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1
Second Law of Thermodynamics
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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 low 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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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 low 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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Piston cylinder arrangement is an example of a heat engine..
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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
1
input Requiredoutput DesiredePerformanc
in
out,netth Q
W
in
outin
QQQ
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Thermal efficiency
Thermal Efficiency
< 100 %1 L
H
,net outth
H
WQ
H L
H
Q QQ
QH= magnitude of heat transfer between the cycledevice and the H-T medium at temperature TH
QL= magnitude of heat transfer between the cycledevice 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 0.4100th
Automobile Engine 20%Diesel Engine 30%Gas Turbine 30%Steam Power Plant 40%
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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? I 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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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 5-1: Net Power Production of a Heat Engine
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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:
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High-temperature Reservoir at TH
Low-temperature Reservoir at TL
QH
QL
WRefQL = 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 RefrigeratorThe efficiency of a refrigerator is expressed in term of the coefficient of performance (COPR).
Desired outputRequired inputRCOP
,
1
1
L L
Hnet in H L
L
Q QQW Q QQ
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Heat Pumps
High-temperature Reservoir at TH
Low-temperature Reservoir at TL
QH
QL
WHP
QH = W + QL
Objective
Read to parts of pp 259 and 260
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Heat Pump
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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 outputRequired inputHPCOP
,
1
1
H H
Lnet in H L
H
Q QQW Q QQ
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Relationship between Coefficient of Performance of a Refrigerator (COPR) and a Heat Pump (COPHP).
,
,
net in LH HHP
net in H L H L
W QQ QCOPW Q Q Q Q
, 1net in LHP R
H L H L
W QCOP COPQ 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!Read pp 262
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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.
Carnot’s PrincipleNo heat engine can be more efficient than a
reversible heat engine when both engines work between the same pair of temperature H
and C.Isothermal Process: the temperature of the
system and the surroundings remain constant at all times. (q=-w)
Adiabatic: a process in which no energy as heat flows into or out of the system. (∆U=w)
Carnot cycle
four stage reversible sequence consisting of 1 .isothermal expansion at high temperature T2 2 .adiabatic expansion 3 .isothermal compression at low temperature T1 4 .adiabatic compression
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The Carnot Cycle
Process 1-2:Reversible isothermal heat addition at high temperature, TH > TL, to the working fluid in a piston-cylinder device that does some boundary work.
Process 2-3:Reversible adiabatic expansion during which the system does work as the working fluid temperature decreases from TH to TL.
Process 3-4:The system is brought in contact with a heat reservoir at TL < TH and a reversible isothermal heat exchange takes place while work of compression is done on the system.
Process 4-1:A reversible adiabatic compression process increases the working fluid temperature from TL to TH
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You may have observed that power cycles operate in the clockwise direction when plotted on a process diagram. The Carnot cycle may be reversed, in which it operates as a refrigerator. The refrigeration cycle operates in the counterclockwise direction.
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Carnot Principles
The second law of thermodynamics puts limits on the operation of cyclic devices as expressed by the Kelvin-Planck and Clausius statements. A heat engine cannot operate by exchanging heat with a single heat reservoir, and a refrigerator cannot operate without net work input from an external source.
Consider heat engines operating between two fixed temperature reservoirs at TH > TL. We draw two conclusions about the thermal efficiency of reversible and irreversible heat engines, known as the Carnot principles.
(a)The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs.
th th Carnot ,
(b) The efficiencies of all reversible heat engines operating between the same two constant-temperature heat reservoirs have the same efficiency.
As the result of the above, Lord Kelvin in 1848 used energy as a thermodynamic property to define temperature and devised a temperature scale that is independent of the thermodynamic substance.
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The following is Lord Kelvin's Carnot heat engine arrangement.
Since the thermal efficiency in general is
thL
H
1
For the Carnot engine, this can be written as
th L H L Hg T T f T T ( , ) ( , )1
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Considering engines A, B, and CQQ
1
3
1
2
2
3
This looks like
f T T f T T f T T( , ) ( , ) ( , )1 3 1 2 2 3
One way to define the f function is
f T T TT
TT
TT
( , ) ( )( )
( )( )
( )( )1 3
2
1
3
2
3
1
The simplest form of is the absolute temperature itself.
f T T TT
( , )1 33
1
The Carnot thermal efficiency becomes
th revL
H
TT, 1
This is the maximum possible efficiency of a heat engine operating between two heat reservoirs at temperatures TH and TL. Note that the temperatures are absolute temperatures.
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These statements form the basis for establishing an absolute temperature scale, also called the Kelvin scale, related to the heat transfers between a reversible device and the high- and low-temperature heat reservoirs by
TT
L
H
L
H
Then the QH/QL ratio can be replaced by TH/TL for reversible devices, where TH and TL are the absolute temperatures of the high- and low-temperature heat reservoirs, respectively. This result is only valid for heat exchange across a heat engine operating between two constant temperature heat reservoirs. These results do not apply when the heat exchange is occurring with heat sources and sinks that do not have constant temperature.
The thermal efficiencies of actual and reversible heat engines operating between the same temperature limits compare as follows:
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Reversed Carnot Device Coefficient of Performance
If the Carnot device is caused to operate in the reversed cycle, the reversible heat pump is created. The COP of reversible refrigerators and heat pumps are given in a similar manner to that of the Carnot heat engine as
COP QQ Q Q
QT
T T TT
RL
H L H
L
L
H L H
L
1
1
1
1
COP QQ Q
TT T
TT
TT
HPH
H L
H
L
H
L
H
H L
H
L
H
L
1
1
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Again, these are the maximum possible COPs for a refrigerator or a heat pump operating between the temperature limits of TH and TL.
The coefficients of performance of actual and reversible (such as Carnot) refrigerators operating between the same temperature limits compare as follows:
A similar relation can be obtained for heat pumps by replacing all values of COPR by COPHP in the above relation. Example 6-2
A Carnot heat engine receives 500 kJ of heat per cycle from a high-temperature heat reservoir at 652oC and rejects heat to a low-temperature heat reservoir at 30oC. Determine
(a) The thermal efficiency of this Carnot engine.(b) The amount of heat rejected to the low-temperature heat reservoir.
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a.
th revL
H
TT
KK
or
,
( )( )
. .
1
1 30 273652 273
0 672 67 2%
TT
KK
Q kJkJ
L
H
L
H
L
( )( )
.
( . )
30 273652 273
0 328
500 0 328164
b.
QL
WOUT
QH
TH = 652oC
TL = 30oC
HE
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Example 6-3
An inventor claims to have invented a heat engine that develops a thermal efficiency of 80 percent when operating between two heat reservoirs at 1000 K and 300 K. Evaluate his claim.
th revL
H
TT
KK
or
,
.
1
1 3001000
0 70 70%QL
WOUT
QH
TH = 1000 K
TL = 300 K
HE
The claim is false since no heat engine may be more efficient than a Carnot engine operating between the heat reservoirs.
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Example 6-4
An inventor claims to have developed a refrigerator that maintains the refrigerated space at 2oC while operating in a room where the temperature is 25oC and has a COP of 13.5. Is there any truth to his claim?
QL
Win
QH
TH = 25oC
TL = 2oC
R
COP QQ Q
TT T
KK
RL
H L
L
H L
( )( ).
2 27325 2
1196
The claim is false since no refrigerator may have a COP larger than the COP for the reversed Carnot device.
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Example 6-5
A heat pump is to be used to heat a building during the winter. The building is to be maintained at 21oC at all times. The building is estimated to be losing heat at a rate of 135,000 kJ/h when the outside temperature drops to -5oC. Determine the minimum power required to drive the heat pump unit for this outside temperature.
21 oC
HP-5 oC
QLost
Win
QL
The heat lost by the building has to be supplied by the heat pump.
QH
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Q Q kJhH Lost 135000
COP QQ Q
TT T
KK
HPH
H L
H
H L
( )( ( ))
.
21 27321 5
1131
Using the basic definition of the COP
COPQ
W
W QCOP
kJ h hs
kWkJ s
kW
HPH
net in
net inH
HP
, /. /
.
,
,
135 0001131
13600
1
3316
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