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Second Law of
Thermodynamics
6CHAPTER
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
 The amount of EE is
equal to the amount of
energy transferred to
the room.
Example 2
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
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
• 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
 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
 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
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
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
Steam power plant is another
example of a heat engine..
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
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
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
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
6.2. Statements of the
Second Law of
Thermodynamics
16
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
 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
 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
High-temperature Reservoir at TH
Low-temperature Reservoir at TL
QH
QL
W
RefQL = QH - W
Objective
Refrigerators
21
An example of a Refrigerator
and a Heat pump ..
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
Heat Pumps
High-temperature Reservoir at TH
Low-temperature Reservoir at TL
QH
QL
W
HPQH = W + QL
Objective
24
Heat Pump
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
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
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
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
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
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
 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
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.

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Second Law of Thermodynamics Explained

  • 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
  • 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.