Thermo Chap V

7/28/2019 Thermo Chap V

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2/26

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.


3/26

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


4/26

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


5/26

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


8/26

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


9/26

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


10/26

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


12/26

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


14/26

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


15/26

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.


16/26

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


17/26

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


18/26

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


19/26

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


20/26

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


26/26

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

Thermo Chap V

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