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UNIT – II

PART – A 1. State the Kelvin – Planck statement of second law of thermodynamics. Kelvin –

Plank states that it is impossible to construct a heat engine working on cyclic process, whose only purpose is to convert all the heat energy given to it an equal amount of work.

2. State Clausius statement of second law of thermodynamics. It states that heat can flow from hot body to cold body without any external aid but heat cannot flow from cold body to hot body without any external aid.

3. Write the two statements of the Second law of thermodynamics. Kelvin-Planck statement: It is impossible to construct an engine working on an cyclic process which converts all the heat energy supplied to it into equivalent amount of useful work. Clausis statement Heat cannot flow from cold reservoir to hot reservoir without any external aid. But can flow from reservoir to cold reservoir without any external aid. 4. State Carnot’s theorem. No heat engine operating in a cyclic process between two – fixed temperatures can be more efficient that a reversible engine operating between the same temperature limits.

5. What are the Corollaries of Carnot theorems? i. All the reversible engines operating between the two given thermal reservoirs with

fixed temperature have the same efficiency ii. The efficiency of any reversible heat engine operating between two reservoirs is

independent of the nature of the working fluid and depends only on the temperature of the reservoirs.

6. Define – PMM of second kind. Perpetual motion of second kind draws heat continuously from single reservoir and converts it into equivalent amount of work. Thus it gives 100% efficiency.

7. What is difference between a heat pump and refrigerator? Heat pump is a device which operating in a cycle process, maintains the temperature of a hot body at temperature higher that the temperature of surrounding. A refrigerator is a device which operating in a cycle process maintains the temperature of a cold body temperature lower than the temperature of the surrounding.

8. What is mean by heat engine? A heat engine is a device which is used to convert the thermal energy into mechanical energy.

9. Define the term COP. Coefficient of performance (COP) is defined as the ratio of heat extracted or rejected to work input

Heat extracted rejectedCOP

Work input

10. Write the expression for COP of a heat pump and a refrigerator. COP for heat pump

2HP

2 1

THeat rejectedCOMP

Work input T T

COP for refrigerator

1ref

2 1

THeat extractedCOP

Heat input T T

11. Why Carnot cycle cannot be realized in practice?

i) In a Carnot cycle all the four processes are reversible but in actual practice there

is no process is reversible. ii) There are two processes to be carried out during compression and expansion.

For isothermal process the piston moves as fast as possible. This speed variation during the same stroke of the piston is not possible.

iii) It is not possible to avoid friction between moving parts completely

12. Name two alternative methods by which the efficiency of a Carnot cycle can be increased. i) Efficiency can be increases as the higher temperature T2 increases ii) Efficiency can be increases as the lower temperature T1 decreases

13. Why a heat engine cannot have 100% efficiency? For all the heat engines there will be a heat loss between system and surroundings.

Therefore we can’t convert all the heat input into useful work.

14. When will be the Carnot efficiency maximum? Carnot cycle efficiency will be maximum, when the temperature is O K (i.e. zero degree Kelvin)

15. What are the processes involved in Carnot cycle? Carnot cycle consist of

i. Reversible adiabatic compression ii. Reversible isothermal heat addition iii. Reversible adiabatic expansion iv. Reversible isothermal heat rejection

a. .

PV Diagram T-S Diagram Process 1 - 2: Isentropic Compression Process 2 – 3: Isothermal heat addition Process 3 – 4: Isentropic expansion and Process 4 – 5: Isothermal heat rejection 16. Write the expression for efficiency of the Carnot cycle.

Carnot = 2 1

2

T T

T

17. Is the second law independent of first law? Explain. Yes. The second law is independent of first law. The second law speaks about the quality of energy. 18. Define entropy. Entropy is an index of unavailability or degradation of energy. 19. Define change of entropy. How entropy is compared with heat transfer and absolute

temperature. The measure of irreversibility when the energy transfer takes place within the system or

1

2 3

4

1

2

3

4

between system and surrounding is called a change of entropy. It is simple known as unaccounted heat loss.

20. Define the terms source, sink and heat reservoir. Source: The part where the heat to be rejected to absorbing or work developing device is called source Sink: The part which receives heat from work absorbing or working developing device is called sink. Reservoir: The part which supplies or receives heat continuously without change in its temperature is called as reservoir.

21. Why the performances of refrigerator and heat pump are given in terms of C.O.P. and

not in terms of efficiency? The performance of any device is expressed in terms of efficiency for work developing machines. But heat pump and refrigerator are work absorbing machines. So, the performance of those devices based on C.O.P. only. 22. Comment on the statement “The entrophy of universe tends to be maximum. If the entropy of universe tends to be maximum the irreversibility will be more due to friction between moving parts. 23. Write down the equation for Carnot C.O.P of a heat pump which works between two

heat reservoirs of temperature T1 and T2 if T1>T2.

Carnot C.O.P. of heat pump = 1 2

1

T T

T

24. What is meant by principle of increase of entropy? For any infinitesimal process undergone by a system, change in entropy, dS>dQ/T

For reversible dQ = 0, hence, dS=0 For irreversible ds>0

So the entropy of an insolated system would never decrease. It will always increase and remains constant if the pressure is reversible is called as principle of increase of entropy.

25. What do you mean by “Calusius inequality”? It is impossible for a self acting machine working in a cyclic process unaided by any external agency to convey heat from a body at a low temperature to a body at a higher temperature.

27. Explain briefly clausius inequality. dQ

0T

is known as inequality of clausius

If 1. dQ

0T

, the cycle is reversible.

2. dQ

0T

, the cycle is irreversible and possible

3. dQ

0T

, the cycle is impossible (Violation of second law).

28. For compression process between same and states, which work will be more, reversible or irreversible. Irreversible work will be more in the compressor. Generally for compression, the actual work given will be higher than the calculated work (Wrev). 29. A heat pump pumps 10MJ/KW whr to the high temperature reservoir. What is the C.O.P.?

C.O.P. = Heat Supplied

Work input

310 102.78

3600

30. Find the entropy of universe when 1000 KJ of heat is transferred from 800K to 500K.

Entropy of universe, univ

1 2

Q QS

T T

1000 1000

800 500

0.75KJ/K

31. Give the expressions to find change in entropy during constant pressure and polytropic process. Show on T-S diagram.

For constant pressure process,

22 1 p

1

TS S S mC In

T

For polytropic process,

2 22 1 PIn

1 1

2 22 1 vIn

1 1

T PS S S m C RIn

T P

or

T VS S S m C RIn

T V

32. Explain the term “Reversibility”.

If the process traces the same path when it is reversed is called as reversibility. 33. Can entropy of universe ever decrease? Why? Entropy of universe can not ever decrease. It will be remains constant or will increase due to irreversibility. 34. What is the essence of the second law of thermodynamics?

1. To know the feasibility of process 2. To know about the quality of energy

35. If Carnot engine efficiency is 50%. Find C.O.P. of Carnot refrigerator working between same temperatures.

1 2

1

2

1

2 1

1 2

T TH.E. 50%

T

T1 0.5

T

T T0.5 2

T T

COP of refrigerator 2 1

11 2

2

T T

TT T1

T

1

2 1

1

36. Define the term absolute entropy.

The change entropy of the system with respect to ambient conditions or any other standard reference condition is known as absolute entropy.

PART – B 1. Two heat engines operating on Carnot cycle are arranged in series. The first engine A

receives heat at 927C and rejects heat at a constant temperature T2. The second engine B

receives the heat rejected by the first engine, and in turn rejects heat to a reservoir at 27C. Calculate the temperature T2, in degree Celsius, for the situation where (a) the work output of the two engines are equal and (b) the efficiency of two engines are equal.

System : Two Carnot engines A & B operating in series between 927C and 27C

Known : T1 = 927C = 1200 K

T3 = 27C = 300 K To find : Intermediate temperature T2 when

(a) WA = WB

(b) A = B Diagram :- Analysis : case (a)

WA = WB Since it has been already proved that total work output from any number Carnot

Reservoir at 1200 K

Engine A

Engine B

Reservoir at 300 K 3 (a)

WA

WB

Q2 T2

Q3

engines operating in series is equal to that of a single Carnot engine operating between the same reservoirs.

1

3001

1200A BQ W W

Since WA + WB

1

21 1

2

2

2

3001 2

1200

2 11200

1200 29002

1200 1200

1200 9001200

2 1200

750

A

H

H L

Q W

T TQ Q

T T

T

T

T K

Case (b) When

2

2

12

2

2

3001 1

1200

(1200 300)

600

A B

T

T

T

T K

2. Two Carnot refrigerators are arranged in series. One receives 300 KJ/cycle from a heat source at 300K. The heat rejection from this refrigerator serves as the heat input to a second refrigerator, which delivers its output heat to reservoir maintained at 1000 K. If the two refrigerators have the same COP, determine

a) Heat rejection to the 1000 K reservoir b) The intermediate temperature between the two refrigerators in Kelvin and c) COP of the two refrigerators. System : Two Carnot refrigerators operating in series Known : Q3 = 300 KJ/Cycle T3 = 300 K T1 = 1000 K

To find : (a) Q1 – heat rejection to the reservoir at 1000 K (b) T2 – Intermediate temperature in K (c) COP

Diagrams: (b) (COP)Ref1 = (COP)Ref2

3 2

2 1 1 2

2

2 2

2

2 2 2

2

2

300

300 1000

300 1000 300 300

300 1000

547.7

T T

T T T T

T

T T

T T T

T

T K

(a) 2 2

2 1 2

547.7

1000 547.7

Q T

W T T

2

2

22

1 2

1 2 2 2

1.211

1.211

0.826

1.826

Q

W

QW

W Q

Also Q Q W Q

Reservoir 1 @ 1000 K

Q1

W2

W1

Q2 T2

Ref2

Ref1

Q3

Reservoir 3 @ 300

Where Q2 = Q3 + W1

33

2

2

1.211

1.211

QQ

Q

W

Ref2 Ref1since COP = COP

1

1

1

11

1.211

300 1.826

547.7 /

Q 1.826 547.7

1000.14 /

Q

KJ Cycle

Q KJ Cycle

(c) COPref1 = COPRed2 = 1.211

3. A Carnot engine receives 90KJ from a reservoir at 627C. It rejects heat to the

environment at 27C. One-fifth of its work output is used to derive a Carnot refrigerator.

The refrigerator rejects 60 KJ to the environment at 27C Find:

a) The work output of the engine. b) The efficiency of the heat engine c) The temperature of the low temperature reservoir for the refrigerator in degree

Celsius. d) The COP of the refrigerator.

System : A cyclic heat engine operating a cyclic refrigerator both working on Carnot cycle.

Known :

Heat engine Refrigerator

TH = 627C = 900 K

TL = 27C = 300 K Qin = 90 KJ

WRef = 1

5 Wheat engine

Qout = 60 KJ TH = 300K

To find : (a) Wheat engine

(b) heat engine

(c) TL refrigerator (d) COP Refrigerator

Diagrams :-

1

5heat engineW

Analysis : (a) Wheat engine = 1 Lin

H

TQ

T

3001 90

900

60 KJ

(b) Heat engine = 1 L

H

T

T

3001

900

0.667 (66.7%)

(c) Wref =

1

5Heat engineW

Source at 900K Environment at 300 K

90 KJ 60 KJ

HE Ref

Sink at 600 K Refrigerator space at TL Ref

4

5 Wheat engine

R Re

ReRe

160

5

12

60 12

48

out ef f

inL

fH L f

KJ

Q Q W

KJ

QT

T T W

in Ref

484

300 12

4(300 )

240

L

L

L L

L

T

T

T T

T K

d) COPRef = Re

in

f

Q

W

48

412

4. An irreversible heat pump is designed to remove heat from the atmosphere at 7C and to

supply 43,200 KJ/hr of heat to a constant temperature reservoir kept at 52C. The heat pump is of COP 80% of the maximum possible between the two reservoirs. Power required

running the heat source kept at 1000 K and the reservoir at 52C which is receiving heat

from the heat pump. Taking the efficiency of the heat engine at 70% of supplied to the 52C reservoir and also the heat extracted by the heat engine from the reservoir at 1000 K.

System : An irreversible heat pump driven by an irreversible engine.

Known : Qout HP = 43,200 KJ/hr

43,200

3,600

12 /

52 273 325

7 273 280

1000

52 273 325

H HP

L HP

H HE

L HE

KJ

S

KJ S

T K

T K

T K

T K

To find : (a) Qsupplied to 52% reservoir (b) Qin for the heat engine

Description

Analysis : (a) COPHP = 0.8 HP

HP HP

H

H L

T

T T

3250.8

325 280

5.78

5.78

5.78

HP

HP

out

HP

HP

Out

HP

QAlso COP

W

QW

12

5.78

2.08 /KJ s

3250.7 1

1000

0.4725

2.08

0.4728

HE

HE

HE

HEHE HE HP

in

HPin

HE

WAlso and W W

Q

WQ

h

Source at 1000 K

HE

HP

Qin HE

WHE

WHP

Qout HP

QOut HE

ab

Reservoir at 52C

Reservoir at

7C

= 4.40-2.08 = 2.322 KJ/s

HE HEin outQ Q

Hence the total heat supplied to the

Reservoir kept at 52°C

= 15 + 2.322 = 17.322 KJ/s Or

4.40 /

15.840 /

HEinQ

KJ S

Or

KJ hr

Hence extracted by the heat engine

from the reservoir at 1000K

5. A heat engine operates between the maximum and minimum temperature of 671C and

60C respectively with an efficiency of 50% of the appropriate Carnot efficiency. It drives a

heat pump which uses river water at 4C to heat a block of flats in which the temperature

difference of 10C exists between the working and the river water on the one hand, and the required room temperature on the other hand, and assuming the COP of heat pump to be 50% of the ideal COP that can be obtained under the same working conditions, find the heat input the engine per unit heat output from the heat pump. Why is direct heating thermodynamically more wasteful?

System : A cyclic heat engine operating a cyclic heat pump with their efficiency / COP have

been defined in terms of ideal efficiency/COP.

Known : For heat engine TH = 671 + 273 = 944K TL = 60 + 273 = 333 K

HE = 0.5Carnot For the heat pump

Tsource = 4C

Tsink = 20C

A temperature difference of 10C required on either side.

Diagram Analysis :

. .

1

333 1

944

0.647

0.5 0.647

0.5 0.647

0.323

LCarnot

H

H E

T

T

COPHP = H

H L

T

T T

Where TH = Tsink + 10C = 20 + 10

= 30C = 303 K

TLTSource-10C = 4-10

= -6C COPHP = 0.5COPIdeal

671C 20C

Q1

HE HP

T = 10C

W

Q2

T = 10C

Q3

60C

Q4

4

4

1

3030.5

30 ( 6)

3030.5 4.2

36

1

1 0.238

4.2

0.238 0.736

0.323

HP

HE

When Q KJ

QW

COP

KJ

WQ

Result: Per unit head output from the heat pump, 0.736 KJ of heat is to be given to the heat engine.

Comment: It can be understood from the result that with the help of 0.736 KJ. 1KJ is

supplied to the conditioned space whereas in direct heating whatever the quantity required

it has to be directly supplied. Moreover direct heating results in degradation of energy.

6. A reversible engine works between three thermal reservoirs. A, B and C. The engine absorbs an equal amount of heat from the thermal reservoirs A and B kept at temperature TA and TB respectively and rejects heat to the thermal reservoir C kept at temperature TC. The efficiency of the engine, which works between the two reservoirs A and C. prove that

(2 1) 2(1 )A A

n C

T T

T T

System : A reversible heat engine operating between three reservoirs A, B and C

Known : 1. Heat supplied to the engine from reservoirs A and B are equal.

2. Efficiency of the given engine is equal to times the efficiency a reversible engine operating between reservoirs A and C.

To prove : (2 1) 2(1 )A A

B C

T T

T T

Diagram : Analysis : As the given heat engine is reversible

0

1 1

CA B

A B C

A B

CA

A B C

C C C

A A B

QQ Q

T T T

Since Q Q

QQ

T T T

Q T Tor

Q T T

Also it is given that

1 1

1 12

C C

A B A

C C

A A

Q T

Q Q T

Q T

Q T

C

A

QSubstituting for we get

Q

11

2

2 1

C C C

A B A

C C C

A B A

T T T

T T T

T T T

T T T

TA TB

TC

HE QA QB

QC

Wnet

Multiplying A

C

T

T by we get,

(2 1) 2(1 )A A

B C

T T

T T

7. A solar powered refrigerator receives heat from a solar collector at Th and rejects heat to the atmosphere at Ta, and removes heat from a space at Tc. The three heat transfers are Qh,

Qa and Qc respectively. Derive an expression for the minimum ratio of C

A

Q

Q, in terms of the

three temperatures. If Th = 400K, Ta = 300K, Tc = 270K and Qc = 10KW, what is the minimum Qa? If the collector captures 0.2KW/m2, what is the minimum collector area required? System : A refrigerator driven by solar heat Known : Tn = 400K Ta = 300K Tc = 270K Qc = 10 KW

20.2 /CQKW m

A

To find : (i) An expression for h

c

Q

Q

(ii) To find the minimum are of the collector Diagram

Solar Collector Tn

Atmosphere Ta

Qh

Qa

Ref

Qc

Cold Space Tc

Analysis : (1) The ratio represents the performance index of the refrigerator which will be minimum if the processes are reversible. Therefore,

0h C A

n C A

Q Q Q

T T T

Dividing by QC and rearranging, we get

1 1 1h a

C h C C a

Q Q

Q T T Q T

From energy balance Qh + QC = Qa Therefore

1 1 1

1 1 1 1

h h C

C h C C C a

h h

C h C C a a

Q Q Q

Q T T Q Q T

Q Q

Q T T Q T T

1 1 1 1

/

( /

h

C a h C a

a c a ch

C h a a h

h h a C

C C h a

Q

Q T T T T

T T T TQ

Q T T T T

Q T T T

Q T T T

The expression represents the ratio of h

C

Q

Q

in terms of the three temperatures.

(ii) Substituting the numerical values for Th, Ta, Tc and Qc, we get 400 300 270

10 270 400 300hQ

2

4.4

0.2

h

h

Q kW

Q kwGiven

A m

24.4 22

/ 0.2h

h

QTherefore A m

Q A

Minimum collector are required is 22m2.

8. It is required to maintain a house at a temperature of 17C where the ambient

temperature falls to 0C in winter. A device working on reversed Carnot cycle is proposed to

maintain the temperature of the room. It heat transfer across the walls and roof is estimated as 2000 KJ per degree temperature difference between inside and outside, estimate the power required?

If the same device is used to cool the room during summer by supplying the same amount of power, what is the maximum outside temperature upto which we can maintain

the room temperature at 17C? Case (i) System : A heat pump working on reversed Carnot cycle Known : TL = 273 K TH = 300 K

2000hQ KJ

T h

Diagram Analysis :

1

1

2000( )

2000 17

2000 17

3600

9.44 /

HP

in

Q T

KJ

h

KJ

s

KJ S

QCOP

W

where

(Room)T1

HP

(Atmosphere) T2

1Q

2Q

inW

HHP

H L

TCOP

T T

300

17.65300 273

9.44 0.535

17.65

Lin

HP

QW

COP

KW

Case (ii)

System : A refrigerator working on reversed Carnot cycle.

Known : TL = 300K

0.535inW KW

Q2 = 2000 (TH – TL) To find : TH Diagram : Analysis : It is given that

2

2

2000 ( )

2000 ( 300)( )

3600

H L

H

Q T T

TQ a

As reversed Carnot cycle is being followed

inW

(Atmosphere) TH

Ref

(Room) TL

Q1

Q2

2

2

300( )

300

L

in H L

H

Q TCOP

W T T

Q bT

From equations (a) and (b) we get

2

2000 ( 300) 300 0.535

3600 ( 300)

300( 300)

2000 0.535 3600

317 (44 )

H

H

H

H

T

T

T

T K C

9. An insulated rigid vessel is divided into two chambers of equal volumes. One chamber contains air at 500 K and 2 Mpa. The other chamber is evacuated. If the two chambers are connected d, what would be the entropy change.

System : Closed system Process : Unresisted expansion Known : T1 = 500K

P1 = 2 103 kPa To find : Entropy change Diagrams: Analysis :

2 22 1

1 1

2 22 1

1 1

( )

P

V

T PS S C In R In or

T P

T VS S C In R In

T V

After expansion air will occupy the entire volume of the container.

V2 = 2V1 Also 1W2 = 0 since it is an unresisted expansion Q12 = 0 since the vessel is insulated

Air Vacuum Air

Initial State Final State

Applying the first law of thermodynamics

1 2Q U W

Therefore U = 0 For air

2 1

2 1

( ) 0

. .

VCM T T

i e T T

Hence S2 – S1 = CV in 2 2

1 1

T V

R InT V

= 0.287 In 1

2

2V

V

= 0.199 KJ/kgK Comment: Though the process is adiabatic entropy increases as the process involving unresisted expansion is an irreversible process. It also proves the fact that.

( ) 0dQ

Ds or dsT

10. An adiabatic chamber is partitioned into two equal compartments. On one side there is

oxygen at 860kPa and 14C. On the other side also, there is oxygen, but at 100 kPa and

14C. The chamber is insulated and has a volume of 7500 cc. The partition is abruptly removed. Determine the final pressure and the change in entropy of the universe. System : Closed Process : Adiabatic Mixing Known :

Subsystem I Subsystem II

Fluid Oxygen Oxygen

Initial pressure 850 kPa 100 kPa

Initial temperature 14C 14C

Initial Volume 7500

2

cc

7500

2cc

Diagrams: Analysis : Here the energy interaction is taking place only between the two fluids and therefore the energy lost by one fluid should be equal to the energy gained by the other fluid. Taking tF as the final temperature we get

O2

850 kPa

14C

O2

100 kPa

14C

O2

Initial State Final State

M1C1 (t1 – tF) = m2C2 (tF – t2) Since the same fluid is stored in both the systems at the same temperature C1 = C2 and

t1 = t2 = 14C

Therefore the final temperature will also be 14C

After removing partition total mass of oxygen is occupying the entire 7500cc at 14C. Hence the final pressure can be computed as given below :

1

1 1

1m

PV

RT

Mass of oxygen

in the subsystem1

6850 3750 10

8.314287

32

0.0427 kg

2

2 2

2

6

100 3750 10

8.314287

32

0.00503

m

PV

RT

kg

Mass of oxygen

in the subsystem 2

To find the final pressure

1 2

1 2

1 2

( )

8.314(42.7 5.03) 287

32

7.5

475

F F

F

rF

r

F

System

P Vm m

RT

m m RTP

V

P

kPa

S S S

= m1

2

1 1 2 2

f ftF

V V

V VTTC In R In m C In R In

T V T V

3

8.314 8.3140.0427 2 0.00503 2

32 3

8.596 10

0

8.596

Surroundings

Universe

In In

KJ

K

S

JS

K

11. Two vessels, A and B each of volume 3m4 may be connected by a tube of negligible

volume. Vessel A contains air at 0.7 Mpa, 95C while vessel B contains air at 0.35 Mpa,

205C. Find the change of entropy when A is connected to B by working from the first principles and assuming the mixing to be complete and adiabatic. System : Closed Process : Adiabatic mixing Known :

Properties Subsystem A Subsystem B

Fluid Air Air

Pressure 0.7 MPa 0.35 MPa

Volume 3 m3 3 m3

Temperature 95C 205C

Diagrams : After Mixing Analysis : Since the energy interaction is taking place only between the two fluids energy

lost by one fluid is equal to the energy gained by the other fluid.

Subsystem A Air @ 0.7 MPa3m3

95C

Subsystem B Air @ 0.35 MPa3m3

205C

Before Mixing

Subsystem A

Air @ 2

2

P

T

Subsystem B Air @ P2 T2

After Mixing

QA = QB Taking t2 as the final temperature after mixing ma Ca (t2 – t1a) = mb Cb (t1b – t2). Since in both A and B the same fluid is stored, Ca = Cb Also ma

1

700 3

0.287 368

19.9

A A

A A

p V

R T

kg

mb = 1

B B

B B

p V

R T

2 2

2 2

2 2

2

350 3

0.287 478

7.65

19.9 ( 95) 7.65 (205 )

2.6 ( 95) 205

2.6 205 2.6 95

125.6

kg

t t

t t

t t

t C

Entropy Change = SA + SB

2 2

1 1

125.6 273 619.9 0.717 0.287

95 273 3

5.08

A A V

A A

T VS m C In R In

T V

In In

KJ

k

2 2

1 1

Bm V

B B

T VS C In R In

T V

125.6 273 67.65 0.717 0.287

205 273 3

0.525

In In

KJ

K

5.08 0.525

5.605

0

5.605

sys

surr

universe

S

KJ

K

S

KJS

K

Final pressure P2 = 2

2

mRT

V

(19.9 7.65) 0.287 (125.6 273)

6

525 kPa

12. Air enters a turbine at 400C, 30 bar and velocity 160 m/s. It leaves the turbine at 2 bar,

120C and velocity 100 m/s. At steady state it develops 200 KJ of work per kg of air. Heat transfer occurs between the surroundings and the turbine at an average temperature of 350K. Determine the rate of entropy generation.

System : Open Process : Steady flow Known :

Properties Inlet Outlet

Pressure 30 bar 2 bar

Velocity 160 m/s 100 m/s

Temperature 400C 120C

Ambient temperature = 350 K Work output = 200 KJ/kg Diagram:

160 m/s 30 bar

400C

Ambient @ 350 K

400C

w = 200 KJ/kg of air

2 bar 120C 100 m/s

To find : Rate of entropy generation

Analysis :

( ) ( )surr CV

Rate of entropyS S

generation

2 1( ) ( )CVS m S S

For unit mass 2 2

1 1

( ) CV P

T PS C In R In

T P

393 21.005 0.287

673 30

0.236 /

( ) sursur

sur

In In

KJ kgK

QS

T

Where 2 2

2 12 1)

2000CV

V Vsur Q W m h h

Q

2 2100 160200 1 1.005(393 673)

2000

89.2 /

89.2( )

350

0.255 /

sur

sur

Q KJ kg

S

KJ kgK

Rate of entropy generation = 0.255 – 0.236 = 0.019 KJ/kgK. 13. A turbine operating at steady state receives air at a pressure of p1 = 3.0 bar and temperature of 390K. Air exist the turbine at a pressure of p2 = 1.0 bar. The work developed is measured as 74 KJ/kg of air flowing through the turbine. The turbine operates adiabatically, and changes in kinetic and potential energy between inlet and exit can be neglected. Using ideal gas model for air, determine the turbine the turbine efficiency. System : Open Process : Steady flow Known : P1 = 3.0 bar P2 = 1.0 bar T1 = 390 K Wa =74 J/kg Diagrams:

Air out

1 bar

Air in (1)

W

~

T

s

11

3 bar

2s

2

Analysis : 1 2

1 2

t

s

h h

h h

1 2

12

T T

T T s

for an ideal gas

where

1 1

2 2 2

1 1 1

r r

r rs sT T P

T P P

0.4

1.4

2

2

1390

3

284.9

s

s

T

T K

1 2

1 2

( ) 74

74

1.005

73.63

aP

WC T T

m

T T

K

Hence 1 2

1 2

t

s

T T

T T

73.63

390 284.9

0.7 ( 70%)or

14. A closed system is taken through a cycle consisting of four reversible processes. Details of the processors are listed below. Determine the power developed if the system is executing 100 cycles per minutes.

Temperature (K)

Process Q(KJ) Initial Final

1 – 2 2 – 3 3 – 4 4 – 1

0 + 1000 0 -

300 1000 1000 300

1000 1000 300 300

System : Closed Process : The system is executing cyclic process

Known : Heat transfer in process 12, 23 and 34 and Temperature change in all the process. No. of cycles per minute.

To find : Power developed. Diagrams : Analysis : To find power developed Wnet per cycle must be known. From I Law Wnet = Qnet which can be computed from the following table.

Process Q(KJ) Temperature (K)

S Initial Final

1 – 2 2 – 3 3 – 4 4 – 1

0 1000 0 -

300 1000 1000 300

1000 1000 300 300

0 1000

11000

KJ KJ

k k

0

S41

For a cyclic process = 0

where is any property

s = 0

(i.e) S12 + S34 + S41 = 0

0 + 1 + 0 + S41 = 0

S41 = 1KJ

K

Since the process 4-1 is isothermal

41

41

1300

300

Q

Q KJ

Therefore

T(K)

1000

300

3

4

2

1

S

12 23 34 41

0 1000 0 300

700

netQ Q Q Q Q

KJ per cycle

700 net netW Q KJ

And power developed

sec

100700

60

1166.7

netW Cycle

Cycle

KW

15. Two kilogram of air is heated from 200C at constant pressure. Determine the change in entropy.

System : Open / closed Working : Air fluid Process : Constant pressure heating

Known : 1) t1 = 200C

2) t2 = 500C Diagram :

To find : Change in entropy s Analysis :

t(C)

500

200

1

2 p = C

S

2 2

1 1

2

1

500 273 2 1.005

200 273

0.987 / .

p

P

T PS m C In R In

T P

TmC In

T

In

KJ K

16. A Carnot engine operated between 4C and 280C. If the engine produces 300 KJ of work, determine the entropy change during heat addition and heat rejection.

System : Open / Closed Process : The working fluid is executing Carnot cycle

Known : 1) t1 = 280C

2) t2 = 4C 3) W = 300 KJ Diagram :

To find : 1. s during heat addition

2. s during heat rejection Analysis : In Carnot engine heat is added at constant temperature Therefore

Source @ 280C

Heat Engine

Sink @ 4C

300 KJ

Qin

Qout

1

2

1

T = 1 -

4 273 1 0.499

280 273

1.087 /

in

in

QS

T

WWhere Q

T

KJ K

Therefore Qin = 300

0.499 = 601.1 KJ

1

601.1

(280 273)

inQS

T

2. In Carnot engine heat rejection is also taking place at constant temperature. Therefore

2

601.1 300

out

out in

QS

T

where Q Q W

301.1

301.1

(4 273)

1.087 /

KJ

S

KJ K

Comment:

In a Carnot change two isothermal process and two isentropic process. Therefore s

during heat addition must be equal to S during heating rejection so that.

0ds

which obeys Clausius inequality. 17. Air flows through a perfectly insulated duct. At one section A the pressure and

temperature are respectively 2 bar 200C and at another section B further along the duct

the corresponding values are 1.5 bar and 150C. Which way the air flowing? System : Open Process : Steady flow process Known : 1. P1 = 2 bar

p = 2 bar

t = 200C

Section A Section B

p = 1.5 bar

t = 150C

2. t1 = 200C 3. P2 = 1.5 bar

4. t2 = 150C To find : To know flow direction Diagram : Analysis : This problem cannot be solved by simple application of first law of thermodynamics. Because there is nothing to tell us whether the fluid is expanding from A to B or being compressed from B to A.

However, since the duct is insulted the inference is that there is no heat transfer to or from the environment and therefore there is no change of entropy in the environment. But in any real process change of entropy of the system plus the surroundings must be

positive. In other words SAB > 0.

B BB A p

A A

T PS S C In R In

T P

273 150 1.5 1.005 0.287

273 200 2

0.02966

In In

KJ

kg K

Thus SA > SB and the flow is from B to A. Even though entropy cannot be measured directly it can still be used to find the sense of flow in a well insulated duct given two salient states as above. 18. A certain fluid undergoes expansion in a nozzle reversibly and adiabatically from 500

kPa, 500 K to 100 kPa. What is the exit velocity? Take 1.4 0.287KJ

and RkgK

System : Open Process : Reversible adiabatic expansion Known : 1. Inlet pressure = 500 kPa 2. Inlet temperature = 500 K 3. Exit Pressure = 100 kPa 4. The ratio of specific heats = 1.4

5. Characteristic gas constant = 0.287 KJ

kgK

To find : Exit velocity Diagram

Analysis: applying steady flow energy Equation

Q W m h ke pe

(1) (2)

Flow diagram

therefore 2 2

2 11 2

C Ch h

2

2 p 1 2C 2C T T

Where Cp and T2 unknowns

To find Cp P

1C 1 R

p

RC

1

Substituting and R we get cp = kJ

kgK

It is stated in the problem that the process of expansion reversible. Therefore,

dQ

dsT

Also the process is given as adiabatic. That is

2 1

3p

1 1

22 1

P 1

e

dQ0

T

(or) ds 0

S S 0

PTC In Rin 0

T P

pRT Te In

C p

0.287 100 500 In

1.005 500

315.8 K

Substituting numerical values for T2 and Cp, we get

2C 2x1005 (500 315.08)

608.5 m/s

19. Show from the first principle that, for a perfect gas with constant specific heat capacity expanding polytropically (Pvn = constant) in a non-flow process, the change of entropy can be expressed by

22 1

1

PnS S xIn

1 P

Gaseous methane is compressed polytropically by a piston from 25 and 0.8 bars to a pressure of 5.0 bar. Assuming an index of compression of 1.2, Calcutta the change of entropy and work done, per unit mass of gas. The relative molecular weight of methane is

16 and = 1.3. System : Closed Process : Polytrapic (pVn = C) Known : 1. T1 = 298 K 2. P1 = 80 kPa 3. P2 = 500 kPa 4. n = 1.2 5. M = 1.6

6. = 1.3 To find:

1. 1W2 – Work done

2. S – change inentropy Analysis : a) To prove

22 1

1

Pn RS S x xIn

1 n P

From first law of Thermodynamics

Q12 = 1w2 + U

2 2 1 1v 2 1

2 1

v 2 1

v 2 1v 2 1

v 2 1

v 2 1

v 2 1

p v p vC T T

n 1

R T TC T T

n 1

C ( 1)T TC (T T )

n 1

11 C (T T )

n 1

1 n 1C (T T )

n 1

nC (T T )

n 1

In differential for

v

ndQ C dT

n 1

for a polytropic process

Therefore ds = dQ

T

v

n dTC

n 1 T

Upon integration we get

22 1 v

1

TnS S C In

1 T

From the process relation

n 1

n2 2

1 1

T P

T P

Substituting for 2

1

T

T we get

n 1

n2

2 1 v

1

22 1 v

1

PnS S C In

1 P

Pn n 1S S C x xIn

n 1 n P

We know that

R = Cp - Cv

R = Cv ( - 1) 22 1

1

Pn RS S x xIn

n 1 n P

Cv = R

1

Substituting for Cv we get (2) Work done

2 11 2

p vW

n 1

2 1R(T T )

n 1

Where T2 = T1

n 1

n2

1

p

p

=

0.2

1.25298

0.8

= 404.45 K Substituting numerical values

1 2

8.3142(404.45 298)

16W

1.2 1

= -276.6 kJ/kg. (3) Change in entropy

22 1

1

pn RS S x xIn

1 n p

1.2 1.3 8.314/14 5 In

1.3 1 1.2 0.8

kJ 0.2645

kg K

Comment: The negative sing in work indicates that work is given into the system. The negative sign in entropy change indicates that there is heat rejection by the system to the ambient during compression.

20. In a refrigerant condenser superheated vapour of ammonia enters steadily at 1.4 Mpa,

70C. It leaves the condenser at 20C. At 1.4 Mpa condensation begins and ends at 36.28C.

Cooling water enters condenser at 10C and leave 15C. Determine.

a) The amount of heat rejected per kg of ammonia vapour condensed for the given inlet and exit conditions.

b) Mass of water to be supplied for each kg of ammonia vapour condensed. c) The change in specific entropy of ammonia d) The entropy generation per kg of ammonia

Take Cpvapour = 2.9 kJ/Kg K, Cpliquid = 4.4 KJ/KgK and latent heat of evaporation of ammonia at 1.4 Mpa is 1118 KJ/Kg. Also represent the process in a T-S. Diagram: System : Open Process : Steady flow process

Know : T1 = 70C P1 = 1.4 Mpa

T2 = 20C

Tw1 = 10C

Tw2 = 15C To find : a) the mount of heat rejected per kg of ammonia vapour condensed for the given inlet and exit conditions. (b) mass of water to be supplied for each kg of ammonia vapour condensed. (c) the change in specific entropy of ammonia (d) the entropy generation pr kg of ammonia. Diagrams : Schematic diagram

Analysis : (a) Heat rejected per kg of ammonia Q12 = Q12a + Q2a – 2b + Q20.2

p vapour 1 2 p liquid 2b 2

Latent heatC t t C t t

of evaporation

2.9 (70 36.28) 1118 4.4 (36.28 20)

1287.42 kJ/kg.

(b) Water flow rate required per kg of ammonia.

p water(mC T) 1287.42

1287.42 m

4.186 (15 10)

kg of water 61.51

kg of ammonia

(c) Change in Specific entropy of ammonia

1 2a 2a 2b 2b 2

2a 2pvapour p liquid

1 2a 2b

S S S

T TLatent heatC xIn C x In

T T T

36.28 273 ( 1118) (20 273) 2.9 In 4.4 x In

70 273 (36.28 273) 36.28 273

kJ 4.153

kg

(d) water ammoniaS S

Where owater p

m

TS mC In

T

288

61.51 4.186 In283

4.509

Substituting the values we get

universeS 4.509 ( 4.153)

kJ 0.356

kg of ammonia

Comment: As heat is removed from ammonia its entropy decreases where entropy of water increases as it receives heat. But total entropy change will be positive as heat is transferred through finite temperature difference. 21. The specific heats of gas are given by Cp = a + kT and Cv = b+kt, where a b and k are constants and T is in K. Show that for an isentropic of this gas. Tb Va-b ekT = constant System : Closed Process : Isentropic Known : 1. Cp = a + kT 2. Cv = b + kT To prove : Tb Va-b e kT = constant for an isentropic process. Proof : for a gas Cp – Cv = (a + kT) – b (b+kT) (or) R = a-b For an isentropic process Ds = 0

(or) v

dT dvC R 0

T v

Substituting for Cv and R

dT dv(b KT) (a b) 0

T v

Upon integration Bin T + KT + (a – b) Inv = constant Taking antilog TbeKT

Va-b = constant.

22. Calculate the entropy change of the universe as a result of the following process:

(a) A metal block of 0.8 kg mass and specific heat capacity 250J/kgK is placed in lake at 8C

(b) The same block, at 8C, is dropped from a height b of 100 m into the lake.

(c) Two such blocks, at 100C and 0C, are joined together. Case (a) System : A metal block Process : Cooling the metal block by dipping it in a lake. Known : 1. Initial temperature of the block (T1) = 100 + 273 = 373 2. Final temperature of the Block (T2) = 8 + 273 = 281 K 3. Mass of the metal block (m) = 0.8 Kg. 4. Specific heat capacity of the metal block.

J(C) 250

kgK

To find : Entropy change of the universe. Diagram : Analysis :

(d) universe surroundingsS S

Where Swater = mCIn 2

1

T

T

2810.8 250 In

373

J56.6

K

surroundings

surroundings

surroundings

sur sys

2 1

surroundings

QS

T

WhereQ Q

m C (T T )

0.8 250 (281 373)

18400 J

18.400S

281

65.4

1

8K

Substituting the values we get

Ssurroundings = -56.6 + 65.48 = 8.84 J/K Comment: As discussed earlier the entropy change of the universe is positive. The reason is the irreversible is positive. The reason is the irreversible heat transfer through finite temperature difference from the metal block to the lake. Case (b) System : A metal block Process : Falling of the metal block into the ake and reaching equilibrium. Known : 1. Initial temperature = 281 K 2. Final Temperature = n281 K 3. Initial Height = 100m 4. Mass of the metal block (m) = 0.8 kg 5. Specific heat capacity of the metal block (C) = 250 j/kgK.

Where Ssystem = 0, as the system is at the same temperature at both the initial and final state.

surroundings

surroundings

surroundings

surroundings system

surroundings

surroundings

QS

T

S E

=mgh

=0.8 9.81 100=784.8J

784.8S 2.79J/K

281

S 2.79J/K

Comment: increase in entropy of the universe indicates that there is a irreversibility or degradation of energy. That is the high grade potential energy is converted low grade energy during this process. Case (c) Systems : Two metal blocks Process : Two metal blocks are brought in thermal

contact and allowed to reach equilibrium Known : Initial temperature of the blocks T1a =373K T1b = 273 K To find : Entropy change of the universe Diagrams: (a) (b) (a) (b) Initial State Final State

universe a b

2a n

1

2a

1

S S S

TAnalysis : where S mCI

Ta

T S mCIn

Tb

To find T2 Qa = -Qb

2 ta 2 1

1a 1b2

m T T m T Tb

T TT

2

373 273323K

2

1000C 00C T2 T2

2 2universe

2

T TS mc In In

Tfa T1b

323 = 0.8 250 In

373 273

J =4.85

K

Comment: In this process also the heat transfer through finite temperature difference makes the process irreversible which in turn results in increase in entropy of the universe. 23. Find the maximum work developed when air expands in piston-cylinder assembly from an initial state of 600 kpa and 150 kPa and 500C. Also find the availability at the initial and final states, max useful work and change in availability. Assume T0 = 300K, P0=100kPa. Worked Fluid : Air System :Closed Process : a polytropic process Known : P1 = 600kPa P2 = 150kPa T1 = 423 K T2 = 323 K To find: Max work, availability at the initial and final states, change in availability and max useful work

600 Kpa 423 K Air

Analysis: a) Max work = Wrev = (U1-U2) – T0(S1-S2)

1 1v 1 2 0 p

2 2

T PC T T T C In RIn

T P

423 6000.717 (423-323) - 300 1.005In 0.287In

323 150

109.74kJ/kg.

b. Availability at the initial state

150 Kpa 323 K Air

1 1 0 0 1 0 0 1 0

01 1 1v 1 0 0 p

1 0 0 0

U U P V V T S S

RTRT T PC T T P C In R In

P P T P

0.287 423 0.287 3000.717 423 300 100

600 100

423 600300 1.00In R 73kJ/kg

300 100

b. Availability at the final state

1 2 0 o 2 0 0 2 0

02 2 2v 2 0 0 P

2 0 0 0

U U P V V T S S

RTRT T PC T T P C In R In

P P T P

0.287 323 0.287 3000.717(323 300) 100

150 100

323 1501.003 In 0.287 In 4.83kJ/kg

300 100

d. Change in availability

1 2

73 4.83

68.17kJ/ kg

e. Maximum useful work = change in availability = 68.17 kJ/kg 24. Determine the available energy of 80kg of water at 1000C. Temperature of the surroundings is 150C. System: Closed Process: (To determine the available energy) water is assumed to be cooled at constant pressure to a dead state. Known: 1. Mass of Water m = 80kg 2. Temperature of water = 1000C 3. Atmospheric Temperature = 150C To find : Available Energy

Diagram: Analysis : From the T-S diagram it is evident that AE = Area under AB – Area BCDE = Q – TB(SD – SC)

AA B B

B

AA B B

B

TmC T T T mCxIn

T

Tmc T T T In

T

373

80 4.186 100 15 288In288

3522.3kJ

Comment : to determine the available portion of total energy content, the system is assumed to undergo an imaginary process in which it brought to dead state that is state at which comes in equilibrium with the surroundings. 25. Consider the transfer of 1000kJ of heat from a reservoir at 1200K to 5 kg of a gas initially at 100kPa and 500 K in a closed tank. For the gas, Cv=0.8 kJ/kg K throughout the temperature range involved. The lowest temperature in the surroundings is 300K. Determine how much of heat removed from the reservoir is available and unavailable and how much of that absorbed by the gas is available and unavailable. Case (i) Working Fluid: Air Case (i) Working Fluid: Air System A reversible heat engine is assumed to receive the heat from the reservoir at 1200K and rejecting heat to 300 k sink. Known : Quality of heat extracted = 100kJ

1000C

B

150C

UAE

AE E

C S D

To find : Available and unavailable part of this 1000kJ Diagrams:

AE – Available Energy, UAE – Un Available Energy Analysis a. Available Energy = Area ABCD in the Ts diagram.

= (1200 – 300) S

Qwhere s=

1200

10000.833kJ/K

1200

AE (1200 300) 0.833

750kJ

b. Unavailable Energy = 300 x s =300x 0.833 = 250 kJ Alternatively : UAE = Q – AE = 1000 – 750 = 250kJ

Reservoir @ 1200K

Rev

HE

Reservoir @ 1200K

AE

JAE

S

T(K)

1200

Available Energy

D Unavailable

Energy

E F

B A

300

Case (ii) System : Closed Process : 1000kJ of heat received by the system at constant volume Known : P1 = 100kPa T = 500K Cv = 0.8kJ/kg T0 = 300K To find : Available and unavailable parts of the 1000 kJ given to the gas Diagrams: Initial state Final state Analysis : a. AE = Area ABCE b. UAE = Area CDEF

2 2v 2 1

1 1

2

1

2

v 2 1

T VWhere s=m C In R In Since V V

T V

T5 0.8 In

T

To find T : Consider First Law

Q-W = U since dv = 0

U=mC T T 1000

5 Kg of gas 100 kPa 500@

5 Kg of gas @ V2 = V1

UAE D

E

AE

A

B

C

F

T(K)

500K

2 1

v

1000T T

mC

1000500

5 0.8

750

750s 5 0.8 In 1.62kJ/K

500

Hence UAE=300 1.62

=486.6kJ

AE=1000-486.6=513.4kJ

26. Helium enters an actual turbine at 300 kpa, 3000C, and to 100kpa, 1500C, heat transfer to the atmosphere at 101.3 kpa, 250C amounts to 7.0kj/kg. Calculate the entering stream availability, the leaving steam availability, and the maximum work. For helium Cp = 5.2kJ/kgk and. Molecular weight = 4.003 kg/kg.mol System: open Process: Steady flow process Worked Fluid: helium Known: P1 = 300 kpa P2 = 100 kpa T1 = 473 k T2 = 423 k Qcv = -7.0KJ/kg P0 = 101.32 kpa T0 = 298k To find: Availability at the inlet and outlet, maximum work Diagrams: Analysis: Availability at the inlet

Qcv=7.0kJ/kg

100kpa 1500C

in

300 kpa 2000C

1 0 0 1 2

1 1p 1 0 0 p

2 0

V H H T S S

T PC T T T C in R in on unit mass basis

T P

423 100=5.2 423-298 298 5.2 in 2.077 in

298 101.32

650 298 1.848

99.1kJ/kg

maximum 1 2W

866 99.1

766.9kJ/kg

27. An ideal gas having a constant pressure specific heat of 1.6 KJ/kgk and a molar mass of 30 KJ/kg. Mol, initially at a pressure of 6.7 Mpa and a temperature of 4250C, under goes a steady Flow process until it reaches a pressure of 1.3 MPa and a temperature of 1500C. If the environment is at 100KPa and 250C, find the reversible work that would be obtained in this process: System : Open Process : Steady flow process Working Fluid: An ideal gas Known : M =30kg/kg.mol Cp =1.6kj/kgK P1=6.7mPa P2 = 1.3 < Pa T1 = 4250C T2 = 1500C T0 = 250C P0 = 100 kPa To find: Reversible Work Diagrams: Analysis:

P0 = 100KPa

T0 = 250C

P2 =100KPa T2 = 1500C

In

6.7 MPa 4250

Wrev=(h1-h2) – T0 (S1 – S2)

1 1p 1 2 0 p

2 2

T PC T T T C In R In

T P

8.314 8.314where R= 0.277kJ/kgK

M 30

425 273 6.71.6 425 150) 298 1.6 In 0.277 In

150 273 1.

336.6kJ/kg

28. In an adiabatic mixing chamber, 80 kg of water at 1000C are mixed with 50 kg of water at 600C. Determine the decrease in available energy due to mixing. The ambient temperature = 1500C. System : There are two streams of water mixing adiabatically in a mixing chamber Known : 1. Temperature of stream 1 = 1000C 2. Temperature of steam 2 = 600C 3. Mass of stream 1 = 80 kg 4. Mass of stream 2 = 50 kg 5. Ambient temperature = 150C Diagram: Analysis: Decrease in available energy is the difference in the total available energy before mixing and after mixing Total available energy Available Energy of Stream 1

before mixing Available Energy of stream 2

1 11 1 a a 2 2 a a

a a

T Tm C T T T in m C T T T in

T T

373 83380 4.186 100 15 288 in 50 4.186 60 15 288 in

288 288

4189.4kJ

MIXING

CHAMBER

80 Kg 1000C

50kg 600C

130kg T3 = ?

Available energy after mixing

31 2 3 a a

a

Tm m C T T T in

T

Where T3 is the temperature after mixing which can be obtained through energy balance. Q1=Q2 m1C(T1 – T2)=m2C(T3-T2) m1T1+m2T2 = (m1+m2)T3

3

80 373 50 333T

80 50

357.6K

Therefore AE after mixing = 357.6

80 50 4.186 357.6 288 288 in288

=39951.3kJ Decrease in available energy due to mixing =4189.4 – 3951.8 = 237.6kJ Comment: Mixing is an irreversible process and hence available energy decreases due to mixing. 29. Oxygen gas is throttled from 5 bars and 270C to 1 bar through a well insulated valve. Determine the reversible work and irreversibility Take T0 = 288K. System : open Process : Steady flow process Known : 1. Inlet pressure P1 = 500 kPa 2. Inlet temperature T1 = 300 K 3. Exit pressure P2 = 100 kPa To find : 1. Reversible work (Wrev) 2. Irreversibility (1) Diagram:

O2

5 bar

270 C

Analysis:

rev 1 2 0 1 2W h h T S S

For a throttling process h1 = h2 and T1 = T2

1 10 p

2 2

10

2

T PT C in R in

T P

PT R in

P

rev

8.314 500288 in

32 100

120KJ/kg

Irreversibility (1) = W 120KJ/kg

30. Air enters an adiabatic nozzle, operating at steady state with negligible velocity. At the inlet, the pressure is 180 kpa, and the temperature is 650C. The mass flow rate is 0.15 kg/s, and the exit pressure is kpa. If the exit velocity is 300 m/s, pressure is 100 kpa. If the exit velocity 300 m/s, determine:

a. The isentropic efficiency of the nozzle b. The exit temperature c. The irreversibility of the process d. The exit area of the nozzle

Assume T0 = 250C and P0 = 100 kPa. System : Open Process : Steady flow process Working Fluid : Air (ideal gas) Known: P0 = 100 kPa P1 = 180 kPa

P2 = 100 kPa T1 = 650C C2 = 300 m/s T1 = 338 K T0 = 298 K

Diagrams:

To find:

1. istentropic 2. T2 3. I 4. A2

Analysis: As shown in the h – s diagram 1 – 22 Isentropic 1 – 2 Actual Since P2 = P2s

1

2s 2s

1 1

1

21 1

1

0.4

1.4

2

T P

T P

PS T

P

100S 338

180

285.75K

180kPa 650C in

100kPa

S

h

1

2

2s

Also from I law thermodynamics

2 2

2 11 2

2 2

2 1p 1 2

2 2

2 12 1

p

Q W m h ke pe

h ke

C Ch h

2

C CC T T

2

C CT T

2C

300

3382 1.005

T2 = 293.2K or T2 = 20.20C

1 2isentropic

1 2s

p 1 2

p 1 2s

h ha.

h h

C T Tsince air is an ideal gas

C T T

338 - 293.2=

338 285.75

= 0.86 T2 = 293.2k or b) Exit temperature

T2 = 20.2C

c) Irreversibility = T0 [(S2 – S1)sys+Ssurr] since the flow is adiabatic

2 20

1 1

293 1000.15 298 1.005 in 0.287

338 180

1.153 /

p

T PmT C in R in

T P

in

kJ S

d) Exit area

2

2 2

22

2

-4 2 2

m =

P

RT

0.15 =

100300

0.287 293.2

=4.21 10 4.21

mA

PC

C

m or cm

31. Carbon dioxide gas is contained in a 1.0 m3 tank initially at 1.2 bar 300k. The temperature is increased to 400 k by supplying heat from a reservoir at 500k. The ambient conditions are 1.0 bar and 300k. find the irreversibility. Take CP = 1.043kJ/kgk. And Cv =0.854kJ/kg System : Closed Worked : CO2 – ideal gas Process : Constant volume heating Known : P1 = 1.2 bar T2 = 400k T1 = 300k Tres – 500k T0 = 300k P0 = 100kpa Cp = 1.043 kJ/kgk Cv = 0.854 kJ/kgk To find : Irreversibility Diagram: Analysis: I advance volume is remaining constant

02 0 1 21 2 0 1

1R

TS S QR

TU U P V V T

Reservoir

500K

Tank V = 1.0m3 Carbon dioxide

02 21 2 0

1 1

1 1

1

p

R

V 2 1

T in 1

T

8.314 C

8.314 =

44

=0.1889kJ/kgk

120 1m=

0.1889 300

= 2.12kg

Q

=W

=mC

1=2

V P R

R

v

system

act

TVmC T T mT C Rm Q

V T

Vm

RT

R also Cmoleculareweight

Q

U

T T

hence

.12 0.854 300-400 2.12 0.854

300 3002.12 0.854 400 300 1

400 400

181 156 72.4

47.4

in

kJ

32. The inlet and exit conditions of a working fluid expanding in a turbine are listed below.

Heat is rejected by the fluid at an average temperature of 200C whereas the work developed is 650kJ/kg. Neglecting the changes in kinetic and potential energies and

assuming T0 = 25C and P0 = 100kpa find: a) The heat transfer b) The reversible work c) The irreversibility

Properties P mpa T C Hkj/kgk Skj/kgk

Inlet 2.5 400C 3240 7.0148

Outlet 50kpa 81.33 2530 7.2688

System : Open Process : Steady flow Working : Not known

Known : Exit and inlet conditions W = 650kJ T0 = 298K P0 = 100kpa TRejection = 473K Diagrams: S = 7.0148 KJ/kgk S1 = 7.0147kJ/kgk

Analysis : Q – W = [ ]m h KE PE

on unit mass basis

Q = W+h = 650 + (2530 – 3240) = -60 kJ/kg Heat is rejected by the system to the surroundings b) Reversible work

0

01 0 0 1 0 Re

Re

1

1

2983240 2530 298 7.0158 7.688 60 1

473

763.2 /

in out

j

g

Tq

T

Th h T S S q

T

kJ kg

c) Irreversibility

763.2650

113.2 /

rev actW W

kJ kg

33. A lead storage battery of the type used in an automobile is able to deliver 5.2 mj of electrical energy. This energy is available for starting the car.

Suppose we wish to use compressed air for doing an equivalent amount of work in starting

the car. The compressed air is to be stored at 7 mpa 25c. What volume of tank would be required to have the compressed air have an availability of 5.2mj? System : Closed existing in a thermodynamicstate Working fluid : Air (Ideal gas) Known : Pressure and Temperature P = 7Mpa T = 25+273 = 298K And the required total energy To find : Volume of the air required Analysis : Following steps can be adopted to solve the problem

1. Finding the availability of 1 kg of air @ 7MPA and 298K 2. Computing Mass air required for the required 5.2MJ 3. Calculating the volume required from the mass.

Step 1: Availability @ 7MPa and 298K

0 0 0 0 0 0

00 0 0

0 0 0

0

in

T=T

0.287 298 0.287 298 7000100 298 0287

7000 100 100

279.2 /

v P

u u p V V T S S

RTRT T PC T T P T C Rin

P P T P

Since

In

kJ kg

Step 2: Mass required to supply 5.2MJ

35.2 10

279.2

18.625kg

Step 3: Required volume

3

18.625 0.28 298

7000

0.222

mRTV

P

m

34. At a certain location the temperature of the water supply is 15C. Ice is to be made form this water supply by the process shown in the following figures. The final temperature of ice

is -10C, and the final temperature that is used as cooling water in the condenser is 30C? What is the minimum work required to produce 1000kg of ice?

System: A refrigerator that converts water at 15C into ice at - 10C by pumping the heat from it and reject the heat to the remaining water increasing its temperature from 15 to

30C Known:

1. Inlet temperature of water Tla = 288K and Tib = 288k

2. Exit temperature of ice T2a = 263k 3. Exit temperature of condenser T2b=303k

To find: Wmin Analysis: From energy balance across the refrigeration we get Q1 = Q2 +W where

2

1

heat1000 288 273 273 263

of fusion

= 1000 4186 288-273 335 2.09 273 263

= 418690kJ

Q 4.186 30 15

Pwater pice

cw

LatentQ C C

m

The term mcw is the mass of condenser cooling water which is an unknown its value corresponding to the minimum work can be obtained as follows. For the work to be minimum heat rejection to the condenser cooling water and formation of ice must be reversible processes. Thus the entropy change of the universe is zero.

0

0

universe

a b

S

S S

Where

22731000

273 273

273 335 2631000 4.186 2.09

288 273 273

apw picea

la

TheatS C in C in

T

in in

1529.0

3034.186

288

0.212

cwb

cw

kJ

k

S m in

m

Substituting the results we get -1529.0+0.212mcw = 0

(or) mcw =1529

0.212

=451722kJ Wmin = Q1-Q2 =4517222.4 – 4186990 =33032kJ =33.03kJ 35. The exhaust from a gas turbine at 1.12 bar, 800k flows steadily into a heat exchanger which cools the gas to 700k without significant pressure drop. The heat transfer from the gas heat an air flow at constant pressure, which enter the heat exchanger at 470k. The mass

flow rate of air is twice that of gas and the surroundings are at 1.03 bar, 20. Determine a) The decrease in availability of the exhaust gases, and b) The total entropy production per kg of gas what arrangement would be necessary to make the process reversible and how much would this increases the power output of plant per kg of turbine gas? Take Cp for exhaust gas as 1.08 and for air 1.05 kJ/kgk. Neglect heat transfer to the surroundings and the changes in kinetic and potential energy System : open Process : steady flow Known

1) Inlet temperature of the gas Tg1 = 800k 2) Exit temperature of the gas Tg2 = 700k 3) Inlet temperature of the air Ta1 = 470k 4) mair = 2mg 5) Temperature of the surroundings T0 = 293k 6) Cpg = 1.08kj/kgk 7) Cpg = 1.05ij/kgk

To find:

a) Decrease in availability of the exhaust gas b) Entropy production c) Means of making the process reversible and the corresponding increase in power

output.

Diagram:

Analysis:

a) Decrease in availability of exhaust gas

g1 g2

1 2 0 1 2

h h T S S

g1

pg 1 2 0

g2

TC T T T In

T

= 1.08 ( 800-700 ) – 293 x In 800

700

= 66 KJ

kg

b) Entropy production

gas air

s S

Where

g2

psgasg1

TS C xIn

T

= 1.08 x In 700

800

= -0.1442 KJ

kgk

a2a p2air

a1

TS m C xIn

T

For each kg of the exhaust gas, heat transferred to air

= 1 x pg g1 g2C x T T

= 1x1.08 x In 800 - 700

= 108 KJ

kg

This heat is fully transferred to air therefore,

a pa g1 g2 m C T T 108

a2

108T 470

2x1.05

521k

Substituting Ta2, we get

air

521S 2x1.05xin

470

= 0.216 KJ

kg of gas k

Entropy production

= 0.2164 – 0.1442

= 0.216 KJ

kg of gas k

Entropy production = 0.2164 – 0.1442

=0.216 KJ

kg of gas k

c) In order to make the process of heat transfer reversible a heat engine must be

introduced as shown below: As all the heat are not given to air the exit temperature of air can no longer be 521 k the actual exit temperature must be such that the entropy change of universe is zero

mathematically

gas air

S universe = 0

That is s s 0

g2 g1

pg pa

g1 g2

T TC in C in 0

T T

1.05 x In a2T 8001.08xin

470 700

Ta2=503k Work delivered form the heat engine is the increase output which can be obtained from the energy balance. That is W = Q1 – Q2 = mg Cpg ( Tg1-Tg2) –mgCpg ( Tg1-Tg2) = 1 x 1.08 ( 800 – 700 ) – 2 x 1.05 ( 503 – 470 )

= 38.7 KJ

kg of gas

36. An inventor claims to have developed an efficient hot engine which would have a heat source at 10000 C and reject heat to a sink at 500 C and gives a efficiency of 90% Justify his claim is possible or not. Given data:

0

H

0

L

T 1000 c = 1273k

T 50 C 323k

n 90% Solution: According to Carnot theorem reversible engine gives maximum efficiency than other heat engine. Maximum efficiency,

h Lmax

H

T T 1273 323n 0.476 74.6%

T 1273

Maximum efficiency ( 74.6% ) is less than proposed engine efficiency ( 90% ). Therefore his claim is impossible. Ans. Result:

Inventor’s claim is impossible. 37. An inventor claims that his propose engine has the following specification: Power developed = 50 kW Fuel burnt = 3 kg/hr Calorific value of the fuel = 75000 KJ/kg Temperature limits = 270 C and 6270C Find out whether it is possible or not. Given data: P = 50kW Fuel burnt = 3 kg/hr C.V. of fuel = 75,000 kj/kg

Temperature limits, 0

L

0

H

T 27 C 300k

T 627 C 900k

Solution: Heat supplied to the engine = Fuel burnt x C.V = 3 x 75000 = 225000KJ/hr QS=62.5kJ/sec=62.5kW Work done, p = 50kW From Carnot theorem cannot engine gives maximum efficiently than any other engine.

H Lcarnot max

H

T TCarnot efficiency, n n

T

900 300

0.66 66.6%900

Efficiency of inventor’s claim engine,

actual

s

workdone W 50n 0.8 80%

Heat supplied Q 62.5

here the inventor’s claim engine has the higher efficiency than maximum engine efficiency which is possible. Result:

Inventor’s claim is impossible

38. Determine whether the following cases represent the reversible irreversible or impossible heat engines:

i) 900 kW heat are rejected ii) 560 kW of heat rejected iii) 108 kW of heat rejected

In each case the engine is supplied with 1120 kJ/sec of heat. The source and sink temperature are maintained at 560 k and 280 k. Given data:

L R

H R

s R

T 280 k i)Q 900kw

T 560 ke ii)Q 560kw

Q 1120KJ/sec iii)Q =108kw

Solution: From Carnot theorem, cannot engine gives maximum efficiency than any other engine. Maximum efficiency or Carnot efficiency,

H Lmax

H

T T 560 280n 0.5 50%

T 560

Case(i)

s R1

s

1 max

Q Q 1120 900n 0.196 19.6%

Q 1120

n n The engine is possible heat engine.

Case (ii)

s RH

s

Q Q 1120 560n 0.5 50%

Q 1120

n1=nmax

Therefore, it is reversible heat engine because by II law all the reversible engines have same efficiency. Case (ii)

s Rm

s

m max

Q Q 1120 108n 0.90 90%

Q 1120

n n it is impossible

Result: Case (i) is possible heat engine 1 max n n

Case (ii) is reversible heat engine 1 max n n

Case(iii) is impossible heat engine m maxn n

39. A heat engine of 30% efficiently drives a heat pump of COP = 5. The heat is transferred both form engine and the heat pump to circulating water for heating building in winter. Find the ratio of heat transfer to the circulating water from the heat pump to the heat transfer to the circulating water from the heat engine. Given data: Nhe = 30% COP OF H.P = 5 To find:

s

s

Q?

Q

QR

'

RQ

Qs

'

sQ

Solution: Where,

'

sQ heat transfer to the circulating water from heat pump.

H.E.

H.P.

sQ - heat transfer to the circulating water from heat engine.

For Heat Engine For Heat pump

s RH.E

s

Q Qn 0.3

Q

s

s R

QCOP 5

Q Q

s R sQ Q 0.3Q s

s R

QQ Q

5

R R RQ Q 0.3Q 0.7Q

R

s

Q0.7

Q

ss R

QQ Q

5

s R s sW Q Q Q 0.7Q s R

1Q 1 Q

5

sW 0.3Q s R0.8Q Q

s

wQ

0.3 R

s

Q0.8

Q

R s sW Q Q Q 0.8QS'

sw 0.2Q

s

wQ

0.2

s

s

Q w 0.3 1.5

0.2Q 0.2w

0.3

Result:

Ratio s

s

Q1.5

Q

40. A reversible heat engine operating between reservoirs at 900k and 300k drives a reversible refrigerator operating between reservoirs at 300k and 250k. The heat engine receives 1800kJ heat from 900k reservoir. The net output from the combined engine refrigerator is 360kJ. Find the heat transferred to the refrigerator and the net heat rejected to the reservoir at 300 k. Given data : T2 = 900 k

T1=300 k T3= 250 k Qs1 = 1800 KJ To find Qs2 =? QR1+QR2=? W1 W2

H.E H QR.1 QR2 Solution : Maximum efficiency of heat engine

H L 2 1max

H 2

T T T Tn

T T

1max

s1

900 3000.66 66.6%

900

wn

q

1 s1 maxW Q xn 1800x0.66 1200KJ

1 s1 R1 R1 s1 1W Q Q Q Q W 1800 1200

R1Q 600kJ

3Lref

H L 1 3

TTCOP

T T T T

T2 = 900 T3 = 250k

H.E H.P

.

T1 = 300 k

250

5300 250

R2ref

s2 R2

QAlso, COP

Q Q

R2

2

Q5 ----------- (2.11)

W

It is given that W1-W2=360kJ W2=W1-360=1200-360 W2=840kJ From equation (2.11)

s2

R2

Q5

40

Q 5X840 4200kJ

Heat transfer to refrigerator R2Q 4200kJ

we know that, W2 = Qs1 – QR2 840 = Qs2 – 4200 = 5040kJ Ans. Net heat transferred the reservoir at 300k = QR1+ Qs1 = 600+5040kJ Ans. Result:

i) Heat transfer to the refrigerator QR2=4200kJ ii) Heat rejection to the reservoir at 3000C=5640kJ

41. A carnot heat engine cycle works at maximum and minimum temperature of 10000C and 400C respectively. Calculate thermal efficiency and work one if Qs

= 1010kJ. Given data: TH = 10000C = 1273k

Tt = 400C=313 k Qs = 1010kJ To find: n, W Solution: Thermal efficiency,

H L

H

T Tn

T

1273-313 =

1273

= 75.4% Ans.

s

s

wn

Q

W nxQ 0.754x1010

W = 761.54kJ Result: n = 75.4% W = 761.54kJ 42. A Carnot heat engine receives heat from 600o source. The efficiency of the engine is 59%. Find the amount of heat supplied and heat rejected per kW of work output. Given data: n = 59% TH = 6000C W = 1kW To find: Qs = QR Solution:

s

s

w W 1n Q 1.695 kw

Q n 0.59

s R

R s R

R s

R

w = Q Q

Q Q Q

Q = Q w 1.695 1

Q 0.695kW

Results:

1. Heat supplied, Qs = 1.695 kW 2. Heat rejected, QR = 0.695 kW

43. The temperature in a domestic refrigerator is to be maintained at 100C. The ambient air temperature is 300C, If the heat leaving through the refrigerator is 3 kW, determine the least power necessary to pump out this heat continuously. TL = 100c = 263 k TH=300C = 303 k Qs = kw To find: Power (w) Qr R W Qs

Solution: A refrigerator removes heat from refrigerator at the same rate at which the heat leaks from it.

TH = 300C

Ref

TL = -100C

+-For reversible engine, condition of minimum power requirement.

sR

L H

LR s

H

R

s R

QQ

Q T

TQ xQ

T

263Q x3 2.6kW

303

w = Q Q 3 2.6 0.396kW

Result: Least power necessary to pump out this heat continuously, W = 0.396kW 44. Two-Carnot engines A and B are operated in series. The first one (a) receives heat at 870k and rejects to a reservoir at temperature T. The second engine (B) receives to a heat reservoir at 300k. Calculate the intermediate temperature T in 0C between two heat engines for the following cases. (iv) The work output of the two engines are equal and (v) The efficiencies of the two engines are equal. Give Data

TH = 870k TL = 300k Case (i) WA = WB

Case (ii) A = B To find: T in C for (i) & (ii) case

Solution:

Let QSA and QRA – Heat supplied and rejected to H.E.-A QSB and QRB --- Heat supplied and rejected to H.E.-B.

Case (i) WA = WB

2

2 870 300

1170585

2

312

A A B B

A A B B

A A A B A B

A A B

A S R B S R

S R S R A B

S R R R R S

R S R

H L

W Q Q and W Q Q

Q Q Q Q W W

Q Q Q Q Q Q

Q Q Q

T T T Q is proportional to T

T k

t C

Where, T is the intermediate temperature between H.E.-A and h.e.-b. Case (ii)

870(2.12)

870A A

A A

A B

S RAA

S S

Q QW T

Q Q

300(2.13)

870 300

870

B B

B B

S RBB

S S

Q QW T

Q Q T

T T

T

2

,

(870 ) 870 ( 300)

870 870 261000

A B equating above two equation

T T T

T T T

2 261000

510.88

237.88

T

T K

T C

Result :

(i) If WA = WB, the temperature T = 312C

(ii) If A = B, the temperature T = 237.88C 45. An office room which was heated by electric resistance heater consumer 1200 kW-hr of electrical energy in a winter month. Instead of this heater if the same office room is heated by a heat pump which is having 20% of COP of the ideal Carnot pump. The room

temperature is 24C while surrounding is at 0C. If heat supplied from the surrounding by the heat pump is 0.65 kJ, determine COP and money saved per month. Assume Cost of Electricity is Rs.1.75 kW/hr. Given data: Power consumed by electric heater = 1200kW/hr COP of H.P. = 20% of COP of ideal Carnot pump

TL = 0C =273k

TH = 24C = 297 Heat supplied by the Hp, Qs = 0.65 kW Cost of Electricity = Rs. 1.75/kW-hr To find :

(i) COP of heat pump (ii) Money saved per month

Solution: Electricity charge by using heater = 1200 × 1.75 = Rs.2100 per month. COP of Carnot pump

29712.375

297 273H

H L

T

T T

COP of Heat pump = 20% of COP of Carnot pump

20

12.375 2.475100

COP of heat pump = sup

Heat plied

Work done

Work done = sup 0.65

. . 2.475

Heat plied

COP of H P

W = 0.26kW Power required by heat pump in kW-hr = work done × 3600 = 0.26 × 3600 = 945.45 kW-hr Electricity charge for running heat pump = 945 × 1.75 = Rs. 1654.50 Money saved per month = 2100 - 1654.50 = Rs. 445.50 Result : COP of heat pump = 2.475 Money saved per month = Rs.445.50 46. Briefly Explain CLAUSIUS INEQUALITY. Clausius inequality states that “when a system undergoes a cyclic process, the summation of dQ

Taround a closed cycle is less than or equal to zero.

Consider an engine operating between two fixed temperature reservoirs TH and TL. Let dQs. units of heats be sullied at temperature TH and dQR units of heat be rejected at temperature TL during a cycle.

Thermal efficiency, s R

s

dQ dQ

dQ

Thermal efficiency of any reversible engine working on the same temperature limit is given by

Thermal efficiency for reversible engine = H L

H

T T

T

The efficiency of an actual engine cycle must be less than that of a reversible cycle. Since no engine can be more efficient than that of a reversible engine. Hence,

s R H L

s H

dQ dQ T T

dQ T

0

H

R L

s H

sR

L

sR

L H

dQ T

dQ T

dQdQ

T T

dQdQ

T T

for the entire cycle, dQ

0T

This equation is known as Clausius inequality. It provides the criterion of the reversibility of a cycle. If

dQ

0T

, the cycle is reversible

dQ

0T

, the cycle is irreversible and possible

dQ

0T

, the cycle is impossible.

Since the cyclic integral dQ

T is less than zero in a cycle, the cycle violates the

second law of thermodynamics. So, it impossible. We can apply the equality to the Carnot cycle since it is reversible cycle. Then the equation

becomes, dQ

0T

.

47. An inventor claims that his new engine will develop 3 kW for a heat addition of 240

kJ/min. The highest and the lowest temperature of the cycle are 1527C and 32C respectively. Would you agree his claim? Use Clausius inequality method. Given data:

W – 3kW = 30 60 = 1800 kJ/min Q1 = 240 kJ/min

T1 = 1527C

T2 = 327C To find : Agreement his claim = ? Solution: By clausius inequality,

1 2

1 2

Q QdQ

T T T

But, Work out, W = Q1 – Q2 1800 = 240 –Q2

Q2 = -1560kJ / min = 0

1 2

1 2

Q Q 240 15602.47 kJ/min.

T T 1800 600

Here dQ

0T

.

So, we agree his claim. Result: Inventor claim is accepted. 48. Entropy is an index of unavailability of degradation of energy. Heat always flows from hot body to cold body and thus becomes lesser value available. This unavailability of energy is measured by entropy. It is an important thermodynamics property of the working

substance = fp

L

TC In

T

By entropy principle

Entropy of universe, Suniv

f fP P

H H

2

fP

H L

f

2

fP

1 2

T TC In C In 0

T T

TC In 0

T T

T to be minimum

TC In should be equal to zero

T T

2

fP

1 2

2

fP P

1 2

TC In 0

T T

TC In 0& C 0

T T

2

f

1 2

2

f

1 2

f 1 2

TIn Inl

T T

T1

T T

T T T

Maximum work, Wmax = P 1 2 1 2C (T T 2 TT

= 2

P 1 1 2C T 2 T T

Wmax = 2

1 2T T

49. 1.6 Kg of air compressed according to the law pV1.3C form pressure of 1.2 bar and

temperature of 20C to a pressure of 17.5 bar. Calculate (a) the final volume and temperature (b) work done (c) heat transferred and (d) change in entropy. Given data: M = 1.6 kg

PV1.3 = C n = 1.3

P1 = 1.2 bar = 120 kN/m2

T1 = 20C = 293 K P2 = 17.5 bar = 1750 kN/m2 To find :

V2, T2, W, Q, and S Solution: From general gas equation.

1 1 1

311

1

1.3 1.3

1 1 2 2

1

1.32 1

1 2

1

1.3

2

3

2

p V mRT

mRT 1.6 0.287 293V 1.121m

P 123

p V p V

V p

V p

120V 1.121

1750

V 0.1427m

From polytropic process relation.

n 1

n2

2 1

1

1.3 1

1.3

PT T

P

1750293

120

543.82 K

Work done, W = 1 1 2 2p V p V

n 1

120 1.121 1750 0.1427

W1.3 1

W 384.02 kJ

Heat transfer.

nQ Wx

1

1.4 1.3Q 384.02

1.4 1

Q 96kJ

Change in entropy, S = mR in

1.121 543.83

1.6 0.718 In0.1427 293

S = 1.657 kJ/K. Result:

(i) Final Volume, V2 = 0.1427m3 (ii) Final Temperature, T2 = 543.826 K (iii) Work done, W = - 384.02 kJ (iv) Heat transfer, Q = -96 kJ

(v) Change in entropy S = 1.657 kJ/K.

50. Air expands from 11 bar at 550C to a pressure of 3 bar adiabatically. Determine temperature at the end of expansion and work done. Find also the change in entropy. Given data : P1 = 11 bar = 10 kN/M2

T1 = 550C = 550 = 273= 823 K P2 = 3 bar = 300 kN/m2

Process : adiabatic To find :

T2, W and S Solution:- The p, V & T relation for adiabatic process

1

2 22

1 1

1 1.4 1

1.42

1

1

2

T pT

T p

p 300 T 823

p 1100

T 567.78k

Work done, 1 1 2 2 1 2p V p V mR(T T )W

1 1

1 0.287 (823 567.78)

1.4 1

183.12kJ/kg.

Change in entropy S = 0 for adiabatic process. Result :

(i) Final temperature T2 =567.78 K

(ii) Change in entropy S = 0 (iii) Work done, W = 183.12 kJ/kg.

51. 1 Kg of air is compressed according to the law p. V1.25 = c from 1 bar and 15C to 17 bar. Calculate the change in entropy. Cp = 1.005 kJ/Kg K and Cv = 0.72 kJ/Kg K. Given data : M = 1 Kg pV1.25 = C; n = 1.25 P1 = 1 bar = 100 kN/m2 P2 = 17 bar = 1700 kN/m2

T1 = 15C = 288 K Cp = 1.005 kJ/kg K Cv = 0.072 kJ/kg K R = Cp – Cv = 1.005 – 0.72 = 0.285 kJ/kg K. Solution : From polytropic relation

n 1

n2

2 1

1

1.25 1

1.25

2 2

pT T

p

1700T 288 T 507.55 K

100

Change in entropy (S)

1 2p

2 1

p TS mRIn mC In

p T

1 507.55S 1 0.285 e In 1 1.005 In

17 288

S 0.238 kJ/k

Result :

Change in entropy S=-0.238kJ/K. 52. One kg of air in a closed system initially at 50C occupying a volume of 0.3 m3 undergoes a constant pressure heating process to 1000C. There is no work other than pdV work. Find the work transfer heat transfer and the entropy change of the gas. [Madras Univ. Apr’95] Given data:

M=1kg T1=50C V1=0.3m3 T2=1000C Solution: Work transfer, W=P(V2-V1)=mR(T2-T1)=1x0.287(100-5) [Assume for air, R=0.287kJ/kg K]=27.26kJ Heat transfer, Q=mCp(T2-T1)=1x1.005(100-5)=95.48kJ

2p

1

373 in entropy, S=mC 1 1.005 In

278

=0.295 KJ/K

TChange In

T

53. Ten grams of water at 200 C is converted into ice at -10C at constant atmospheric pressure. Assuming specific heat of liquid water to remain constant at 4.2 j/g K and that of ice ot be half of this value and taking the latent heat of fusion of ice at 00C to be 335 J/g. Calculate the total entropy change of the system Cp of ice = 2.093 J/gK.

[Madras Univ. Oct-95] m=10g Tw=200C Tice=-100C Cp1=4.2J/gk Hfg=335J/g Solution: Heat absorbed from the water to form 10g of ice. Q=Heat absorbed from the liquid + latent heat + heat absorbed from the solid phase.

Q=mwCpw(T1-0)+hfg+miceCpice(0-T2) =10x4.2(20-0)+335+10x2.093(0-(-10)) =4399.3J

system

4399.3Entropy change of atmosphere S

273

=16.12J/K.

Q

T

Entropy change of the system from 200 to 00C

01 w pw

1

S =m C

27310 4.2 In

293

=2.97 /K

TIn

T

Entropy change of the system from 0 to 100C

01 w pice

1

S =m C

26310 2.093 In

293

=0.781 /K

TIn

T

Total entropy change S=S1+S2+Ssystem

=-2.97-0.781+16.12 =12.369 J/K 54. A constant volume chamber of 0.3m3 capacity contains 1 kg of air at 560C. Heat is transferred to the air until the temperature is 1000C. Find the work transfer, heat transfer and the change in internal energy enthalpy and entropy. [Madras Univ.Oct’95] Given data: V1=0.3 m3 M=1kg T1=560C T2=1000C Solution: Work transfer, W=0 for constant volume process Heat transfer = change in internal energy.

Q=U=mCv(T2-T1) =1 x 0.718(100-56) =31.59kJ

Change in enthalpy H=U=31.59KJ

21 v

2

change in entropy, S =mC

3731 0.718 In

329

=0.09kJ/K /K

TIn

T

55. 10 kg water 900C mixes with 2.5kg of water at 200C under adiabatic condition. Find the final temperature and entropy generation. Given data:

M1=10kg T1=900C M2=2.5kg T2=200C To find:

1. Tf=?

2. S=? Solution:

1 1 1 2 2 2

1 1 2 2

f

0

f

water=4.187kJ/kg K.

10 4.187 363 2.5 4.187 293T

1.4.187 2.5 4.187

=349K

T 76

f

p

m C T m C TT

m C m C

C of

C

1 1 2 2

1 2

Entropy generation, S=m C m C

349 34910 4.187 In 2.5 4.187 In

363 293

=0.184kJ /K

f fT TIn In

T T

Result:

1. The final temperature, Tf=760C

2. Entropy generation, S=0.184kJ/K 56. In a certain heat exchange, 45 kg/min of water is to be heated from 600C to 1150C by hot gases, which enter the heat exchanges at 225 and flow at the rate of 90kg/min. compute the net change of entropy. Assume specific heat for water and gases as 4.18 and 1.045kJ/kg K. Given data: Mw=45kg/min Tw1=600C Tw2=1150C Tg1=220C Mg=90 kg/min.

To find:

Net change of entropy, S=? Solution: By energy balance Heat gained by the water = Heat given out by the gas.

2 1 1 2

2

2

2

1

g

45 4.18(115 60) 90 1.045 225

115 .

Now change in entropy of gas, S

90 115 = 1.045

60 225

s

s

u p w w g p g g

g

g

g

g P

g

m C T T m C T T

T

T K

Tm C In

T

In

=1.052kJ/K.

2

1

wNow change in entropy of water, S

45 388 = 4.18

60 333

=0.479 kJ/K.

w

w

w P

w

Tm C In

T

In

Net change in entropy, S=Sg+Sw =-1.052+0.479 =0.573 kJ/K. Result:

Net change in entropy, S=-0.573kJ/K. 57. A perfect gas is contained in a cylinder and undergoes a controlled expansion according to the law p-A=A+BV, where A and B constants. P be the pressure in kgf/cm2 and v the volume in m3. The initial and final pressure are 8.4 kgf/cm2 and 2.8 kgf/cm2 and the corresponding volumes are 0.056 m3 and 0.168 m3.

Assume =1.39, R=0.28 kJ/kg K. The initial temperature is 150C. Determine (a) work done by the gas (b) heat transferred during the process in magnitude and direction (c) change in entropy per kg of gas during expansion (d) maximum value of internal energy per keg reckoned from 00C. Given data: P= A+BV

P1 =8.4kgf/cm2 P2=2.8 kgf/cm2 V1=0.056 m3 V2 = 0.168 m3. T1 = 150C.

=1.39 & R=0.28 kJ/kg.K. To find:

(i) Work done, W=? (ii) Heat transfer, Q=?

(iii) Change in entropy, S=?

(iv) Change in internal energy, U=? Solution: P=A+BV ------------------------(2.31) At p1-8.4 kgf/cm2 7 V1=0.056m3 8.4=A+0.056 B ----------------(2.32) At p2 = 2.8 kgf/cm2 & V2 =0.168m3 2. 8=A+0.168 B----------------(2.33) Subtracting eqn. (2.32) from (2.33) -5.6=0.112B B=-50

A=112

The equation (2.32) becomes p=11.2-50V

2

1

2

1

2

1

v

v

v 0 168

v 0 056

v 0 168

2

v 0 056

22

Work done, W= 100

= 11.2 50

50 =100 11.2V-

2

=100 11.2 .168-0.056 25[(0.168) 0.055

pdv

V dv

V

=62.72kJ.

1 1

1

p of the gas, m=

8.4 100 0.056 =

0.28 288

=0.583kg

VMass

RT

2

mRT know that, V= (2.34)[ ]

p

(2.34) (2.31)

0 (2.35)

We pV mRT

substituting in

BmRTp A

p

p Ap BmRT

For getting maximum temperature, differentiate the eqn. (2.35) and equate to zero. 2pdp – Adp + BmRT =0

20

2 0

/ 2

11.25.6 .

2

dT p A

dp BmR

p A

p A

p bar

From eqn(2.31) 5.6=112-50 V V=0.112m3 At p=5.6 bar & V=0.112m3

max

5.6 100 0.112

0.583 288

=384.2K

pVT

mR

Maximum Internal energy,

Umax=Cv(Tmax-T0)

max 01

0.28(384.2 273)

1.39 1

79.84 / .

[ transfer, Q=work transfer [ for pV=C]

RT T

kJ kg

Heat

= 62.72kJ

2 2v

1 1

1 2

2

1

[ is supplied into the system]

Entropy, S=C my gas.

For T

In

0.1680.28 In

0.056

0.308 kJ/kg K.

Heat

T VIn Rin for

T V

T

VS R

V

Result:

(i) Work done, W= 62.72kJ (ii) Heat transfer, Q=62.72kJ

(iii) Entropy, S=0.308kJ/kg K

(iv) Max internal energy, Umax=79.84kJ/kg. 58. As single stage air turbine is to operate with air inlet pressure and temperature of 1 bar and 60k. During the expansion the turbine loses are 20kJ/kg to the surroundings which is at 1 bar and 300k, for 1 kg.of mass flow rate determine (i) decrease in availability (ii) maximum work (iii) the irreversibility. Given data: P1=1bar T1=600K Q=-20kJ/kg P2=1bar T2=300K M=1kg. To find:

(i) Decrease in availability 1-2=? (ii) Maximum work, Wmax=? (iii) Irreversibility, l=?

Solution: For any process, change in entropy

2 22 1 p

1 1

S=S -S =m C I

300 11 1.005 In 0.287 In

600 1

0.697 kJ/ K.

T PIn R n

T P

Decrease in availability,

1 2 1 2 1 2

1 2 1 2 =

=1 1.005 600-300 300 0.697

=510.6kJ/kg

o

p o

m h h T S S

m C T T T S S

Maximum work – decrease in availability Wmax = 510.6kJ/kg From SFEE, H1+Q=h2+W W=m[(h1-h2)+Q] =m[Cp(T1-T2)+Q] =1x[1.005 9600-300)-20] W=281.5kJ/kg Irreversibility, I=wmax-W =510.6-281.5 =229.1kJ/kg

Decrease in availability, 1-2=510.6kJ/kg Maximum work, Wmax =510.6kJ/kg Irreversibility, I=229.1kJ/kg 59. In a steam generator, the steam generating tubes receive heat from hot gases passing over the oxide surface evaporating water inside the tubes. Flue gas flow rate is 20kg/s with an average specific heat f 1.04kJ/kg K. The gas T decreases from 6500 C to 4000C while generating steam at 3000C water enters the tube as a saturated liquid and leaves with a quality of 90%. Assume environment temperature as T0=270C. Determine the water flow rate. Availability of hot fluid and cold fluid, irreversibility and second law efficiency. Given data: M1 =20kg/s C1=1.04kJ/kg K T1=6500C T2=4000C T3 = 3000C (saturated liquid) X4=0.9 T0=270C To find:

1. m2=?

2. 1=?

3. 2=?

4. L=?

5. II =? Solution: For heat exchanger energy balance, Heat given out by the flue gas=Heat gained by the steam Corresponding to 3000C, read hf 7 hfg, Sf and Sfg H3=hf=1345kJ/kg,hfg=1406kJ/kg S3 =sf = 3.255kJ/kg k,sfg =2.453kJ/kg K

h4=hf+x4hfg =1345+0.9 x 1406 2610.4KJ/kg s4=sf +x4sfg =3.255+0.9x2.453 =5.46kJ/kg K m1C1(T1-T2)=m2(h4-h3) 20x1.04(650-400)=m2(2610.4-1345) m2=4.11kg/s Availability of hot fluid,

1 1 1 2 0 1 2

11 2 p

2

923 =20[1.40(650-40)-300x 1.40In

6T3

T3228.88 s -s =C In

T

pm C T T T s s

kW

Availability of cold fluid,

2 2 1 2 0 1 2

1 2

II

=4.11[ 2610.4-1345 -300 5.467 3.255

2473.39

,

=3228.88-2478.39

=755.49KW

AvailSecond law efficiency,

pm C T T T s s

kW

Irreversibility I

ability of cold fluid

Availability of Hot fluid

2473.39 = 100 76.6%

3228.88

Result:

1. Mass of stem, m2=4.11 kg/s

2. Availability of cold fluid 2=2473.39kW

3. Availability of Hot fluid 1=3228.88kW 4. Irreversibility Hot fluid, I=755.49kW

5. Second law efficiency,II =76.7% 60. Steam flows in a pipeline at 1.5 Mpa. After expanding to 0.1 Mpa in a throttling calorimeter, the temperature is found to be 1200C. Find the quality of steam in the pipeline and also calculate availabilities, irreversibility and second law efficiency. Assume T0=250C Given data: P1=1.5Mpa, and p2 = 0.1 Mpa T2 =1200C To find:

1. x1=?

2. 1=?

3. 2=? 4. I=?

5. II=? Solution: Corresponding to p2=1 bar & T2=1200C from the super heated Table, read H2=2716.2kJ/kg. S2=7.4606kJ/kg K For throttling process H1=h2 H1=2716.2kJ/kg But at p1 = 15bar, read hf hfg, sf &Sfg H1=844.89kJ/kg Hfg=1947.3kJ/kg Sf=2.3315kJ/kg Sfg=6.448kJ/kg H1=hf+x1hg 2716.2 =844.89+x1(1947.3) x1=0.963 s1=sf+x1sfg s1=2.3315+0.963(6.4448) =8.538kJ/kg K

Availability at inlet, 1=h1=T0s1 =2716.2-298 x 8.538 =171.876 kJ/K

Availability at outlet, 2 =h2-T0s2 =2716.2-298 x 7.4606 = 492.94 kJ/kg

Irreversibility, I=2-1 =492.94-171.876 =321.064kJ/kg or =T0 (s1-s2) =298(8.538-7.4606) =321.065kJ/kg

1II

2

law efficiency,

171.876 = 100

492.94

=34.867%

out

in

ASecond or

A

Result:

1. Quality steam, X1=0.963

2. Availability of inlet, 1=492.94kJ/kg

3. Availability of Outlet 2=171.876kJ/kg 4. Irreversibility, I=321.065kJ/kg

5. Second law efficiency, II=34.87% 61. An inventor claims to have developed a refrigerating unit which maintains the refrigerated space at -60 while operating in a room where temperature is 270C and has COP 8.5. find out whether his claims is correct or not. Given data: Inventor’s refrigerating unit. TH=270C TL=-60C COP=8.5 To find: Decision = ? Solution:

1

H

inventor

T of carnot =

T

-6+273 =

27-(-6)

=8.09

here, COP

His claim is not correct.

L

carnot

COPT

COP

Result: His claim is not correct because COPcarnot is less than the inventor’s claimed COP. 62. A heat engine is used to drive a heat pump. The heat transfer from the heat engine and from the heat pump is used to heat the water circulating through the radiators of building. The efficiency of the heat engine is 27% and COP of the heat pump is 4. (i) Draw the neat diagram of the arrangement and (ii) evaluate the ratio of heat transfer to the circulating water to the heat transfer to the heat engine. Given data:

H.E =27% COPH.P=4 To find:

1 2

1

?R R

S

Q Q

Q

Solution:

1 1.

1

1

1

1

1

1 1

.

R2

S1 1

R2

S1 1

done

supplied

0.27 1

0.73

0.73 (2.38)

supplied

done

Q =

Q

Q 4= (2.39)

Q

s RH E

S

R

s

R

S

R s

H P

R

R

Q QWork

Heat Q

Q

Q

Q

Q

Q Q

HeatCOP

Work

Q

Q

R2

S1 S1

R2

S1

R2 S1

(2.38) (2.39)

Q 4=

Q 0.73 4=Q

Q =

0.27Q

Q 108Q (2.40)

Substituting in

Total heat supplied to the water, Q= 1RQ + 2RQ

1 1

1

1

1 1

0.73 1.08

1.81

1.81

=1.81

s s

s

s

s s

Q Q

Q

QQ

Q Q

Result: The ratio of heat transfer to the circulating to the heat transfer to the heat engine,

1 2

1

1.81R R

S

Q Q

Q

63. A carnot heat engine takes heat from an infinite reservoir at 5500C. Half of the work delivered by the engines used to run generator and the other half if used to run heat pump which takes heat at 2750C and rejects it at 4400C. Express the heat rejected at 4400C by the heat pump as % of heat supplied to the engine at 5500C. If the operation of the generator is 500kW, find the heat rejected per hour by the heat pump at 4400C. Given data: T1=5500C T2=2750C T4=4400C Wg=500kw To find:

2

1

R2

1. ?

2. rejected, Q ?

R

S

Q

Q

heat

Solution: For cannot heat engine

1 1

1 2

11 1

2

1

R1

823 =

548

=1.502Q (2.41)

S R

S R

R

Q Q

T T

TQ Q

T

Q

Work done by the heat engine, W=Qs1-QR1 =1.502 QR1-QR1 =0.502 QR1

1g

R1

0.502 input, W

2

=0.251 Q

RQGenerator

Work input to the heat pump. WHP=0.251 QR1 Heat rejected by the heat pump, QR2 = Qs2 + WHP = Qs2+0.251 QR1---------(2.42)

2 2

2 4

22 2

4

2

2 2

s2

For reverse heat pump,

543 =

713

0.77

Substitutinge Q (2.42)

S R

S R

R

S R

Q Q

T T

TQ Q

T

Q

Q Q

in

2 2 1

2 1

2 1

2

1

0.77 0.251

1 =0.77 0.251

1.502

0.727

72.7%

R R R

R s

R s

R

s

Q Q Q

Q Q

Q Q

Q

Q

The generator power input, Wg=500kw 0.251 1RQ =500

1RQ =1992.03kW

Qs1=1.502 1RQ

=1.502 1RQ

=2992.03 kW

2 0.727 2992.03

=2175.21kW

2175.21 =

1000

=7830.74MJ/hr

RQ

Result:

1. Heat rejected by the heat pump 2

1

72.7%R

s

Q

Q

2. Heat rejected by the heat pump if the generator has 500kW, 1RQ =7830.74 MJ/hr.

64. A heat engine operates between a source at 6000C and a sink at 600C. Determine the least rate of heat rejection per kW net output of the engine. Given data: TH=6000C=273+600=873K TL=600C=273+60=333K To find: QR/net output=? Solution: Take W=1kW. At the least rate of heat rejection, the efficiency should be maximum. Carnot efficiency of heat engine,

.

H.E

873 333

873

=61.86%

1

0.616

H LH E

H

carnot

s

s

T T

T

W

Q

Q

1.616

=1.616-1

=0.616-1

=0.616/kW of the output

s

R s

Q

Q Q W

Result: QR/ net output = 0.616/kW

65. 5 kg of air at 2 bar and 30C is compressed to 24 bar pressure according to the law

p.V1.2= Constant. After compression air is cooled at constant volume to 30C. Determine, (i) Volume and temperature at the end of compression, (iii) Change in entropy during compression, (iii) Change in entropy during constant volume cooling. Take Cp=1.005 kJ/kg K, Cv= 0.718 kj/kg K, Cv=0.718 kJ/kg K. Given data: M=5 kg P1=2 bar

T1=30C P2=24 bar pV1.2= Constant

T3=30C Cp=1.005kJ/Kg K Cv=0.718 kJ/kg K To find:

1. V2=? 2. T2=? 3. S2-S1=? 4. S3-S2=?

Solution: According to the law, pV1.2= C

p1V1=p2V2 From characteristic gas equation , pV=mRT

11

1

3

1.2 1.2

1 2 2

5 0.287 (303) =

200

=2.17 m

From p

mRTV

p

V p V

1.2

12 1

2

1

12

3

2 = 2.17

24

=0.274 m

pV V

p

Similarly,

1.2 1

1.22 2

1 1

0.2

1.2

2

24 T 303

2

=458.47K

=185.47 C

T p

T p

Change in entropy during compression,

2 2

1 1

2 22 1

1 1

2 1

458.47 0.2745 0.718 5 0.287I

303 .217

=1.483kJ/K

Process 2-3 is a constant volume process.

p

v

T pmC In mRIn

T p

T VS S mC In mRIn

T V

S S In n

33 1

2

in entropy, S

303 =5 0.287In

458.47

=1.487kJ/K.

v

TChange S mC In

T

Result;

1. Final volume at the end o compression, V2=0.274m3

2. The corresponding temp, T2=185.47C. 3. Change of entropy during compression S2-S1-1.483 kJ/K 4. Change of entropy during constant volume, S3-S2=1.487kJ/K

66. 0.2kg of air at 1.5 bar and 27C is compressed to a pressure o 15 bar according to the

law PV1.25= Constant. Determine work done on or by air, eat flow to or from the air, increase or decrease in entropy. Given data: M=0.2kg P1=1.5bar

T1=27C=300K P2=15bar pV1.25=C To find:

W, Qs,S=? Solution: By general gas equation P1V1=mRT1

11

1

3

1 1.25 1

1.252

2

1

1 1

1.251

2

2

3

0.2 0.28 300

150

=0.148m

15300

1.5

=475.5k

1.5V 0.148

15

=0.018195m

n

n

n

mRTV

p

PT

P

p

p

Work done,

2 2 1 1

5 5

1

-1.5 10 0.1148 15 10 0.018195 =

1.25 1

=40.29kJ

p V pVW

n

Heat transfer

1.4 1.2540.29

1 1.4 1

=15.1088kJ

nQ W

change in entropy

2 2

1 1

0.01148 475.50.2 0.287 0.2 0.718

0.018195 300

v

V TS mRIn mC In

V T

In in

=0.17187kJ/K Result:

(i) Work done=40.29kJ (ii) Heat transfer=15.1088kJ (iii) Change of entropy=0.17187kJ/kgK

67. A domestic food freezer maintains a temperature of -15C. the ambient air is at 30C. If heat leads into the freezer at a continuous rate of 1.75kJ/s, what is the least power necessary to pump the heat out continuously? Given data:

TL=15C=273-15=258K

TH=30C=273+30=303K Qs=1.75kW To find: Least power. W=? Solution:

Carnot COP=258

303 258

=5.733

L

H L

T

T T

Actual COP of refrigerator = sQ

W

For minimum power required to pump the heat, Carnot COP = Actual COP

5.733=1.75

W

W=0.305 kW Result: Least power necessary to pump heat, W=0.305kW

68. Find the change in entropy of 1 kg of ice which I heated from ink - 5C to 0C. It melts

into water at 0C.

CPice = 2.093kJ/kgK. The pressure during heating is maintained at 1 atm constant. Latent heat of fusion of rice =334.96kJ/kg. Given data: M=1kg

Tice=5C=-5+273=268 K

T0=0C=0+273=273K Cpice=2.093 kJ/kg K L=334.96 kJ/kg To find

Change in entropy, S=/ Solution: Heat released by 1kg of ice, Q=mCpice (T0-Tice)+mL =1x 2.093 x (273-268)+1x 334.96 =345.425kJ Entropy change,

273

0268

( )

273 1 334.961 2.093

268 273

1.266 /

fusionice

pice

S S S

dT mLmC

T T

In

kJ K

Result:

Change in entropy ,S=1.266kJ/K 69. Three identical bodies of A, B and C constant heat capacity are at temperatures of 300,300 and 100K. A heat engine is operated between A and B and a heat pump working as refrigerator is operated between B and C. The heat pump is operated by the output of heat engine. If no work or heat supplied from outside , find the highest temperature to which any one of the body can be raised by the operation of heat engine or refrigerator. Solution; A heat engine and a refrigerator are operated between temperature limits 300,300 and 100 k.

Let Tf be the final temperature of bodies A and B. Tf be the final temperature of body C and C be the final heat capacity of these three identical bodies.

(S)univ > 0 -------(2.43)

(S)univ = (S)A +(S)B +(S)c +(S)H.E +(S)H.P

But (S)H.E and (S)Ref=0

(S)univ=(S)A+(S)B+(S)C >0

From (2.43). (S)A+(S)B +(S)c > 0

' 2

1

2 '

0 300 100 300

0300 300 100

f f fp

f f

T T T TCIn CIn CIn S C In

T

T TCIn

For minimum value of Tf (S)univ=0

2 '

6

2 '

6

.0

3 10

.1

3 10

f f

f f

T TCIn

T TCIn In

Since In 1 =0, the above equation becomes

HP

W=QS1-QR1

QS1

Tf A

300k

C

300k

HE

B 300K

Tf

2 , 6

s1 1 2

2 1

'

f

9 10 (2.44)

also Q

(300 ) ( 100) ( ' 100)

[ ( )]

300 100 ' 300

T 700 2 (2.45)

f f

R R

f f f

f f f

f

T T

and Q Q

C T C T C T

Q mC T T

T T T

T

From (2.44) and (2.45), Tf

2 (200-2 x Tf)=9 x 106 Tf=300K Tf=700 -2 x Tf =700-2 x 300 =100 K Result: The maximum temperature can be raised for 100K body and 300 K of B.

70. A reversible heat engine operates between a source at 800C and a sink at 30C. What I the least rate of heat rejection per kW network output of the engine? Given data:

TH=800C =273+800=1073K

TL=30C =273 +30 = 303 K To find: Least rate of heat rejection / KW of network output, Q=? Solution;

, 1

303 =1-

1073

Lcarnot

H

TCarnot efficiency

T

=0.7176 Work Output, W =1kW W=Qs-QR 1-Qs-QR Qs=1+QR ----------(2.46) But efficiency of the engine

,

1

0.7176 1

0.7176 11

0.3935

R

s

R

s

R

R

R

Q

Q

Q

Q

Q

Q

Q kW

Result; Least rate of heat rejection , QR=0.3935 kW

71. One kg o ice at -5C is exposed to the atmosphere which is at 20C. the ice melts and comes into thermal equilibrium with the atmosphere (i) Determine the entropy increase of the turbine. (ii) what is the minimum amount of work necessary to convert the water back

to ice at - 5C? Assume Cp for ice as 2.093 kJ/kg K and the latent heat o fusion of ice as 333.3 kJ/kg. Given data:

T1=5C=273-5=268K

Ta=20C=273+20=293K Cpi=2.093 kJ/kg K L=333.3 kJ/kg To find:

1. Entropy increase of universe, (S)univ =? 2. Minimum amount of work, Wmin=? Solution:

Heat absorbed by air from atmosphere, (Q)= Heat absorbed in solid phase+ Latent heat+ eat absorbed in liquid phase

0 0

pw

( ) ( ) ( )

Assuming 1 kg and C 4.187 / K

Q=1 2.093(0-(-5))+1 333.3+1 4.187(20-0)

=427.535 kJ

pi i pW aQ mC T T mL mC T T

m kJ kg

Entropy change of atm.

( )

425.535 = 146 /

293

atm

QS

T

KJ K

Entropy change of system

(S)system=(S)ice +(S)fusion +(S)liquid

273 293

268 273

273 1 333.3 2931 2.093 1 4.187

268 273 273

1.556 /

pi pw

dT mL dTmC mC

T T T

In In

KJ K

Entropy of universe,

(S)univ =(S)sys +(S)atm =1.556-1.46 =0.096 kJ/K If water is to be converted back to ice using reversible refrigerator, heat to be removed from water.

Q=427.535 kJ

Now, (S)sys =1.556kJ/K

But, (S)atm =atm

Q W

T

(S)ref = 0 as the refrigerant operates in a cycle.

(S)univ =(S)sys +(S)ref +(S)atm 0 -1.536+0+

min

445.908

445.908 -427.535

So, W 28.373 KJ

atm

Q WkJ

T

W

Result:

1. Entropy increase of universe, (S)univ =0.096 kJ/K 2. Minimum amount of work, Wmin =28.373 kJ

72. Air is closed vessel of fixed volume 0.15m3, exerts pressure of 12bar at 250C. If the vessel is cooled so that the pressure falls to 3.5 bar, determine the final temperature, heat

transfer and change of entropy. Given data:

V1 = 0.15m3 P1 = 12bar = 1200 kN/m2 T1 = 2500C = 273 + 250 = 523 K p2 = 3.5bar = 350 kN/m2 To find:

1. Final temperature, T2 = ? 2. Heat transfer, Q = ?

3. Entropy change, S = ? Solution: From ideal gas equation

1 1

1

p Vm

RT

1200 0.151.2kg

0.287 523

For constant volume process,

22 1

1

pT T

p

3.5523 152.54K

12

Heat transfer, Q = mC(T2 – T1) = 1.2 x 1.005 (152.54 – 523) ( Cp of air = 1.005 kJ/kgK) = -446.78 kJ (-ve sign indicates that the heat is rejected from the system)

2 2p

1 1

2

1

T PEntropy change, S=mC In mR In

T P

T mCuIn forV C

T

152.54 3.51.2 1.005 In 1.2 0.287In

523 12

1.06kJ/K

Result:

1. Final temperature, T2 = 152.54 K 2. Heat transfer, Q = -446.78kJ

3. Entropy change, S = -1.06 kJ/K 73. Two reversible heat engines A and B are arranged in series. A rejecting heat directly to B. Engine receives 200kJ at a temperature of 4210C from hot source, while engine B is in communication with a cold sink at a temperature of 4.4oC. If the work output of a A is twice that of B, find: i. The intermediate temperature between A and B, ii. The efficiency of each engine, and iii. The heat rejected to the cold sink. Given data: TH = 4210C = 421 + 273 = 694 K TL = 4.40C = 4.4 + 273 = 277.4 K Qs1 = 200kJ WA = WB To find: i. The intermediate temperature between A and B, T = ?

ii. The efficiency of each engine, A and B = ? iii. The heat rejected to the cold sink, QR2 = ? Solution: Work output from engine A, WA = QS1 – QR1 = 200 – QR1

For reversible heat engine,

s1H

R1

R1

QT2.47

T Q

694 200So,

T Q

TH TL

QR1 = 0.288T --------(2.48) So, WA = 200 – 0.288 T But WB = 100 – 0.144T --- (2.49) ( 2WA = WB) and also WB = Qs2 – QR2 = 0.288T – QR2 ------ (2.50) ( Qs1 – Qs2) Equating equations (2.49) and (2.50) 100 – 0.144T = 0.288T – QR2

QR2 = 0.432T -100 -------(2.51) Similarly, for reversible engine B,

2

2

1

2 1

2

2

s

L R

R

s R

R

R

QT

T Q

QTSo, Q Q

277.4 Q

0.288T 0.288T

Q 0.432T 100

T 416.42K

OR T = 143.420C So, QR1 = 0.288 x 416.42 = 119.93 kJ

WA=2WB A

B

QR1

QS2

WB

QR2

and QR2 = 0.432 x 416.42 – 100 = 79.89 kJ

1

1

R

A

S

QEmergency of engine A, 1

Q

119.93 1 40.04%

200

2

2

2 1

s

B

R

s R

QEfficiency of engine A, 1

Q

79.89 = 1- Q Q

119.93

= 33.39%

Result:

i. The intermediate temperature between A and B, T = 143.420C

ii. The efficiency of each engine, A = 40.04% and B = 33.39% The heat rejected to the cold sink, QR2 = 79.89kJ

unit ii a - wordpress.com cycle efficiency will be maximum, when the temperature is o k (i.e. zero...

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