Department of
Mechanical Engineering
ME 322 – Mechanical Engineering
Thermodynamics

The Second Law of Thermodynamics
Entropy is a balanced quantity. Therefore,
Entropy transported into a system
-
Entropy transported out of a system
+
Entropy produced within a system
=
Entropy gained within a system
2
S
T
S
T
+
S
P
=
S
G
S
T
+
S
P
=
S
G
S
P
Total Entropy Form
Entropy Rate Form
S
G
Something to think about ... We have already seen that entropy production occurs due to irreversibility. But, how is entropy transported across a system boundary?

3
The Second Law of Thermodynamics
First, consider a closed system. For this system, we know that the Second Law is,
S
T
+
S
P
=
S
G
  dS
T
+ dS
P
= dS
G
Q
Recall that Clausius discovered entropy analyzing a Carnot cycle as a closed system.
His discovery was, dS

 dQ
T

 rev
T
H
T in this expression is the boundary temperature where Q is being transferred.
T
L
W

4
The Second Law of Thermodynamics
For a reversible process, we know that the entropy production is zero. Therefore, for a reversible process, the Second Law says, dS
T
= dS
G
Q W
Now, consider Clausius’ discovery again, dS
 dQ

 T  rev
For the closed system that we are analyzing, it is clear that heat and work cross the system boundary. Therefore, we can say that, dS
T
=
 dQ

 T  rev
and dS
G
= dS sys

5
The Second Law of Thermodynamics
We have just discovered that that the entropy is transported into the system by heat , but NOT by work!
Q W
For any closed system process, we can write,
 dQ

 T 
+ dS
P
= dS sys
This expression can now be integrated between any two states,

1
2
 dQ

 T 
+

1
2 dS
P
=

1
2 dS sys

6
The Second Law of Thermodynamics
Analyzing each integral ...

1
2
 dQ

 T 
+

1
2 dS
P
=

1
2 dS sys

1
2
 dQ

 T 
=
 k
Q k
T k
Assumption: The system boundary is isothermal.
The summation sign accounts for all heat transfer.

1
2 dS sys
=
S
2
-
S
1
= 
2
s
1


1
2 dS
=
P
???
The entropy change of the system.
This is the integral of the entropy production in an irreversible process. But ... how do we evaluate the integral???

7
The Second Law of Thermodynamics
Entropy production is caused by irreversibilities. Consider two piston-cylinder assemblies. They are identical in every way except that one of them operates reversibly and the other is irreversible.
Questions ...
Q
W rev reversible
Q
W irr irreversible
1.
Which system has zero entropy production?
2.
Which system delivers more work?
3.
Is work a property?
4.
Is entropy production a property?

8
The Second Law of Thermodynamics
Entropy production is NOT a property. It is a function of path, just like heat and work!
 dQ

 T 
+ dS
P
= dS sys
Now, we can integrate the entropy production term!

1
2 dS
P
=
S
P ,12
> 0 means that the process is irreversible
= 0 means the process is reversible
< 0 means the process is impossible

9
The Second Law of Thermodynamics
For a closed system, the Second Law of Thermodynamics can be written as,
S
T
+
S
P
=
S
G
 k
Q k
T k
+
S
P ,12
= 
2
s
1

This can be written on a per unit mass basis by dividing both sides of the equation by the mass of the system,
 k

Q / m k

T k
+
S
P ,12 m
=  s
2
s
1
  k q k
T k
+ s
P ,12 s
2 s
1

10
The Second Law of Thermodynamics
Incorporating the net entropy transport into the system due to the mass flows gives the complete form of the Second Law of
Thermodynamics!
 k
Q k
T k
+  i m s i i
 e m s e e
+
S
P
= dS sys dt
Notice that if the system is closed,
 k
Q k
T k
+
S
P
= dS sys dt
 k
Q k
T k
+
S
P ,12
= 
2
s
1


11
Conservation of Mass – The Continuity Equation i
 m i
 e m e
= dm sys dt
Conservation of Energy – The First Law of Thermodynamics i
 m i

 h i
+
V
2 i g
2 c
+ g g c z i


 e m e

 h e
+
V
2 e g
2 c
+ g g c z e


= dE sys dt
The Entropy Balance – The Second Law of Thermodynamics
 k
Q k
T k
+ i
 m s i i
 e m s e e
+
S
P
= dS sys dt

12
Special Application – Closed Systems
Consider a closed system that is reversible and adiabatic.
The second law for such a system is,
 k
Q k
T k
0
+
S
P ,12
0
= 
2
s
1

 s
2
s
1
 =
0
 s
2
= s
1
We have just learned that a reversible and adiabatic process for a closed system is isentropic!
Notice: Reversible does not mean isentropic
Adiabatic does not mean isentropic
Reversible and adiabatic means isentropic!

13
Special Application – Open Systems
Consider an open system that is operating in steady state and is reversible and adiabatic with one flow in and one flow out. The second law for such a system is,
0
 k
Q k
T k
+  i m s i i
 e
0 m s e e
+
S
P
= dS sys dt
0
 i
s e
 =
0
 s i
= s e
We have just learned that a reversible and adiabatic process for a system of this type is isentropic!
Entropy is the gate keeper of the Second Law!

14
c p
We previously derived the following expressions for an ideal gas from the Gibbs Equations, s
2
- =
1

T
1
T
2 c v dT
T
+
R ln v
2 v
1 s
2
- =
1

T
1
T
2 c p dT
T
-
R ln
P
2
P
1
If the gas is undergoing a process where the heat capacity can be assumed to be constant, s
2
- =
1 c p ln
T
2
T
1
-
R ln
P
2
P
1
Consider a case where the process is also isentropic. Then, c p ln
T
2
T
1
-
R ln
P
2
P
1
=
0

15
c p
Algebra time ...
c p ln
T
2
T
1
-
R ln
P
2
P
1
=
0
 c p ln
T
2
T
1
=
R ln
P
2
P
1
R

T
2

P
2

T
1
 P
1
 c p
 ln
T
2
=
R P p ln
T c P
1 1
2
Rearrange the exponent,
R c p
= c p
c v c p
=
 c p
/
 c v

 c / c p v c
 v
/ c v

= k
-
1 k k
 c p c v

16
c p
Substitution gives,
T
2

P
2
 k k
-
1
T
1
 P
1

If we consider an isentropic process with the other
D s equation,
1
k s
2 1
0 c v ln
T
2
T
1
+
R ln v
2 v
1

T v
2 2
T v
1 1
Therefore, for an ideal gas with constant heat capacity undergoing an isentropic process,
T
2
=


P
2


T
1
 P
1
 k
-
1 k
= v
2
 
1
k

17
c p
But wait ... there’s more!!
T
2
=


P
2


T
1
 P
1
 k k
-
1
= v
2
 
1
k



P
P
1
2

 k k
-
1
= v
 
1
k
Rearranging ...
k
-
1
P
2 k v
1
1
k = k
-
1
P
1 k v
2
1
k k
-
1
P
2 k v
2 k
-
1 = k
-
1
P
1 k v
1 k
-
1
 k
-
1
 
 k k
-
1

 =
 k
-
1
 
 k k
-
1


P v k
2 2
=
Pv
1 1 k
Does this look familiar?

18
P
P
2
1
=

 v v
1
2

 n
T
T
2
1
=
 
 
1
n
T
T
2
1
=
 
 
1
k
Pv n = constant v
2 v
1
=


P
P
1
2


1/ n
T
T
2
1
=


P
P
2
1


  n
T
T
2
1
=


P
P
2
1


 k
 k
Any fluid model
Ideal Gas Model
Ideal Gas Model – Isentropic process, constant heat capacity n k / (ideal gas with constant heat capacity - isentropic process) p v n
=
0 (isobaric) n
=  
(isochoric) n
=
1 (ideal gas, isothermal)