Files 2-Lectures Ch05a 2nd Law

8/11/2019 Files 2-Lectures Ch05a 2nd Law

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Second Low of

Thermodynamics


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Heat always flows from

high temperature to lowtemperature.

So, a cup of hot coffee

does not get hotter in acooler 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.

Example 1


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The amount of EE is

equal to the amount ofenergy 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 thedirection of a process.

Therefore we need another law (thesecond law of thermodynamics) todetermine the direction of a process.


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Thermal Energy Reservoir

If it supplies heat then itis called a source.

It is defined as a body to which and fromwhich heat can be transferred without achange in its temperature.

If it absorbs heat then itis called a sink.


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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).


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We all know that doing work on the water willgenerate heat.

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 thework gained by the shaft.

Heat Engines


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Previous example leads to the concept ofHeat Engine!.

We have seen that work always convertsdirectly and completely to heat, butconverting 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 thisheat to work.

They reject the remainingwaste heat to a low-temperature sink.

They operate on (athermodynamic) cycle.

High-temperature

Reservoir at TH

Low-temperature

Reservoir at TL

QH

QL

WHE


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Piston cylinder arrangement isan 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 thetransfer of heat from a high-temperature body to a low-temperature body.

A thermodynamic cycle involves a fluid to and from which heat istransferred 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 anotherexample of a heat engine..


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Thermal efficiency

Thermal Efficiency

< 100 %

in

out

Q

Q= 1

inputRequired

outputDesiredePerformanc =

==

in

out,net

thQ

Win

outin

QQQ


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Thermal efficiency

Thermal Efficiency

< 100 %1 L

H

Q

Q=

,net out

th

H

W

Q

= = H L

H

Q Q

Q

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


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thermal efficiency can not

reach 100%

Even the Most Efficient Heat

Engines Reject Most Heat asWaste Heat

Even the Most Efficient Heat

Engines Reject Most Heat asWaste 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=15kJ

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 toreject 85 kJ

A heat- engine cycle

cannot be completed

without rejectingsome 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.

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 thatoperates on a cycle to receive heat

from a single reservoir and producea 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 apower 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 ofthe notes leads to the concept of

Refrigeratorand 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

W

RefQL = 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 output

Required inputR

COP =

,

1

1

L L

Hnet in H L

L

Q Q

QW Q Q

Q

= = =


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Heat Pumps



QH

QL

WHP

QH

= W + QL

Objective

Read to parts ofpp 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 output

Required inputHPCOP =

,

1

1

H H

Lnet in H L

H

Q Q

QW 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 QCOP

W Q Q Q Q

+

= = =

,1

net in LHP R

H L H L

W QCOP COP

Q Q Q Q= + = +

1HP R

COP 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 cycleand 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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QH + QL

QL

W = QH

RefHE

QH

Net QIN = QL

Net QOUT = QL

HE + Ref

Equivalence of the Two

Statements


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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 Machines

Any device that violates the first or second law is

called a perpetual motion machine

If it violates the first law, it is a perpetual motionmachine of the first type (PMM1)

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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The second law of thermodynamics statethat 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 iscalled the reversible process.


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