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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


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


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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