusion:
For constant-pressure process, we can use :
 Qin  Wother  Wb  U
or

Qin  Wother  H 2  H1
The electrical work done by the surroundings is (the heater is inside the system),
We  VIt = 120 (0.2) 300 = 7200 J = 7.2 kJ
At state 1 (300 kPa, sat. vapor)
Qin  Wother  H 2  H1 = mh2  h1 
h1  hg ,300kPa = 2724.9 kJ/kg
Qin = – 3.7 kJ (heat loss)
Wother = – We = – 7.2 kJ
– 3.7 – (– 7.2) = 0.025 h2  2724.9
At state 2 (300 kPa, h = 2864.9) superheated steam
h2 = 2864.9 kJ/kg
T2 = 200 oC
2
8
Example
A rigid tank is divided into two equal parts by a partition. Initially, one side of the tank
contains 5 kg of water at 200 kPa and 25 oC, and the other side is evacuated. The partition is
then removed, and the water expands into the entire tank. The water is allowed to exchange
heat with the surroundings until the temperature in the tank returns to the initial value of 25 oC.
Determine (a) the volume of the tank, (b) the final pressure, and (c) the heat transfer for this
process.
P, kPa
Evacuated
partition
1
.
200
.
.
200 kPa
25 oC
5 kg
partition
removed
5 kg
25 oC
2
1
2
(a)
State 1 (200 kPa, 25 oC) is compressed liquid
At state 1, volume occupied by water is :
Total volume of tank = 2 V1 = 2 (0.005) = 0.01 m3
v
v1  v f , 25C = 0.001003  0.001 m3/kg
V1  mv1 = 5 (0.001) = 0.005 m3
V 0.01

= 0.002 m3/kg
m
5
At 25 oC, vf = 0.00103 m3/kg, vg = 43.340 m3/kg.
Since vf < v2 < vg state 2 is mixture. Thus P2 = Psat, 25C = 3.1698 kPa
(b) At state 2, v 2 
(c) At state 1, u  u f , 25C = 104.83 kJ/kg
At state 2, x 2 
v2  v f
v fg

0.002  0.001
= 2.3 x 10-5
43.34  0.001
u 2  u f  x 2 u fg = 104.83 + (2.3 x 10-5)(2304.3) = 104.88 kJ/kg
The work term is zero, Wb and other forms of work are not present.
0
Q  W  mu2  u1  = 5 (104.88 – 104.83) = 0.25 kJ
The +ve sign indicates heat is transferred to the water during the process.
9
4.3
Specific Heats
cv = specific heat at constant volume is “the energy required to raise the temperature of a unit
mass of a substance by one degree as the volume is maintained constant”.
cp = specific heat at constant pressure is “the energy required to raise the temperature of a unit
mass of a substance by one degree as the pressure is maintained constant”
cp > cv because for the expansion process, work must be supplied to the system.
 u 
 cv   
………… (kJ/kg.K or kJ/kg.oC)
 T  v
 h 
 cp   
….……… (kJ/kg.K or kJ/kg.oC)
 T  P
cp and cv are properties, since they are expressed in terms of other properties.
4.3.1 Internal Energy, Enthalpy, and Specific Heats of Ideal Gases
It has been demonstrated experimentally that for an ideal gas,
For an ideal gas,
Definition of enthalpy,
Thus,
u = u(T).
Pv = RT
h = u + Pv
h = u + RT
Since R is constant and u = u(T)
h = h(T)
Since u and h depend only on temperature for an ideal gas, cp and cv also depend on
temperature only. Thus, for ideal gases :
 du 
cv  

 dT  v
and
 dh 
cp  

 dT  P
or
du  cv dT
dh  c p dT
The change in internal energy or enthalpy for an ideal gas during a process from state 1 to state
2 is obtained by integrating these equations, to give :
2
u  u 2  u1   cv dT
…….. ( kJ/kg )
1
2
h  h2  h1   c p dT
……… ( kJ/kg )
1
Although cv and cp are temperature dependent, using average specific heats simplifies the
calculations :
u 2  u1  cv ,avg T2  T1 
h2  h1  c p ,avg T2  T1 
10
For an ideal gas,
h = u + RT
dh = du + RdT
can be differentiated to give
Using the specific heats, c p dT  cv dT  RdT
c p dT
Dividing by dT,
dT

cv dT RdT

dT
dT
cp – cv = R
gives
Note : cp and cv can be used for any process and any system !
Example
Air at 300 K and 200 kPa is heated at constant pressure to 600 K. Determine the change in
internal energy per unit mass using the average specific heat value (Table A-2b).
Average temperature = (300 + 900)/2 = 450 K
At this temperature from table, cv = 0.733 kJ/kg.K
Thus,
u 2  u1  cv T2  T1  = 0.733 (600 – 300) = 220 kJ/kg
Example
An insulated rigid tank initially contains 0.7 kg of helium at 27 oC and 350 kPa. A paddle
wheel with a power rating of 0.015 kW is operated within the tank for 30 min. Determine (a)
the final temperature and (b) the final pressure of the helium gas. Take cv as 3.1156 kJ/kg.oC.
P, kPa
He
0.7 kg
27 oC
350 kPa
Wsh
0.7 kg
T2
P2
paddle
27
1
2
o
C
350
V
(a)
The amount of paddle-wheel work done on the system is
Wsh  Wsh t = 0.015 (30 x 60) = 27 kJ
The system is insulated
Q=0
Qin  Wout  mu 2  u1   mcv T2  T1 
0 – (– 27) = 0.7 (3.1156) (T2 – 27)
(b) At state 1, m 
Thus,
P1 P2

T1 T2
P1V
RT1
and at state 2, m 
T2 = 39.4 oC
P2V
RT 2
P 
 350 
P2  T2  1   39.4  273
 = 364.5 kPa
 27  273 
 T1 
Note : In an ideal gas equation, P and T are absolute quantities !
11
Example
A piston-cylinder device initially contains 0.5 m3 of nitrogen at 400 kPa and 27 oC. An electric
heater within the device is turned on an allowed to pass a current of 2 A for 5 min from a 120
V source. Nitrogen expands at constant pressure, and a heat loss of 2800 J occurs during the
process. Determine the final temperature of nitrogen. Take R = 0.297 kJ/kg.K and cp = 1.039
kJ/kg.K, for nitrogen gas.
Electrical work done on nitrogen, We = VI t = 120 (2) (5 x 60) = 72 000 J = 72 kJ
PV
4000.5
At state 1, m  1 1 
= 2.245 kg
RT1 0.297300
Qin = – 2800 J = – 2.8 kJ
For constant-pressure process, Qin  Wother,out  H 2  H 1  mh2  h1  = mc p T2  T1 
– 2.8 – (– 72) = 2.245 (1.039) (T2 – 27 )
T2 = 56.7 oC
Example
A piston-cylinder device initially contains air at 150 kPa and 27 oC. At this state, the piston is resting
on a pair of stops, and the enclosed volume is 400 L. The mass of the piston is such that a pressure of
350 kPa is required to move it. The air heated until its volume doubles. Determine (a) the final
temperature, (b) the work done by the air, and (c) the total heat transferred to the air. For air, R = 0.287
kJ/kg.K and cv = 0.823 kJ/kg.K.
.
.
.
.
.
Heating
150 kPa
400 L
27 oC
1
m
(a)
P1V1 P3V3

RT1
RT3
Heating
V=C
350 kPa
400 L
P=C
350 kPa
800 L
2
3
PV
350800
= 1400 K
T3  T1 3 3  300
P1V1
150400
(b) From 1 to 2, Wb = 0. From 2 to 3, W23 is boundary work for a constant-pressure process.
Thus total boundary work done by the system is
Wtotal,out  W12  W23  0  P2 V3  V2 
= 350 kPa (0.8 m3 – 0.4 m3) = 140 kJ
(c)
… +ve sign indicates work done by system
P1V1
1500.4
= 0.697 kg

RT1 0.287300
Process 1–2
………. W12 = 0 for constant volume process
Q12  W12  U 2  U1
Q23  W23  U 3  U 2
Process 2–3
Adding, gives Q12  Q23  W12  W23  U 3  U 1 or
Qtotal,in  Wtotal,out  mu 3  u1   mcv T3  T1 
m
Qtotal,in  140  0.697(0.823)(1400  300)
Qtotal,in = 771 kJ ……….. +ve sign indicates heat is transferred to system.