Second Law of Thermodynamics Explained

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Second Law of
Thermodynamics
6CHAPTER

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 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

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

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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.

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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.

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• 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

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 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

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 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:

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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

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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

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

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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

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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

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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%

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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
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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.

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 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.

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 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
!?

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

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An example of a Refrigerator
and a Heat pump ..

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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
= = =
− −

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Heat Pumps
QH
QL
W
HPQH = W + QL
Objective

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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
= = =
− −

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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= +

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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!

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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

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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:

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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

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 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.

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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.

Second Law of Thermodynamics Explained

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