PdfdCopy of 2nd Law
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Transcript of PdfdCopy of 2nd Law
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1
Second Law ofThermodynamics
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Example 12
Heat always flows from
high temperature to lowtemperature.
So, a cup of hot coffee does
no ge o er n a coo erroom.
Yet, doing so does notviolate the first low as longas the energy lost by air isthe same as the energygained by the coffee.
Room at
25 C
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Example 2 3
The amount of EE is equal
to the amount of energytransferred to the room.
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It is clear from the previous examples that..4
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 secondlaw of thermodynamics) to determine thedirection of a process.
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Thermal Energy Reservoir5
It is defined as a body to which and fromwhich heat can be transferred without achange in its temperature.
If it supplies heat then itis called a source.
If it absorbs heat then itis called a sink.
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6
Some obvious examplesare solar energy, oilfurnace, atmosphere,lakes, and oceans
Anotherexample is two-
phase systems,
and even the air in a room if theheat added or absorbed is smallcompared to the air thermal
capacity (e.g. TV heat in a room).
Air
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Heat Engines 7
We all know that doing work on the water will.
However transferring heat to the liquid will notgenerate work.
Yet, doing so does not violate the first low as longas the heat added to the water is the same as the
work gained by the shaft.
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8 Previous example leads to the concept ofHeat Engine!.
We have seen that work always convertsdirectly and completely to heat, butconvertin heat to work re uires 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-temperaturesource.
High-temperature
Reservoir at TH
QH
heat to work.
They reject theremaining waste heat toa low-temperature sink.
They operate on (a
thermodynamic) cycle. 9
Low-temperature
Reservoir at TL
QL
W
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Piston cylinder arrangement is anexample of a heat engine..
10
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Difference between Thermodynamicand Mechanical cycles11
A heat engine is a device that operates in a thermodynamic cycle
and does a certain amount of net positive work through thetransfer of heat from a high-temperature body to a low-temperature body.
ermo ynam c cyc e nvo ves a u o an rom w c ea stransferred while undergoing a cycle. This fluid is called theworking fluid.
Internal combustion engines operate on a mechanical cycle (the
piston returns to its starting position at the end of eachrevolution) but not on a thermodynamic cycle.
However, they are still called heat engines
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Steam power plant is another exampleof a heat engine..
12
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Thermal efficiency13
Thermal Efficiency
inputRequired
outputDesiredePerformanc =
< 100 %
in
out
Q
Q= 1
==
in
out,net
thQ
Win
outin
QQQ
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Thermal efficiency14
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
Thermal Efficiency
< 100 %1H
Q
Q=
,net out
th
H
W
Q = = L
H
Q Q
Q
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thermal efficiency can not reach 100%15
Even the Most Efficient Heat
Engines Reject Most Heat asWaste Heat
Even the Most Efficient Heat
Engines Reject Most Heat asWaste Heat
40
.100th
Automobile Engine 20%
Diesel Engine 30%
Gas Turbine 30%
Steam Power Plant 40%
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Can we save Qout? 16 Heat the gas (QH=100
kJ)
Load is raised=>W=15 kJ How can you go back
to et more wei hts
(i.e. complete thecycle)? By rejecting 85 kJ Can you reject it to the
Hot reservoir? NO What do you need? I need cold reservoir
to reject 85 kJ
A heat- engine cycle
cannot be completed
without rejectingsome heat to a low
temperature sink.
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17Heat 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.
Example 5-1: Net Power Production of a Heat
Engine
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The Second Law of Thermodynamics:Kelvin-Plank Statement (The first)18
The Kelvin-Plank statement:
It is impossible for any devicethat operates on a cycle to receive
produce a net amount of work.
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19 It can also be expressed as:
No heat engine can have athermal efficiency of 100%, or asfor a power plant to operate, the
working fluid must exchange
ea w e env ronmen aswell as the furnace. Note that the impossibility of having a 100%
efficient heat engine is not due to friction orother dissipative effects.
It is a limitation that applies to bothidealized and the actual heat engines.
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20
Example 1 at the beginning of thenotes leads to the concept ofRefrigerator and Heat Pump..
Heat can not be transferred from lowtemperature body to high temperature oneexcept with special devices.
These devices are called Refrigerators andHeat Pumps
Heat pumps and refrigerators differ in
their intended use. They work the same. They are characterized by the following:
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Refrigerators
21High-temperature Reservoir at TH
QHW
RefQL = QH - W
Low-temperature Reservoir at TL
QL
Objective
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An example of a Refrigerator and a Heat pump ..
22
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Coefficient of Performance of a
RefrigeratorThe efficiency of a refrigerator is expressed in term of
the coefficient of performance (COPR).
Desired output
,
1
1
L L
Hnet in H L
L
Q Q
QW Q Q
Q
= = =
23
Required inputR
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Heat Pumps24
High-temperature Reservoir at TH
QH
Objective
Low-temperature Reservoir at TL
QL
WHP
H=
L Read to parts ofpp 259 and 260
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Heat Pump
25
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Coefficient of Performance of a
Heat PumpThe efficiency of a heat pump is expressed in term of the
coefficient of performance (COPHP).
Desired out ut
,
1
1
H H
Lnet in H L
H
Q Q
QW Q QQ
= = =
26
Required inputHP=
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27
Relationship between Coefficient of
Performance of a Refrigerator(COPR)
and a Heat Pump (COPHP).
,net in LH HHP
W QQ QCOP
+
= = =
,
1net in L
HP R
H L H L
W Q
COP COP Q Q Q Q= + = +
1P RCOP COP = +
,net in H L H L
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28
The second Law of Thermodynamics:
Clausius Statement
The Clausius statementis
expressed as follows:
It is impossible to construct a
ev ce a opera es n a cyc eand produces no effect other
than the transfer of heat from
a lower-temperature body to a
higher-temperature body.
Both statements are negative
statements!
Read pp 262
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Equivalence of the Two Statements29
High-temperature Reservoir at TH
QH + QLQH
Net QOUT = QL
Low-temperature Reservoir at TL
QL
W = QH
RefHE
Net QIN = QL
HE + Ref
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30
Example (5-5): Heating a House by a Heat Pump
A heat pump is used to meet the heating requirements ofa 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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Perpetual Motion Machines31
Any device that violates the first or second
law is called a perpetual motion machine
If it violates the first law, it is a perpetualmo on mac ne o e rs ype 1
If it violates the second law, it is a perpetualmotion machine of the second type (PMM2)
Perpetual Motion Machines are not possible
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32 The second law of thermodynamics state
that no heat engine can have an efficiencyof 100%.
Then one may ask, what is the highestefficiency that a heat engine can possibly
have.
Before we answer this question, we need
to define an idealized process first, whichis called the reversible process.
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