Chapter 6
ENERGY ANALYSIS OF CLOSED
SYSTEM
(1st Law of Thermodynamics)
The first law of thermodynamics is an expression of the conservation of energy
principle. Energy can cross the boundaries of a closed system in the form of heat or
work. Energy transfer across a system boundary due solely to the temperature
difference between a system and its surroundings is called heat.
Work energy can be thought of as the energy expended to lift a weight.
Closed System First Law
A closed system moving relative to a reference plane is shown below where z is the
elevation of the center of mass above the reference plane and is the velocity of the
center of mass.
Heat

V
Closed
System
Work
z
Reference Plane, z = 0
For the closed system shown above, the conservation of energy principle or the
first law of thermodynamics is expressed as
2
or
Ein  Eout  Esystem
According to classical thermodynamics, we consider the energy added to be net heat
transfer to the closed system and the energy leaving the closed system to be net
work done by the closed system. So
Qnet  Wnet  Esystem
Where
Qnet  Qin  Qout
Wnet  (Wout  Win )other  Wb
2
Wb   PdV
1
Normally the stored energy, or total energy, of a system is expressed as the sum of
three separate energies. The total energy of the system, Esystem, is given as
3
E = Internal energy + Kinetic energy + Potential energy
E = U + KE + PE
Recall that U is the sum of the energy contained within the molecules of the system
other than the kinetic and potential energies of the system as a whole and is called
the internal energy. The internal energy U is dependent on the state of the system
and the mass of the system.
For a system moving relative to a reference plane, the kinetic energy KE and the
potential energy PE are given by
mV 2
KE   mV dV 
V 0
2
V
PE  
z
z 0
mg dz  mgz
The change in stored energy for the system is
E  U  KE  PE
where
U  mu 2  u1 

1
m V22  V12
2
PE  mg  z 2  z1 
KE 

u2 and u1(get from
thermodynamic property
relations)
4
Now the conservation of energy principle, or the first law of thermodynamics for
closed systems, is written as
Qnet  Wnet  U  KE  PE
If the system does not move with a velocity and has no change in elevation, the
conservation of energy equation reduces to
Qnet  Wnet  U
where ΔU = U2 – U1
We will find that this is the most commonly used form of the first law.
Closed System First Law for a Cycle
Since a thermodynamic cycle is composed of processes that cause the working fluid
to undergo a series of state changes through a series of processes such that the final
and initial states are identical, the change in internal energy of the working fluid is
zero for whole numbers of cycles. The first law for a closed system operating in a
thermodynamic cycle becomes
Qnet  Wnet  U cycle
Qnet  Wnet
5
Moving Boundary Work
Work is energy expended when a force acts through a displacement. Boundary
work occurs because the mass of the substance contained within the system
boundary causes a force(automobile engines), the pressure times the surface
area, to act on the boundary surface and make it move. This is what happens
when steam, the “gas” in the figure below, contained in a piston-cylinder device
expands against the piston and forces the piston to move; thus, boundary work is
done by the steam on the piston. Boundary work is then calculated from
6
Since the work is process dependent, the differential of boundary work Wb
 Wb  PdV
is called inexact. The above equation for Wb is valid for a quasiequilibrium(also called a quasi-static process, is closely approximated by
real engines when piston moves at low velocities) process and gives the
maximum work done during expansion and the minimum work input during
compression. In an expansion process the boundary work must overcome
friction, push the atmospheric air out of the way, and rotate a crankshaft.
Wb  Wfriction  Watm  Wcrank
2
  ( Ffriction  Patm A  Fcrank )ds
1
7
To calculate the boundary work, the process by which the system changed
states must be known. Once the process is determined, the pressure-volume
relationship for the process can be obtained and the integral in the boundary
work equation can be performed. For each process we need to determine the
functional relationship between P and V during the process
P  f (V )
So as we work problems, we will be asking, “What is the pressure-volume
relationship for the process?” Remember that this relation is really the forcedisplacement function for the process.
The boundary work is equal to the area under the process curve plotted on the
pressure-volume diagram.
The quasi-equilibrium expansion process
described is shown on a P-V diagram. On
this diagram, the differential area dA is equal
to PdV, which is the differential work. The
total area A under the process curve 1-2 is
obtained by integration
2
2
Area  A   dA   PdV
1
1
8
Note from the above figure:
P is the absolute pressure and is always positive.
When dV is positive, Wb is positive.
When dV is negative, Wb is negative.
Since the areas under different process curves on a P-V diagram are different,
the boundary work for each process will be different. The next figure shows that
each process gives a different value for the boundary work.
9
Example
A piston-cylinder device contains 0.05 m3 of a gas initially at 200 kPa. At this
state, a linear spring that has a spring constant of 150 kN/m is touching the
piston but exerting no force on it. Now heat is transferred to the gas, causing the
piston to rise and to compress the spring until the volume inside the cylinder
doubles. If the cross-sectional area of the piston is 0.25 m2 , determine
(a) the final pressure inside the cylinder, (b) the total work done by the gas.
10
Assumption 1 The expansion process is quasi-equilibrium. 2 The spring is linear
in the range of interest.
Analysis A sketch of the system and the P-V diagram of the process are shown
above.
(a) The enclosed volume at the final state is


V2  2V1  2 0.05m3  0.1m3
Then the displacement of the piston (and of the spring) becomes
V 0.1  0.05m3
x

 0.2m
A
0.25m 2
The force applied by the linear spring at the final state is
F  kx  150kN / m0.2m  30kN
The additional pressure applied by the spring on the gas at this state is
F
30kN
P 
 120kPa
2
A 0.25m
11
Without the spring, the pressure of the gas would remain constant at 200 kPa
while the piston is rising. But under the effect of the spring, the pressure rises
linearly from 200 kPa to
200 + 120 = 320 kPa
(b) An easy way of finding the work done is to plot the process curve. From
fig above the area under the process curve (a trapezoidal) is determined to
be
W  area 
200  320kPa 0.1  0.05m3 
2
1kJ 
 13kJ

3 
1
kPa
.
m


12
Some Typical Processes of Closed System
Constant volume
If the volume is held constant, dV = 0, and the boundary work equation
becomes
P
1
2
V
P-V diagram for V = constant
If the working fluid is an ideal gas, what will happen to the temperature of the
gas during this constant volume process?
Q12  U
Q12  U 2  U1
Q12  mu2  u1 
kJ/kg
kJ
13
For ideal gas
Q12  mu2  u1 
Q12  mCv T2  T1 
14
EXAMPLE 9-2
A rigid tank contains air at 500 kPa and 150°C. As a result of heat transfer to
the surroundings, the temperature and pressure inside the tank drop to 65°C
and 400 kPa, respectively. Determine the boundary work done during this
process.
A sketch of the system and the P-V diagram of the process are shown
below.
2
Wb   PdV  0
1
This is expected since rigid tank has a constant volume and dv = 0.
Therefore, there is no boundary work done during this process. That is,
the boundary work done during a constant-volume process is always zero. 15
Example 9-3
A tank contains nitrogen at 27C. The temperature rises to 127C by heat transfer to
the system. Find the heat transfer and the ratio of the final pressure to the initial
pressure.
System: Nitrogen in the tank.
2
System
boundary
P
T1=
27C
Nitrogen gas
T2=127C
1
P-V diagram for a constant
volume process
V
Property Relation: Nitrogen is an ideal gas. The ideal gas property relations
apply. Let’s assume constant specific heats. (You are encouraged to rework
this problem using variable specific heat data.)
Process: Tanks are rigid vessels; therefore, the process is constant volume.
Conservation of Mass:
m2  m1
16
Using the combined ideal gas equation of state,
PV
PV
2 2
 1 1
T2
T1
Since R is the particular gas constant, and the process is constant volume,
V2  V1
P2 T2 (127  273) K


 1333
.
P1 T1
(27  273) K
Conservation of Energy:
The first law closed system is
Ein  Eout  E
Qnet  Wnet  U
For nitrogen undergoing a constant volume process (dV = 0), the net work is
(Wother = 0)
17
Using the ideal gas relations with Wnet = 0, the first law becomes (constant
specific heats)
The heat transfer per unit mass is
18
Constant pressure
P
2
1
V
P-V DIAGRAM for P = CONSTANT
If the pressure is held constant, the boundary work equation becomes
W12  P  dV  PV2  V1   mPv2  v1 
V2
V1
Heat Transfer
Q12  U 2  U1  W12
Q12  U 2  U1  PV2  V1 
Q12  U 2  PV2   U1  PV1 
19
Now, we know that h = u + pv or H = U + pV. Hence
Q12 = H2 – H1
For ideal gas
Q12  mh2  h1 
Q12  mC p T2  T1 
20
Example 2-13
Air undergoes a constant pressure cooling process in which the temperature
decreases by 100C. What is the magnitude and direction of the work for this
process?
System:
P
2
System
Boundary
1
Wb
Air
V
Property Relation: Ideal gas law, Pv = RT
Process: Constant pressure
Work Calculation: Neglecting the “other” work
21
The work per unit mass is
wnet ,12 
Wnet ,12
m
 (0.287
 R(T2  T1 )
kJ
kJ
)(100 K )  28.7
kg  K
kg
The work done on the air is 28.7 kJ/kg.
Note: The UNIVERSAL GAS CONSTANT Ru is 8.31447 kJ/kmol.K
Substance
R, kJ/kg.K
Air
0.2870
Nitrogen
0.2968
Helium
2.0769
22
Constant temperature, ideal gas
If the temperature of an ideal gas system is held constant, then the equation of
state provides the pressure-volume relation
PV  mRT  cons tan t  C
Then, the boundary work are
V2
V2
V1
V1
Wb   PdV   C
dV
C
V
ln V 
V2
V1
V 
V 
 C ln  2   P1V1 ln  2 
 V1 
 V1 
or
or
 P1 
Wb  mRT ln  
 P2 
V2 
Wb  P2V2 ln  
 V1 
Note: The above equations are the result of applying the ideal gas assumption for
the equation of state. For real gases undergoing an isothermal (constant
temperature) process, the integral in the boundary work equation would be done
23
numerically.
Heat Transfer
Energy balance to this case is applied:
U1 + Q = U2 + W
For a ideal gas
U1 = mcvT1 and U2 = mcvT2
As the temperature is constant
U1 = U2
Substituting in the energy balance equation,
Q=W
Thus, for a ideal gas, all the heat added during a constant temperature
process is converted into work and the internal energy of the system
remains constant
24
Example 2-12
Three kilograms of nitrogen gas at 27C and 0.15 MPa are compressed
isothermally to 0.3 MPa in a piston-cylinder device. Determine the
minimum work of compression, in kJ.
System: Nitrogen contained in a piston-cylinder device.
Process: Constant temperature
2
P
System
Boundary
1
Nitrogen
gas
Wb
V
P-V DIAGRAM for T = CONSTANT
Since, under the given conditions, nitrogen is an ideal gas, and we use the
ideal gas equation of state as the property relation.
PV  mRT
25
Work Calculation:
For an ideal gas in a closed system (mass = constant), we have
m1  m2
PV
PV
1 1
 2 2
RT1 RT2
Since the R's cancel, we obtain the combined ideal gas equation. Since T2 = T1,
V2 P1

V1 P2
26
The net work is
Wnet ,12  0  Wb,12  184.5 kJ
On a per unit mass basis
wnet ,12
Wnet ,12
kJ

 615
.
m
kg
The net work is negative because work is done on the system during the
compression process. Thus, the work done on the system is 184.5 kJ, or
184.5 kJ of work energy is required to compress the nitrogen.
27
Example 2-15
Air is expanded isothermally at 100C from 0.4 MPa to 0.1 MPa (in a pistoncylinder device). Find the ratio of the final to the initial volume, the heat
transfer, and work.
System: Air contained in a piston-cylinder device, a closed system
Process: Constant temperature
System
boundary
1
P
2
Air
Wb
T = const.
V
P-V diagram for T= constant
28
Property Relation: Assume air is an ideal gas and use the ideal gas property
relations with constant specific heats.
PV  mRT
u  CV (T2  T1 )
Conservation of Energy:
Ein  Eout  E
Qnet  Wnet  U
The system mass is constant but is not given and cannot be calculated;
therefore, let’s find the work and heat transfer per unit mass.
29
Work Calculation:
Conservation of Mass: For an ideal gas in a closed system (mass = constant), we
have
m m
1
2
PV
PV
1 1
 2 2
RT1 RT2
Since the R's cancel and T2 = T1
V2 P1 0.4 MPa


4
V1 P2 01
. MPa
30
Then the work expression per unit mass becomes
The net work per unit mass is
wnet ,12  0  wb ,12
kJ
 148.4
kg
Now to continue with the conservation of energy to find the heat transfer. Since T2 =
T1 = constant,
U12  mu12  mCV (T2  T1 )  0
So the heat transfer per unit mass is
31
qnet
qnet  wnet
Qnet

m
 u  0
qnet  wnet
kJ
 148.4
kg
The heat transferred to the air during an isothermal expansion process
equals the work done.
32
The Polytropic Process
The polytropic process is one in which the pressure-volume relation is given
as
PV n  constant
The exponent n may have any value from minus infinity to plus infinity depending on the process. Some of the more common values are given below.
Process
Constant pressure
Constant volume
Isothermal & ideal gas
Adiabatic & ideal gas
Exponent n
0

1
k = CP/CV
Here, k is the ratio of the specific heat at constant pressure CP to specific
heat at constant volume CV. The specific heats will be discussed later.
The boundary work done during the polytropic process is found by
substituting the pressure-volume relation into the boundary work equation.
The result is
33
For an ideal gas under going a polytropic process, the boundary work is
34
Notice that the results we obtained for an ideal gas undergoing a polytropic
process when n = 1 are identical to those for an ideal gas undergoing the
isothermal process.
Heat Transfer
Q12  U 2  U1   W12
Relationship P,v and T for a polytropic process between two ideal
gas states:
P1  v2 
  
P2  v1 
T1  v2 
  
T2  v1 
T1  P1 
  
T2  P2 
n
n 1
n 1
n
35
Problem 2-16: An air in a closed system undergoes a process from the initial state
p = 3 bars, and V = 0.1 m3 to the final state V = 0.2 m3 according to the relation
pV 1.2  Const.
Plot the process on the p-V diagram. Determine (a) the final pressure of the air,
and (b) the work done during this process in kJ.
Solution:
p1
1
p
p2
2
V2
V
V1
(a)The process is known as a polytropic process with n = 1.2 and the pressure at
the final state can be obtained as follows:
1.2
1 1
pV
 p2V
1.2
2
 Const.
36
Or
p2  p1 (
V1 1.2
0.1
)  3( )1.2  1.31 bars
V2
0.2
Answer
(b) The work during the polytropic process is given by
p2V2  p1V1 [(1.31)(0.2)  (3)(0.1)] x105
W

kJ
3
1 n
(1  1.2) x10
 19 kJ
The work done is a positive work ( 19 kJ) for expansion, therefore the work
is done by the system.
37
Specific Heats and Changes in Internal Energy and Enthalpy for Ideal Gases
Before the first law of thermodynamics can be applied to systems, ways to calculate
the change in internal energy of the substance enclosed by the system boundary
must be determined. For real substances like water, the property tables are used to
find the internal energy change. For ideal gases the internal energy is found by
knowing the specific heats. Physics defines the amount of energy needed to raise
the temperature of a unit of mass of a substance one degree as the specific heat at
constant volume CV for a constant-volume process, and the specific heat at
constant pressure CP for a constant-pressure process. Recall that enthalpy h is the
sum of the internal energy u and the pressure-volume product Pv.
h  u  Pv
In thermodynamics, the specific heats are defined as
(1)
(2)
38
cp > cv (cp involves additional work)
For and ideal gas the internal energy is a function of the temperature only. That is
Using the definition of enthalpy and the equation of state of an ideal gas, we
have
Since R is constant and u = u(T), it follow that the enthalpy of an ideal gas
is also a function of temperature only:
39
Since u and h depend only on temperature for ideal gas, the specific heat cv and
cp also depend, at most, on temperature only.
For ideal gases, the partial derivatives in Eqs. (1) and (2) can be replaced by
ordinary derivatives.
(3)
(4)
The change in internal energy or enthalpy for an ideal gas during a
process from state 1 to state 2 is determined by integrating these
equations:
u2  u1  Cv T2  T1 
h2  h1  C p T2  T1 
kJ/kg
kJ/kg
40
Relation between CP and CV for Ideal Gases
Using the definition of enthalpy (h = u + Pv) and writing the differential of enthalpy, the
relationship between the specific heats for ideal gases is
h  u  Pv
dh  du  d ( RT )
C P dT  CV dT  RdT
C P  CV  R
where R is the particular gas constant. The specific heat ratio k (fluids texts often use
 instead of k) is defined as
CP
k
CV
41
The Systematic Thermodynamics Solution Procedure
When we apply a methodical solution procedure, thermodynamics problems are
relatively easy to solve. Each thermodynamics problem is approached the same way
as shown in the following, which is a modification of the procedure given in the text:
Thermodynamics Solution Method
1. Sketch the system and show energy interactions across the boundaries.
2. Determine the property relation. Is the working substance an ideal gas or a real
substance? Begin to set up and fill in a property table.
3. Determine the process and sketch the process diagram. Continue to fill in the
property table.
4. Apply conservation of mass and conservation of energy principles.
5. Bring in other information from the problem statement, called physical constraints,
such as the volume doubles or the pressure is halved during the process.
6. Develop enough equations for the unknowns and solve.
42