CBB 2024 Chapter 4-Second Law of Thermo V2

CHAPTER 4

THE SECOND LAW OF THERMODYNAMICS

CONTENTS

5 - 1 The 2nd law & thermal energy reservoirs

5 - 2 Heat engines

5 - 3 Refrigerators & heat pumps

5 - 4 Perpetual-motion machines

5 - 5 The carnot cycle

5 - 6 The thermodynamic temperature scale

5 - 7 The carnot heat engine

5 - 8 The carnot refrigerator & heat pump

Objectives• Introduce the second law of thermodynamics.

• Identify valid processes as those that satisfy both the first and

second laws of thermodynamics.

• Discuss thermal energy reservoirs, reversible and irreversible

processes, heat engines, refrigerators, and heat pumps.

• Describe the Kelvin–Planck and Clausius statements of the second

law of thermodynamics.

• Discuss the concepts of perpetual-motion machines.

• Apply the second law of thermodynamics to cycles and cyclic

devices.

• Apply the second law to develop the absolute thermodynamic

temperature scale.

• Describe the Carnot cycle.

• Examine the Carnot principles, idealized Carnot heat engines,

refrigerators, and heat pumps.

• Determine the expressions for the thermal efficiencies and

coefficients of performance for reversible heat engines, heat

pumps, and refrigerators.

INTRODUCTION TO THE SECOND LAW

A cup of hot coffee does

not get hotter in a cooler

room.

Transferring

heat to a wire

will not

generate

electricity.

Transferring

heat to a

paddle wheel

will not cause

it to rotate.

These processes

cannot occur

even though they

are not in violation

of the first law.

Processes occur in a

certain direction, and not

in the reverse direction.

A process must satisfy both

the first and second laws of

thermodynamics to proceed.

MAJOR USES OF THE SECOND LAW

1. The second law may be used to identify the direction of processes.

2. The second law also asserts 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. The second law provides the necessary means to determine the

quality as well as the degree of degradation of energy during a process.

3. The second law of thermodynamics is also used in determining the

theoretical limits for the performance of commonly used engineering

systems, such as heat engines and refrigerators, as well as predicting the

degree of completion of chemical reactions.

THERMAL ENERGY RESERVOIRS

Bodies with relatively large thermal

masses can be modeled as thermal

energy reservoirs.

A source

supplies

energy in the

form of heat,

and a sink

absorbs it.

• A hypothetical body with a relatively large thermal energy capacity (mass x

specific heat) that can supply or absorb finite amounts of heat without

undergoing any change in temperature is called a thermal energy reservoir,

or just a reservoir.

• In practice, 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 masses.

HEAT ENGINES

Work can always

be converted to

heat directly and

completely, but the

reverse is not true.

Part of the heat

received by a heat

engine is

converted to work,

while the rest is

rejected to a sink.

The devices that convert heat to

work.

1. They receive heat from a high-

temperature source (solar energy,

oil furnace, nuclear reactor, etc.).

2. They convert part of this heat to

work (usually in the form of a

rotating shaft.)

3. They reject the remaining waste

heat to a low-temperature sink

(the atmosphere, rivers, etc.).

4. They operate on a 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.

5-1 The 2nd law & thermal energy reservoir

2nd law of processes occur in a certain direction, not

thermodynamics in just any direction.

E.g. a cup of coffee does not get hotter in a cooler room & transferring of heat to a wire will not generate electricity.

1st law places no restriction on the direction of process

A process will not occur until it satisfies both 1st & 2nd law.

Thermal energy hypothetical body that can absorb or

reservoir/Heat reject finite amounts of heat reservoir isothermally

E.g. oceans, lakes, rivers, atm. air & industrial furnace

A high T heat reservoir from which heat is transferred -heat source

A low T heat reservoir to which heat is transferred - heat sink

Heat transfer from industrial sources to env. is of major concern.

Work reservoir - sufficiently large system in stable equilibrium to which & from which finite amounts of work can be transferred adiabatically without any change in its P. Thermodynamic cycle - when the system undergoes a series of

processes & then returns to its original state, so that the properties of the system at the end of the cycle are the same as its beginning.

5-2 Heat engines Work can be converted directly & completely, but converting heat to

work requires heat engines. Characteristic:1) Receive heat from high-temperature

source.2) Convert part of heat to work

(rotation shaft)3) Reject remaining waste heat to low T

sink4) Operate in a cycle.

A steam power plant

A portion of the work output

of a heat engine is consumed

internally to maintain

continuous operation.

Steam power plant as a heat engine operating in a thermodynamic cycle:

The net work output of power plant: Wnet,out = Wout - Win (kJ)

Can be analyzed as a closed system undergoing a cycle, U = 0, therefore net output of the system:

Wnet,out = Qin - Qout (kJ)

Thermal efficiency

Some heat engines perform better

than others (convert more of the

heat they receive to work).

Schematic of

a heat engine.

Even the most

efficient heat

engines reject

almost one-half

of the energy

they receive as

waste heat.

Can we save Qout?

A heat-engine cycle cannot be completed without

rejecting some heat to a low-temperature sink.

In a steam power plant,

the condenser is the

device where large

quantities of waste

heat is rejected to

rivers, lakes, or the

atmosphere.

Can we not just take the

condenser out of the

plant and save all that

waste energy?

The answer is,

unfortunately, a firm

no for the simple

reason that without a

heat rejection process

in a condenser, the

cycle cannot be

completed.

Every heat engine must waste

some energy by transferring it to a

low-temperature reservoir in order

to complete the cycle, even under

idealized conditions.

Thermal efficiency, hth

index of performance of a work-producing device or a heat engine.

defined as the ratio of the net work output to the heat input.

hth = Desired result

Required input

for heat engines, the desired result is the net work done & the input is the heat supplied to make the cycle operate.

Thermal efficiency is always less than 1 or less than 100%.

where

Wnet,out = Wout - Win

Qin Qnet

Applying 1st law to cyclic heat engine:

in

outnet

thQ

W ,h

UWQ outnetinnet ,,

innetoutnet QW ,,

outinoutnet QQW ,

The cycle thermal efficiency may be written as

or

Example:

A steam power plant produces 50 MW of net work while burning fuel to produce 150MW of heat energy at the high temperature. Determine the cycle thermal efficiency and the heat rejected by the cycle to the surroundings.

Wnet,out = QH - QL

QL = QH - Wnet,out

= 150 MW - 50 MW

= 100 MW

in

outth

Q

Q1h

H

Lth

Q

Q1h

H

outnet

thQ

W ,h

%3.33333.0150

50

MW

MW

Example 5-17, Cengel

A 600-MW steam power plant, which is cooled by a nearby river, has athermal efficiency of 40%. Determine the rate of heat transfer to theriver water. Will the actual heat transfer rate be higher or lower thanthis value? State why?

Solution:

The rate of heat rejection is to be determined & the result is to be compared to the actual case in practice.

Assumptions: (1) The plant operates steadily (2) The heat losses from the working fluid at the pipes and other components are negligible.


An automobile engine consumes fuel at a rate of 28 L/h and deliver 60 kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and a density of 0.8 g/cm2, determine the efficiency of this engine.

Solution:

Thermal efficiency of the engine is to be determined as power output & fuel consumption rate of a car are given.

Assumptions: The car operates steadily

The Second Law of

Thermodynamics:

Kelvin–Planck Statement

A heat engine that violates the

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.

No heat engine can have a thermal

efficiency of 100 percent, or as for a

power plant to operate, the working fluid

must exchange heat with the

environment as well as the furnace.

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 the

idealized and the actual heat engines.

Statements of the 2nd Law:

Kelvin-Planck Statement:

It is impossible for any system to operate in a thermodynamic cycle & deliver a net amount of work to its surroundings while receiving energy by heat transfer from a single thermal reservoir.

The statement does not rule out the possibility of a system developing a net amount of work from a heat transfer drawn from a single reservoir. It only denies this possibility if the system undergoes a thermodynamic cycle.

NO!

System

undergo

thermo.

cycle

Analysis of Kelvin-Planck statement:

(1) From 1st Law:

Wcycle = Qcycle

the net work done by the system undergoing a cycle = the net heat transfer to the system. If Wcycle is -ve, then Qcycle is -ve.

if a net amount of energy is transferred by work to the system during the cycle, then an equal amount of energy is transferred by the heat from the system during the cycle.

(2) From 2nd law:

direction of energy transfer.

a system undergoing a cycle from a single reservoir cannot deliver a net amount of work to its surroundings. Therefore Wnet,cycle cannot be +ve.

But the statement not rule out the possibly that the net work is zero.

Therefore Wcycle 0 (single reservoir )

MAXIMUM POSSIBLE EFFICIENCY IS LESS THAN 100 %

hth 100%

Clausius Statement:

It is impossible for any system to operate in such a way that the sole result would be an energy transfer by the heat from a cooler to a hotter body.

The statement does not rule out the possibility of transferring energy by heat from a cooler body to a hotter body.

‘Sole result’ - when a heat transfer from cooler to hotter body occur, there must be some other effect within the system, surrounding or both.

COP

Q Hot

metal

Yes bar No

Cold Q

REFRIGERATORS AND HEAT PUMPS• The transfer of heat from a low-

temperature medium to a high-

temperature one requires special

devices called refrigerators.

• Refrigerators, like heat engines,

are cyclic devices.

• The working fluid used in the

refrigeration cycle is called a

refrigerant.

• The most frequently used

refrigeration cycle is the vapor-

compression refrigeration cycle.

Basic components of a

refrigeration system and

typical operating conditions.

In a household refrigerator, the freezer compartment

where heat is absorbed by the refrigerant serves as

the evaporator, and the coils usually behind the

refrigerator where heat is dissipated to the kitchen

air serve as the condenser.

5-3 Refrigerators & pumps

nature process: heat flows from high T to low T

reverse process: heat from low T to high T refrigerators

refrigerators & heat engines are cyclic device. The most frequently used - vapor compression refrigeration cycle (a compressor, a condenser, an expansion valve & an evaporator)

Refrigerators: How it works

Refrigerant enters the compressor as a vapor & is compressed to condenser pressure (800 kPa & 60C). It leaves the compressor at a relatively high temperature & cools down & as it flows through the coils of condenser(800 kPa & 30C) by rejecting heat to the surrounding medium.

It then enters a capillary tube (expansion valve) where its pressure & temperature drops drastically due to throttling effect.

The low-temperature refrigerant(120 kPa & -25C) then enters the evaporator, where it evaporates by absorbing heat from the refrigerated space.

The cycle is completed as the refrigerant( 120 kPa & -20C) leaves the evaporator & reenters the compressor.

freezer compartment where heat is picked up by refrigerant serves as evaporator.

coils where heat is dissipated serve as condenser.

The objective of a refrigerator is to

remove QL from the cooled space.

QL is the magnitude of heat removed

from the refrigerated space at TL

QH is the magnitude of heat rejected

to the warm env. At TH.

Wnet,in is the net work input to the

refrigerator.

Coefficient of Performance

The objective of a refrigerator is to

remove QL from the cooled space.

The efficiency of a refrigerator is expressed

in terms of the coefficient of performance

(COP).

The objective of a refrigerator is to remove

heat (QL) from the refrigerated space.

Can the value of COPR be

greater than unity?

Coefficient of performance (COP)]

The efficiency of a refrigerator is expressed in terms of coefficient of performance (COPR)

COPR = desired output = QL

required input Wnet,in

Apply 1st law of cyclic refrigerator,

COP relation can be expressed as

value of COPR can be greater than unity, that is the amount of heatremoved from the refrigerated space can be greater than the amountof work input.

LHinnetin

cycleinHL

QQWW

UWQQ

,

00

1/

1

LHLH

LR

QQQQ

QCOP

Heat

Pumps

The objective

of a heat

pump is to

supply heat

QH into the

warmer

space. The work

supplied to a

heat pump is

used to extract

energy from the

cold outdoors

and carry it into

the warm

indoors.

for fixed values of QL and QH

Can the value of COPHP

be lower than unity?

What does COPHP=1

represent?

Heat pumps

device that transfer heat from low T to high T.

Objective: to maintain a heated space at

a high T by absorbing heat from low T

source.

COPHP = desired output = QH

required input Wnet,in

can also be expressed as

under the same conditions,

HLLH

HHP

QQQQ

QCOP

/1

1

1 RHP COPCOP

When installed backward,

an air conditioner

functions as a heat pump.

• Most heat pumps in operation today have a

seasonally averaged COP of 2 to 3.

• Most existing heat pumps use the cold outside air

as the heat source in winter (air-source HP).

• In cold climates their efficiency drops considerably

when temperatures are below the freezing point.

• In such cases, geothermal (ground-source) HP

that use the ground as the heat source can be

used.

• Such heat pumps are more expensive to install,

but they are also more efficient.

• Air conditioners are basically refrigerators whose

refrigerated space is a room or a building instead

of the food compartment.

• The COP of a refrigerator decreases with

decreasing refrigeration temperature.

• Therefore, it is not economical to refrigerate to a

lower temperature than needed.

Energy efficiency rating (EER): The amount of heat removed from the

cooled space in Btu’s for 1 Wh (watthour) of electricity consumed.

The Second Law of Thermodynamics:

Clasius Statement

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.

It states that a refrigerator cannot operate unless

its compressor is driven by an external power

source, such as an electric motor.

This way, the net effect on the surroundings

involves the consumption of some energy in the

form of work, in addition to the transfer of heat

from a colder body to a warmer one.

To date, no experiment has been conducted that

contradicts the second law, and this should be

taken as sufficient proof of its validity.

A refrigerator that

violates the Clausius

statement of the second

law.


An air conditioner removes heat steadily from a house at rate of 750 kJ/min while drawing electric power at a rate of 6 kW. Determine

(a) The COP of this air conditioner (2) The rate of heat transfer to the outside air.

Solution:

Assumption: The air conditioner operates steadily.


Determine the COP of a heat pump that supplies energy to a house at a rate of 8000 kJ/h for each kW of electric power it draws. Also, determine the rate of energy absorption from the outdoor air.

Solution:

Assumption: The heat pump operates steadily.

PERPETUAL-MOTION MACHINES

A perpetual-motion machine that

violates the first law (PMM1).

A perpetual-motion machine that

violates the second law of

thermodynamics (PMM2).

Perpetual-motion machine: Any device that violates the first or the second

law.

A device that violates the first law (by creating energy) is called a PMM1.

A device that violates the second law is called a PMM2.

Despite numerous attempts, no perpetual-motion machine is known to have

worked. If something sounds too good to be true, it probably is.

5-4 Perpetual-motion machines

A process cannot take place unless its satisfies both 1st & 2nd law of thermodynamics.

Any device that violates either law is called perpetual-motion machine

device violates the 1st law, it is a perpetual-motion machine of the 1st kind (PMM1)

device violates the 2nd law, it is a ……………..of the 2nd kind (PMM2)

Reversible processes A reversible process is a quasi-equilibrium, or quasistatic process with a more restrictive requirement.

Internally reversible quasiequilibrium process, which once having taken place, can be reversed & in doing so leave no change in the system. This says nothing about what happens to the surroundings about the system.

Totally or externally quasiequilibrium process, which once

reversible process having taken place, can be reversed and in

the system or surroundings.

Irreversible process process that is not reversible.

All real processes are irreversible & occur because of:

1. Friction

2. Unrestrained expansion of gasses

3. Heat transfer through a finite T difference

4. Mixing of 2 different substances

5. Hysteresis effects

6. I2R losses in electrical circuits

7. Any deviation from a quasistatic process

5-5 The carnot cycle

Heat engines are cyclic devices & the working fluid returns to its initial state at the end of each cycle.

Work is done by the working fluid during 1 part of the cycle & on the working fluid during another part.

The difference between these two = net work delivered by heat engine.

The net work, cyclic efficiency can be max. by require least amount of work & deliver the most - using reversible processes.

Carnot cycle, proposed by Sadi Carnot in 1824.

The theoretical heat engine that operates on Carnot cycle is called the Carnot heat engine.

Composed of 4 reversible processes - 2 isothermal & 2 adiabatic & can be executed either in a closed or a steady-flow system.

Consider a closed system that consists of a gas contained in an adiabatic piston-cylinder device.

The insulation of the cylinder head may be removed to bring the cylinder into contact with reservoir to provide heat transfer.

P - V diagram of the Carnot cycle

(1) Reversible isothermal expansion (process 1-2, TH = constant)

At state 1, temperature of gas is TH & cylinder head is in contact with the source at temperature TH.

The gas is allowed to expend but T of the gas tends to decrease . Heat flows from the reservoir into the gas , rising the gas T so that T is kept constant to TH.

This is a reversible process as the T difference between the gas & reservoir never exceeds a differential amount dT.

The process continuous until reaches position 2.

Amount of heat transferred to the gas during this process is QH.

(2) Reversible adiabatic expansion (process 2-3, T drops from TH to TL)

At state 2 the reservoir is removed & replaced by insulation, so the system becomes adiabatic.

The gas continuous to expand slowly until temperatures drop from TH

to TL (state 3).

The piston is assumed frictionless & the process to be quasi-equilibruim, so that the process is reversible as well as adiabatic.

(3) Reversible isothermal compression (process 3-4, TL = constant)

At state 3, the insulation at the cylinder head is removed & brought into contact with a sink at temperature TL.

Now the piston is pushed inward by an external force, doing work on the gas. As the gas is compressed, its temperature tends to rise but once it rises to infinitesimal amount dT, heat flows from the gas to the sink, causing the T to drop to TL. Thus the gas T is maintained at TL.

This is a reversible process as T difference between gas & sink never exceed dT. It continuous until the piston reaches state 4.

The amount of heat rejected from the gas is QL.

(4) Reversible adiabatic compression (process 4-1, temperature rises from TL to TH)

At state 4, the low temperature reservoir is removed & insulation is put back on the cylinder head. The gas is compressed in a reversible manner.

The gas returns to initial state (state 1)

The temperature rises from TL to TH during the reversible adiabatic compression process, which complete the cycle.

From the P - V diagram:

area under the process curve = boundary work for quasi-equilibrium(internally reversible) processes

area under curve 1-2-3 = work done by the gas during the expansion

area under curve 3-4-1 = work done on the gas during compression

area enclosed by path cycle 1-2-3-4-1 = difference between the two & represents the net work done during the cycle.

The reversed carnot cycle

The Carnot heat-engine cycle described is a totally reversible cycle.

All processes that can be reversed becomes the ‘Carnot refrigeration cycle’.

The cycles remain the same except direction of Q & W is reversed.

Heat in the amount of QL is absorbed from low T reservoir

Heat in the amount of QH is rejected to a high T reservoir & work input of Wnet,in is required to accomplish.

The Carnot principles

The 2nd law of thermodynamics puts limits on the operation of cyclic device as expressed by the Kelvin-Planck & Clausius statements.

Considering heat engines operating between two fixed temperature reservoir at TH TL. Two conclusions about the thermal efficiency of reversible & irreversible heat engines known as 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.

hth hth, Carnot

(b) The efficiencies of all reversible heat engines operating between same two constant temperature heat reservoir have the same efficiency.

= HE

High-temperature reservoir

at TH

1 2 3

Irrev Rev Rev

hth,1hth,2 hth,2=hth,3

Low-temperature reservoir

at TL

Diagram of Carnot principles

These 2 statements can be proved by demonstrating that the violation of either statement results in the violation of the 2nd law of thermodynamics.

To prove the 1st statement, consider 2 heat exchanges operating between the same reservoir. By referring to figure (a),1 engine is reversible & other is reversible.

In violation of 1st Carnot principle, we assume hth, irrev hth,rev

Now, let the reversible heat engine be reversed & operate as a refrigerator. It will receive a work input Wrev & reject heat to the high-temperature reservoir.

Since the refrigerator is rejecting heat QH to the high-temperature reservoir & irreversible heat engine is receiving the same amount of heat from this reservoir. Net heat exchange = 0

Now consider the refrigerator & irreversible engine together in figure (b), producing a net work

Wirrev - Wrev exchanging heat

with a single reservoir (violation

Kelvin-Planck statement 2nd law.

Therefore, our initial assumption

that hth, irrev hth,rev is incorrect.

Conclusion: No heat engine can be

more efficient than a reversible

heat engine operating between

the same reservoir.

5-6 The thermodynamic temperature scale

Definition: A temperature scale that is independent of the properties of the substances that are used to measure temperature

From 2nd Carnot principle:

(a) hrev engine is independent of the working

fluid employed & its properties

(b) the way the cycle is executed

(c) or the type of reversible engine used

Thermal efficiency of reversible heat

engine is a function of reservoir T only

hth,rev = g(TH,TL)

or

since hth = 1 - QL/QH

All Reversible heat engines operating between the same two reservoirs have the same efficiency (the Second Carnot principle)

),( LH

L

H TTfQ

Q

Since the thermal efficiency in general is

hth = 1 - QL/QH

For the Carnot Engine, this can be written

as

Considering engines A, B, & C:

This look like

One way to define the f function is

HLHLth TTfTTg ,1, h

3

2

2

1

3

1

Q

Q

Q

Q

Q

Q

),(),(),( 322131 TTfTTfTTf

)(

)(

)(

)(

)(

)(),(

3

1

3

2

2

131

T

T

T

T

T

TTTf

The simplest form of is the absolute Temperature itself.

The Carnot thermal efficiency becomes

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

These statements form the basis for establishing an absolute temperature scale, also called the Kelvin scale, related to the heat transfer between a reversible device & the high and low temperature heat reservoirs by

where TH/TL are the absolute temperatures of the high & low temperature heat reservoirs.

Only valid when heat engine operating between 2 constant temperature heat reservoir.

3

131 ),(

T

TTTf

H

Lrevth

T

T1,h

H

L

revH

L

T

T

Q

Q

5-7 The carnot heat engine

Carnot heat engine hypothetical heat engine that operates on the reversible Carnot cycle

Thermal efficiency of any heat engine, reversible or irreversible:

where QH = heat transferred to the heat engine from a high T reservoir at TH

QL = heat rejected to a low T reservoir at TL.

Efficiency of a Carnot engine or any reversible heat engine, becomes

The above relation is known as the Carnot efficiency. It is the highest efficiency a heat engine operating between the 2 thermal energy reservoirs at temperatures TL & TH can have.

All irreversible (i.e actual) heat engines operating between these limits (TL & TH) will have lower efficiencies.

H

Lth

Q

Q1h

H

Lrevth

T

T1,h

An actual heat engine cannot reach this maximum theoretical efficiency value because it is impossible to completely eliminate all the irreversibilities associated with the actual cycle.

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

hth,rev irreversible heat engine

hth = hth,rev reversible heat engine

hth,rev impossible heat engine


A Carnot heat engine receives 650 kJ of heat from a source of unknown temperature and rejects 200 kJ of heat to a sink at 17C. Determine (a) the temperature of the source (b) the thermal efficiency of the heat engine.

Solution:

Assumption: The Carnot heat engine operates steadily.

Carnot Steam power cycle

(Process 1-2)

As water flows through the boiler, a change of phase from liquid to vapor at constant temperature TH occurs at a result of heat transfer from hot reservoir.

As temperature remains constant, pressure also remains constant during the phase change.

(Process 2-3)

The steam exiting the boiler expands adiabatically through the turbine & work is developed. In this process temperature decreases to temperature of the cold reservoir TL & there is an accompanying decrease in pressure.

(Process 3-4)

As the steam passes through the condenser, heat transfer to cold reservoir occurs & some vapor condenses at constant TL. Temperature & pressure remains constant as water passes through the condenser.

(Process 4-1)

Pump - receives two-phase liquid-vapor mixture from the condenser & returns it adiabatically to the state of boiler entrance.

During this process, work input is required to increase the pressure & temperature increases from TL to TH.

The thermal efficiency of actual heat engines can be maximized by supplying heat to the engine at the highest possible temperature (limited by material strength) & rejecting heat from the engine at the lowest possible temperature.

5 - 8 The carnot refrigerator & heat pump

A refrigerator or a heat pump that operates on the reversed Carnot cycle is called Carnot refrigerator or a Carnot heat pump.

The COP reversible or irreversible is given by:

and

where QL = amount of heat absorbed from the low-temperature medium

QH = amount of heat rejected to the high-temperature medium

1

1

L

HR

Q

QCOP

H

LHP

Q

QCOP

1

1

The COPs of all reversible refrigerators or heat pumps:

replace heat transfer ratio by ratio of the absolute temperature

of the high and low temperature reservoirs:

and

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

The COP of actual & reversible (such as Carnot) refrigerators operating between the same temperature limits compare as follows:

COPR,rev irreversible refrigerator

COPR = COPR,rev reversible refrigerator

COPR,rev impossible refrigerator

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

1

1,

L

HrevR

T

TCOP

H

LrevHP

T

TCOP

1

1,


A Carnot refrigerator operates in a room in which the temperature is 25C. The refrigerator consumes 500 W of power when operating and has a COP of 4.5. Determine (a) the rate of heat removal from the refrigerated space (b) the temperature of the refrigerated space.

Solution:

Prob. 5.75 M&S:

Two kg of water execute a Carnot power cycle.

During the isothermal expansion, the water is

heated until it is a saturated vapor from an

initial state where the pressure is 40 bar and

the quality is 15%. The vapor then expands

adiabatically to a pressure of 1.5 bar while

doing 491.5 kJ/kg of work.

a) Sketch the cycle on p-v diagram

b) Evaluate the heat and work for each process,

in kJ.

c) Evaluate the thermal efficiency.

Carnot cycle: 5.61, Moran & Shapiro

One-half kilogram of water executes a Carnot power cycle.During the isothermal expansion, the water is heated until it is asaturated vapor from an initial state where the pressure is 15bar and the quality is 25%. The vapor then expandsadiabatically to a pressure of 1 bar while doing 403.8 kJ/kg ofwork.

(a) Sketch the cycle on p-v diagrams

(b) Evaluate the heat & work for each process, in kJ

(c) Evaluate the thermal efficiency

Prob

A Carnot refrigeration cycle executed in a closed

system in the saturated liquid-vapor mixture region

using 0.96 kg of R-134a as the working fluid. It is

known that the maximum absolute temperature in the

cycle is 1.2 times the minimum absolute temperature,

and the net work input to the cycle is 22 kJ. If the

refrigerant changes from saturated vapor to saturated

liquid during the heat rejection process, determine the

minimum pressure in the cycle.

P max=?

Prob

Consider a Carnot heat pump cycle executed in a steady

flow system in the saturated liquid-vapor mixture region

using R-134A flowing at rate of 0.264 kg/s as the working

fluid. It is known that the maximum absolute temperature in

the cycle is 1.25 times the minimum absolute temperature,

and the net power input to the cycle is 7 kW. If the

refrigerant changes from saturated vapor to saturated liquid

during the heat rejection process, determine the ratio of the

maximum to minimum pressures in the cycle.

CBB 2024 Chapter 4-Second Law of Thermo V2

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Transcript of CBB 2024 Chapter 4-Second Law of Thermo V2