ae 2202 thermodynamics- 2 mark questions

8/13/2019 AE 2202 THERMOdynamics- 2 Mark Questions

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Notes prepared by M.Sundaresan.

AERO-ENGINEERING THERMODYNAMICS

A. Two mark question and answers.

1.0. State the Kelvin-Planck statement of second law of thermodynamics.Ans. Kelvin-Planck states that it is impossible to construct a heat engine working on a cyclicprocess, whose only purpose is to convert all the heat energy given to it in an equal amount of

work.

It essentially means that while the heat is getting converted to work you need a second bodywhich is at a lower temperature through which part of the heat should also pass. That is why heat

is considered a lower grade of energy compared to work.

2.0 State Clausius statement of the second law of thermodynamics.Ans. It states that heat can flow from a cold body to a hot body only with external aid or work

added in a cyclic process. But heat can easily pass from a hot body to a cold body without any

external aid and in the process does work3.0 State Carnots theorem.Ans. No heat engine operating in a cyclic process between two fixed temperatures can be more

efficient than that of a reversible engine operating between the same temperature limits.

4.0 What are the corollaries of Carnot theorem ?Ans. i) All the reversible engines operating between the two given thermal reservoirs with

the fixed temperatures have the same efficiency.

.ii) The efficiency of any reversible heat engine operating between two reservoirs isindependent of the nature of the working fluid and depends only on the temperature of the

reservoirs.

5.0 Define PMM of the second kind.Ans. Perpetual motion machine of the second kind draws heat continuously from a singlereservoir and converts it into equivalent amount of work. Thus it gives 100% efficiency. This is

impossible.

6.0 What is the difference between a heat pump and refrigerator ?Ans. Heat pump is a device which operating in a cyclic process maintains the temperature of a

hot body at a temperature higher than that of the surroundings

A refrigerator is a device operating in a cyclic process maintains the temperature of acold body at a temperature lower than the temperature of the surroundings.

7. What is meant by a heat engine.?

Ans. A heat engine is a device, which is used to convert the thermal energy int mechanicalenergy.

8. Define the term COP ?Ans. Co-efficient of performance is defined as the ratio of heat extracted or rejected to workinput.

COP = Heat extracted /rejected

Work input.

9. Write the expansion for COP of a heat pump and refregirator.Ans

COP for a heat pump


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COPHP=inputwork

rejectedHeat

...

... =

)(21

1

TT

T

COP for a refrigerator

COPref = inputwork

extractedHeat

...

... = )(

21

2

TT

T

10 Why Carnot cycle cannot be realized in practice ?Ans. (i) In a Carnot cycle, all the four processes are reversible but in actual practice there is noprocess that is reversible.

(ii) There are two processes to be carried out during compression and expansion. For

isothermal process, the piston moves slowly and for adiabatic process the piston moves as fastas possible. This speed variation during the same stroke of the engine is not possible.

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

11. Name two distinct methods by which the efficiency of a Carnot cycle can be

increased.

Ans (i) Efficiency can be increased as the higher temperature T2 increases.(ii) Efficiency can be increased as the lower temperature T1 decreases.

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

Therefore we cannot convert all the heat input into useful work.

13. When the Carnot cycle efficiency will be the maximum?.Ans. Carnot cycle efficiency is maximum when the initial temperature is 0K.

14. What is the process involved in Carnot cycle ?

Ans. Carnot cycle consists of the following processes.(i) reversible adiabatic compression.(ii) Reversible isothermal heat addition.(iii) Reversible adiabatic compression.(iv) Reversible isothermal heat rejection.

15. Sketch the p-v and T-s diagram for Carnot cycle.

16. Is the second law independent of the first law ?Ans. Yes , the second law of thermodynamics independent of the first law. The second law

speaks about quality of energy.

17. Define Entropy ?

w

s

T

V

p


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Ans. The entropy of an assembly in a state of thermal equilibrium is a thermodynamic

property, in principle related through equations of state to other thermodynamic properties. Thechange of entropy is defined only for reversible processes stated as :

dsT

dq=

2

1

.

Hence in a closed isolated system the entropy change 0ds . This aspect relates to the second

law of thermodynamics in dictating the direction and flow of heat in a workable engine or heatpump. The property of the entropy namely specific entropy has been arrived on the basis that

equilibrium values are easily and quickly achieved though the various molecules are in random

and disorderly motion and are in combination as various meta states in a system. Once theassembly are system is in a highly probable macro-state though disorderly, the probability of

returning to a low probable macro-state is very small.

18. Define the terms source sink and heat reservoir.

Ans. Source. The part from where heat is rejected to a colder body for doing work, orthe part from where heat is generated for doing work is called as the source,

Sink. The part which receives heat from work absorbing or work developing device is called

sink.

Reservoir.The part which supplies or receives heat continuously without change in its

temperature is called a reservoir.

19. Why is the performance of refrigerator and heat pump are given in terms of COP

and not in terms of efficiency ?Ans. The performance of any device is expressed in terms of efficiency for work developing

machines.. But for Heat pump and refrigerator are work absorbing machines. So the performance

of these devices are based on COP only.

20. What is meant by principle of increase of entropy ?Ans. For any infinitismal process undergone by a system, change in entropy,

( )T

dQds

For a reversible, 0=dQ and hence 0=dS

For irreversible, 0>dS

So the entropy of a system would never decrease, it will always increase and remains constant if

the pressure is reversible and is called as principle increase of entropy.

21. What is meant by Clausius inequality?

Ans. It is impossible for a self acting machine working in a cyclic process unaided by anyexternal agency to convey heat from a body at a lower temperature to a body at a higher

temperature,

22. Explain briefly the Clausius inequality.Ans.

. 0 TdQ

is known as inequality of Clausius


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If 0= TdQ

, the cycle is reversible

0 TdQ

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

23. For the compressible process between the same two states, which work will be more

reversible or irreversible ?Ans. Irreversible work will be more in the compressor. Generally for compression, the actual

work given will be higher than the calculated work (Wrev).

24. A heat pump pumps 10MJ/KWhr to the high temperature reservoir. What is the COP

?

Ans.COP= Heat supplied / work output.

78.23600

1010 3=

25. Find the entropy of the universe where 1000kj of heat is transferred from800K to

500K.

Ans.

Entropy change of universe = KKJT

Q

T

Q/75.0

500

1000

800

1000

21

=+

=+

26. Give the expressions for change in entropy during constant pressure and polytropic

process. Show on T-s diagramAns. For the constant pressure process,

1

212 ln

T

TmCSSS P== .

For polytropic process,

]lnln[1

2

1

2

12p

pR

T

TCmSSS P ==

or

]lnln[

1

2

1

2

12v

vR

T

TCmSSS v +== (adiabatic S =0 ).

27. Explain the term reversibility.Ans If the process traces the same path during the process when reversed is called as

reversibility. And the entropy change is zero.

28. Can the entropy of the universe decrease? why?Ans Entropy of the universe cannot decrease. It will be constant or will increase due to

irreversibility.

T

2T

2n

2s


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29. What is the essence of the second law of thermodynamics?Ans. 1.0 To know the feasibility of process.

2.0 To know about the quality of energy.30. If the Carnot engine efficiency is 50%. Find COP of Carnot refrigerator working

between same temperatures

Ans.

Efficiency (HP) =1

21

T

TT =0.5

22

1=

T

T

COP of refrigerator 112

1

1

1

2

121

2=

=

=

T

TTT

TAns.

31. Define the term absolute entropy.Ans. The change in entropy of the system with respect to ambient conditions or any other

standard reference condition is known as absolute entropy.

32. Define a system.Ans. A THERMODYNAMIC SYSTEM IS DEFINED AS A DEFINITE SPACE OR AREA WHICH THE STUDYOFENERGY TRANSFER AND ENERGY CONVERSIONS IS MADE.

33. What is a closed and open system.? and an isolated systemAns. In open systems both the mass and energy transfer take place. The open system is oftenspecified within a control volume. e.g. air compressor. The system boundary is determined by

the space that the matter occupies. In a closed system it does not permit mass transfer, but only

the energy transfer takes place. A closed system is also referred to control mass.

The isolated system is not affected by the surroundings. Simply there is no heat, work and mass

transfer. e.g Universe.33. Write the two property equations involving entropy.

Ans.

vdpdhTds

pdvduTds

=

+=

34. Define the steam rate in using the Rankine cycle

SURROUNDINGSSYSTEMBOUNDARY


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Ans. Steam rate =hkW

Kg

WWPT .

3600

where Wt and Wp are the work done by

turbine and pump respectively.

35. Define the heat rate in using the Rankine cycleAns.

heat rate =hkW

kJ

WW

Q

CYCLEPT

i

.

36003600

=

where Qi heat supplied.

36. Define dryness fraction of steam, and total volume of steam.

Ans.gf

f

mm

mX

+= .

gf xvvxv ++= )1( where v f and vg are specific liquid and gas volume respectively.

37. Give the formulae relating temperature , pressure and value for an adiabatic

process.ANS .

1

2

1

1

1

2

1

2

=

=

V

V

P

P

T

T

38. What is meant by one tonne of refrigeration ?

Ans. One tonne of refrigeration is defined as the rate of heat removal from the surroundingsequivalent to the heat required to melt 1 tonne of ice in oe day. If the latent heat of fusion of ice

is taken as 336kj /kg then 1 tonne is equivalent to heat removal at the rate of

91000x336)/24=14000kJ/hr.

Therefore capacity of the refrigeration plant is=1400

3600)41( hhw

PT

VS

1

2

1

2

P 1

P 2

T 1

T 2

V 2 = V 1

V 2 = V 1

C O N S T A N T V O L U M E P R O C E S S ( F O R S T E A M )

==

=

=

2

1

2

1

12

12

T

d u

T

d qSS

uuq

d ud q

21

1

2

2

1

12

1212

ln

)(

T

P

T

P

T

TC

T

d uSS

TTC vouuq

vo

=

==

==


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PT

VS

1

2

1

2

P1

P2

T 2

T2 = T1

P2

REVERSIBLE ISOTHERMAL PROCESS ( STEAM )

W QP1

V2=V1

22111

2

2

1

12

1

21122,12

1212

....;ln

ln....,

...,)(

vpvpandvvR

TdqSS

v

vRTpdvwqandVPVPuu

dwdudqandwuuqSST

====

=====

+=+==

PT

VS

12

1

2

P2=P1

T 1

T 2

REVERSIBLE CONSTANT PRESSURE PROCESS ( FOR STEAM )

1

2

2

1

1

2

2

1

2

1

12

12

12

1212

2

1

ln

)(

)()(..

T

v

T

v

T

TC

T

dh

T

dqSS

hhq

TTCdhdwdudq

TTRvvppdvWWORK

po

po

=

===

=

==+=

===

W

P2=P1

Q


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P T

V

V

P

T

S

FOR AIR

T1

T2

1

2

1

2

S2=S1

P2

P1

P2

P1

2

1

S2=S1

REVERSIBLE ADIABATIC PROCESS

S

P2

P1 P2

P1

FOR STEAM

T1

T2

PT

V

V

P

T

S

S

FOR AIR

T1

T2

1

2P2=P11

2

T2=T1P2

P1

P2

P1

21

W Q

P2=P1

T2=T1

W

Q

CONSTANT PRESSURE

CONSTANT TEMPERATURE


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First law and second law

PHASE RULE.

The number of independent variables with a multicomponent, multiple system is

given by the phase rule. It is also called the Gibbs phase rule. It is expressed by the

equation as:

N= C +2

Where,

N= the number of independent variables.C= the number of components.

= the number of phases present in equilibrium.

For the single component (C=1) Two phase system (=2) system, the number of

independent variables needed is :

N=1-2+2=1

For example, water (C=1) at the triple point has three phases (=3 ) and thus n=0. That is

none of the properties of a pure substance at the triple point can be varied. From this rule

we can know that a pure substance which exists in a single phase (=1) will have two

independent variables (n= 2 ). That means there are two independent intensive properties

required to be specified to fix up the state of the system at equilibrium.

Problem

A reheat cycle operating between 30 and 0.04 bar has a superheat and a reheat

temperature of 450 deg C. The first expansion takes place till the stem is dry saturated and

then reheat is given. Neglecting feed pump work determine the reheat cycle efficiency.

1

2

1

1

1

2

1

2

1212

12

12

)()(

)(..;0

=

=

==

=

===

=

kk

k

vo

V

V

P

P

T

T

TTCuuW

SS

uuWandTdsQ

dudw


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Solution :-given data ;- P1=30 BAR; P2=0.04 BAR, T1=450 deg C and T3=450degC,and

0.12 =x .

T

S

0.04 bar

30 bar

1

2

3

45

6

450 deg C

From steam tables : at 30 bar at 450 deg C : h1= 3344 kJ/kg ; S1= 7.08 Kj/kg K

At 0.04 bar;

1-2 ISENTROPIC PROCESSS1=S2=7.08 Kj/kg K

S2=Sg .. dry saturated steam

P2= Psat at Sg

From steam table : P2 =2.3 bar.

At 2.3 bar :

kgkJh /33441=

kgKkJs /08.71 =

kgKkJskgkJhff

/423.0,/12144

==

kgKkJskgkJhfgfg

/05.8,/243344

==


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3-4 Isentropic processS3=S4=8.306 Kj/kg/K

X4 =0.98

=2505 Kj/kg.

The cycle efficiency :

=

= x100

Efficiency =0.3873 =38.73% Ans/

,/6.27122

kgkJh =

kgKkJskgkJh /306.8,/0.338133

==

053.8423.04444

+=+= xSxSSfgf

243398.04.1214444 +=+=

fgf hxhh

+

+

)()(

)()(

2351

4321

hhhh

hhhh

+

+

)6.27123381()4.1213344(

)25053381()6.27123344(


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AERO -THERMODYNAMICS PROBLEMS.

Problem 1.

Three identical bodies of A,B and C constant heat capacity are at temperatures of 300, 100 and

300K respectively. A heat engine is operated between A and B and a heat pump working as arefrigerator between B and C. The heat pump is operated by the output of heat engine. If no

work or heat is supplied from outside, find the highest temperature to which any one of the body

can be raised by the operation of heat engine or a refrigerator.

fT /

fT

1SQ

11 RS QQW =

1RQ

fT

Let fT be the final temperature of bodies A and B./

fT be the final temperature of body C

and C be the final heat capacity of these three identical bodies.

(S)Universe > 0 (1)

(S)Univ = (S)A + (S)B + (S)C+ (S)HE+ (S)HP.

But (S)HE& (S)HP =0

Therefore (S)Univ = (S)A + (S)B + (S)C > 0

300ln fTC +

100ln fTC +

300ln

/

fTC 0 [ s =CPln1

2

TT ]

C ln)100300300(

/2

ff TT0

For minimum value of Tf (S)Univ =0

B 100K

C 300K

HE HP


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C ln 0)103( 6

/2

=

ff TT

C ln 1ln)103( 6

/2

=

ff TT

Since ln 1=0 the above equation becomes

)2...(..........109 6/2 = ff TT

and also 211 RRS QQQ +=

C(300-Tf ) =C(Tf -100) + C (Tf/ -300)

[ because Q=m C (T2 T1)]

300-Tf

= Tf 100 + Tf/ -300

Tf'=700-2 x Tf (3)

From 2 &3

62 109)2700( = ff TT

KTf 300=

Tf'=700-2 x Tf

=700-2 x 300

=100KThe maximum temperature can be raised for 100K body as 300 K of B

PROBLEM 2.

Two reversible engines A & B are arranged in series. A rejecting heat directly to B engine ,

receives 200 kJ at a temperature of 421deg C from a hot source, while engine B is incommunication with a cold sink at a temperature of 4.4 degC. If the work output of A is twice

that of B, find :

i) The intermediate temperature between A and Bii) The efficiency of each engine, andiii) The heat rejected to the cold sink.

Solution:Work output from engine A,

WA= QS1 - QR1 =200- QR1


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For reversible heat engine,

)1(..........1

1

R

SH

Q

Q

T

T= QS1

WA=2 WB

1

200694

RQT =

QR1

Therefore QR1=0.288t.. (2)

QS2

WBQR2

So WA=200-0.288 T

But WB=100-0.144 T..(3)

And also WB= QS2- QR2 = 0.288T - QR2 . (4) (Because QR1= QS2)

Eliminating (3) and (4)

100-0.144T =0.288T- QR2

QR2 =0.432T 100 .(5)

Similarly, for reversible engine B,

2

2

R

S

L Q

Q

T

T=

So asQ

QT

R

R ......4.277

2

1= QS2 = QR1

2

288.0

RQ

T=

100432.0

288.0

=

T

T. Therefore T=416.42 K or 143.42 deg C..Ans

So QR1 = 0.288 x 416.42=119.93 KJ.

And QR2 =0.432 x416.42 100 =79.89 KJ. Ans

Efficiency of engine A, = 1-1

1

S

R

Q

Q= 1 -

200

93.119 =40.04 %

Efficiency of engine B, = 1-

2

2

S

R

Q

Q= 1 -

93.119

89.79 =33.39%. Ans.

(because QS2 = QR1) )

TH

A

B

TL


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Problem 4: Throttling process:

Steam at 1Mpa and 0.9 dry is throttled to a pressure of 200 kPa. Using steam table, find the

quality of steam and change of entropy. Check your answer using Mollier chart. State whetherthe process is reversible or reversible.

SOLUTION:

Given data :- P1=1 Mpa=10 bar. X=0.9. Throttling process P2= 200 kPa=2 bar.To find : quality of steam and change of entropy.

During the throttling process enthalpy remains constant, i.e. h2=h1.

From steam tables at 1 Mpa or 10 bar:

Quality of steam is wet

kgkJhf

/6.7261= kgkJhfg /6.20131 =

kgKkJsf

/138.21= kgKkJsfg /445.41

=

kgkJhh /8.2574:6.20139.02.76211

=+=

kgkJs

sxsssfgf

/1385.6

445.49.0138.2:

1

1111

=

+=+=

kgkJhkgkJhfgf

/6.2201:/7.50422

==

kgKkJsf

/53.12= kgKkJs

fg/597.5

2=

94.0:

6.22017.50484.2574::

2

221

=

+==

xAns

xhhSince

222 fgf xhhh +=

6.22017.50422

+= xh


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Generally throttling is an irreversible process.

From Mollier chart,

For x1=0.9 at 10 bar line, note the enthalpy of steam,

S1=6.1 kJ/kgK.

Since the throttling process is a constant enthalpy process, draw horizantal line in the

Mollier chart upto 2 bar as shown in figure below. Now read theentropy of final state of

steam,S1= 6.76 kJ/kgK.

Change in entropy = (s2-s1)= 0.66kJ/kgK.

Since s is positive, the process is irreversibleAnswer.

h2=h1

h

Sp. Entropy (s)

1 2

10

bar

2ba

r

X=0.90

X=0.94

X=0.96

Problem 5

Steam turbine receives steam at a pressure of 20 bar superheated at 300 degC. The exhaustpressure is 0.07 bar and expansion takes place isentropically. Using steam steam tables calculate

the following ;

a)Heat supplied assuming that the feed pump supplies water to the boiler at 20 bar.

kgKkJs

sssentropyinchange

kgKkJs

s

/652.0

138.679.6....

/79.6

)59.5(94.053.1

12

2

2

=

===

=

+=


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b) Heat rejected.c)work done.

d) Thermal efficiency.

e) Theoretical steam consumption.

T

ENTROPY (S)

P2=0.07 bar

p1=20 bar

1

2

3

300 C

4

SOLUTION:Given data :- p1=20 bar : T1=Tsup=300deg C ; P2=0.07 bar

To find Qs, Qr, W, Efficiency (Rankine), and SSC.

From super heated steam table at 20 bar and 300 degC,

kgkJh /5.3021 = kgKkJs /77.6

1 =

barattablessteamfrom .07.0........

kgkJhf

/42.1632 = kgkJh

fg/2.2409

2 =

kgKkJs f /559.02 = kgKkJs fg /718.72 =

kgmvf

/001007.0 32 =

8.0

718.7558.077.6

....

2

22121

=

+=

+==

x

x

sxsss

thatknowwe

fgf

kgkJW

ppv

ppvhhWworkpump

P

f

P

/0068.2

)72000(001007.0)(

)(:.

212

34334

=

==

==

)0069.24.163(3025)(:sup21

+=+=pfS

whhQpliedheat

./79.2859 kgkJQ S =

./76.20902.24098.04.1632

2222

kgkJh

hxhhfgf

=+=

+=

AnskgkJ

WhhorQQWdoneworkPRS

......./34.932

37.19277.2859)(......21

=

==

AnskgkJQ

hhhhQrejectedheat

R

fR

..../36.1927

4.16376.2090:..2232

=

===

%6.32326.079.2859

34.932.. ====

SQ

WefficiencyThermal


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Theoretical steam consumption :

IMPORTANT FORMULAE IN STEAM OR PURE SUBSTANCE

PROPERTIES WET STEAM DRY STEAM SUPER HEATED STEAM

ENTHALPY ( h )

KJ/kgfgfwet

hxhh .+= fgfg

hhh += )(sup SSUPPg

TTChh +=

SPECIFIC

VOLUME, (v) in

m3/kg

gwet vxv .=

gdry vv =

S

SUPg

T

Tvv

=

sup

Density () in

kg/m3

wet

wetv

1=

g

gv

1=

sup

1

vwet =

Work done (W)

in KJ/kggWET

vpW .100=

gg vpW .100=

supsup.100 vpW =

Internal energy

in KJ/kgWETwetWET

WhU =

GGG

WhU = SUPSUPSUP

WhU =

Specific Entropy

(s) in KJ/kg.Kfgfwet

SxSS .+= fgfDRY

SSS +=

+=

S

SUP

PSGSUPT

TCSS ln

FOR NON-FLOIW PROCESS.

PROCESSES

WORK DONE HEAT TRANSFER

CONSTANT VOLUME W=0 )()(111sup212 g

vxpvphhQ =

CONSTANT

PRESSURE

W=p(v2-v1) Q =h2 h1

CONSTANT

TEMPERATURE

W=p(v2-v1) Q= h2 - h1

HYPERBOLIC

=

1

2

11 ln v

vvpW )(ln 12

1

2

11 hhv

vvpW +

=

ISENTROPIC W=U1- U2 Q= 0

POLYROPIC

1

2211

=

n

vpvpW )(

1

)(12

2211 hhn

vpvpnQ +

=

hrkWkgW

=== /86.334.932

36003600


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FOR FLOW PROCESS.

SYSTEMS WORK DONE HEAT TRANSFER

BOILER W=0 Q=h2-h1

TURBINE W=h2-h1 Q=0

CONDENSOR W=0 Q=h1 h2

NOZZLE W=0 Q=0

APPLICATION OF STEADY FLOW ENERGY EQUATIONS TO VARIOUS

ENGINEERING SYSTEMS.

1.0 BOILER OR STEAM GENERATOR.12

hhQ = .in KJ.

2.O CONDENSOR.

21 hQh =+

2.0 NOZZLE.( )2

1122)(2 VhhV +=

The expansion of the fluid in a nozzle is treated as reversible adiabatic or isentropic.

( )21212

)(2 VTTCVP

+=

+

=

2

1

1

1

2

112(2 V

P

PTTCV

P

as..

1

1

2

2

=

P

PT

2

1

1

1

2

1212 V

P

PTCV

P +

=

m/sec


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20

TURBINE.

Turbine is a device which converts potential energy of working fluid into mechanical work.

The turbine is fully insulated. Therefore there is no heat transfer (Q=0).

Whh +=21

AIR COMPRESSOR.

Whh =21 ..FOR NO HEAT LOSS

WhQh =21 ..IF Q IS REJECTED TO SURROUNDINGS.


21/56

21

THROTTLING PROCESS.

When a liquid expands through a minute aperture orifice or a slightly opened valve, the

process is called throttling process.

Q=0, W=0, AND Enthalpy is constant i.e. h1=h2

Relation between specific heats at constant volume and pressure.

The ratio between specific heat

V

P

C

C IS DENOTED BY GAMMA, i.e.

=

=

VP

V

P

CC

C

C

PROBLEM 6.

A refrigerator works on an ideal vapor compression refrigeration cycle, uses R-12 as theworking fluid. The minimum and maximum pressure of the cycle is 0.15 Mpa and 0.9Mpa

respectively. If the mass flow rate of the refrigerant is 0.045 kg/sec; determine (a) the rate of heat

removal from the refrigerated space and the power input to the condenser. (b)the rate of heatrejection to the environment and (c) COP of the refrigerant.

SOLUTION:

111 ,, UVP122

,, UVP


22/56

22

In an ideal vapour compression refrigeration cycle the compression process is isentropic and therefrigerant nters the compressor as a saturated vapour at the evaporator pressure and leaves the

condenser as saturated liquid at the condenser pressure.

T

ENTROPY (S)

P2=0.15 MPa

p1=0.9 MPa

2

14

3

Q2

Q1

IRREVERSIBLE PROCESS

BECAYSE OF

THROTTLING h3=h4

Win

The enthalpy of the refrigerant at all four states is obtained from the refrigerant R-12TABLES.

At P1 =0.15Mpa,

kgKkJSSkgkJhhgg

/7089.0:/64.17811

====

At P1 =0.9 Mpa,

kgKkJSSandkgkJh /7089.0..:/43.210122

===

At P3 =0.9 Mpa,

kgkJthrottlinghhandkgkJhhf

/93.71:..:/93.71343

====

The rate of heat removal and the power input to the compressor respectively is given by :


23/56

23

)(............).....(12

.

41

.

hhWandhhQ mm INL ==

kWQL

8.4)93.7164.178(1

.

045.0 ==

kWWIN

43.1)64.17843.210(045.0.

==

(b) The rate of heat rejected from the refrigerant to the environment is :

kWhhm 33.6)93.7143.210(045.0)32( ==

Also QH is given by :

kWWin

LH QQ 23.643.180.4

.

=+=+=

( c ) COP= AnsW

Q

IN

L ...36.343.1

80.4==

PROBLEM :7In an aircraft engine compressed air at 3 bar and 450K enters a combustion chamber in which heat is added at

constant pressure to the air by the combustion of the fuel.Air is thus heated to a temperature of 1250K.It then enters

a turbine with a negligible velocity . It expands in the turbine until its temperature falls to 100K.The velocity of air

leaving the turbine is 50m/sec.Air then enters a convergent divergent nozzle wherein it expands until its temperature

drops to 800K.Flow through both turbine and nozzle may be considered as reversible adiabatic. Determine :

(a)Heat added in the co0mbustion chamber/kg of air.(b)Work done in the turbine per kg of air.(c)Velocity of air leaving the nozzle.(d)Pressure of air leaving the turbine and at the exit from the nozzle.

Combustion

chamber

Air

turbine

1

3

2

Compressed

air at 3 bar ,

450K

3 bar,

1250K

1000K, 50 m/sec

nozzle

v4

w


24/56

24

(a) combustion chamber.

Heat added to air kgkJTTChhQPO

/804)4501250(0035.1)()(1212 ====

(b) Turbine :

work done in the turbine

QVV

hhW +

+= )2

()(2

3

2

2

32 as Q=0

kgkJV

TTCWPO

/2512000

50)10001259(0035.1

2)(

22

3

32

===

note that the kinetic terms are negligible.

( c ) Velocity of air leaving nozzle.

2

343

2

3434)(2)(2 VTTCVhhV

PO +=+=

sec/63650)8001000(5.100322

4 mV =+=

(d) Pressure of air leaving the turbine is given by:

barT

TPP

k

k

374.11250

10000.3

4.0

4.1

1

2

3

23 =

=

=

(e) pressure of air at exit of the nozzle is given similarly by:

barT

T

PP

k

k

63.01000

800

374.1

4.0

4.1

1

3

4

34 =

=

=

PROBLEM 8

An aircraft is flying at a speed of 1000km.p.h. The ambient air pressure and temperature is 0.35bar and 15 deg C. respectively. What is the pressure of air entering the compressor after inlet


25/56

25

diffuser section, and what is the pressure ratio required for compressor for the pressurization ofthe aircraft?

Solution:

Speed of aircraft

sec/8.2773600

10001000 mV ==

from the SSSF conditions for the diffuser.

2

2

1

12

Vhh +=

Fuel flow

Combustion chamberDiffuser intake

Exhaust gases from exit

Ambient air

Turbine

Turbo jet engine

1 2

3 4

5 6

T

S

2

1

3

4

5

6

COMPRESSOR TURBINE

DIFFUSER

COMB CHAMBER

NOZZLE


26/56

26

2

2

1

12

VTCTC

POPO +=

Whence the temperature of air leaving inlet diffuser,

KC

VTT

PO

5.2965.382585.10032

8.2772582

22

112 =+=

+=+=

barT

TPP

k

k

57.0258

5.29635.0

4.0

4.1

1

1

2

12 =

=

=

Pressure ratio of compression required for cabin pressurization:

78.157.0

01325.1== Answer

PROBLEM : 9

An engine operates on the air standard cycle. The compression ratio is 18. The pressure and

temperature at the start of compression process are 100kPa and300K.The heat added is 1800

KJ/kg of air. Determine the maximum pressure and temperature in the cycle, the thermal

efficiency and the mean effective pressure.

State 2:

v

T

1

3

4

2

S=CONSTANT

MAX

QH

QL


27/56

27

Kv

vTT 3.953)18(300 4.0

1

2

1

.12 ==

=

.5720)18(100 4.12

1.12 kPav

vPP ==

=

state :3

KTT

TTChhQQPOH

3.2744);3.953(0035.11800

)(

33

2323322

==

===

Maximum temperature and pressure of the cycle.

Tmax=T3=2744.3K

Pmax=P2 =p3=5720 kPa.State 4 :

kgmP

RTv /1377.0

5720

)3.2744(287.0 3

3

3

3 ===

kgmP

RTvv /861.0

100

)300(287.0 3

1

1

14 ====

Kv

vTT 3.1318861.

1377.03.2744

4.01

4

3

.34 =

=

=

kPav

vPP 4.439

861.

1377.05720

4.1

4

3

.34 =

=

=

Heat rejected:

kgkJTTCUUQQVOL

/731)3.1318300(7165.0)(414114 =====

Net work :

kgkJQQWLHNET

/10697311800 ===

Thermal efficiency:

%)4.59(594.01800

1069

1

==

H

NET

theorQ

W

Mean effective pressure:


28/56

28

( ).1419

)18/86.0(861.0

1069

21

kPavv

Wp NET

mean =

=

=

1.0 PROBLEM 10.

Consider the entropy change of an ideal gas following reversible path :

1-a (isothermal) and a-2 (constant volume )

& 1-b (adiabatic & b-2 (constant volume )

as shown below. Show that the path combination produces the same entropy.

For path 1-a-2:

1

2

2

1

2

1

1ln.:lnlnT

TCSSand

v

vR

v

vRSS

VOa

a

A ===

So that by adding we get;

1

2

1

2

12lnlnT

TC

v

vRSS

VO+= .(A)

For path 1-b-2:

B

VOBBTTCSSadiabaticQasSS 2

21ln)(0....0 ===

For adiabatic process ])(ln[. 1

1

2

1

2

2

1

1

1

=

=

v

v

T

TCSS

v

v

T

TVOB

b

B

1

2

1

2 ln)1(lnv

vKC

T

TC

VOVO +=

p

v

1

2

b

a

T=C

Q=0

V=C


29/56

29

1

2

1

2 lnlnv

vR

T

TC

VO += (B) EXPRESSIO SAME AS (A)

)1(..1...... === KCRieC

R

C

CRCCas

VO

VOV

P

VP

2.0 REVERSIBLE POLYTROPIC PROCESS FOR AN IDEAL GAS.

2N

REVERSIBLE

POLYTROPIC

WORK

T

S

P2

P1

2T

2S

Q

T2=T1

ISOTHERMAL N=1

REVERSIBLE

ADIABATIC

N=K

)/ln(

)/ln(

loglogloglog

......

.....0

......

12

21

2111

22

11

vv

ppn

vnpvnp

pvCvpvpalso

v

pn

dv

dp

v

dvn

p

dp

Cpvhavewe

nnn

n

=

+=+

===

==+

=

)2.....(....................1

1

11

..

...

1122

21

1122

1

111

1

122

21

1

1

1

2

2

1

2

1

21

2211

n

vpvpW

n

vpvp

n

vvpvvpW

n

vvC

v

dv

CpdvWHence

v

vp

v

vp

v

cpalso

nnnnnn

n

n

n

n

n

n

=

=

==

=

==

===


30/56

30

Substituting

=

=

1)(

1

....1

1

211

21

1

2

1

1

2 n

nn

p

p

n

vpW

p

p

v

v

3.0 PROBLEM 11.

Air expands irreversibly from 3 bars, 200 deg C, to 1.5 bars and 105deg C.Compute the entropychange/kg of air.

Solution.

Although the process is irreversible, entropy being a property change can be computed by

integrating along a reversible path between the two specified states.

4.0 PROBLEM 12.

An insulated tank is divided into equal parts by volume. One part contains Argon gas at 30 degC and 5 bar. The

other part is vacuum. The partition between these two parts is broken and the gas fills the whole tank. Determine

final pressure and entropy change per kg of Argon gas.

Solution.

Considering it as an ideal gas, we have :

[ ]

21

12

12

12

1221

)1

(

)()1(

)()1

(

)(.....

1

Wn

nkq

TTn

nkCq

TTN

RC

Wuuqnow

TTn

RW

VO

VO

=

=

+=

+=

=

kgKkJ

P

PR

T

TCSS

PO

/0323.03

5.1ln278.0

473

378ln0035.1

lnln1

2

1

2

12

==

=


31/56

31

Entropy change ( choosing the reversible path)

kgKkJv

vR /1441.02ln

40

344.8ln

1

2=== as molecular weight of Argon gas =40

5.0 PROBLEM 13

A Single stage reciprocating air compressor has a swept volume of 2000cm3and runs at 800rpm.

It operates on a pressure ratio of 8, with a clearance of 5 % of the swept volume. Assume NTP

room conditions and at inlet (p=101.3 kpa, t=15 deg C), and polytropic compression andexpansion with n=1.25; calculate (a)indicated power (b) volumetric efficiency c) mass flow rate

d) FAD (free air delivery ) e) isothermal efficiency f)the actual power needed to drive thecompressor, if mechanical efficiency is 0.85.

SUCTION AT CONSTANT TEMPERATURE T1.

P1=101.3kpa;P2=8P1=810.4kpa; T1=288K, VS=2000cm3

V3==VC=0.05*VS=100 cm3.

V1=VC+VS=2100 cm3.

barpv

vpp

processansionfreeforuuasTT

5.22

1

...exp.........,..

1

2

1

12

1212

==

=

==

V

P

1

3

4

2

PVN =C

MAXP2

P1

VC VS


32/56

32

( ) 325.11

3

1

4

3

444335281008. cmV

P

PVVPVP

nnn

==

==

V1-V4=2100-528=1572 cm3

=

1)(1

1

1

2

411

n

n

P

PVVP

n

nW OR

=

1)41(1

1

1

2

n

n

P

Pmm

n

nW

( ) JW nn

41118101572103.101251.0

25.1 163=

=

NOTE : If the number of stages is N then work done =Nx W

Indicated power= wattsdoneworkrpminspeed

60

......

Indicated power kW

60

10800411 3 5.46 Kw

(b) volumetric efficiency = %6.782000

1001572=

c) mass of compressed air per cycle :3

63

1093.1288287

101572103.101

=

==

RT

pVm

THEREFORE MASS FLOW RATE= min/54.18001093.13 kg=

d) FAD (free air delivery ) min/26.1800101572 36 m=

e)

kWPowerInput

Jp

pVVPW

ISOTHERMAL

44.685.0

47.5....

%6.806047.5

800331.0

3318ln101572103.101ln)(63

1

2

4111

=

=

=

===


33/56

33

6.0 EFFICIENCY OF COMPRESSOR

The efficiency of a compressor is defined the ratio of isothermal work done to the actual work

done by the compressor.

Efficiency of a compressor =COMPCOMPC

T

W

V

VVP

W

P

PVP

W

W 2

111

1

211 lnln

===

The work done by the compressor

=

1

1

1

1

1n

n

X

C

P

P

n

nRTNW

Where N=number of stages of compression and Px is the intermediate pressure.

The optimum pressure Px= 31 PP for two stages. For number of stages the optimum pressure is

given by

N

S

DX

P

P

P

P1

1

= Where Pd=delivery pressure and Ps=suction pressure and Px=intermediate

pressure.

7.0 PROBLEM 14

A two stage air compressor with perfect inter-cooling takes in air at 1 bar pressure at 27deg C .

The law of compression in both the stages is tconsPV tan3.1 = .The compressed air is delivered

at 9 bar from the HP cylinder to an air reservoir. Calculate per kg of air a) the minimum work

done and b) the heat rejected to the intercooler.

SOLUTION.

The minimum work required is a two stage compressor is given by :-

=

11

1

1

1n

n

XC

P

P

n

nRTNW


34/56

34

( )

= 133.0

300287.03.123.1

3.01

..as Px=3

=26x0.287x100x0.287=214.16kJ/kg

2886.13 3.13.0

1

1

2

1

2==

=

n

n

P

P

T

T

therefore T2=386.56 K

Heat rejected to the intercooler =1.005(386.56-300)=86.99kJ/kgAns

8.0 PROBLEM 15.GAS POWER CYCLE OTTO CYCLE.

An engine working in the Otto cycle is supplied with air at 0.1mPa, 35 deg C. The compression

ratio is 8. Heat supplied is 2100 KJ/kg.Calculate the max pressure and temperature of the cycle,

the cycle efficiency and the mean effective pressure.(for air Cp=1.005, Cv==0.718, andR=0.287kJ/kgK).

( )

( )

AnsMPaMEPPPVVP

kgkJQW

MPaPPT

VP

T

VP

MPaPVV

PPKTT

KTkgkJTTCQ

KTV

V

T

T

kgmV

MMM

CYCLENET

MAX

MAX

V

.........533.1774.0

5.1186)...11.0884.0()(

/5.1186565.02100

426.9708

363837.1............

873.1........37.188...................3633

2925718.0

21004.708.........:/2100)(

4.7083083.2.......:3.28

/11.08

884.0

21

1

3

2

22

3

33

2

4.1

2

1

1

23

3231

2

4.0

1

2

1

1

2

3

2

======

===

====

===

===

====

====

=

==

V

P

1

3

42

S= C

CPv =

OTTO

CYCLE

T

S

1

3

4

2Q1

Q2

WC

WE

3

1

11

2

1

4.01

1

2

1

1

844.0100

308287.0,......8

%5.56....565.03.2

11

8

11

11

4.1...8,../2100

/1001.0

30835273

mP

RTV

V

V

orr

rkgkJQ

mkNMPaP

KT

k

cyc

k

=

===

====

===

==

=+=


35/56

35

9.0 PROBLEM 16

A diesel engine has a compression ratio of 14 and cut off takes place at 6% of the stroke. Find

the air standard efficiency.

10.0 PROBLEM 16.

v

P

1

3

4

2

S=CONSTANT

MAX

QH

QL

CPV =

%5.60605.078.0

24.1248.01

178.1178.1

141

4.111

1

1111

78.1.......:78.1

78.0)14(06.0)(06.0.........:14

4.1

4.0

1

2

3

23

2222123

2

1

==

=

=

===

=====

c

c

k

diesel

c

k

r

r

r

rV

VRATIOOFFCUTVV

VVVVVVVV

Vr


36/56

36

A turbojet aircraft flies with a velocity of 300 m/sec at an altitude where the air is at 0.35 bar and40 degC. The compressor has a pressure ratio of 10, and the temperature of the gases at the

turbine inlet is 1100deg C. Air enters the compressor at a rate of 50 kg/sec. Estimate a) the

temperature and pressure of the gases at the turbine exit. b) the velocity of gases at the nozzleexit and c) the propulsive efficiency of the cycle.

For isentropic flow of air in the diffuser,

3 45

6

T

S

2

1

3

4

5

6

DIFFUSER

COMB

CHAMBER

NOZZLE

12

COMPRESSOR TURBINE

Q1

Q2

PRODUCT

GAS

P=C

MKTTP

PT

kPaPrP

kPamkNT

TPP

KTC

VTT

VTTC

VVhhWQ

p

P

P

6.536......:)10(8.277

.6.64776.6410

76.649233

8.277/35)(

8.277...:005.12

10300233

2

2)(0

2

3

4.1

4.0

2

1

2

3

3

23

4.0

4.1

21

1

2

12

2

322

1

12

2

1

12

2

1

2

2

122112

==

=

===

=

==

=

+=

+=

=

+=


37/56

37

b) for isentropic expansion of gases in the nozzle:

Neglecting the KE of gas at nozzle inlet,

[ ]21

6561000][2 = TTCV

P

[ ] sec/4.1029sec/)(2 21

656 mmhhV =

=

c) The propulsive efficiency of a turbo-jet engine is the ratio of the propulsive power developed

Wp to the total heat transfer to the fluid.

[ ]

[ ]

Ans

MWhhwQ

MW

VVVWWAIRCRAFTINTEXITP

%........7.25257.0026.42

806.10

026.42)66.5361373(005.150)(

80.103003004.102050

341

===

===

==

=

11.0 JET PROPULSION

EEAIREfuelair APPVmVmmT +++=

)()( &&&

Where T =thrust in Newtons,air

m& andfuel

m& are mass flow of air and fuel respectively.

.69.311

1373

12.11146.647

2.111478.27766.5361373

...:

5

5.3

4

1

4

5

5

2345

55235423

kPaP

PT

TP

KTTTT

TTTThhhh

WWTC

=

=

=

=+=+=

==

=

12.59669.311

352.1114

286.01

5

6

56 =

=

=

P

PTT


38/56

38

V .velocity of free stream air or aircraft velocity

EV velocity of jet exhaust

EP ..pressure at nozzle exit

P .free stream pressure or static pressure at the altitude where aircraft is flying.

12 ROCKET MOTOR.

The specific Impulse of the rocket motor is given by :

2

1

1

)(11

2

=

O

Eo

SPP

PTRI

where Isp=specific impulse;oT =chamber temperature

O

E

PP is the ratio of the nozzle exit

pressure to the rocket chamber pressure, R is the gas constant, and = ratio of the specific

heat for the combustion gases taken as 1.4 for adiabatic flow.

The mass flow through the nozzle is given by :-

+

=

+

1

1

1

2

RT

AP

mO

O

&

Where A is the area of throat cross section.

13.0 STEADY STATE STEADY FLOW ENERGY EQUATIONS.SSSFE.

ddWvvpuw

ddQvvpuw ++=+++ )

2()

2(

2

2

2222

2

1

1111.is energy flow per unit time

dm

dWvvpu

dm

dQvvpu ++=+++ )

2()

2(

2

2

222

2

1

111for energy flow per unit mass

In differential form :-


39/56

39

dWgdzVdVdhdQ +++=

Where

For closed system the work done by the gas is pdv while for flow systems the work done is

given by vdp .

14.0 PROBLEM 17.

NOZZLE FLOW FOR A GAS FOR A POLYTROPIC PROCESS.

Since this is a flow system i.e. not a closed system the work done should be a vdp term. Anozzle flow analyses is taken as an example to make things more clearer.

A system above is steady flow nozzle ideally insulated device so that no heat transfer takes

place with the surroundings, while the flow takes place through it. A gas expands through the

nozzle following a reversible polytropic law Cpv = . There is no change in PE but the

pressure drop from 15 bar to 1.5 bar and the specific volume increases from 0.1 m3to 0.6m3.

vdpdhTds

vdpdhdq

vdpdqdh

vdppdvdudh

pvdudh

pdvdudQ

=

=

+=

++=

+=

+=

12

V1=70 m/S

.v1=0.1m3/kg

P1=15 bar

V1=? m/S

.v1=0.6m3/kg

P1=1.5 bar


40/56

40

If the entrance velocity to the nozzle is 70 m/sec determine the exit velocity.

We have;

dWgdzVDVdhdQ +++=

here Q=0 ; W=0 as no external work is added and the potential energy is =0 as z is in the same

datum.Therefore,

== 2

2

1

2

2

2

1

VVVdVVdV

15.0 Reversible adiabatic flow of a fluid with area change.Consider reversible adiabatic flow of a fluid in a passage of varying cross sectional area.

The following equations can be written for the flow :-

Continuity equation : tconsmVA tan== &

Differentiating we have

0=++V

dV

A

dAd

SSSF equations (neglecting PE changes )

( )2211

2

1

1

.......

vpvpnn

processpolytropicaindoneworkvdp

VdVvdp

VdVvdpdqdq

VdVvdppdvdudq

=

=

=

++=

+++=

( )

smV

JV

JVV

/5.7244900106.22

106.22

70

106106.033.46.0105.11.010153.0

3.1

2

5

2

5

22

2

5555

2

1

2

2

=+=

=

===


41/56

41

dpdhTds

VdVdh

=

== 0

Therefore combining SSSF equations with the property relation we get;

dp

VdV

1=

substituting for , dV ,from the continuity equations

2V

dpd

V

dVd

A

dA

+==

+= 1

2

2 dp

dV

V

dp

A

dA

Since the flow is reversible adiabatic ,

2

2

2

M

VC

d

dp==

Substituting the same in the above we obtain

( )22

1 MV

dp

A

dA=

.(1)

For nozzles always dp 0.

Hence on closer analyses of the above equation we can see the following :-

FOR NOZZLES :-

For subsonic portion of the nozzle the nozzle cross section converges i.e. when M1.0.

FOR DIFFUSERS:

It is the reverse of nozzles i.e. at the convergent portion diverges or M>1.0 and converges

for supersonic flow.


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EXAMPLE E7-3 A 4-cylinder SI engine with a cylinder displacement of 0.5 L and aclearance volume of 62.5 mL is running at 3000 RPM. At the beginning of the compression

process, air is at 100 kPa and 20oC. The maximum temperature during the cycle is 1800

K. Employing the cold-air standard Otto cycle, determine (a) the power developed by theengine, (b) the thermal efficiency, and (c) the MEP. (d) What-if-Scenario:How would

the answer in part (b) change if the maximum temperature were raised to 2200 K? (e)How would the answer in part (b) change if the ideal gas model were used? [Manual Solution][TEST Solution]

Answers:(a) 30.7 kW, (b) 58.6%, (c) 614 kPa, (d) 58.5%, (e) 53.5%


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EXAMPLE E7-5 An air standard Diesel cycle has a compressionratio of 18. The heat transferred to the working fluid per cycle is

2000 kJ/kg. At the beginning of the compression process thepressure is 100 kPa and the temperature is 25oC. Employing theperfect gas (PG) model for the working fluid, determine (a) the

pressure at each point in the cycle, (b) the cut-off ratio, (c) thethermal efficiency, (d) the net work per unit mass, and (e) the

MEP. (f) What-if-Scenario:How would the thermal efficiencychange if the heat transfer were 1500 kJ/kg? [Manual Solution][TESTSolution]

Answers:(a) 5728 kPa, 487 kPa, (b) 3.1, (c) 57.8%, (d) 1153 kJ/kg, (e) 1430 kPa, (f)60.6%

EXAMPLE E7-6 The Diesel cycle described in Example E7-5is

modified into a Dual cycle by breaking the heat addition process

into two equal halves so that 1000 kJ/kg of heat is added atconstant volume followed by 1000 kJ/kg of heat addition at

constant pressure. Determine (a) the temperature at each point inthe cycle, (b) the thermal efficiency, (c) the net work per unit

mass, and (d) the MEP. (e)What-if-Scenario:How would thethermal efficiency change if 75% of the total heat transfer took

place at constant volume? [Manual Solution][TEST Solution]

Answers:(a) T2 = 947 K, T3 = 2340 k, T4 = 3335 K, T5 = 1209 K, (b) 67.3%, (c) 1346kJ/kg, (d) 1666 kPa, (e) 61.1%


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EXAMPLE E8-2Air enters the compressor of a 50 MW simplegas turbine at 100 kPa, 298 K. The pressure ratio of the

compressor is 11 and the turbine inlet temperature is 1500 K.

Using the cold-air standard Brayton cycle to model the powerplant, determine (a) the mass flow rate, (b) the thermalefficiency, and (c) the back work ratio. (d)What-if-Scenario:

How would the efficiency and mass flow rate change if the

compression ratio is varied within the range 4-50? [ManualSolution][TEST Solution]

Answers:(a) 11.61 kg/s, (b)


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EXAMPLE E9-2 A steam power plant operates on an ideal Rankine

cycle. Superheated steam flows into the turbine at 3 MPa and 400oCwith a flow rate of 100 kg/s and exits the condenser at 50oC as

saturated water. Determine (a) the net power output, (b) the thermal

efficiency, and (c) the quality of steam at the turbine exit. (d)What-if-scenario:How would the efficiency change if the condensertemperature can be lowered to 30oC? [Manual Solution][TEST

Solution]Answers:(a) 108,179 kW, (b) 33.2%, (c) 91.3%, (d) 36.5%


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10-1-7 A refrigerator uses R-12 as the working fluid and operates

on an ideal vapor compression refrigeration cycle between 0.15 MPa

and 1 MPa. If the mass flow rate is 0.04 kg/s, determine (a) thetonnage of the system, (b) compressor power, and (c) the COP. (d)What-if-scenario:How would the answer in part (c) change if R-

12 was replaced with R-134a, a more Environmentally benign

refrigerant? [Manual Solution][TEST Solution]Answers:(a) 1.165 tons, (b) 1.36 kW, (c) 3.01, (d) 3.33


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Notes prepared by M.Sundaresan.

EXAMPLE E10-5 A two-stage cascade refrigeration plant uses R-22

as the working fluid in both the stages. The lower cycle operates

between the pressure limits of 120 kPa and 380 kPa and the toppingcycle has a condenser pressure of 1200 kPa. The heat exchangerthat couples the two cycles requires a minimum temperature

difference of 5oC between the heating and the heated streams. If the

mass flow rate in the lower cycle is 0.08 kg/s, determine (a) themass flow rate in the upper cycle, (b) the cooling capacity, and (c)

the COP. (d) What-if-Scenario:How would the COP change if a

single cycle operated between 1200 kPa and 120 kPa?[Manual Solution]Answers:(a) 0.11 kg/s, (b) 4.53 tons, (c) 2.7, (d) 2.58


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ae 2202 thermodynamics- 2 mark questions

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