thermo chap v
TRANSCRIPT
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Before we formulate the second law we need to clear few concepts.
THERMAL ENERGY RESERVOIRS
These are Bodies or Systems , which have large thermal energies.
THESE CAN ABSORB OR SUPPLY LARGE ANMOUNTS OF HEAT
WITHOUT ANY TEMPERATURE CHANGE WITHIN THEIR SYSTEM.Examples are
A Thermal Reservoir which absorbs a large amount of heat is called a
HEAT SINK
A Thermal Reservoir which SUPPLIES a large amount of heat is called a
HEAT SOURCE
We now look at the concept of Heat Engines
We had seen earlier that it is possible to convert one form of energy to another
form.
It is possible to convert Work into Heat : Friction
It is however with difficulty that we are able to convert Heat into Work. We can
do so by using specialized devices called as HEAT ENGINES.
HEAT ENGINES
a. Receive heat from a High Temperature Source
b. Convert part of Heat into work
c. Reject part of Heat to a Low temperature Sink
d. Operate in a cycle.
We thus represent the Heat Engine in the following manner.
ATMOSPHERE
SEA
CAN ABSORB LARGE AMOUNT OF HEATTRANSFER WITHOUT ANY TEMPERATURECHANGE.
IndustrialFurnace
Sun
CAN SUPPLY A LOT OF HEAT WITHOUT ANYTEMPERATURE CHANGE.
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The best example of a Heat Engine is a Steam Power Plant.
Here QNET = Qin Qout = WNET = Wout - Win
Sometimes we use the term Heat Engine for devices which do not operate in
Thermodynamic cycle. For example a petrol engine
Work
Qin
Qout
Heat Engines usually have a fluidfrom which heat either taken orremoved during the cyclic process.This is called the WORKING FLUID
BOILER
CONDENSOR
PUMPTURBINE
Qin
Qout
WORKIN
Cool airand fuel
Hot gases inexhaust
Combustionin cylinder
The working fluid does nothave the same propertiesas the inlet after the end ofprocess
WORK
OUT
Work toFlywheel
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Another such cycle is the jet engine
In fact all devices which use internal combustion are not satisfying the strict
definition of Heat engines
The Steam Power plant has four individual components ie
BOILER, TURBINE , CONDENSOR AND PUMP. All are open systems and do
not operate in a cycle individually.
But when they are combined together they form a closed system as no mass
enters or leaves the system.
So for a closed system which operates in a cycle
NET NET
in out out in
Q W or
Q Q W W
=
=
For a Heat Engine WNet is a very important commodity.
THERMAL EFFICIENCY
We now look at some means of determining the performance of Heat Engines.
In a Heat Engine Qout is the energy which is not utilized and is rejected. But
Qout is necessary to get a cycle , so our effort is to minimize it.
Now performance of a Heat Engine is determined by its ability to reduce this
Qout as mush as possible. So THERMAL EFFRICIENCY is very much related
to the quantity of Heat rejected.
Cool air
Compressor
The working fluid does not have the same properties as the inlet afterthe end of process. Here also the cycle is not achieved.
CombustionChamber
Turbine
Work given tocompressor Fuel flow to
C.C
Shaft Workfrom Turbine
Hotgases
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In general terms the performance of any system is given by
Re
Desired Actual OutputPerformance
quired Input=
For Heat Engines Performance Criteria is Thermal Efficiency
1
Net
in
Net in out out Th
in in in
WNet Work outThermal Efficiency or
Heat put in Q
W Q Q Q
Q Q Q
= =
= = =
Thus the higher the value of Qout the lower the value ofTh
generally for cyclic engines
Qin=QH HEAT FLOW FROM HIGH TEMP RESERVOIR TO CYCLE
and Qout = QL HEAT FLOW FROM CYCLE TO LOW TEMP RESERVOIR
1Net LThin H
W Q
Q Q = = Thermal Efficiency is always less than 1.
For IC Automobile Engines using Petrol it is equal to 25 to 30 %For Diesel engines it is equal to 30 35 %
Gas Turbine and Steam Power Plants it is equal to 45 50 %
Our effort is to keep QL as low as possible. We still need to have it to have a
proper cycle. We now see a few examples.
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SECOND LAW OF THERMODYNAMICS
We have now understood the concept of Energy Reservoirs and Heat Engines
We will now try to understand the Second Law by using these concepts.
There are two ways of expressing the Second Law , we will at this moment
express it in relation to Heat Engines.KELVIN-PLANCK STATEMENT
IT IS IMPOSSIBLE FOR ANY DEVICE THAT OPERATES IN A CYCLE , TO
RECEIVE HEAT FROM A SINGLE RESERVOIR AND PRODUCE NET
AMOUNT OF WORK
This can be represented as
Since efficiency is paramount for devices which convert one form of energy toanother so we will now look at some other devices which do so. We know now
.Net
th
in
WFor Heat Engines Let us see other devices
Q =
A. Water Heater. Converts Electrical Work into Heat. Here however
some heat is lost in the pipes so the heaters efficiency is defined as
Heat given to WaterEfficiencyEnergy given by heater
=
Work
QH
NOT POSSIBLE
This implies that it is absolutelymandatory for any Heat engine to rejectheat.
It also implies that NO HEAT ENGINECAN HAVE EFFICIENCY OF 1This shall be seen later.
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B. Combustion efficiency. A combustion device converts Chemical
energy into heat energy. The amount of Heat produced by the fuel
depends upon the Heating Value of Fuel, which is defined as
Amount of Heat released when unit mass of fuel is completely burnt
at Room Temperature and the products are then cooled to RoomTemperature Now this definition can be represented as
We have two values of Heating Value of Fuel
Higher Heating Value of Fuel.(HHV) when H2O in the products is in liquid
state. This has a high value because Heat is given off as H2O condenses.
Lower Heating Value of Fuel.(LHV) when H2O in the products is in vapor
state.So now the combustion efficiency is defined as
Heat Released
Heating ValueCombustion
=
Electrical Power Output.
Mechanical Power InputC Generator Efficiency =
Mechanical Power Output.
Electrical Power InputD Motor Efficiency =
Amount of Light given in LUMENS.
Amount of Energy consumed in WATTSE Lighting Efficiency=
CombustionChamber
One kg ofFuel at
Room Temp.
Air at RoomTemp.
Combustion products at RoomTemp.
GenerallyRoomTemperatureis taken as25oC
HHV or LHV
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Sometimes we have several devices linked together so we define what is
termed as Overall Efficiency = Product of individual efficiencies
So if we have the following arrangement
( ) ( ) ( )
*
*
Overall Comb Th Generator
eH H L
H H LFuel
e
Fuel
X X
WQ Q QX X
Q Q QHHV m
W
HHV m
=
=
=
We will now look at two other energy conversion devices which is used a
lot in our environment. These are referred to as REFRIGERATORS AND
HEAT PUMPS.
REFRIGERATORS AND HEAT PUMPS
Now we do understand that heat flows high temperatures to low
temperatures naturally. What if we want to transfer heat from low
temperature to a high temperature. Now this cannot be done naturally , but
it is possible to develop a device which can do this by specific means. So
lets look at a device which also operates in a cycle but operates in a
manner totally opposite to the Heat Engine
HeatEngine
Combustion Chamber
Generator
Fuel and Air
QH
QL
Work Electrical Work
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In fact a refrigeration process is a combination of four components which
operate together as a cycle. Lets look at one
The processes are
a. Condenser. ( 1-2)Change of phase from compressed vapor to
compressed liquid.
b. Expansion valve. (2-3)Pressure reduction to give Low pressure and
Low Temperature Refrigerant
c. Evaporator. (3-4)Isothermal Change of phase from Liquid to Vapor
d. Compressor. (4-1)Compression to High Pressure vapor.
this goes through a cycle and can be shown to be reverse cycle of a Heat
Engine as
Low Temperature Reservoir
High Temperature Reservoir
QL
QH
Win
Device
If the device is required to removeheat from Low Temperature Device(Q
L) , then it is called
REFRIGERATOR
If the device is required to give heatto High Temperature Device(Q
H) , then it is called HEAT PUMP
Condensor
Evaporator
EXPANSION VALVECOMPRESSOR
QH Heat Given toHigh Temperature
SHAFTWORKIN
QL
Heat removed from
Low Temperature
800 Kpa30oC
120 Kpa-25oC
800 Kpa60oC
120 Kpa-25oC
12
3
4
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CO-EFFICIENT OF PERFORMANCE
The performance of Refrigerators and heat pumps is assessed by an index
called as Co-efficient of Performance (COP). This is defined as
Ref
HP
1
Re1
1
1
L L
Hin H L
L
H H
Hin H L
L
Q QDesired outputCOP so COP
Qquired input W Q Q
Q
Q QCOP
QW Q Q
Q
= = = =
= = =
COP can be greater than 1 depending upon the value of Win
Ref
Ref1
H LHP
H LHP
Q QNow COP and COP
W W
Q Qso COP COP for the same device
W
= =
= =
Air Conditioners are also a type of refrigerator where the Low Temperature
is a room or a building.
An A/C can be used as a Heat Pump if we reverse it.
Air Conditioners and refrigerators are also evaluated by another index
called as ENERGY EFFICIENCY RATING. (EER) where
( ) Ref
( ' )1 3.412
1 3.412
LQ In BTU s
EER now Wh BTU
Watt Hour of electrical energy consumedso if COP then EER COP
= =
= =
QH
QL
Work
If QH
is important then we call it
Heat Pump.
If QL
is important then we call it
Refrigerator
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Most A/Cs have EER of 8-12. Also generally EER and COP of
refrigerators decrease with decreasing TL. So freezers have Low COP.
CLAUSIUS STATEMENT OF SECOND LAW
This statement relates to Heat Pumps and Refrigerators. The statement isIT IS IMPOSSIBLE TO CONSTRUCT A DEVICE THAT OPERATES IN A
CYCLE , AND PRODUCES NO EFFECT OTHER THAN THE TRANSFER
OF HEAT FROM A LOW TEMPERATURE BODY TO A HIGH
TEMPERATURE BODY.
Both Statements of Second Law govern the conversion of energies.
EQUIVALENCE OF TWO STATEMENTS
Both Statements are equivalent and either of them can be used to express
the Implications of Second Law.
Supposing we have a combination of Heat Engine and Refrigerator
operating between the same two reservoirs. And supposing the heat
engine violates the second law
QH
QL
To do this we need to put energy into
the cyclic device , other than heat.
It is only possible to transfer heatfrom HTR to LTR naturally.
The reverse requires input of Work
High TEMPERATURE
RESERVOIR
Low TEMPERATURERESERVOIR
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The combination will look like this
We had first evolved the PMM1 from First law . And now we have PMM2
from Second law
A device which behaves like PMM1 or PMM2 is not possible
Any device must satisfy First and Second Laws completely.
Read Pages 271-273
( QH
+ QL
)
QL
High TEMPERATURE RESERVOIR
Low TEMPERATURE RESERVOIR
QH
Work = QH
Heat Engine
Violates theSecond Law
QL
QL
The combination is a violation of theClausius Statement.
THUS A VIOLATION OF K-PSTATEMENT IS A VIOLATION OFCLAUSIUS STATEMENT and
Vice Versa
High TEMPERATURERESERVOIR
Low TEMPERATURERESERVOIR
W
PMM1 QH
W
PMM2
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REVERSIBLE AND IRREVERSIBLE PROCESSES
Second Law states that no Heat Engine can have 100 % efficiency. Well if
this is not possible , then what could be the highest efficiency which can be
achieved. To answer this we need to study the concept of REVERSIBLE
and IRREVERSIBLE Processes.Lets look at a few process;-
a. Cooling of water. The water will lose heat to air. We cannot force the
heat to go back to the water. In fact we will have to use some energy
source to heat the water back.
b. Stopping of a car by applying the brakes. When we apply brakes , the
friction generated by the brake pads cause the car to stop. In the
process the brakes get heated. Now we cannot apply heat to cold
brakes and expect the wheel to rotate. We will have to apply some other
energy source to rotate the wheel.
Now this means that to reverse the process some external energy has to be
put in. This energy has to come from surroundings. So what could be a
reversible process. Lets first define it.
REVERSIBLE PROCESS IS A PROCESS THAT CAN BE REVERSED
WITHOUT LEAVING ANY TRACE ON THE SURROUNDINGS.
This is only possible, if the process retraces its path when the process
reverses.
An irreversible process on the other hand will not reverse along the same pathto reach its original state , and hence will require energy or give energy tocome back to its original state. Here ENET is not equal to 0
These will be shown as
P
VE
out
Ein
Because the process goes from 1 to 2 and
then reverses along the same path to gofrom 2-1, so ENET
= 0
2
1
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You must understand that REVERSIBLE PROCESS is an ideal concept , and
does not occur in nature , but they are the standard to which we compare the
real processes in nature.
IRREVERSIBILITIES reduce the efficiency of devices , hence to achieve
better design and performance we need to lower the factors which cause
irreversibilities. Lets investigate the factors which cause irreversibilities.They
can be several. Some of the major ones are :-
a. FRICTION
b. UNRESTRAINED EXPANSION OR COMPRESSION
c. MIXING OF TWO GASES
d. HEAT TRANSFER ACROSS A FINITE TEMPERATURE
DIFFERENCE
e. ELECTRICAL RESISTANCE
f. INELASTIC DEFORMATION
g. CHEMICAL REACTIONS
FRICTION occurs when bodies are in motion. If a body moves in one
direction, a part of the kinetic energy is converted into heat by friction. The
heat generated goes into the surrounding; hence the process is irreversiblebecause the heat cannot be taken back.
We try to reduce friction as much as possible , so as to reduce irreversibility
but we cannot eliminate friction.
Sometimes we even need friction to do some work. When we walk it is friction
which allows us to do so. It is very difficult to walk on slippery surfaces. A car
moves because of friction between the tires and road surface.
P
V
The process goes from 1 to 2 and thencomes back to original state by twodifferent paths. The reverse path dependsupon the degree of irreversibility presentin the process. Here E
NETwill more be 0
2
1
AB
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UNRESTRAINED EXPANSION AND FAST COMPRESSION
We have earlier seen the concept of QUASI-STATIC PROCESS. ( A process
which is so slow that property change is uniform in the system)
Lets look at a Piston-Cylinder arrangement
Now if we have unrestrained expansion , then that is also IRREVERSIBLE
FAST PROCESSES ARE HIGHLY IRREVERSIBLE
SLOW PROCESSES ARE LESS IRREVERSIBLE
However because of human nature to do processes in short time we tend to
do most processes fast , and cause irreversibilities.
Frictionless
Very Slow Frictionless movement of Pistoninwards will keep the properties uniform
Very Fast Frictionless movement of Piston inwardswill cause a pressure buildup ( Compression Wave),at the face of the piston. This pressure wave willoppose movement , and more work has to be doneto move the piston.
If we move the piston in slowly , a Compression wave is not set up and lesswork will be required to move the piston down.
Now if we move the piston outwards in a fast manner , then pressure atpiston face reduces very fast , and we will get less PdV is given to thesystem. A slow movement will give more work.
Gas Vacuum
Membrane
Now if the membrane is removed ,the gas will fill the vacuumquickly. No work is given out ofthe system.
So to reverse the process wehave to put work into system tocompress the gas.
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HEAT TRANSFER
Whenever we have Heat Transfer , across a finite temperature difference, we
will have an Irreversible Process.
ALL HEAT TRANSFER PROCESSES ARE IRREVERSIBLE
ELECTRIC RESISTANCE WORK
INELASTIC DEFORMATION
CHEMICAL REACTIONS
System at5 oC
Surroundingat 25 oC
Q
Heat will go from system to thesurroundings.
System at
5o
C
Surroundingat 25 oC If we want to reverse the
process then we willhave to refrigerate , thusgetting external work
from the surroundings.The result will be energyloss of thesurroundings.
Refrigerator
QL
QH
Work
+_
QAs the current flows through the wire , heatis generated and goes to the surroundings.We cannot put heat into the system andexpect the wire to generate electricity.
Once the bar has deformed , then we
need some external force to bring itback to its original shape.
A BA+B
PRODUCTSTo reverse and get Aand B we have to getwork from surroundings
REACTANTS
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INTERNALLY AND EXTERNALLY REVERSIBLE PROCESS
A reversible process is an ideal conception. It is the best process between two
states.
Now if a process is reversible then it is called INTERNALLY REVERSIBLEif there are no irreversibilities within the system. Here the forward path of the
process is the same as the reverse path.
Now if a process is such that there may be irreversibility in the system , but
there will be no irreversibility outside the system then this process is called
EXTERNALLY REVERSIBLE
Such is possible when Heat Transfer between system and surrounding is at
same temperatureA TOTALLY REVERSIBLE PROCESS IS ONE WHICH HAS NO
IRREVERSIBILITY EITHER IN THE SYSTEM OR IN THE
SURROUNDINGS.
We generally call such a process a Reversible process.
For such a process there is no heat transfer, friction, Non quasi
process, Chemical reaction , Inelastic deformation etc.
System with noirreversibility P
V
2
1
System withirreversibility
No irreversibility in the surroundings
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CARNOT CYCLE
A cycle which is totally made up of reversible processes, is called a Reversible
Cycle . This is an ideal concept , but it is the best cycle possible. We will now
look at a special cycle which is totally reversible ,called as CARNOT CYCLE.
A Carnot cycle involves four processes,a. Reversible Isothermal Expansion
b. Reversible Adiabatic Expansion
c. Reversible Isothermal Compression
d. Reversible Adiabatic Compression
If we use an ideal gas with a piston cylinder device then it is executed as
We can show it on P V as well as T V diagram
HEAT
SOURCE
QH T
His
Const.
212 3Q
inis
ZeroReversible IsothermalExpansion. St 1 to 2
Reversible AdiabaticExpansion St 2 to 3
HEAT
SINK
QL
TLis
Const.
34 1 4Q
OUTis
ZeroReversible IsothermalCompression. St 3 to 4
Reversible AdiabaticCompression St 4 to 1
3
4
2
1P
V
TH
TLQ = 0
Q = 0
2
34
1T
V
QL
WNet
QH
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This cycle can also be done in an open system as long as we have the four
essential processes.
Such a cycle whether closed or open is called a CARNOT HEAT ENGINE
because it produces work from heat input.
REVERSED CARNOT CYCLE
Since a Carnot Cycle is made up of 4 Reversible processes , so it can be
reversed. And we have
This cycle is called a REVERSED CARNOT CYCLE. This can also be
achieved by both open and closed systems.
BOILER
CONDENSOR
TURBINEPUMP
21
4
3
QH
at TH
QL at TL
Q2-3
is zeroQ4-1
is zero
3
4
2
1P
V
TH
TLQ = 0
Q = 0
23
41
T
V
QL
WNet
QH
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THE CARNOT PRINCIPLES
We have seen the KELVIN-PLANCK and CLAUSIUS STATEMENTS , as
they govern Heat engines ,Refrigerators/Heat Pumps. We can now make two
strong conclusions from these statements:-
A. The efficiency of an Irreversible Heat Engine is always less then theefficiency of a Reversible Heat Engine operating between the same
two reservoirs.
B. The efficiencies of all Reversible Heat Engines operating between the
same two reservoirs are the same.
THIS CAN BE PROVED BY SHOWING THAT THE VIOLATION OF EITHER
STATEMENT IS VIOLATION OF SECOND LAW.
Let us say that we have an IRR Heat Engine which has a higher efficiency
than a REV heat engine. Both operate between the same reservoirs and take
the same amount of QH ,
IRR REV
IRR REV
IRR REV
L L
Now if
then W W
and Q Q
>
>
. Then A BA B L LW W and Q Q>
Most engines have an efficiency below 40 %. So if we have an engine
operating between 750oK and 300oK the carnot efficiency will be 0.6 = 60%
So the engine is fairly close to 40 %.
FOR A CARNOT ENGINE TH
INCREASES IF TH INCREASES and also INCREASES IF TL DECREASES
So we make an effort in Actual engines to have operation betweenHighest Possible temp. Source( Limited by Material Strength)
Lowest possible temp Sink ( Limited by the cooling medium)
IRRA IRR
B
QH
= 1000 W
QL= 900W
WA
= 100 WW
B= 180 W
QH
= 800 W
QL
=620 W
HIGH TEMPERATURE RESERVOIR at TH
= 400 oK
LOW TEMPERATURE RESERVOIR at TL
= 300 oK
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CARNOT REFRIGERATOR AND HEAT PUMP
Based on our definition of Absolute Temperature we can say that
Re .
1 1 1 1
1 1 1 1
L Hf H P
H H L LH L H L
L L H H
Q QCOP and COP
Q T Q T Q Q Q Q
Q T Q T
= = = = = =
These are the highest COP of Refrigerator/heat Pump operating between two
temperature limits . So we can say that
Re
Re
Re
1Re
1
1
1
1
1
f
H
L
fH
L
f
H
L
COP versible refrigerator T
T
COP Irreversible refrigerator T
T
COP Not possibleT
T
=
1Re
1
1
1
1
1
HP
L
H
HP
L
H
HP
L
H
COP versible Heat Pump
TT
COP Irreversible Heat PumpT
T
COP Not PossibleT
T
=
Lets look at a few problems.
Also read pages 291-294