Chapter 6

2nd Law of Thermo-Dynamics

What is 1st Law of thermodynamics ?

Violation of 2nd Law

This discussion leads us to a conclusion that processes proceed in a certain direction and not in the reverse direction.

Draw back in the first law is that id does not place a restriction on the direction of a process, but satisfying the first law does not ensure that the process can actually occur.

The second law of thermodynamics addresses this inadequacy.

A process cannot occur unless it satisfies both the first and the second laws of thermodynamics.

5

The use of the second law of thermodynamics also tells us that energy has quality as well as quantity. The first law is concerned with the quantity of energy and the transformations of energy from one form to another with no regard to its quality. Preserving the quality of energy is a major concern to engineers, and the second law provides the necessary means to determine the quality as well as the degree of degradation of energy during a process.

Quality and Quantity

Some Basic Concepts

Thermal Energy (Heat) reservoir

A heat reservoir is a sufficiently large system in stable equilibrium to which and from which finite amounts of heat can be transferred without any (significant) change in its temperature.

A high temperature heat reservoir from which heat is transferred is sometimes called a heat source. A low temperature heat reservoir to which heat is transferred is sometimes called a heat sink.

i.e. large bodies of water such as oceans, lakes, and rivers as well as the atmospheric air can be modeled accurately as thermal energy reservoirs because of their large thermal energy storage capabilities or thermal masse

A two-phase system can be modeled as a reservoir also since it can absorb and release large quantities of heat while remaining at constant temperature.

Another familiar example of a thermal energy reservoir is the industrial furnace as it is kept at constant temperature.

Work reservoir

A work reservoir is a sufficiently large system in stable equilibrium to which and from which finite amounts of work can be transferred adiabatically without any change in its pressure.

8

Heat Engine

Consider arrangements shown in figure (a) & (b). The mechanical work done by the shaft, is first converted to the internal energy of the water. This energy may then leave the water as heat.

What if we reverse this process ?

Transferring heat to the water will cause the shaft to rotate ?

Surely NO.

So we can say that work can be converted to heat directly and completely, but converting heat to work requires the use of some special devices. These devices are called heat engines.

a b

Heat Engine

A heat engine is a thermodynamic system operating in a thermodynamic cycle to which net heat is transferred and from which net work is delivered.

The system, or working fluid, undergoes a series of processes that constitute the heat engine cycle.

The following figure illustrates a steam power plant as a heat engine operating in a thermodynamic cycle.

Heat engines and other cyclic devices usually involve a fluid to and from which heat is transferred while undergoing a cycle. This fluid is called the working fluid.

,net out out inW W W

,net out in outW Q Q

Basic characteristics of Heat Engine are:

• They receive heat from a high-temperature source (solar energy, oil furnace, nuclear reactor, etc.).

• They convert part of this heat to work (usually in the form of a rotating shaft).

• They reject the remaining waste heat to a low temperature sink (the atmosphere, rivers, etc.).

• They operate on a cycle.

Heat Engine

Thermal Efficiency

We know

Qout represents the magnitude of the energy wasted in order to complete the cycle. But Qout is never zero; thus, the net work output of a heat engine is always less than the amount of heat input. That is, only part of the heat transferred to the heat engine is converted to work. The fraction of the heat input that is converted to net work output is a measure of the performance of a heat engine and is called the thermal efficiency ηth .

,net out in outW Q Q

The thermal efficiency is the index of performance of a work-producing device or a heat engine and is defined by the ratio of the net work output (the desired result) to the heat input (the costs to obtain the desired result).

th Desired Result

Required Input

thnet out

in

W

Q ,

where

W W W

Q Q

net out out in

in net

,

Here the use of the in and out subscripts means to use the magnitude (take the positive value) of either the work or heat transfer and let the minus sign in the net expression take care of the direction.

Continue to next slide

For heat engines, the desired output is the net work output, and the required input is the amount of heat supplied to the working fluid. Then the thermal efficiency of a heat engine can be expressed as:

Network out put

Total Heat inputth

Q W U

W Q

W Q Q

net in net out

net out net in

net out in out

, ,

, ,

,

The cycle thermal efficiency may be written as

thnet out

in

in out

in

out

in

W

Q

Q Q

Q

Q

Q

,

1

Now apply the first law to the cyclic heat engine

The thermal efficiency of the above device becomes

thL

H

Q

Q 1

Cyclic devices such as heat engines, refrigerators, and heat pumps often operate between a high-temperature reservoir at temperature TH and a low-temperature reservoir at temperature TL.

15

Can we save Qout ?

Is it possible to transfer the 85 kJ of excess heat at 90°C back to the reservoir at 100°C for later use?

16

Second Law Statements

Kelvin-Planck statement of the second law

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.

The Kelvin-Planck statement of the second law of thermodynamics states that no heat engine can produce a net amount of work while exchanging heat with a single reservoir only. In other words, the maximum possible efficiency is less than 100 percent.

th < 100%

Heat engine that violates the Kelvin-Planck statement of the second law

17

Heat Pump

A heat pump is a thermodynamic system operating in a thermodynamic cycle that removes heat from a low-temperature body and delivers heat to a high-temperature body. To accomplish this energy transfer, the heat pump receives external energy in the form of work or heat from the surroundings.

While the name “heat pump” is the thermodynamic term used to describe a cyclic device that allows the transfer of heat energy from a low temperature to a higher temperature, we use the terms “refrigerator” and “heat pump” to apply to particular devices.

Refrigerator is a device that operates on a thermodynamic cycle and extracts heat from a low-temperature medium.

The heat pump also operates on a thermodynamic cycle but rejects heat to the high-temperature medium.

18

Refrigerator as a heat pump operating in a thermodynamic cycle.

19

For the heat pump acting like a refrigerator or an air conditioner, the primary function of the device is the transfer of heat from the low- temperature system.

For the refrigerator the desired result is the heat supplied at the low temperature ( heat removed from space) and the input is the net work into the device to make the cycle operate.

COPQ

WRL

net in

,

Coefficient of Performance, COP

The index of performance of a refrigerator or heat pump is expressed in terms of the coefficient of performance, COP, the ratio of desired result to input. This measure of performance may be larger than 1, and we want the COP to be as large as possible.

COP Desired Result

Required Input

20

Now apply the first law to the cyclic refrigerator.

( ) ( )

,

Q Q W U

W W Q QL H in cycle

in net in H L

0 0

and the coefficient of performance becomes

COPQ

Q QRL

H L

For the device acting like a “heat pump,” the primary function of the device is the transfer of heat to the high-temperature system. The coefficient of performance for a heat pump is

COPQ

W

Q

Q QHPH

net in

H

H L

,

Note, under the same operating conditions the COPHP and COPR are related by

COP COPHP R 1

21

Clausius statement of the second law

The Clausius statement of the second law states that 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.

Or energy from the surroundings in the form of work or heat has to be expended to force heat to flow from a low-temperature medium to a high-temperature medium.

Thus, the COP of a refrigerator or heat pump must be less than infinity.

COP

Heat pump that violates the Clausius statement of the 2nd law

COPQ

WRL

net in

,

22

Heat Pump and Air Conditioner Ratings

Heat pumps and air conditioners are rated using the SEER system. SEER is the Seasonal adjusted Energy Efficiency Rating. The SEER rating is the amount of heating (cooling) on a seasonal basis in Btu/hr per unit rate of power expended in watts, W.

The heat transfer rate is often given in terms of tons of heating or cooling. One ton equals 12,000 Btu/hr = 211 kJ/min.

23

The food compartment of a refrigerator is maintained at 4°C by removing heat from it at a rate of 360 kJ/min. If the required power input to the refrigerator is 2 kW, determine (a) the coefficient of performance of the refrigerator and (b) the rate of heat rejection to the room that houses the refrigerator.

(a) The coefficient of performance of the refrigerator is

That is, 3 kJ of heat is removed from the refrigerated space for each kJ ofwork supplied.

(b) The rate at which heat is rejected to the room that houses the refrigeratoris determined from the conservation of energy relation for cyclic devices,

We see that both the energy removed from the refrigerated space as heat and the energy supplied to the refrigerator as electrical work eventually show up in the room air and become part of the internal energy of the air. This demonstrates that energy can change from one form to another, can move from one place to another, but is never destroyed during a process.

24

Second Law Statements

Kelvin-Planck statement of the second law

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.

In other words, the maximum possible efficiency is less than 100 percent. th < 100%

The Clausius statement of the second law states that 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.

Clausius statement of the second law

Thus, the COP of a refrigerator or heat pump must be less than infinity.

COP

25

A violation of either the Kelvin-Planck or Clausius statements of the second law implies a violation of the other. Assume that the heat engine shown below is violating the Kelvin-Planck statement by absorbing heat from a single reservoir and producing an equal amount of work W. The output of the engine drives a heat pump that transfers an amount of heat QL from the low-temperature thermal reservoir and an amount of heat QH + QL to the high-temperature thermal reservoir. The combination of the heat engine and refrigerator in the figure acts like a heat pump that transfers heat QL from the low-temperature reservoir without any external energy input. This is a violation of the Clausius statement of the second law.

Equivalent

26

Perpetual-Motion Machines

Any device that violates the first or second law of thermodynamics is called a perpetual-motion machine. If the device violates the first law ( creating energy), it is a perpetual-motion machine of the First kind (PMM1). If the device violates the second law, it is a perpetual-motion machine of the second kind (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;-

•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.

•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.

28

Reversible Processes

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 path to reach its original state , and hence will require energy or give energy to come back to its original state. Here ΔENET is not equal to 0

Eout

Ein

Because the process goes from 1 to 2 and then reverses along the same path to go from 2-1, so ΔENET = 0

29

Irreversible Process

An irreversible process is a process that is not reversible. All real processes are irreversible. IRREVERSIBILITIES reduce the efficiency of

devices , hence to achieve better design and performance we need to lower the factors which cause irreversibilities : Following are some major factor

FrictionUnrestrained expansion of gases

Heat transfer through a finite temperature difference

Mixing of two different substancesHysteresis effects

I2R losses in electrical circuitsInelastic Deformation

Any deviation from a quasi-static process

A

B

30

31

32

Internally reversible process

The internally reversible process is a quasi-equilibrium process, which, once having taken place, can be reversed and in so doing leave no change in the system. This says nothing about what happens to the surroundings about the system. ( reversibilities occur only within boundaries of the system) Quasi equilibrium process

Externally reversible process

The externally reversible process is a process, which, once having taken place, can be reversed and in so doing leave no change on surroundings.

What is Totally Reversible ????

33

The Carnot Cycle

Scientists and researchers remained eager to produce the most efficient practical engine since ever. We now that heat engines operate in cycle i.e. the working fluid of a heat engine returns to its initial state at the end of each cycle.

Work is done by the working fluid during one part of the cycle and on the working fluid during another part.

The difference between these two is the net work delivered by the heat engine. The efficiency of a heat-engine therefore greatly depends on how the individual processes that make up the cycle are executed. The net work, thus the cycle efficiency, can be maximized by using processes that require the least amount of work and deliver the most, i.e. using reversible processes

34

The Carnot Cycle

French military engineer Nicolas Saadi Carnot (1769-1832) was among the first to study the principles of the second law of thermodynamics. Carnot was the first to introduce the concept of cyclic operation and devised a reversible cycle that is composed of four reversible processes, two isothermal and two adiabatic. His design looks like following on P-v & T-v diagram :

3

4

2

1

P

v

TH

TL Q = 0

Q = 0

2

34

1

T

v

QL

WNet

35

36

The Carnot Cycle

Process 1-2:Reversible isothermal heat addition at high temperature, TH > TL, to the working fluid in a piston-cylinder device that does some boundary work.

Process 2-3: Reversible adiabatic expansion during which the system does work as the working fluid temperature decreases from TH to TL.

3

4

2

1TH

3

4

2

1TH

Q = 0

P

v

37

The Carnot Cycle

Process 3-4: The system is brought in contact with a heat reservoir at TL < TH and a reversible isothermal heat exchange takes place while work of compression is done on the system.

Process 4-1:A reversible adiabatic compression process increases the working fluid temperature from TL to TH

3

4 2

1TH

TL Q = 0

3

4

2

1TH

TL

Q = 0

Q = 0

38

You may have observed that power cycles operate in the clockwise direction when plotted on a process diagram. The Carnot cycle may be reversed, in which it operates as a refrigerator. The refrigeration cycle operates in the counterclockwise direction.

39

We know that area under the process curve is equal to the boundary work. Network during Carnot Cycle is area enclosed between 1-2-3-4. Larger this area larger is the net work done.

Now assume we try to save the energy lost (QL) (removing condenser or heat sink) .

Then we will not encounter point 4 in the cycle. Hence , there will be no net work done.

40

Carnot Principles

The second law of thermodynamics puts limits on the operation of cyclic devices as expressed by the Kelvin-Planck and Clausius statements. A heat engine cannot operate by exchanging heat with a single heat reservoir, and a refrigerator cannot operate without net work input from an external source.

Consider heat engines operating between two fixed temperature reservoirs at TH > TL. We draw two conclusions about the thermal efficiency of reversible and irreversible heat engines, known as the Carnot principles.

(a) The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs.

th th Carnot ,

(b) The efficiencies of all reversible heat engines operating between thesame two reservoirs are the same

41

Most of our Temperature scales are based on some property of a substance like thermal expansion.A THERMODYNAMIC TEMPERATURE SCALE IS INDEPENDENT OF THE PROPERTIES OF THE SUBSTANCE. Such a scale has been derived from the Carnot Principles i.e. The Second One

Since the thermal efficiency in general is

thL

H

Q

Q 1

Here the efficiency is independent of the properties of the material used but is dependent upon the Heat transfer of each engine. Now the efficiencies of each engine will be the same because they operate between the same reservoirs.

( , ), ( , )Lth L H H L

H

Qg T T T T

Q f

THERMODYNAMIC TEMPERATURE SCALE

Since energy reservoirs are characterized by their temperatures, we can write:

42

Considering engines A, B, and C

Q

Q

Q

Q

Q

Q1

3

1

2

2

3

This looks like

f T T f T T f T T( , ) ( , ) ( , )1 3 1 2 2 3

One way to define the f function is

f T TT

T

T

T

T

T( , )

( )

( )

( )

( )

( )

( )1 32

1

3

2

3

1

Above equation can be satisfied by several forms of (T). Lord Kelvin used simply (T)=T where T is the absolute temperature itself.

f T TT

T( , )1 3

3

1

The Carnot thermal efficiency becomes

th revL

H

T

T, 1

This is the maximum possible efficiency of a heat engine operating between two heat reservoirs at temperatures TH and TL. Note that the temperatures are absolute temperatures.

43

1

273.16

This scale is known as Kelvin Scale and temperatures are known Absolute Temperatures. The relation between Celsius and absolute scale is given by

The magnitude of oK is defined as of the Temperature interval between absolute Zero and Triple point of Water.

273.16oK C

So for a reversible device the heat transfers between the high- and low-temperature heat reservoirs is written as :

Q

Q

T

TL

H

L

H

Then the QH/QL ratio can be replaced by TH/TL for reversible devices, where TH and TL are the absolute temperatures of the high- and low-temperature heat reservoirs, respectively.

44

The thermal efficiencies of actual and reversible heat engines operating between the same temperature limits compare as follows:

th revL

H

T

T, 1

Now as per our definition absolute zero will be achieved when

01 1 1.00L L L L

ThH H H H

Q T Q Twhich will mean

Q T Q TNow this is not possible so Absolute Zero is not achievable

THFOR A CARNOT ENGINE INCREASES IF TH INCREASES and

also INCREASES IF TL DECREASES

45

Reversed Carnot Device Coefficient of Performance

If the Carnot device is caused to operate in the reversed cycle, the reversible heat pump is created. The COP of reversible refrigerators and heat pumps are given in a similar manner to that of the Carnot heat engine as

COPQ

Q Q QQ

T

T T TT

RL

H L H

L

L

H L H

L

1

1

1

1

COPQ

Q Q

QQ

QQ

T

T T

TT

TT

HPH

H L

H

L

H

L

H

H L

H

L

H

L

1

1

46

Again, these are the maximum possible COPs for a refrigerator or a heat pump operating between the temperature limits of TH and TL.

The coefficients of performance of actual and reversible (such as Carnot) refrigerators operating between the same temperature limits compare as follows:

A similar relation can be obtained for heat pumps by replacing all values of COPR by COPHP in the above relation.

47

a. th revL

H

T

T

K

K

or

,

( )

( )

. .

1

130 273

652 273

0 672 67 2%

Q

Q

T

T

K

K

Q kJ

kJ

L

H

L

H

L

( )

( ).

( . )

30 273

652 2730 328

500 0 328

164

b.

QL

WOUT

QH

TH = 652oC

TL = 30oC

HE

A Carnot heat engine receives 500 kJ of heat per cycle from a high-temperature heat reservoir at 652oC and rejects heat to a low-temperature heat reservoir at 30oC. Determine

(a) The thermal efficiency of this Carnot engine.(b) The amount of heat rejected to the low-temperature heat reservoir.

48

Example 6-3

An inventor claims to have invented a heat engine that develops a thermal efficiency of 80 percent when operating between two heat reservoirs at 1000 K and 300 K. Evaluate his claim.

th revL

H

T

T

K

Kor

,

.

1

1300

10000 70 70%

QL

WOUT

QH

TH = 1000 K

TL = 300 K

HE

The claim is false since no heat engine may be more efficient than a Carnot engine operating between the heat reservoirs.

49

Example 6-4

An inventor claims to have developed a refrigerator that maintains the refrigerated space at 2oC while operating in a room where the temperature is 25oC and has a COP of 13.5. Is there any truth to his claim?

QL

Win

QH

TH = 25oC

TL = 2oC

R

COPQ

Q Q

T

T T

K

K

RL

H L

L

H L

( )

( )

.

2 273

25 2

1196

The claim is false since no refrigerator may have a COP larger than the COP for the reversed Carnot device.

50

Example 6-5

A heat pump is to be used to heat a building during the winter. The building is to be maintained at 21oC at all times. The building is estimated to be losing heat at a rate of 135,000 kJ/h when the outside temperature drops to -5oC. Determine the minimum power required to drive the heat pump unit for this outside temperature.

21 oC

HP-5 oC

QLost

Win

QL

The heat lost by the building has to be supplied by the heat pump.

QH

51

Q QkJ

hH Lost 135000

COPQ

Q Q

T

T T

K

K

HPH

H L

H

H L

( )

( ( ))

.

21 273

21 5

1131

Using the basic definition of the COP

COPQ

W

WQ

COP

kJ h h

s

kW

kJ skW

HPH

net in

net inH

HP

, /

. /.

,

,

135 000

1131

1

3600

1

3 316

52

A gasoline engine produces 20 hp using 35 kW of heat transfer from burning fuel. What is its thermal efficiency and how much power is rejected to the ambient?

Conversion : 20 hp = 20 × 0.7457 kW = 14.91 kW

Efficiency: ηTH = Wout/QH = 14.91/35 = 0.43

Energy equation: QL = QH - Wout = 35 – 14.91 = 20.1 kW

53

LQ

net, in

R

90 kJ/min500 kJ/min

COP 1.8L

QW 0.83 kW

net, in90 50

H LQ Q W 140 kJ / min

5-52 A household refrigerator with a COP of 1.8 removes heat from the refrigerated space at a rate of 90 kJ/min. Determine (a) the electric power consumed by the refrigerator and (b) the rate of heat transfer to the kitchen air.

The COP and the refrigeration rate of a refrigerator are given. The power consumption and the rate of heat rejection are to be determined.

Analysis (a) Using the definition of the coefficient of performance, the power input to the refrigerator is determined to be

(b) The heat transfer rate to the kitchen air is determined from the energy balance,

54

5-57 When a man returns to his well-sealed house on a summer day, he finds that the house is at 32°C. He turns on the air conditioner, which cools the entire house to 20°C in 15 min. If the COP of the air-conditioning system is 2.5, determine the power drawn by the air conditioner. Assume the entire mass within the house is equivalent to 800 kg of air for which cv 0.72 kJ/kg · °C and cp 1.0 kJ/kg · °C.

Assumptions 1 The air conditioner operates steadily. 2 The house is well-sealed so that no air leaks in or out during cooling. 3 Air is an ideal gas with constant specific heats at room temperature.Properties The constant volume specific heat of air is given to be cv = 0.72 kJ/kg.C.

Analysis

kW3.07 2.5

kW 7.68

COPRinnet, LQ

W

Using the definition of the coefficient of performance, the power input to the air-conditioner is determined to be

55

5-86 A heat engine is operating on a Carnot cycle and has a thermal efficiency of 55 percent. The waste heat from this engine is rejected to a nearby lake at 60°F at a rate of 800 Btu/min. Determine (a) the power output of the engine and (b) the temperature of the source.

Btu/min 1777.8Btu/min 800

155.01th HHH

L QQQ

Q

hp23.1 Btu/min 977.8Btu/min 1777.80.55thoutnet, HQW

L

H

L

H

T

T

Q

Q

rev

R1155.6 R 520Btu/min 800

Btu/min 1777.8

rev

L

L

HH T

Q

QT

60°F

TH

HE

800 Btu/min

The Carnot heat engine operates steadily. Analysis (a) The rate of heat input to this heat engine is determined from the definition of thermal efficiency,

Then the power output of this heat engine can be determined from

(b) For reversible cyclic devices we have

Thus the temperature of the source TH must be

56

Assignment

6-74

6-80

6-86

6-88

A car engine takes atmospheric air in at 20C, no fuel, and exhausts the air at –20C producing work in the process. What do the first and the second laws say about that?

57

58