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Introduction To
The Second Law ofThermodynamics
Understanding of
Thermal EfficiencyPrepared by
PM Muhammad Abd RazakFKM UiTMPP
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The Second Law of Thermodynamics
The 2nd law of thermodynamics is a natural law thatstates
that
processes can occur in a certain direction, not in justany
directionGases expand from a high pressure to a low pressure.Heat
flows from a high temp. to a low temperature.
No heat engine is able to convert completely all theheat
supplied into work output and there must besome heat rejection at a
lower temperature than the
source.
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Heat Engine
An energy conversion system which:
operates in a thermodynamic cycle
operates between two heat reservoirs where
net heat is transferred
net work is delivered.
Heat (Thermal) Reservoir a sufficiently large system in stable
equilibrium has finite amounts of heat can be transferred without
any
change in its temperature. high temperature heat reservoir : a
heat source. low temperature heat reservoir : a heat sink .
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Example of a heat engine : Steam Plant
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Thermal Efficiency, th the index of performance of a heat engine
defined by the ratio of the net work output to the heat input
th =Desired Result
Required Input
thnet out
in
W
Q=
,W W W
Q Q
net out out in
in net
, =
where
thermal efficiency is 0
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Applying the 1st Law of Thermodynamics
Q W U
W Q
W Q Q
net in net out
net out net in
net out in out
, ,
, ,
,
=
=
=
0 for cyclicprocess
th net outin
in out
in
out
in
W
Q
Q Q
Q
Q
Q
=
=
=
,
1
Then
thL
H
Q
Q= 1or
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Refrigerator & Heat Pump operates in a thermodynamic cycle
absorbs heat from a low temperature body and
delivers heat to a high temperature bodymust receives external
energy (work or heat) from the
surroundings.refrigerator: extracts heat from low-temperature
media.heat pump : rejects heat to the high-temperature media.
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Coefficient of Performance, COP
index of performance of a refrigerator & heat pump is
interms of the coefficient of performance, COP,
the ratio of desired result to input larger than 1 and theCOP to
be as large as possible.
COP =Desired Result
Required Input
For a refrigerator or an air conditionerHeat is transfered from
the low temperature reservoir.
ThenCOP Q
WR
L
net in
=,
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( ) ( )
,
Q Q W U
W W Q Q
L H in cycle
in net in H L
= == =
0 0
COPQ
Q QR
L
H L
=
From the 1st Law Equation
Then
For a heat pumpheat is transfered to the high temperature
system, then
COPQ
W
Q
Q QHP
H
net in
H
H L
= =,
We can also show that COP COPHP R= + 1
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The Carnot Cycle Sadi Carnot (1769-1832) was among the first to
study the
principles of the 2nd law of t/dynamics on cyclic operations
devised a reversible cycle composed of four reversible
processes:
two isothermal and two adiabatic.
A vapour cycle
Process 1 2 : Reversibleadiabatic expansion (in turbine).
System produces work, Wout-
The working fluid temperaturedecreases from T
Hto T
L.
Process 2-3 : reversibleisothermal heat rejection QL (in a
condenser)
T
TH
TL
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Process 3-4: reversible adiabaticcompression (in a
compressor)
system receives work input, Win
working fluid temperature
increases from TL to THProcess 4-1: reversible isothermal
heat addition, QH (in a boiler)
Note that the Carnot power cycle operates in the
clockwise direction when plotted on a
process diagram.(T-v, P-v, T-s) for a refrigerator & heat
pump, theCarnot cycle is reversed, the cycleoperates in the counter
clockwise
direction.A gas cycle
QH
QL
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Again, the thermal efficiency is
thL
H
Q
Q= 1
For a reversible heat engine, the energy transfer ratio QL/QH
canbe replaced by ratio of absolute temp TL/TH
th revL
H
TT
, = 1
This is the maximum possible efficiency of a heat
engineoperating between two heat reservoirs at constant
temperatures THand TL.
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The Kelvin scale, relates the heat transfers in a reversible
devicebetween the high and low-temperature heat reservoirs
atconstant temperature as
Q
Q
T
T
L
H
L
H= LL
H
H
T
Q
T
Q
=
Summarising all heat in and heat out
o
o
oi
i
i T
Q
T
Q == = 11 =in out
T
Q
T
Q
= 0T
QFor a cyclic process
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The term Q/T depends only on the initial & final states,not
on the process. Thus it is a point function or propertydefined as
entropy, s
Then
revTQds
= =
2
1
12TQss
1 2
Q = T ds [kJ/kg]Then
Q12 = T(s2 s1)
[kJ/kgK]
