chapter 04 the second law of thermodynamics (pp 82-99)

CH4 SECOND LAW OF THERMODYNAMICS 82

CHAPTER 04 THE SECOND LAW OF

THERMODYNAMICS

4-1 EFFICIENCY OF A HEAT ENGINE

Example 4-1

A diesel engine performs J2200 of mechanical work and

discards J4300 of heat each cycle.

(a) How much heat must be supplied to the engine in each

cycle?

(b) What is the thermal efficiency of the engine?

Solution

(a) JWQQ OUTIN 650022004300 =+=+=

(b) %3434.06500

2200or

Q

W

IN

===η

Example 4-2

An engine with %20 efficiency does J100 of work in each

cycle.

(a) How much heat is absorbed in each cycle?

(b) How much heat is rejected in each cycle?

Solution (a) The efficiency of a heat engine is given by

INQ

W=η

JQ

Q ININ 500

20.0

100===

η

(b) JWQQ INOUT 400100500 =−=−=


Example 4-3

An engine absorbs J100 and rejects J60 in each cycle.

(a) What is its efficiency?

(b) If each cycle takes s5.0 , find the power output of this

engine in watts.

Solution (a) The efficiency of the heat engine is defined as

%4040.0100

60100or

Q

QQ

Q

W

IN

OUTIN

IN

=−

=−

==η

(b) Wt

QQ

t

WP OUTIN 80

5.0

60100=

−=

−==

Example 4-4

Find the efficiency of a Carnot engine working between the

temperatures C0225 and C

025 . B.U. B.Sc. (Hons.) 1983A

Solution The efficiency of a Carnot engine is given by

%1001 ×

−=

H

L

T

Tη

%40%100273225

273251 =×

+

+−=η

Example 4-5

What is the maximum efficiency of a steam engine that

utilizes steam from a boiler at K480 and exhausts at

K373 . K.U. B.Sc. 2003


%1001 ×

−=

H

L

T

Tη

%3.22%100480

3731 =×

−=η


Example 4-6

A heat engine performs 200 joules of work and at the same

time rejects 3000 joules of heat to the cold body. Calculate

the efficiency. B.U. B.Sc. (Hons.) 1989A


%1001 ×

−=

IN

OUT

Q

Qη

%25.6%1002003000

30001 =×

+−=η

Example 4-7

A Carnot engine is operated between two heat reservoirs at

temperatures of K400 and K300 . If the engine receives

1200 joules of energy from the reservoir at K400 in each

cycle, how many joules of energy does it reject to the

reservoir at K300 ? B.U. B.Sc. (Hons.) 1986A


−=

−=

IN

OUT

H

L

Q

Q

T

T11η

IN

OUT

H

L

Q

Q

T

T=

JQT

TQ OUT

H

LOUT 900)1200(

400

300=

=

=

Example 4-8

A steam engine, operating between C0115 and C

060 ,

develops hp2500 . How much heat must be supplied if its

overall efficiency is 10 percent? What is the maximum

efficiency that an ideal heat engine can have when it

operates between the stated temperatures? (1 hp = 746 W).

B.U. B.Sc. 1999A


Solution

Let INQ be the supplied heat, then

ININ QQofW 1.0%10 ==

INQ1.07462500 =×

JQIN

710865.11.0

7462500×=

×=

The efficiency of ideal heat engine is given by

%2.14142.0273115

2736011 or

T

T

H

L =+

+−=

−=η

Example 4-9

A heat engine absorbs kJ4.52 of heat and exhausts kJ2.36

of heat per cycle. Calculate the efficiency and the work done

by the engine per cycle. P.U. B.Sc. 2003, 2009


%1001 ×

−=

IN

OUT

Q

Qη

%9.30%1004.52

2.361 =×

−=η

The work done by the engine per cycle is

kJQQW OUTIN 2.162.364.52 =−=−=

Example 4-10

A Carnot’s engine takes 61036.3 × joules of heat from a

reservoir at C0527 and gives some heat to sink at C

027 .

What is the efficiency of engine? How much work does it

perform in joules? K.U. B.Sc.2006

Solution The efficiency of heat engine is given by

−=

H

L

T

T1η


%5.62625.0273527

273271 or=

+

+−=η

The work done by the engine is

JQW IN

66 101.2)1036.3)(625.0( ×=×== η

Example 4-11

A Carnot engine extracts J240 of heat from a high

temperature reservoir during each cycle. It rejects J100 of

heat at reservoir at C015 .

(i) How much work does the engine do in one cycle?

(ii) What is its efficiency?

(iii) What is the temperature of the hot reservoir?

K.U. B.Sc. 2001

Solution (i) The work done by the engine is given by

JQQW OUTIN 140100240 =−=−=

(ii) The efficiency of the given Carnot engine is

%3.58%100240

1001%1001 =×

−=×

−=

IN

OUT

Q

Qη

(iii) The efficiency of Carnot engine in terms of

temperatures of reservoirs is given by

%1001 ×

−=

H

L

T

Tη

10027315

13.58 ×

+−=

HT

HT

2881583.0

100

3.58−==

418.0583.01288

=−=HT

CorKTH

0418691418.0

288==


Example 4-12

A heat engine operates between two reservoirs at K300

and K500 . During each cycle it absorbs cal200 of heat

from the hot reservoir.

(a) What is maximum efficiency of the engine?

(b) Determine the maximum work the engine performs

during each cycle. B.U. B.Sc. 2008A

Solution (a) The efficiency of heat engine is given by

−=

H

L

T

T1η

%4040.0500

3001 or=−=η

(b) The heat absorbed by the engine from hot reservoir in

each cycle is

JJcalQIN 2.837)186.4)(200(200 ===

The work done by the engine in each cycle is

JQW IN 88.334)2.837)(40.0( === η

Example 4-13

A Carnot engine absorbs heat from a reservoir at a

temperature of C01000 and rejects heat to a cold reservoir

at a temperature of C00 . If the engine absorbs 1000 joules

from the high temperature reservoir, find heat rejected and

efficiency of heat engine. B.U. B.Sc.(Hons.) 1988A, 1989A

Solution Now

H

L

IN

OUT

T

T

Q

Q=

JQT

TQ IN

H

LOUT 214

)2731000(

)1000)(2730(=

+

+=

=

The efficiency of heat engine is given by


%1001 ×

−=

H

L

T

Tη

%6.78%1002731000

27301 =×

+

+−=η

Example 4-14

In a Carnot cycle, the isothermal expansion of an ideal gas

takes at C0140 and isothermal compression at C

025 . During

the compression, J2090 of heat energy is transferred to the

gas. Calculate

(i) The work performed by the gas during isothermal

expansion.

(ii) Heat rejected from the gas during isothermal

compression.

(iii) Work done on the gas during isothermal compression.

K.U. B.Sc. 2002, 2005

Solution

(a) JQW IN 2090−=−= The negative sign indicates that the

work is done by the gas.

(b) The efficiency of Carnot engine is given by

%1001 ×

−=

H

L

T

Tη

%8.27%100273140

373251 =×

+

+−=η

Now %1001 ×

−=

IN

OUT

Q

Qη

)100(2090

18.27

−= OUTQ

−==

20901278.0

100

8.27 OUTQ

722.0278.012090

=−=OUTQ


JQOUT 1509)2090)(722.0( ==

(c) Work done on the gas during isothermal compression

JQW OUT 1509==

Example 4-15

A Carnot cycle has efficiency %75 working between

highest and lowest temperatures. What is the lowest

temperature of its high temperature is C0927 ?

B.U. B.Sc. 2007S


−=

H

L

T

T1η

273927

1100

75

+−= LT

1200

175.0 LT−=

25.075.011200

=−=LT

CKTL

027300)1200)(25.0( ===

Example 4-16

A Carnot engine has an efficiency of %22 . It operates

between heat reservoirs differing in temperature by C075 .

Find the temperature of the reservoirs.

K.U. B.Sc. 2004, 2008

Solution

Let LT and HT be the temperatures of cold and hot reservoirs

(i.e. sink and source) respectively, then the efficiency of Carnot

engine is given by

H

L

T

T−= 1η

Now 22.0%22 ==η


KTT LH )75( +=

Hence

75

75

75122.0

+

−+=

+−=

L

LL

L

L

T

TT

T

T

75

7522.0

+=

LT

755.1622.0 =+LT

5.585.167522.0 =−=LT

CorKTL

0726622.0

5.58−==

CorKTT LH

0683417526675 =+=+=

Example 4-17

A Carnot heat engine with an efficiency of 0.20 operates

between two temperatures that differ from each other

by K100 . What are the temperatures between which the

cycle operates?


H

L

T

T−= 1η

Now 20.0=η

KTT LH )100( +=

Hence

100

100

100120.0

+

−+=

+−=

L

LL

L

L

T

TT

T

T

100

10020.0

+=

LT

1002020.0 =+LT

802010020.0 =−=LT

KTL 40020.0

80==


KTT LH 500100400100 =+=+=

The given engine operates between 400 K and 500 K.

Example 4-18

The exhaust temperature of a heat engine is C0220 . What

must be the high temperature if the Carnot efficiency is to

be %36 ?

Solution

Now H

L

T

T−= 1η

HT

2732201

100

36 +−=

HT

493136.0 −=

64.036.01493

=−=HT

CorKTH

049777064.0

493==

Example 4-19

In one cycle of operation a heat engine takes in J2200 of

heat and performs J620 of work.

(a) What is the efficiency of this engine?

(b) How much heat is exhausted in each cycle?

(c) If the engine completes a cycle each s033.0 , then at

what rate is heat added, heat exhausted, work done?

Solution (a) The efficiency of heat engine is defined as

%2828.02200

620or

Q

W

IN

===η

(b) The heat exhausted in each cycle is

JWQQ INOUT 15806202200 =−=−=

(c) kWWt

QP IN

IN 7.661067.6033.0

2200 4 =×===


kWWt

QP OUT

OUT 9.471079.4033.0

1580 4 =×===

kWWt

WP 8.181088.1

033.0

620 4 =×===

Example 4-20

The low temperature reservoir of a Carnot engine is at

C010 and has an efficiency of %50 . How much the

temperature of the high temperature reservoir is increased

to measure the efficiency to %50 ? K.U. B.Sc. 2007


−=

H

L

T

T1η

HT

27310150.0

+−=

50.050.01283

=−=HT

KTH 56650.0

283==

Let 1η be the new efficiency of heat engine with new source

temperature as *

HT , then *1 1

H

L

T

T−=η

*

283160.0

HT−=

40.060.01283

*=−=

HT

KTH 5.70740.0

283* ==

The desired increase in temperature of hot reservoir is

KTTT HH 5.1415665.707* =−=−=∆


Example 4-21

A heat engine utilizes a heat source at C0580 and has a

Carnot efficiency of 29 percent. To increase the efficiency to

35 percent, what must be the temperature of the heat

source?

Solution The temperature of the cold reservoir i.e. sink is calculated as

under

H

L

T

T−= 1η

273580

129.0+

−= LT

71.029.01853

=−=LT

KTL 606)853)(71.0( ==

Let 1η be the new efficiency of heat engine with new source

temperature as *

HT , then

*1 1

H

L

T

T−=η

*

606135.0

HT−=

65.035.01606

*=−=

HT

CorKTH

0* 65993265.0

606==


Example 4-22

A Carnot heat engine operates between a reservoir at

K950 and has heat transfer of J400 exhausted to another

reservoir at K300 during each cycle. What is the amount of

work done by the heat engine during each cycle?

Solution

Now IN

OUT

H

L

Q

Q

T

T=

JQT

TQ OUT

L

HIN 1267)400(

300

950=

=

=

JQQW OUTIN 8674001267 =−=−=

The given heat engine performs 867 J of work in each cycle.


4-2 COEFFICIENT OF PERFORMANCE OF A REFRIGERATOR ENGINE

Example 4-23

The low temperature of a freezer cooling coil is C015− and

the discharge temperature is C030 . What is the maximum

theoretical coefficient of performance?

Solution The coefficient of performance is defined as

LH

L

TT

TK

−=

733.5)27315()27330(

)27315(=

+−−+

+−=K

Example 4-24

A heat pump acting as a refrigerator is used to heat a house.

The temperature outside the house is C010− and the

interior to be kept at C022 . Find the maximum coefficient of

performance of heat pump. P.U. B.Sc. 2005, 2006

Solution The coefficient of performance of heat pump acting as a

refrigerator is given by

LH

L

TT

TK

−=

Now KKCTL 263)27310(100 =+−=−=

KKCTH 295)27322(220 =+==

KTT LH 32263295 =−=−

Hence 218.832

263==K

Example 4-25

What is the coefficient of performance of a refrigerator that

operates with Carnot efficiency between temperatures

C03− and C

027 ?


The coefficient of performance of heat pump acting as a

refrigerator is given by

LH

L

TT

TK

−=

Now KKCTL 270)2733(30 =+−=−=

KKCTH 300)27327(270 =+==

KTT LH 30270300 =−=−

Hence 930

270==K

Example 4-26

A household refrigerator, whose coefficient of performance is 4.7, extracts heat from the cooling chamber at rate of 250

J per second. How much work per cycle is required to

operate the refrigerator? P.U. B.Sc. 2002

Solution The coefficient of performance K is defined as

W

Q

QQ

QK

L

LH

L=

−=

cycleJK

QW

L/2.53

7.4

250===

Example 4-27

A refrigerator does 153 J of work to transfer 568 J of heat

from its cold compartment.

(a) Calculate the refrigerator’s coefficient of performance.

(b) How much heat is exhausted to the kitchen?

Solution (a) The coefficient of performance K is defined as

71.3153

568===

W

QK

L

(b) Heat exhausted to kitchen will be

JWQL 721153568 =+=+


Example 4-28

Apparatus that liquefies helium is in a laboratory at K296 .

The helium in the apparatus is at K4 . If mJ150 of heat is

transferred from helium, find the minimum amount of heat

delivered to the laboratory.

Solution The coefficient of performance is defined as

LH

L

TT

TK

−=

21037.14296

4 −×=+

=K

Now

K

QQWQQ

L

LLH +=+=

JQL 1.111037.1

11)10150(

2

3 =

×+×=

−

−


ADDITIONAL PROBLEMS

(1) A Carnot engine is operated between two reservoirs

at temperatures C0227 and C

0127 . Evaluate the

efficiency of the engine. B.U. B.Sc.(Hons.) 1988S

(2) A heat engine performs 1000 joules of work and at

the same time rejects 4000 joules of heat energy to

the cold reservoir. What is the efficiency of the

engine? B.U. B.Sc.(Hons.) 1991A

(3) A heat engine receives heat 120 joules and the work

done by it is 90 joules. Calculate the efficiency in

percentage. B.U. B.Sc. 1997A

(4) A heat engine performs 2000 joules of work and at

the same time rejects 6000 joules of heat energy to

the cold reservoir. What is the efficiency of the

engine? B.U. B.Sc. 1998A

(5) A Carnot engine whose heat source is at C0127 takes

100 calories of heat at this temperature in each cycle

and gives up 80 calories to the heat sink. Find the

temperature of heat sink. B.U. B.Sc. 1988S

(6) A Carnot engine absorbs heat from a reservoir at a

temperature of C0100 and rejects heat to a cold

reservoir at a temperature of C00 . If the engine

absorbs 1000 joules from the high temperature

reservoir, find heat rejected and efficiency of heat

engine. B.U. B.Sc. 1986S

(7) What is the maximum efficiency of a steam engine

that utilizes steam from a boiler at K480 and

exhausts at K373 ? K.U. B.Sc. 2003

(8) A heat engine performs 200 joules of work and at the

same time rejects 300 joules of heat to the cold body.

Calculate its efficiency. B.U. B.Sc.(Hons.) 1989A

(9) Calculate the efficiency of a Carnot engine working

between the temperatures C0927 and C

027 .

P.U. B.Sc. 1989


(10) A Carnot engine’s working substance is water. It

uses steam at C0300 and condenses to water at C

040 .

What is the maximum theoretical efficiency of this

engine? B.P.S.C. 1995

(11) A Carnot engine operates between the temperatures

K850 and K300 . The engine performs J1200 of work

each cycle, which takes s25.0 . Calculate its efficiency

and its average power. What are the rates of heat input

and heat exhaust per cycle? F.P.S.C. 2009

Answers

(1) 20 % (2) 20 % (3) 75 % (4) 25 %

(5) 320 K (6) JQOUT 732= , %8.26=η (7) 22.3 %

(8) 6.25 % (9) 75 % (10) 37 %

(11) 65 %, 4800 W, 13600 W and 8800 W

chapter 04 the second law of thermodynamics (pp 82-99)

Documents

heat pump acting

high temperature reservoir

utilizes steam

carnot engine

carnot engine

hlt ttk

hot reservoir

isothermal compression