The document discusses the second law of thermodynamics. It states that heat always flows spontaneously from hotter to colder bodies and not vice versa. This is because processes proceed in a certain direction and not the reverse direction. The second law is needed to determine the direction of spontaneous processes since the first law does not restrict direction. Two common statements of the second law are then presented: 1) It is impossible for a heat engine to receive heat from a single reservoir and produce work without rejecting some heat. 2) It is impossible to transfer heat from a cold to a hot body without external work. Examples of heat engines, refrigerators, and heat pumps are provided to illustrate these principles.

1. 1
Second Law of
Thermodynamics
6CHAPTER

2. 2
 Heat always flows from
high temperature to low
temperature.
 So, a cup of hot coffee
does not get hotter in a
cooler room.
 Yet, doing so does not
violate the first law as long
as the energy lost by air is
the same as the energy
gained by the coffee.
Room at
25° C
!?
Example 1

3. 3
 The amount of EE is
equal to the amount of
energy transferred to
the room.
Example 2

4. 4
It is clear from the previous
examples that..
 Processes proceed in certain direction
and not in the reverse direction.
 The first law places no restriction on the
direction of a process.
• Therefore we need another law (the
second law of thermodynamics) to
determine the direction of a process.

5. 5
Thermal Energy Reservoir
• If it supplies heat then it
is called a source.
• It is defined as a body to which and from
which heat can be transferred without a
change in its temperature.
• If it absorbs heat then it
is called a sink.

6. 6
• Some obvious examples
are solar energy, oil
furnace, atmosphere,
lakes, and oceans
• Another
example is two-
phase systems,
• and even the air in a room if the
heat added or absorbed is small
compared to the air thermal
capacity (e.g. TV heat in a room).
Air

7. 7
 We all know that doing work on the water will
generate heat.
 However transferring heat to the liquid will not
generate work.
 Yet, doing so does not violate the first law as long
as the heat added to the water is the same as the
work gained by the shaft.
Heat Engines

8. 8
 Previous example leads to the concept of
Heat Engine!.
 We have seen that work always converts
directly and completely to heat, but
converting heat to work requires the use of
some special devices.
 These devices are called Heat Engines and
 can be characterized by the following:

9. 9
Characteristics of Heat
Engines..
 They receive heat from
high-temperature source.
 They convert part of this
heat to work.
 They reject the remaining
waste heat to a low-
temperature sink.
 They operate on (a
thermodynamic) cycle.
High-temperature
Reservoir at TH
Low-temperature
Reservoir at TL
QH
QL
WHE

10. 10
Difference between Thermodynamic
and Mechanical cycles
 A heat engine is a device that operates in a thermodynamic cycle
and does a certain amount of net positive work through the
transfer of heat from a high-temperature body to a low-
temperature body.
 A thermodynamic cycle involves a fluid to and from which heat is
transferred while undergoing a cycle. This fluid is called the
working fluid.
 Internal combustion engines operate on a mechanical cycle (the
piston returns to its starting position at the end of each
revolution) but not on a thermodynamic cycle.
 However, they are still called heat engines

11. 11
Steam power plant is another
example of a heat engine..

12. 12
Thermal efficiency
Thermal Efficiency
< 100 %
in
out
Q
Q
−=1
inputRequired
outputDesired
ePerformanc =
==
in
out,net
th
Q
W
η
in
outin
Q
QQ −
High-temperature
Reservoir at TH
Low-temperature
Reservoir at TL
QH
QL
WHE

13. 13
Thermal efficiency
Thermal Efficiency
< 100 %1 L
H
Q
Q
= −
,net out
th
H
W
Q
η = H L
H
Q Q
Q
−
QH= magnitude of heat transfer between the cycle
device and the H-T medium at temperature TH
QL= magnitude of heat transfer between the cycle
device and the L-T medium at temperature TL

14. 14
thermal efficiency can not
reach 100%
Even the Most Efficient Heat
Engines Reject Most Heat as
Waste Heat
40
0.4
100
thη= =
Automobile Engine 20%
Diesel Engine 30%
Gas Turbine 30%
Steam Power Plant 40%

15. 15
Heat is transferred to a heat engine from
a furnace at a rate of 80 MW. If the rate
of waste heat rejection to a nearby river
is 50 MW, determine the net power
output and the thermal efficiency for
this heat engine.
<Answers: 30 MW, 0.375>
Example 6-1: Net Power Production of a Heat
Engine

16. 6.2. Statements of the
Second Law of
Thermodynamics
16

17. 17
The Second Law of Thermodynamics:
Kelvin-Plank Statement (The first)
 The Kelvin-Plank statement:
It is impossible for any device that
operates on a cycle to receive heat
from a single reservoir and produce
a net amount of work.

18. 18
 It can also be expressed as:
No heat engine can have a thermal
efficiency of 100%, or as for a
power plant to operate, the working
fluid must exchange heat with the
environment as well as the furnace.
 Note that the impossibility of having a 100%
efficient heat engine is not due to friction or
other dissipative effects.
 It is a limitation that applies to both idealized
and the actual heat engines.

19. 19
 Example 1 at the beginning of the notes
leads to the concept of Refrigerator and
Heat Pump..
 Heat can not be transferred from low
temperature body to high temperature one
except with special devices.
 These devices are called Refrigerators and
Heat Pumps
 Heat pumps and refrigerators differ in
their intended use. They work the same.
 They are characterized by the following:
Room at
25° C
!?

20. 20
High-temperature Reservoir at TH
Low-temperature Reservoir at TL
QH
QL
W
RefQL = QH - W
Objective
Refrigerators

21. 21
An example of a Refrigerator
and a Heat pump ..

22. 22
Coefficient of Performance of a
Refrigerator
The efficiency of a refrigerator is expressed in term of
the coefficient of performance (COPR).
Desired output
Required input
RCOP =
,
1
1
L L
Hnet in H L
L
Q Q
QW Q Q
Q
= = =
− −

23. 23
Heat Pumps
High-temperature Reservoir at TH
Low-temperature Reservoir at TL
QH
QL
W
HPQH = W + QL
Objective

24. 24
Heat Pump

25. 25
Coefficient of Performance of a
Heat Pump
The efficiency of a heat pump is expressed in term of the
coefficient of performance (COPHP).
Desired output
Required input
HPCOP =
,
1
1
H H
Lnet in H L
H
Q Q
QW Q Q
Q
= = =
− −

26. 26
Relationship between Coefficient of
Performance of a Refrigerator (COPR)
and a Heat Pump (COPHP).
,
,
net in LH H
HP
net in H L H L
W QQ Q
COP
W Q Q Q Q
+
= = =
− −
,
1net in L
HP R
H L H L
W Q
COP COP
Q Q Q Q
= + =+
− −
1HP RCOP COP= +

27. 27
The second Law of Thermodynamics:
Clausius Statement
The Clausius statement is
expressed as follows:
It is impossible to construct a
device that operates in a cycle
and produces no effect other
than the transfer of heat from
a lower-temperature body to a
higher-temperature body.
Both statements are negative
statements!

28. 28
High-temperature Reservoir at TH
Low-temperature Reservoir at TL
QH + QL
QL
W = QH
RefHE
QH
Net QIN = QL
Net QOUT = QL
HE + Ref
Equivalence of the Two
Statements
Consider the HE-RF combination shown below

29. 29
Example (6-2): Heating a House by a Heat Pump
A heat pump is used to meet the heating requirements of
a house and maintain it at 20oC. On a day when the
outdoor air temperature drops to -2oC, the house is
estimated to lose heat at rate of 80,000 kJ/h. If the heat
pump under these conditions has a COP of 2.5,
determine (a) the power consumed by the heat pump and
(b) the rate at which heat is absorbed from the cold
outdoor air.
Sol:

30. 30
6.3. Perpetual Motion Machines
 Any device that violates the first or second law is
called a perpetual motion machine
 If it violates the first law, it is a perpetual motion
machine of the first type (PMM1)
 If it violates the second law, it is a perpetual
motion machine of the second type (PMM2)
 Perpetual Motion Machines are not possible

31. 31
 The second law of thermodynamics state
that no heat engine can have an efficiency of
100%.
 Then one may ask, what is the highest
efficiency that a heat engine can possibly
have.
 Before we answer this question, we need to
define an idealized process first, which is
called the reversible process.

32. 32
Can we save Qout?
 Heat the gas (QH=100
kJ)
 Load is raised=> W=15
kJ
 How can you go back to
get more weights (i.e.
complete the cycle)?
 By rejecting 85 kJ
 Can you reject it to the
Hot reservoir? NO
 What do you need?
 We need cold reservoir
to reject 85 kJ
A heat- engine cycle
cannot be completed
without rejecting
some heat to a low
temperature sink.