ch 6a 2nd law

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1 Second Law of Thermodynamics 6 CHAPTER

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1

Second Law of

Thermodynamics

6CHAPTER

2

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 law as long

as the energy lost by air is

the same as the energy

gained by the coffee.Room at

25° C

!?

Example 1

3

The amount of EE is

equal to the amount of

energy transferred to

the room.

Example 2

4

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.

5

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.

6

• Some obvious examples

are solar energy, oil

furnace, atmosphere,

lakes, and oceans

• Another

example is two-

phase systems,

• and even the air in a room if the heat added or absorbed is small compared to the air thermal capacity (e.g. TV heat in a room).

Air

Thermal Pollution- disrupt marine life!

7

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 law as long as the heat added to the water is the same as the work gained by the shaft.

Heat Engines

8

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:

9

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

10

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

11

Steam power plant is another

example of a heat engine..

12

Thermal efficiency

Thermal Efficiency

< 100 %

in

out

Q

Q1

input Required

output DesiredePerformanc

in

out,net

thQ

W

in

outin

Q

QQ

13

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

14

thermal efficiency can not

reach 100%

Even the Most Efficient Heat

Engines Reject Most Heat as

Waste Heat

400.4

100th

Automobile Engine 20%

Diesel Engine 30%

Gas Turbine 30%

Steam Power Plant 40%

15

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 6-1: Net Power Production of a Heat

Engine

6.2. Statements of the

Second Law of

Thermodynamics

16

17

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.

18

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.

19

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:

20

High-temperature Reservoir at TH

Low-temperature Reservoir at TL

QH

QL

W

RefQL = QH - W

Objective

Refrigerators

21

An example of a Refrigerator

and a Heat pump ..

22

Coefficient of Performance of a

RefrigeratorThe efficiency of a refrigerator is expressed in term of

the coefficient of performance (COPR).

Desired output

Required inputRCOP

,

1

1

L L

Hnet in H L

L

Q Q

QW Q Q

Q

23

Heat Pumps

High-temperature Reservoir at TH

Low-temperature Reservoir at TL

QH

QL

W

HP

QH = W + QL

Objective

24

Heat Pump

25

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 Q

Q

26

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 RCOP COP

27

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!

28

High-temperature Reservoir at TH

Low-temperature Reservoir at TL

QH + QL

QL

W = QH

RefHE

QH

Net QIN = QL

Net QOUT = QL

HE + Ref

Equivalence of the Two

Statements

Consider the HE-RF combination shown below

29

Example (6-2): Heating a House by a Heat Pump

A heat pump is used to meet the heating requirements of

a 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:

30

6.3. 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 motion machine of the first type (PMM1)

If it violates the second law, it is a perpetual motion machine of the second type (PMM2)

Perpetual Motion Machines are not possible

31

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.