Aero Engineering Thermo Dynamics

8/12/2019 Aero Engineering Thermo Dynamics

1/224


2/224

It is defined as the amount of heat energy re+uired raising or lowering the temperature of unit massof the substance through one degree when the pressure kept constant. It is denoted by C p.

;. De+i e s0e i+i heat a0a it" at $ sta t $*)me.

It is defined as the amount of heat energy re+uired to raising or lowering the temperature of unitmass of the substance through one degree when volume kept constant.

A0#. ,6617

Thermodynamic property is any characteristic of a substance which use used to identify the state ofthe system and can be measured, when the system remains in an e+uilibrium state.

11. H$( %$ "$) *assi+" the 0#$0e#t"'

Thermodynamic property can be classified in to two types.

%. Intensive or Intrinsic and'. !tensive and !trinsic property

1,. De+i e I te si e a % E/te si e 0#$0e#ties. ?MU>O t. =9@ O t. =A0#. ==@ A0#.,661

S*.N$. I te si e P#$0e#ties E/te si e P#$0e#ties

%. Independent on the mass of the system. ependent on the mass system.'. If we consider part of the these properties If the consider part of the system it will have a lesser value

:EL TECH D#. RR & D#. SR TECHNICAL UNI:ERSITY

'


3/224

remain same e.g. #ressure, temperature,specific volume, etc.

e.g. Total energy, Total -olume, weight, etc.

12. What %$ "$) ) %e#sta % !" e )i*i!#i)m $+ a s"stem'

When a system remains in e+uilibrium state, it should not under go any changes to its own accord.

18. Whe a s"stem is sai% t$ !e i The#m$%" ami E )i*i!#i)m '4A a U i . A0#.62 4EEE7@ MU>A0#. =< MSU>A0#. =97

When a system is in thermodynamic e+uilibrium, it should satisfy the following three conditions

a. &echanical " #ressure remains constant b. Thermal +uilibrium " Temperature remain constantc. Chemical e+uilibrium " There is no chemical reaction

19. De+i e e#$th *a( a % +i#st *a( $+ the#m$%" ami s. 4A a U i . N$ . 6 A0#.68 4EEE7@ MU>N$ .=2@ A0#. ,661 BRU A0#. =97

/eroth law of thermodynamics states that when two systems are separately in thermal e+uilibrium with athird system, then they themselves are in thermal e+uilibrium with each other.

0irst 1aw of thermodynamics states that when system undergoes a cyclic process net heat transfer ise+ual to work transfer

2 3 W

1;. 4a74i7 State Fi#st La( $+ the#m$%" ami s a % a " t($ $+ its $#$**a#ies. ?A a U i . A0#.624EEE7

/eroth law of thermodynamics states that when two systems are separately in thermal e+uilibrium with a

third system, then they themselves are in thermal e+uilibrium with each other.

0irst 1aw of thermodynamics states that when system undergoes a cyclic process net heat transfer ise+ual to work transfer.

2 3 W

C$#$**a#ies $+ +i#st *a( $+ the#m$%" ami sC$#$**a#" I

There e!ists a property of a closed system such than a change in its value is e+ual to the difference between the heat supplied and the work done any change of state.



4/224

C$#$**a#" II

The internal energy of a closed system remains unchanged if the system is isolated from itssurroundings.

C$#$**a#" II I

$ perpetual motion machine of first kind 4##&5%) is impossible. 1T977 - 3 % ? %@ 7-3@%D?.E< m ;kg mole


@


5/224

Res)*tK&olecular volume, -3@%D?.E< m ;kg mole

,1. A i s)*ate% #i3i% esse* is %i i%e% i t$ t($ 0a#ts !" a mem!#a e. O e 0a#t $+ the esse* $ tai sai# at 16 M0a a % $the# 0a#t is +)**" e a )ate%. The mem!#a e #)0t)#es a % the ai# +i**s the e ti#e

esse*. Is the#e a " heat a % $# ($# t#a s+e# %)#i 3 this 0#$ ess' )sti+" "$)# a s(e#. ?A aU i e#sit" N$ 56 4Ma# h7

0or rigid vessel and unrestrained e!pansionChange in volume d- 3 7Work transfer. W 3 pd- 3 70or insulated vessel, heat transfer, 2 3 7$ccording to the first law of thermodynamics the sum of work transfer is e+ual to the sum of heat transfer. W 3 2 3 7

,,. De+i e the te#m 0#$ essK ?MGU NO:5=9

It is define as the change of state undergone by a gas due to energy flow.

, . De+i e the te#m " *eK ?MGU NO:5=9

When a system under goes a series of processes and return to its initial condition, it is known as cycle.

,2. What is mea t !" $0e a % *$se% " *e'

In a closed cycle, the same working substance will recirculate again and again. In a open cycle, the sameworking substance will e!hausted to the surrounding after e!pansion.

,8. What is mea t !" #e e#si!*e a % i##e e#si!*e 0#$ ess' ?MU A0#5,661 BNU > N$ 5=2

$ process is said to the reversible. It should traces the same path in the reverse direction when the

process is reversed. It is possible only when the system passes through a continuous series of e+uilibriumstate. If a system does not passes through continuous e+uilibrium state, then the process is said to beirreversible.

,9. What is mea t !" 0$i t a % 0ath +) ti$ ' ?MU>O t. ,666

The +uantity which is independent on the process or path followed by the system is known as pathfunctions.

!ample" Feat transfer, work transfer


E


6/224

,;. What is )asi> stati P#$ ess' ?MU>O t. =


7/224

,. P#$ e that the %i++e#e e i s0e i+i heat a0a ities e )a* t$ C 0 C R. ?A a U i . A0#. 64Me h7

Consider a gas heated at constant pressure(o, heat supplied, 2 3 mC p 4T' B T%)Work done, W 3 p4- ' B - %)3 m > 4T ' B T%)Change in internal energy, 8 3 mC v 4T' B T%)

$ccording to the first law of thermodynamics,2 3 W 6 8

(o, mC p 4T' B T%) 3 m> 4T ' B T%)6 mC v4T' B T%) C p 3 r 6 C vC p B cv 3 >

33 State the Ge* i P*a stateme t $+ se $ % *a( $+ the#m$%" ami s' ? MU N$ 5=2@ O t 5=;@O t5 ,666 MGU N$ 5=8@ A0# 5=9

=elvin B #lank states that it is impossible to construct a heat engine working on cyclic process, whoseonly purpose is to convert all the heat energy given to it an e+ual amount of work.

34 State C*a)si)s stateme t $+ se $ % *a( $+ the#m$%" ami s' ?MU>O t5=;@ O t5==@ A0#5,666 MG N$ 5=8

It states that heat can flow from hot body to cold body without any e!ternal aid but heat cannot flowfrom cold body to hot body without any e!ternal aid.

35 W#ite the t($ stateme ts $+ the Se $ % *a( $+ the#m$%" ami s. ? A a U i .A0#56

Ge* i 0*a stateme tK

It is impossible to construct an engine working on an cyclic process which converts all the heat energysupplied to it into e+uivalent amount of useful work.

C*a)sis stateme t

Feat cannot flow from cold reservoir to hot reservoir without any e!ternal aid. Hut can flow fromreservoir to cold reservoir without any e!ternal aid.

36 State Ca# $t5s the$#em.

:o heat engine operating in a cyclic process between two B fi!ed temperatures can be more efficientthat a reversible engine operating between the same temperature limits.



8/224

37 What a#e the C$#$**a#ies $+ Ca# $t the$#ems'

i. $ll the reversible engines operating between the two given thermal reservoirs with fi!ed temperaturehave the same efficiency

ii. The efficiency of any reversible heat engine operating between two reservoirs is independent of thenature of the working fluid and depends only on the temperature of the reservoirs.

38 De+i e PMM $+ se $ % i %.

#erpetual motion of second kind draws heat continuously from single reservoir and converts it intoe+uivalent amount of work. Thus it gives %77 efficiency.39 What is %i++e#e e !et(ee a heat 0)m0 a % #e+#i3e#at$#'

Feat pump is a device which operating in a cycle process, maintains the temperature of a hot body attemperature higher that the temperature of surrounding.

$ refrigerator is a device which operating in a cycle process maintains the temperature of a cold bodytemperature lower than the temperature of the surrounding.

4 What is mea !" heat e 3i e'

$ heat engine is a device which is used to convert the thermal energy into mechanical energy.

41 De+i e the te#m COP.

Coefficient of performance is defined as the ratio of heat e!tracted or reGected to work inputHeat extracted rejected

COPWork input

=

42 W#ite the e/0#essi$ +$# COP $+ a heat 0)m0 a % a #e+#i3e#at$#.

COP +$# heat 0)m02

HP2 1

THeat rejectedCOMP

Work input T T= =

COP +$# #e+#i3e#at$#1

ref 2 1

THeat extractedCOP

Heat input T T= =

43 Wh" Ca# $t " *e a $t !e #ea*i e% i 0#a ti e'


?


9/224

i) In a Carnot cycle all the four processes are reversible but in actual practice there is no process isreversible.

ii) There are two processes to be carried out during compression and e!pansion. 0or 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 completely44 Name t($ a*te# ati e meth$%s !" (hi h the e++i ie " $+ a Ca# $t " *e a !e i #ease%.

i) fficiency can be increases as the higher temperature T ' increasesii) fficiency can be increases as the lower temperature T % decreases

45 Wh" a heat e 3i e a $t ha e 166 e++i ie "'

0or all the heat engines there will be a heat loss between system and surroundings. Therefore we canJtconvert all the heat input into useful work.

46 Whe the Ca# $t e++i ie " (i** !e ma/im)m'

Carnot cycle efficiency is ma!imum when the temperature is O=

47 What a#e the 0#$ esses i $* e% i Ca# $t " *e'

Ca# $t " *e $ sist $+i. >eversible adiabatic compressionii. >eversible isothermal heat additioniii. >eversible adiabatic e!pansioniv. >eversible isothermal heat reGection

48 W#ite the e/0#essi$ +$# e++i ie " $+ the Ca# $t " *e.

Carnot 3 2 12

T T

T

49 Is the se $ % *a( i %e0e %e t $+ +i#st *a(' E/0*ai .

Kes. The second law is independent of first law. The second law speaks about the +uality of energy.

5 De+i e e t#$0".

ntropy is an inde! of unavailability or degradation of energy.51 De+i e ha 3e $+ e t#$0". H$( e t#$0" is $m0a#e% (ith heat t#a s+e# a % a!s$*)te tem0e#at)#e.


D


10/224

The measure of irreversibility when the energy transfer takes place within the system or between systemand surrounding is called a change of entropy. It is simple known as unaccounted heat loss.

52 De+i e the te#ms s$)# e@ si a % heat #ese# $i#.

S$)# eK

The part where the heat to be reGected to absorbing or work developing device is called source

Si K

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

Rese# $i#K

The part which supplies or receives heat

Continuously without change in its temperature is called as reservoir.

53 Wh" the 0e#+$#ma es $+ #e+#i3e#at$# a % heat 0)m0 a#e 3i e i te#ms $+ C.O.P. a % $t ite#ms $+ e++i ie "'

The performance of any device is e!pressed in terms of efficiency for work developing machines. Hutheat pump and refrigerator are work absorbing machines. (o, the performance of those devices based onC.O.#. only.

54 W#ite %$( the e )ati$ +$# Ca# $t C.O.P $+ a heat 0)m0 (hi h ($# s !et(ee t($ heat#ese# $i#s $+ tem0e#at)#e T 1 a % T , i+ T1QT ,.

Carnot C.O.#. of heat pump 3 1 21

T TT

55 What is mea t !" 0#i i0*e $+ i #ease $+ e t#$0"'

0or any infinitesimal process undergone by a system, change in entropy, d ( Ld2 ;T

0or reversible d 2 3 7, hence, d ( 370or irreversible d sL7

(o the entropy of an insolated system would never decrease. It will always increase and remains constantif the pressure is reversible is called as principle of increase of entropy.

56 What %$ "$) mea !" Ca*)si)s i e )a*it" '


%7


11/224

It is impossible for a self acting machine working in a cyclic process unaided by any e!ternalagency to convey heat from a body at a low temperature to a body at a higher temperature.

8;. E/0*ai !#ie+*" *a)si)s i e )a*it".

dQ0

T is known as ine+uality of clausius

If %.dQ

0

T , the cycle is reversible.

'.dQ

0T

, the cycle is irreversible and possible.

dQ0

T , the cycle is impossible 4-iolation of second law).

8


12/224

91. -i e the e/0#essi$ s t$ +i % ha 3e i e t#$0" %)#i 3 $ sta t 0#ess)#e a % 0$*"t#$0i 0#$ ess.Sh$( $ T>S %ia3#am.

0or constant pressure process,2

2 1 p1

TS S S (C )n

T = =

0or polytropic process,

2 22 1 P)n

1 1

2 22 1 #)n

1 1

T PS S S ( C *)n

T P

or

T +S S S ( C *)n

T +

= =

= =

9,. E/0*ai the te#m Re e#si!i*it" .

If the process traces the same path during the process is reversed is called as reversibility .

9 . Ca e t#$0" $+ ) i e#se e e# %e #ease' Wh"'

ntropy of universe can not ever decrease. It will be remains constant or will increase due toirreversibility .

92. What is the esse e $+ the se $ % *a( $+ the#m$%" ami s'

%. To know the feasibility of process'. To know about the +uality of energy

98. I+ Ca# $t e 3i e e++i ie " is 86 . Fi % C.O.P. $+ Ca# $t #e+#i3e#at$# ($# i 3 !et(ee sametem0e#at)#es.

1 2

1

2

1

2 1

1 2

T TH.,. $0-

T

T1 0.$

T

T T0.$ 2

T T

= =

=

= =


%'


13/224

COP $+ #e+#i3e#at$#2 1

11 2

2

T TTT T 1T

= =

12 11

= =

PART B

1. What a#e the Limitati$ s $+ Fi#st La( the#m$%" ami s'

4%) 0irst 1aw of thermodynamics does not specify the direction of flow of heat and work. i.e., whether

the heat flows from hot body to cold or from cold body to hot body.4') The heat and work are mutually convertible the work can be converted fully into heat energy but heatcannot be converted fully into mechanical work. This violates the foresaid statements. $ &achinewhich violates the 0irst law of thermodynamics is known as perpetual motion machine 4#&& 5 N) ofthe first kind which isimpossible.

#&& B % is a machine which delivers work continuously without any input. Thus, the machine violates firstlaw of the thermodynamics.

,. A *$se% s"stem ) %e#3$es a " *e $ sisti 3 $+ t($ 0#$ esses. D)#i 3 the +i#st 0#$ ess@ 26 G $+

heat t#a s+e##e% t$ the s"stem (hi*e the s"stem %$es 96 G $+ ($# i the se $ % 0#$ ess@ 28 G $+($# is %$ e $ the s"stem.a7 Dete#mi e the heat t#a s+e# %)#i 3 the se $ % 0#$ ess.!7 Ca* )*ate the et($# %$ e a % et heat t#a s+e# +$# the " *e.

S$*)ti$ K>(ystem " Closed (ystem#rocess " Cycle consisting of two processes=nown "

To find " 2 '%, 2, W


%


14/224

$nalysis " %2 '% 3 ' W% 6 8 '5%Where ' W% 3 5@E =A.

To find 4 8 '5%)0or cyclic process the algebraic sum of any property of the system must be ero.48 %') 6 48 '%) 3 7 48 %') 3 48) %'3 5 P2 %' B %W ' Q

3 5P@7 B 97Q3 6'7 =A8 '% 3 5@E 6'72 '% 3 5'E=A

' :etwork done 3 4 %W ' 6 ' W%)3 @7 B 'E3 %E =A

:et heat transfer 3 42 %' 6 2 '%)3 97 B @E3 %E =A.

Comment "

The negative sign indicates 2 '% that 'E =A of heat is supplied by the system to the surroundingsduring the process ' to %.

. The +$**$(i 3 %ata #e+e# t$ a *$se% s"stem@ (hi h ) %e# 3$es a the#m$%" ami s " *e $ sisti 3 $++$)# 0#$ esses .

P#$ ess Heat t#a s+e# mi W$# t#a s+e# mia5b

b5cc5dd5a

E7,7775E,777

5%9,777

5@,'77

5','775 ,777

Sh$( that the %ata is $ siste t (ith the +i#st *a( the#m$%" ami s a % a* )*ateK


#rocess 24=A) W4=A)%5''5%

@7unknown

975@E

%@


15/224

a7 Net #ate $+ ($# $)t0)ts is MW!7 E++i ie " $+ the " *e

-i e %ataKa>! 86@666 mi!> 86@666 mi

>% >19@666 mi%>a 6

W %>a 6W !> 2@,66 miW >% >,@,66 miW %>a @666 miT$ Fi %1. W and = ,. '

S$*)ti$ KCyclic heat transfer of the cycle,23E77775E7775%977767 3'D777kA;minsimilarly, cyclic work transfer of the cycle,

W376 @'775''775 777 3'D777kA;min

0rom first law of the thermodynamics 23 WThe gives data is consistent with the first law of thermodynamics.

'D777 :et work output, W3 ; sec

97 3@? . kW

37.@?&W

kJ

Feat supplied, 2 s 3 E7,777kA;min4Q

Taking the positive heat only)

fficiency of the cycle, 3

supplied

'D777 3

E7,777 3E?

s

W Work done Heat Q

=

Res)*tK%. Total work output of the cycle, W37.@?&W

'. fficiency of the cycle, 3E?


%E


16/224

2. A e#tai 3as $+ $*)me 6.2m @ 0#ess)#e $+ 2.8 !a# a % tem0e#at)#e $+ 1 6 6C is heate% i a "*i %e#t$ a = !a# (he the $*)me #emai s $ sta t. Ca* )*ate 4i7 Tem0e#at)#e at the e % $+ the 0#$ ess@ 4ii7the heat t#a s+e#@ 4iii7 ha 3e i i te# a* e e#3" 4i 7 ($# %$ e !" the 3as@ 4s 7 ha 3e i e tha*0" a %4 i7 mass $+ the 3as@ ass)me C 0 1.668 3 G. a % C 6.; 3 G.-i e %ataK: 1 6.2mP 1 2.8!a# 286 N m ,

T 1 1 6 6C 1 6 ,; 26 G P , =!a# =66 N m ,

C 0 1.668 3 G C 6.;1 3 G a %It is 3i e that the 0#$ ess i$ $ sta t $*)me 0#$ essT$ +i %KT , @ @U@ W@H a % m.

S$*)ti$ K4i) Temperature at the end of the process 4T ' )">elation between p.v.T for constant volume process is

% %

' '

'' %

%

D77 @7 ?79@E7

P T P T

P T T k P

=

= =

Heat t#a s+e# 4 7K23mC v4T' 5T%)kA0irst find out the mass of the gas by using characteristic gas e+uation.# %- %3m>T %Where>3C p5Cv3%.77E57.


17/224

4e) Change in nthalpy4F)F3mC v4T' 5T%) 3%.E% ! 7.


18/224

- ' 5- %3?.D?5@.@D 3@.@Dm

4ii) Work transfer"

( )'

%

' %

3%774?.D?5@.@D)

3@@DkA

v

v

W pdv p v v= =

4iii) change in internal energy"

83mC v4T' 5T%)0or constant #ressure process' '

% %

'' %

%

%

%

'- 3 %

-

39'9=

V T V T

V T T

V

=

=

83E!7.


19/224

T$ +i %KT , @ @U&P

S$*)ti$ K4i) 0inal temperature 4T ' )" Hy using p, -*T relation

% %

' '

'' %

%

'

7.% @D?

7.7?ET %

%%

%

'% '

% 7.'DE @D? 3%.77E57.37.'DEkA;kg =

%


20/224

M 6.,8 3P1 1 !a# 166 N m ,

: 1 6. m: , 6.=mT$ +i %K

1. T 1 ',. T , '

. W '2. '5. U '

S$*)ti$ K0rom ideal gas e+uation.#%- % 3 m>T %

=1

100 0.3T

0.2$ 0.2!

0or isothermal processT% 3 T ' 3 @%?.%'=

Work done, W, mp%-, in 12

+p1 orp1#1in

p2 +

0.100 0.3 in 0.3 =

3 '.D9kAFeat absorbed, 2 3 W

3 '.D9kAChange in internal energy, 8 3 7.

Res)*tK%. Work done, W 3 '.D9kA'. Feat absorbed, 2 3 '.D9kG3. Change in internal energy, 8 3 7.


21/224

W$# %$ e@w

S$*)ti$ K( )( )

( )( )

= = =

1

3

(a (en it4 5

+olu(e +

(a ( 1.$+olu(e 1.2 3(

en it4 1 1.1"

-olume 4- %) 3 %.'D m2

1

1

+6ori ot7er(alproce 8W p1+ in

+

0.1!100 1.2 3 in 2$1."k&

1.2 3

=

= = 45ve sign indicates that the work is done on the system)Res)*tKWork done, W 3 5'E%.9kA

=. , 3 $+ 3as at a 0#ess)#e $+ 1.8 !a# $ )0ies $*)me $+ ,.8m . I+ this 3as $m0#ess is$the#ma**" t$1 times the i itia* $*)me. Fi % 4i7 i itia* tem0e#at)#e 4ii7 +i a* tem0e#at)#e 4iii7 ($# %$ e a % 4i 7heat t#a s+e#. Ass)me R 6.,


22/224

W 3 5@% .EkA :ote" Fere, negative sign indicates that the work is done on the system.4iii7 Heat t#a s+e# 4 7K

0rom 0irst law, 2 3 W 6 80or isothermal process, 8 3 72 3 W 3 5@% .EkA

:ote" Fere negative sign indicates that the heat is reGected by the system.

Res)*tKi. Initial and 0inal temperature, T % 3 T ' 3 9E . = ii. Work done 4W) 3 5@% .EkAiii. Feat transfer 42) 3 5@% .EkA

16. 16 3 $+ 3as at 16 !a# a % 266 6C e/0a %s #e e#si!*" a % a%ia!ati a**" t$ 1 !a#. Fi % the ($# %$ ea % ha 3e i i te# a* e e#3". 4Ma $ ma iam S) %a#a a# U i . A0#5=97.

-i e %ataKM 16 3P1 16!a# 1666 N m ,

T 1 2666C 266 ,; 9; G P , 1 !a# 166 N m ,

T$ +i %W a % U

S$*)ti$Hy adiabatic relation,

1

2 2

1 1

1 1

12

T pT p

100T " 3 3 !.$!%1000

=

= =

Work done,( )

( )

1 2(* T TW1

10 0.2! " 3 3 !.$!232 . k&

1. 1

=

= =

Change in internal energy8 3 m ! C v ! 4T ' B T%)

3 %7 ! 7.


23/224

3 5' '


24/224

10

10

0lo9

!000.1$

lo90. 3

=

( )

p

#

p #

p #

# #

#

#

p

1.3!.

C1.3!

C

C 1.3!C

* C C

1.3!C C

0.2 3 C 1.3! 1

0.2 3C 0. 1!k&k9%

0.3!

C 1.3! 0. ! 0. k&' k9%

=

= =

=

= =

=

= =

= =

Hy using >elation between T * - for isentropic process1 1

2 1 12 1

1 2 21.3! 1

2

T + +T T

T + +0.1$

T 2 3 1"0.$ %0. 3

= =

= =

= =

Change in internal energy 4 8)

83mC v4T' B T%)3%.E ! 7.


25/224

i. Work done 4W) 3 %@'.?DkAii. Mas constant 3 7.'< kA;kg =.

iii. >atio of two specific heat p#

C1.3!

C= =

iv. -alues of two specific heat, C p37.DDkA;kg Cv 3 7.


26/224

n 1n

2

1

11.2

2

T p2T p2

"00T 3 3 $$!%

100

=

= =

Change in internal energy.8 3 m ! C v ! 4T ' B T%)3 7.D ! 7.


27/224

F$# 0$*"t#$0i 0#$ ess@ the 0@: a % T #e*ati$

= =

= =

n 1 n 1n n

22 1

1

1.2$ 11.2$

2

T p2 p2T T

T p1 p1

"00T 300 $22.33%

100

W$# %$ eK( )

( )

( )

1 2

u

(* T TW

n 1* =ni#er al a con tant

>a con tant8 *5M (olecular ?ei97t

0.2 k&'k9%

=

=

( ) = =

1 0.2 300 $22.33W

1.2$ 12 ".3 k&

45ve sign indicates that the work is done on the system) Feat transfer, =

= =

nQ W

1

1. 1.2$2 ".3 2.3 !2k&

1. 1

45ve sign indicates that the heat is reGected from the system)Change in enthalpy"

( )( )

p 2 1H ( C T T

1 1.00$ $22.3! 300

233. k&

= = =

Res)*tK

Temperature at the end of the compression, T ' 3E''. ?= Work done on the system, W 3 5'@9. @kAFeat transferred from the system, 2 3 5D'.


28/224

T$ +i %K1. W$# t#a s+e#@ W ',. Heat t#a s+e#@ '

. Cha 3e i i te# a* e e#3"@ U '

S$*)ti$ Kuring poly tropic process,

n 1n

22 1

1

1.3 11.3

PT T

P

300 5300

100 53!".$ %

=

( )

( )

1 2(* T TWork tran fer8 W5n 1

1 .2! 300:3!".$ 5

1.3 1

5:!2.!2k&'k9

r:n7eat tran fer8 Q5 Wr:1

1. :1.3 5 !2.!2

1. :1

Q520. 0$k&'k9.

Hy first law of thermo dynamics,8325W35'7.


29/224

P#$ ess K Stea%" +*$(@ a%ia!atiG $( K

#roperty $t inlet $t e!it:e*$ it" 96 m s =6 m sP#ess)#e 166 Pa 866 PaTem0e#at)#e 26C 1,6 Ca % the mass +*$( #ate 26 3 mi .

Dia3#ams K>

A a*"sis K P QQ W m h ke pe = + + & & &

2 3 7 sincde the flow is adiabatic.

8nless the height from the datum of the inlet and out are mentioned pe 3 7.

Therefore W 3 ( )' '

% '% ' '

C C h h +

( )

( )

' '% '

% '

' '

'

97 D7@7 %77E @7 %'7

'

C C m Cp T T = + = +


'D

InQ = 0

Control

Out

&W

Volume


30/224

9

9

. 79 %7min

. 79 %797

EE.% ;

EE.%

J

J s

KJ s

kW

=

=

= =

C$mme t K (ince the working fluid is an ideal gas h % 5 h' is replaced by C p 4T% B T' ).19. A steam t)#!i e $0e#ates ) %e# stea%" +*$( $ %iti$ s #e ei i 3 steam at the +$**$(i 3 stateKP#ess)#e 18 !a#. S0e i+i $*)me 6.1;m 3 i te# a* e e#3" ,;66 G 3 e*$ it" 166 m s. The e/ha)st$+ steam +#$m the t)#!i e is at 6.1!a# (ith i te# a* e e#3" ,1;8 G 3. s0e i+i 18 m 3 a % e*$ it"

66 m s. The i ta e is m a!$ e the e/ha)st. The t)#!i e %e e*$0s 8 W a % heat *$ss $ e# thes)#+a e $+ t)#!i e is ,6 G 3. Dete#mi e the steam +*$( #ate th#$)3h the t)#!i e.S"stem K O0e S"stemP#$ ess K Stea%" +*$( 0#$ essG $( K

P#$0e#ties At i *et At e/it#ressure %E bar 3 %E %7 #a bar 3 7.% %7 #aInternal energy '


31/224

( ) ( ) ( )

( )

( )

' ' ' '' ' ' % % % ' % ' %

E

E

' '

%4 )

'

'%


32/224

A a*"sis K [ ]Q W m h ke kpe = + + & & &

0low is given as adiabatic hence 7Q =&

:o le will not involve W & Change in potential energy can be neglected in no les and diffusersh 3 ke

' '% '

' '% '

'

'

C C

C C h

= =

$s the inlet velocity is negligible

h ' B h% 3''

'C

3'@


33/224

1C1 ,


34/224

m 6.2, 3 s.T$ +i %KP$(e# $)t0)t W '

S$*)ti$ K(0

' '% '

% % ' '

' '

' '

. D.?% %777.@' '


35/224

11 1

2

+18 1 1 )n

2 +

0.100 0.3

0.332. " .

pWorkdon W mp V In orp v

p

In

k J

= =Feat absorbed, 23W 3 '.D9kAchange in internal energy , 837.

Res)*tK%. Work done, w3 '.D9kA'. Feat absorbed, 23 '.D9kA3. Change in internal energy , 837.

,1. I a stea%" +*$( s"stem a ($# i 3 s)!sta e at a #ate $+ 2 3 s e te# a 0#ess)#e $+ 9,6 N m , at ae*$ it" $+ 66m s. The i te# a* e e#3" is ,166 3 a % s0e i+i $*)me 6. ;m 3. It *ea es the

s"stem at 0#ess)#e $+ 1 6 N m , @ a e*$ it" $+ 186m s@ I te# e e#3" $+ 1866 3 a % s0e i+i $*)me$+ 1.,m 3. D)#i 3 its 0assa3e i the s"stem@ the s)!sta e has a heat t#a s+e# $+ *$ss $+ 6 3 t$ itss)##$) %i 3s. Dete#mi e the 0$(e# $+ the s"stem. State that it is +#$m 4$#7 t$ the s"stem.

-i e %ataKM 2 3 sP 1 9,6 N m ,

C 1 66m sU1 ,166 3: 1 6. ;m 3P , 1 6 m ,

C , 186m sU , 1866 G3: , 1.,m 3

> 6 3T$ +i %KP$(e# $+ the s"stem@ W '

S$*)ti$SFEE


E


36/224


37/224


38/224

' '' %

% %

' '' %

% %

4 )

P

V

T P ! ! C "n R "n or

T P

T V ! ! C "n R "n

T V

=

= +

$fter e!pansion air will occupy the entire volume of the container. - ' 3 '- %

$lso %W ' 3 7 since it is an unresisted e!pansion 2 %' 3 7 since the vessel is insulated

$pplying the first law of thermodynamics

% 'Q # W = +Therefore 8 3 70or air

' %

' %

4 ) 7

. .V C

$ T T

% e T T

==

Fence ( ' B ( % 3 C - in ' '% %

T V

R "nT V +

3 7.'?< In %'

' V V

3 7.%DD =A;kg=

Comment" Though the process is adiabatic entropy increases as the process involving unresisted e!pansionis an irreversible process. It also proves the fact that.

4 ) 7dQ

&s or dsT

> >

,2. A *$se% s"stem is ta e th#$)3h a " *e $ sisti 3 $+ +$)# #e e#si!*e 0#$ esses. Detai*s $+ the0#$ ess$#s a#e *iste% !e*$(. Dete#mi e the 0$(e# %e e*$0e% i+ the s"stem is e/e )ti 3 166 " *es 0e#mi )tes.

Tem0e#at)#e 4G7P#$ ess 4G 7 I itia* Fi a*% B '

' B B @

@ B %

76 %777

75

77%777%777

77

%777%777

7777

(ystem" Closed#rocess " The system is e!ecuting cyclic processG $( K Heat t#a s+e# i 0#$ ess 1,@ , a % 2 a % Tem0e#at)#e ha 3e i a** the 0#$ ess.


?


39/224

:o. of cycles per minute.To find " #ower developed.

Dia3#ams K

$nalysis " To find power developed W net per cycle must be known. 0rom I 1aw W net 3 2 net which can becomputed from the following table.

#rocess 24=A)Temperature 4=) (Initial 0inal

% B '' B

B @@ B %

7%777

75

77%777

%77777

%777%777

7777

7%777

%%777 KJ KJ

k k =7

( @%

0or a cyclic process 3 7where is any property

s 3 74i.e) ( %' 6 ( @ 6 ( @% 3 7 7 6 % 6 7 6 ( @% 3 7

( @% 3 % KJ K (ince the process @5% is isothermal

@%

@%

%77

77

Q

Q KJ

=

= Therefore


D

(K)

%777

300

3

4

2

1

!


40/224

%' ' @ @%

7 %777 7 77


41/224

Res)*t KChange in entropy (357.' ?kA;=.

,9. A i e t$# *aims t$ ha e %e e*$0e% a e++i ie t h$t e 3i e (hi h ($)*% ha e a heat s$)# e at1666 6 C a % #eJe t heat t$ a si at 86 6 C a % 3i es a e++i ie " $+ =6 )sti+" his *aim is 0$ssi!*e $#

$t.

-i e %ataK0

H

0

B

T 1000 c 5 12 3k

T $0 C 323k

=

= =n 0-=

S$*)ti$ K$ccording to Carnot theorem reversible engine gives ma!imum efficiency than other heat engine. &a!imum efficiency,

7 B(ax

H

T T 12 3 323n 0. " ."-

T 12 3 = = = =

Ma/im)m e++i ie " 4 ;2.9 7 is *ess tha 0#$0$se% e 3i e e++i ie " 4 =6 7. The#e+$#e his *aim isim0$ssi!*e. A s.Res)*tKInventorJs claim is impossible.

,;. A i e t$# *aims that his 0#$0$se e 3i e has the +$**$(i 3 s0e i+i ati$ KP$(e# %e e*$0e% 86 WF)e* !)# t 3 h#Ca*$#i+i a*)e $+ the +)e* ;8666 G 3Tem0e#at)#e *imits ,; 6 C a % 9,; 6CFi % $)t (hethe# it is 0$ssi!*e $# $t.-i e %ataKP 86 WF)e* !)# t 3 h#C.:. $+ +)e* ;8@666 J 3

Tem0e#at)#e *imits@0

B

0H

T 2 C 300k

T "2 C 00k

= == =

S$*)ti$ KFeat supplied to the engine 3 0uel burnt ! C.- 3 !


42/224

H Bcarnot (ax

H

T TCarnot efficienc48 n n

T= =

00 300

0."" ""."-00= = =

fficiency of inventorJs claim engine,

actual?orkdone W $0

n 0.! !0-Heat upplied Q "2.$

= = = = =

here the inventorJs claim engine has the higher efficiency than ma!imum engine efficiency which is possible.

Res)*tKInventorJs claim is impossible

,


43/224

Case 4ii7*

(

( (ax

Q Q 1120 10!n 0. 0 0-

Q 1120

n n it i i(po i;le

= = = =

>

Res)*tKCase 4i) is possible heat engine ( )1 (axn n

,=. A i e t$# *aims t$ ha e %e e*$0e% a #e+#i3e#ati 3 ) it (hi h mai tai s the #e+#i3e#ate% s0a eat >9 6 (hi*e $0e#ati 3 i a #$$m (he#e tem0e#at)#e is ,; 6C a % has COP 6

-i e %ataKT L 18 C ,; >18 ,8


44/224


45/224

.

H.,

! 3 2 3! 3

5"". -

2

0.""3 kW

53:2

51kW

=

=

=

=

=

=

H LH E

H

c rnot

s

s

s

! s

T T T

W Q

Q

Q

Q Q W

4b) $t @7 of ideal efficiency,H,

* S

0."" 0.

0.2"$"

2Q0.2"$"

."k?Q Q W

." 2.0 $."kW

= =

==

= = =

,. A #e e#si!*e heat e 3i e $0e#ates !et(ee a s$)# e at


46/224


47/224

for the entire cycle,dQ

0T

This e+uation is known as Clausius ine+uality. It provides the criterion of the reversibility of a cycle.If

dQ0

T= , the cycle is reversible

dQ0

T , the cycle is impossible.(ince the cyclic integral

dQT is less than ero in a cycle, the cycle violates the second law of

thermodynamics. (o, it impossible.

We can apply the e+uality to the Carnot cycle since it is reversible cycle. Then the e+uation becomes,dQ

0T

= .

2. A i e t$# *aims that his e( e 3i e (i** %e e*$0 W +$# a heat a%%iti$ $+ ,26 mi . Thehi3hest a % the *$(est tem0e#at)#e $+ the " *e a#e 18,; C a % , C #es0e ti e*". W$)*% "$) a3#eehis *aim' Use C*a)si)s i e )a*it" meth$%.-i e %ataKW B kW 3 7 97 3 %?77 kA;min2 % 3 '@7 kA;minT% 3 %E'< CT ' 3 '< CT$ +i % K$greement his claim 3

S$*)ti$ KHy clausius ine+uality,

1 2

1 2

Q QdQT T T

= + Hut, Work out, W 3 2 % B 2 '

%?77 3 '@7 B2 '2 ' 3 5%E97kA ; min 3 7

1 2

1 2

Q Q 2 0 1$"02. k&' (in.

T T 1!00 "00+ = =


@


48/224

FeredQ

0T


49/224

(ince it has been already proved that total work output from any number Carnot engines operating inseries is e+ual to that of a single Carnot engine operating between the same reservoirs.

%

77%

%'77 A +Q W W = +

(ince W $ 6 W H

%

'% %

'

'

'

77% '

%'77

' % %'77

%'77 'D77'

%'77 %'77

%'77 D77%'77

' %'77


50/224

=nown" 2 3 77 =A;Cycle T 3 77 = T% 3 %777 =

To find " 4a) 2 % B heat reGection to the reservoir at %777 = 4b) T ' B Intermediate temperature in =

4c) CO#

Dia3#amsK

4b) 4CO#) >ef% 3 4CO#) >ef' '

' % % '

'

' '

'' ' '

'

'

77

77 %777

77 %777 77 77

77 %777

E@


51/224

'

'

''

% '

% ' ' '

%.'%%

%.'%%7.?'9

%.?'9

QW

QW

W Q

A(so Q Q W Q

=

=

== + =

Where 2 ' 3 2 6 W %

'

'

%.'%%

%.'%%

QQ

QW

= +

= =

*ef2 *ef1ince COP 5 COP

%

%

%

%%

%.'%%77 %.?'9

E@


52/224

2 in 3 D7 =A TF 3 77= To find " 4a) W heat engine

4b) heat engine 4c) T1 refrigerator 4d) CO# >efrigerator

Dia3#ams K>

%E heat eng%ne

W

$nalysis " 4a) W heat engine 3 % )

%n H

T QT

77% D7

D7797 KJ

= =

4b) Feat engine 3 % ) H

T T

77%

D77

7.99< 499.< )

=4c) Wref 3

%E Heat eng%ne

W


E'

Source at 00% nvironment at300 %

0 %& "0 %&

H, *ef

Sink at "00 % >efrigerator spaceat TB *ef

W"e#t engine


53/224

> >e

>e>e

%97

E%'

97 %'

@?

out e, ,

%n )

, H ) ,

KJ

Q Q W

KJ

QT T T W

=

==

= =

=

"n !e#

@?@

77 %'

@4 77 )

'@7

)

)

) )

)

T T

T T

T K

= = =

=

d) CO# >ef 3>e

%n

,

QW

@?@

%'= =


54/224

A a*"sis K

%

%

'7774 )

'777 %%3 '77 B 2 >%

0or reversible heat engine,

( )1H

*1

*1

QT2.T Q

" 200So8

T Q

= =

TF

T 1

2 >% 3 7.'??T 555555554'.@?)(o, W $ 3 '77 B 7.'?? THut W H 3 %77 B 7.%@@T 555 4'.@D) 4Q 'W $ 3 W H)

and also W H 3 2 s' B 2 >'3 7.'??T B 2 >' 555555 4'.E7) 4Q 2 s% B 2 s' )

+uating e+uations 4'.@D) and 4'.E7)%77 B 7.%@@T 3 7.'??T B 2>'

2 >' 3 7.@ 'T 5%77 55555554'.E%)

(imilarly, for reversible engine H,


E9

W A52W

AQ

*1QS2

W

Q*2


57/224

( )

2

2

1

2 1

2

2

B *

**

*

*

QTT Q

QTSo8 Q Q

2 . Q

0.2!!T 0.2!!TQ 0. 32T 100

T 1". 2%

=

= =

= =

=

Q

O> T 3 %@ .@'7C(o, 2 >% 3 7.'?? ! @%9.@' 3 %%D.D kAand 2 >' 3 7.@ ' ! @%9.@' B %77

3


58/224

Fi3)#e

0or the heat engine, the heat reGected 2 ' to the panel 4at T ' ) is e+ual to the energy emitted from the panel to the surroundings by radiation of $ is the area of the panel,

2 2

2 2

Q AT

Q %AT ?7ere % i a con tnat=

:ow, 1 21 1

T TWQ T

= =

or 31 2 2 21 2 1 2 2

Q Q %ATW%AT

T T T T T= = = =

( ) ( )3 32 1 2 1 2 2W W

A%% T T % T T T

= =

0or a given W and T %, $ will be minimum when

( ) ( )22 3 3

1 2 2 1 2 22

dA W3T T T T T T 0

dT %

= =

since( )3 2 31 2 2 1 2 2

2

1

T T T 08T T T

T 3 0. $

T

=

= =

UNIT II

-AS AND :APOUR POWER CYCLES & I.C.EN-INES FUNDAMENTALS


E?


59/224

PART A

1. What a#e the ass)m0ti$ s ma%e +$# ai# sta %a#% " *e a a*"sis' 4A.U. OCT ,66,7

I. $ir is the working fluid and it obeys the laws of perfect gas. #v 3 m>T.II. :o change in the mass of the working medium.III. !pansion and compression process are assumed to be adiabatic.I-. :o chemical reaction take place in the cylinder during the cycle.-. $ll the process are reversible processes and form a cycle..

,. F$# the same $m0#essi$ #ati$ a % heat #eJe ti$ @ (hi h " *e is m$st e++i ie t $tt$ Diese* $# D)a*.? A.U. OCT ,66, .

The air standard cycle efficiency of any theoretical cycle is given by 2 r

th 3 % B2 s

Where 2s 3 Feat (upplied2r 3 Feat reGected

We know, r

s

2%

2 =

$ll the cycles have the same 2 r.$s 2 r is constant, U will increase if 2 s increases,2 s for otto, ual and iesel cycles is given by area under curve '5 , '595< and '5E respectively.

$rea under '5 L $rea under '595< L $rea under '5E.Fence otto dual diesel > >


ED


60/224

. De+i e $m0#essi$ #ati$.

It is defined as the ratio of the volume of fluid occupied when the piston reaches the bottom deadcentre position to the volume of the fluid occupied when the piston reaches the top dead centre position.

r c or 4r) 31

2

s c

c

V V V V V + =

2. De+i e mea e++e ti e P#ess)#e.

It is defined as the theoretical constant pressure acting on the piston during the entire power stroke,that will produce the same amount of work as that produced by the actual varying pressure during the entirecycle.

Work done during a cycle in :5m&ep 3 (troke volume in m .

8. What is 0)#e s)!sta e'

#ure substance is a substance which has a fi!ed chemical composition throughout its mass.!amples" Water, nitrogen, carbon dio!ide, and helium. $ pure substance does not have to be of a single

chemical element or compound. $ mi!ture of various chemical elements or compounds also called as a puresubstance as long as the mi!ture is homogeneous.

9. What is sat)#ati$ tem0e#at)#e a % sat)#ati$ 0#ess)#e'


97


61/224

$t a given pressure, the temperature at which a li+uid boils is called the saturation temperature. $t agiven temperature, the pressure at which the li+uid boils is called the saturation pressure. It is also called asvapour pressure.

;. De+i e *ate t hea% $+ a0$#i ati$ . 4MU A0#5,6667

The amount of heat added during heating of water from boiling point or dry saturated stage is calledas latent heat $+ a0$#i ati$ $# e tha*0" $+ a0$#i ati$ $# *ate t heat $+ steam.


62/224


63/224

In the T5s diagram, the region left of the water e!ists as li+uid. In the right of the dry steam line, thewater e!ists as a superheated steam. In between water and dry steam line, the water e!ists as a set steam.Therefore, the dryness fraction lines are represented in these region. The value of various +uantities can bedirectly read from the diagram.

It can be noted from the figure that the water line and steam line are converging with the increase intemperature. $t a particular point, the water is directly converted into dry steam without formation of wetsteam. This point is called C#iti a* P$i t5.

Fi3)#e. Tem0e#at)#e E t#$0" %ia3#am

1=. D#a( the h>s %ia3#am +$# steam a % sh$( a th#$tt*i 3 0#$ ess $ it 4MU>OCT5=;7.

,6. W#ite the +$#m)*a +$# a* )*ati 3 e t#$0" ha 3e +#$m sat)#ate% (ate# t$ s)0e#heat steam$ %iti$ . 4MU>A0# 5==7


9


64/224

supsupntropy of (uperheated steam, ( log g ps c

s

T ! C

T

= +

where( g B entropy of dry steamTsup5(uper heated temperatureTs5(aturated temperatureC ps5(pecific heat of super heated steam

,1. Dete#mi e the $ %iti$ $+ steam $+ , !a# (h$se e t#$0" is 9.,; G 3.4MU>APR5==7

0rom steam Table at ' bar s g3A0#5,6667

0rom steam table at %'7 7C(pecific enthalpy, h g3'


65/224

,8. Dete#mi e (hethe# (ate# at the +$**$(i 3 states is a $m0#esse% *i )i%@ a s)0e# heate% a0$# $# ami/t)#e $+ sat)#ate% (ate# steam. 4a7 1< M0a@ 6.66 m 34!71 66C@,66 Pa.

Case4a)#3%?&pa-37.77 m ;kg0rom steam table, corresponding to %?&pa, read- g37.77


66/224

,=. What a#e the e++e ts $+ $ %e se# 0#ess)#e $ the Ra i e " *e'

Hy lowering the condenser pressure, we can increase the cycle efficiency. The main disadvantageslowering the back pressure increase the wetness of steam isentropic compression of a very wet vapour isvery difficult.

6. A a0$)# " *e i he#e t*" has t($ a% a ta3es $ e# 3as 0$(e# " *e' What a#e the"' 4MU O t5=87

4i) Isothermal heat transfer 4evaporation and condensation) is possible in practice.

4ii) The work ratio is high compared to the gas power cycles.

1. What a#e the *imits $+ ma/im)m a % mi im)m tem0e#at)#es i a steam 0$(e# " *es' 4MU O5=87

The limit of ma!imum temperature of steam is its critical temperatures i.e


67/224

2. Name the %i++e#e t $m0$ e t i steam 0$(e# 0*a t ($# i 3 $ a Ra i 3 " *e.4MUO t5=;7

Hoiler, Turbine, Cooling, Tower or Condenser and #ump

8. Dis )ss the e++e ts $+ steam 0#ess)#e a % tem0e#at)#e at i *et t$ the t)#!i e Ia Ra i e " *e.4MU>O t5=87

$t the higher pressure the heat reGection is less and this results in increase in efficiency. The cycleefficiency does not increase continuously with boiler pressure up to the critical pressure. The increase in

pressure increases the wetness of the steam after e!pansion, which decreases the adiabatic efficiency, andreason is that latest heat decreases at high pressure. The increase the cost of boiler, and turbine and also it

erodes the turbine blades.

Increase in temperature of steam supplied to turbine increases the work done by the turbine and alsoincreases the net cycle efficiency. The efficiency of the superheated cycle continuously increase with

pressure. (uper heating reduces the specific steam consumption. It also increases the dryness fraction ofsteam at the end to the e!pansion. We can reduce the back pressure, which increases the work done.

9. What a#e the e++e ts $ $ %e se# 0#ess)#e $ the Ra i 3 C" *e'

Hy lowering the condenser pressure, we can increase the cycle efficiency. The main disadvantages islowering the backpressure increases the wetness of steam Isentropic compression has to deal with a non5

homogeneous mi!ture of water and stem. Hecause of the large specific volume of li+uid vapour is verydifficult.

;. Wh" Ca# $t " *e a $t !e #ea*i e% i 0#a ti e +$# a0$)# 0$(e# " *es'

The main difficulty to attain the cycle in practice is that isothermal condensation is topped before itreaches to saturated li+uid condition. Therefore the compressor has to deal with a non5homogeneous mi!tureof water and stem. Hecause of the large specific volume of li+uid vapour mi!ture before compression, thecompressor si e and work input have to be a large. The higher power re+uirement reduces the plantefficiency as well as work ratio.


68/224

,oiler

4MU>A0# 5=AP5=87

$t the low reheat pressure the reheat cycle efficiency may be less that the >ankin cycle efficiency.(ince the average temperature during heating will then be low.

21. What a#e the %isa% a ta3es $+ #eheati 3'

The cost of the plant increases due to the reheater and its long connections. It also increases thecondenser capacity due to increased dryness fraction.

2,. List the a% a ta3es $+ #eheat " *e. 4MU>O t5=;7

%. &arginal increase in thermal efficiency'. Increase in work done per kg of steam which results in reduced si e of boiler and au!iliaries for the

same output.. We can prevent the turbine from erosion .

2 . D#a( the a0$)# 0$(e# #eheat " *e $ T>s %ia3#am a % (#ite i%ea* e++i ie " $+ that " *ei te#ms $+ e tha*0" 4MU A0#5=


69/224


70/224

T)#!i e

C$ %e s$#

Fee% (ate# heate#

Fee% 0)m0 Fee% 0)m0

T)#!i e

C$ %e se#

Fee% heate# Fee% heate#

Fee% 0)m0

29. What is the +) ti$ $+ +ee% (ate# heate#s i the #e3e e#ati e " *e (ith !*ee%i 3'4MU>O t 5==7

The main function of feed water heater is to increase the temperature of feed water to the saturationtemperature corresponding to the boiler pressure before it enters into the boiler.

2;. Whe (i** !e the e++i ie " $+ the #e3e e#ati e " *e attai s ma/im)m'

The temperature of the bled steam is appro!imately halfway between the e!treme temperatures of the primary flow cycle.2


71/224


72/224

A sK ;86 #0m.

8;. A** the +$)# $0e#ati$ s i t($ st#$ e e 3i e a#e 0e#+$#me% i )m!e# $+ #e $*)ti$ $+#a sha+t.

A sK $ e

8


73/224

. Feat addition takes place at constantvolume.

. Feat addition takes place at constant pressure

@. Compression ratio is less. It is variesfrom 9 to ?.

@. Compression ratio is more. It variesfrom %' to %?.

92. What is the $the# ame 3i e t$ $tt$ " *e'

A sKConstant volume cycle.

98. What is mea t !" ai# sta %a#% e++i ie " $+ the " *e'

A sKIt is defined as the ratio of work done by the cycle to the heat supplied to the cycle.Work done

fficiency n 3 555555555555555555Feat supplied

99. De+i eK Mea e++e ti e 0#ess)#e $+ a I.C. e 3i e.

A sK&ean effective pressure is defined as the constant pressure acting on the piston during the working

stroke. It is also defined as the ratio of work done to the stroke volume or piston displacement volume.9;. What (i** !e the e++e t $+ $m0#essi$ #ati$ $ e++i ie " $+ the %iese* " *e'

A sK fficiency increases with the increase in compression ratio and vice B versa.

9


74/224

;,. -i e the e/0#essi$ +$# e++i ie " $+ the D)a* " *e.

% =pr 5 %fficiency n 3 % 5 5555555 55555555555555555

4r)r5% 4=5%) 6 y=4p5%)where,r B Compression ratiok B pressure or !passion ratio

p B cut off ratio andy B adiabatic inde!

; . The e++i ie " $+ the D)a* " *e is tha the %iese* " *e a % tha the $tt$ " *e +$#the same $m0#essi$ #ati$.

A sKgreater, less.

;2. What a#e the +a t$#s i +*)e i 3 $+ the D)a* " *e'

A sK%. Compression ratio '.cut off ratio . pressure ratio and @. heat supplied at constant volume andconstant pressure.

;8. The B#a"t$ " *e is mai *" )se% i

A sKMas turbine power plant.

;9. -i e the e/0#essi$ +$# e++i ie " $+ the B#a"t$ " *e.

%fficiency n 3 % 5 5555555555 where > p B pressure ratio.

4> p)y5%

;;. The t($ st#$ e " *e e 3i e 3i es the )m!e# $+ 0$(e# st#$ es as $m0a#e% t$ the +$)#st#$ e " *e e 3i e@ at the same e 3i e s0ee%.

A s Kdouble.

;


75/224

A s K0alse.


76/224


77/224

%$ir standard efficiency, 4 ) 3 % 5

4- c)

% 3 % 5 3 7.E@7?

4- % 3 3 # % % ! %7E

for ideal 9a P + 5 (*T&

for air8 * 5 2! 8 ( 5 1 k9k9%

Assumed

3 7.?@7D m ; kg 7.?@7D

- ' 3 3 7.%'7% m ; kg c4p@5p%)Q P4E75D@) B 94 E. '5%)Q %@.9?Canceling - c 3 3 3 3%%. c 5% ) 4 n5%) 4 95% )4 %.'E5% ) E!7.'E

%' '7- s 3 d ' ! 1 3 ! 3 '.'9 !%7 5' m

@ @ %77 %77

Work done per cycle3 p mvs 3 %%.


89/224

$nd T 3 T @> c XXX.. 4') T

W 3 mC v P T 5T% > c 5 5 6 T%Q XXX.. 4 ) > c

This e!pression is a function of > c when T and T % are fi!ed. The value of W will be ma!imum whendW 3 7d>cdW

3 5 T % 4 5 %) > c 5 T 4 %5 ) > 5 c 3 7d>c

T > c 3 T %> c

T 3 > c'

T%

T > c 3

T%

4b) 0or air 3 %. @

T T %'77 > c 3 3 3 3 E.99

T % T% 77

The air standard efficiency is given by

% % %


?D

1 1

2

2

( 1)

( ) ( )1

2 1

( 1.4 1) 1.25 1.25


90/224

a 3 % 5 3 %5 3 %5 3 7.E or E7 >c 4E.99) 7.@ '

Work done per kg of air is given by the e!pression 4 )

%'77W 3 7. c3 3 3 4@) 7. c 4'.?') 7.9 c 5 6 T % > c %'77

3 .%E %'77 B 77 ! 4'.?') 7.9< 5 6 77 3 D@E kA;kg 4'.?


91/224

4:ote. Feat added will increase by the same percentage).

=. A e 3i e ($# i 3 $ Ott$> " *e i (hi h the sa*ie t 0$i ts a#e 1@,@ a % 2 has )00e# a % *$(e#tem0e#at)#e *imits T a % T1. I+ the ma/im)m ($# 0e# 3 $+ ai# is t$ !e %$ e@ the sh$( that thei te#me%iate a % $)t*et tem0e#at)#e is 3i e !".

T , T 2 T 1 T 4!7 I+ a e 3i e ($# s $ Ott$> " *e !et(ee tem0e#at)#e *imit 12 6 G a % 66 G@ Fi % thema/im)m the$#eti a* 0$(e# %e e*$0e% !" the e 3i e ass)mi 3 the i# )*ati$ $+ ai# 0e# mi )te is 6.2

3.

S$* K 4a7 Usi 3 the e )ati$ 5 $+ the *ast 0#$!*em a % a*s$ the same 0#$ e%)#e T

> c 3T%

T ' 3 T %> 5%c and T @ 33

1c

T !

(ubstituting the value of > c in the above e+uations,

T

T ' 3T % 3 T % T 3 T%T T% T%T T

(imilarly, T @ 3 3 3 T%T T T

T% T%

T ' 3 T @ 3 T %T 'W 3 mC vP 4 T 5T' ) B 4 T@ 5T% ) QT ' 3 T @3 77 ! %@ 7 3 9EE =

W 3 7. " *e is


92/224

Ta e C 0 1.668 GJ 3 GC 6.;1< 3 G

%.77E(ol " % 3 % bar, T % 3 77 = 3 3 %.@

7.T % 7.'?< ! 77 ! %7- % 3 3 3 ?.9% ! %7 5

p% % ! %7E

3 7.?9% m ; kg

T ' 3 T %. > c 3 77 ! ?7.@ 3 77 ! '.'D< 3 9?D.'' =

' 3 %. > e 3 % ! ?%.@ 3 %?. ? - % 7.?9%- ' 3 3 3 7.%7 c ?

Feat added during constant value process '5 is2 s 3 mC v 4T 5T' ) 3 %D77 assume m 3 % kg %D77

T B T' 3 3 '9@9.'@ T 3 E.@9 = 7. c ? 7.@


D'

1

1

1


93/224

??.DE@ 3 3 3 @.?@ bar

> c ? %.@

% %a 3 % 5 3 %5 3 %5 7.@ E 3 7.E9@< or E9 .@<

>c ? 7.@

Feat reGected during constant volume process @5% is

2 r 3 mC v 4T@ 5 T%) 3 7.


94/224

THEORETICAL CYCLE ACTUAL CYCLEE% 5 (uction %' 5 Compression' 5 Constant volume combustion @ 5 !pansion@% 5 (udcen fall in pressure %E 5 !haustTheoretical Work one 3 area %' @% 3 $$ctual Work one 3 area @J' @@J B areaJE%@ 3 $ I5$ '#umping #ower 3 area @JE%@J 3 a '

$ctual Work one V Theoretical work done

THEORETICAL CONDITIONSKI. (uction and e!haust takes place at atmospheric pressure)). (uction and e!haust takes place through %?7 of crank rotation))). Compression and e!pansion takes place through %?7 of crank rotationI-. Compression and e!pansion are isentropic-. The combustion takes place instantaneously at constant volume at the end of compression-I. #ressure suddenly falls to atmospheric pressure at the end of e!pansion

ACTUAL CONDITIONSKI. The suction of mi!ture is possible only if the pressure inside the cylinder is below atmospheric

pressure.II. Hurnt gases can be pushed out only if the pressure of the e!haust gas is above atmospheric

pressure.III. The compression and e!pansion are not isentropicI-. (udden pressure raise is not possible after ignition as combustion takes some time for

completion.-. $ctual pressure raise is less than the theoretical.-I. The pressure increase takes place through some crank rotation.-II. (udden pressure release after opening of e!haust valve is not possible as it takes place through

some crank rotation.


D@


95/224

THEORETICAL AND ACTUAL CYCLES FOUR STROGE DIESEL EN-INE

THEORETICAL CYCLE ACTUALCYCLE

E% 5 (uction %' 5Compression

' 5 Constant pressure combustion @ 5 !pansion@% 5 (udden fall in pressure %E 5 !haust

THEORETICAL CONDITIONS

I. (uction and e!haust takes place at atmospheric pressure)). (uction and e!haust takes place through %?7

of crank rotation

))). Compression and e!pansion takes place through %?7 of crank rotationI-. Compression and e!pansion are isentropic-. The combustion takes place instantaneously at constant volume at the end of compression-I. #ressure suddenly falls to atmospheric pressure at the end of e!pansion

ACTUAL CONDITIONS K

I. The suction of mi!ture is possible only if the pressure inside the cylinder is below atmospheric pressure.

II. Hurnt gases can be pushed out only if the pressure of the e!haust gas is above atmospheric

pressure.III. The compression and e!pansion are not isentropicI-. (udden pressure raise is not possible after ignition as combustion takes some time for

completion.-. $ctual pressure raise is less than the theoretical.-I. The pressure increase takes place through some crank rotation.-II. (udden pressure release after opening of e!haust valve is not possible as it takes place through

some crank rotation.


DE


96/224

1 . C$m0a#e the$#eti a* a % a t)a* " *es $+ , st$ e S.I. a % C.I. e 3i ee#i 3.

THEORETICAL AND ACTUAL CYCLES>TWO STROGE PETROL EN-INE

THEORETICAL CYCLE ACTUAL CYCLE

%' 5 Compression ' 5 Constant -olume Combustion

@ 5 !pansion @% 5 (udden fall in #ressure

%9, 9% 5 !haust E9, 9E 5 Charging and (cavenging

AIR STANDARD EFFICIENCY OF TWO STROGE EN-INES

00 CTI- Compression ratio 3 r e

2

1

11

se c e

c

se e

e

V V r

V

&V L

r

=

=

=

THEORETICAL AND ACTUAL CYCLE>TWO STROGE DIESEL EN-INE


D9


97/224

%' 5 Compression ' 5 Constant #ressure combustion@ 5 !pansion @% 5 (udden fall in pressure

%9, 9% 5 !haust E9,9E 5 Charging and (cavenging

12. I%e ti+" the state $+ H , O +$# the +$**$(i 3 $ %iti$ s.a7 8 !a#@ 66C

!7 8 !a#@ ,==6

C7 8 !a#@ 181.=6C%7 ,86 Pa@ 8 *it 3e7 , M0a@ 666 J 3 s0. E tha*0"+7 , M0a@ 8.


98/224

Miven (p. ntropy is greater than h g and hence it is a superheated steam.f) 0or given pressure( % 3 %.


99/224

3% E%.< kG;kgCase'" Miven #3' bar T3h3pv

When pressure and temperature are known, comparing the given temperature with the saturationtemperature. The state can be identified.

(ince ( given L T saturation the state is superheated vapor referring appendi! $% other properties can be found-3%.7?7 m ' ;mkgF3'?


100/224

' %.E'


101/224

T$ +i %KState $+ steam 4(hethe# %#"@ (et $# s)0e# heate%7

S$*)ti$ K0rom steam table of pressure scale at %' bar, specific volume of dry steam, v g37.%9 '%m3 ;kg, T s3%??oC.(ince vLv g, the steam is in super heated condition.

:ote"-L- g (uper heated-3v g dry steam-V- g Wet steam

We know that

upup 9

upup

9

oup

T

T

0.1 $T T 1!!

0.1"321

T 201.$! C

=

= = =

The steam is super heated to '7% .E?oC

Res)*t K

The state of steam is super heated

1;. Dete#mi e the $ %iti$ $+ steam at a tem0e#at)#e $+ ,,6 $C a % e tha*0" $+ ,;86 3.-i e %ataKT ,,6 $CH ,;86G G3T$ +i %KState $+ steam

S$*)ti$ K0rom steam Table of temperature scale at ''7 oCF f 3D@ .


102/224

?et f

f9

7 7x

7

2 $0 3.0. 3

1!$".2

=

= =Res)*tKThe given steam is wet steam of 7.D< dry.

1


103/224

h3'.9


104/224

3%77


105/224

S$*)ti$ K4i) p3%bar v37.''m ;kg 0rom (.T. at %7bar, v g the steam is super heated.4ii) #3%Ebar t3''E oC 0rom (.T. at %E bar, T s3%D?. oC since T sVTs the stem is super heated .4iii) T3'77 oC h3'


106/224

[ ]

[ ]

[ ]

2 1 2 2 1 1

2 1 1 2 1 1 2

3

( 7 7 / p # p # /

( 7 7 / p # # / p p

10 30$2.1 :!!. / :1000 0.2$! : 0.001010/

*e ultF

1. C7an9e in #olu(e8 + 5 2.$" (

2. C7an9e in #olu(e8 7 5 2!"3 %&3. C7an9e in entrop48 S 5

=

= = =

" .! %&'%

. C7an9e in internal ener948 = 5 2"0" .1 k&

,,. Steam at ,6 !a# a % 96 C e/0a %s i a steam t)#!i e t$ 6.6< !a#. It is the $ %e se% i a$ %e se# t$ sat)#ate% (ate#. The 0)m0 +ee%!a the (ate# t$ !$i*e#. Ass)me@ i%ea* Ra i e " *e

a % %ete#mi eK4i) :et workdone;kg of steam4ii) >ankine efficiency.

S$*)ti$ K-i e K p% 3 '7 bar T sup 3 97 C p b 37.7?E bar.>efer 0ig" .% $ssume, mass flow steam, m3% kg.

Fi3. .1W$# %$ e 3 $+ steamK

0rom steam table at p b 37.7? bar h% 3h f% 3 %< .D =A;kg


%79


107/224

( ) 1 ;p 2 1p p

? 7 710

20:0.0! 5 1. 2 %&'k9

10

= =

= h ' 3h % 6W p 3%< .D6%.DD' 3 %


108/224

Fi3. ., :et work done,

( ) ( )( ) ( )

( ) ( )

3 2 1

*in 3 2

W 7 7 7 7

5 31"0:21!0 1 $.! 2 1 3. !1. k&'k9

W W !1. 0.32 1 or 32. 1-

@ 7 7 31"0 1 $.! 2

= =

= = = = :ote" There will difference in answers by solving the problem by calculations with the help of steam tables

and by solving with the help of &ollierJs diagram because of the errors involved in reading the value ofenthalpies from 4h5() diagram.

, . A steam t)#!i e $+ a 0$(e# 0*a t $0e#ati 3 $ i%ea* Ra i e " *e@ #e ei es steam at ,6 !a#@66C at the #ate $+ 3 s a % it e/ha)sts at 6.1!a#. Dete#mi e the +$**$(i 3.

4i7 Net 0$(e# $)t0)t 4ii7 Steam #ate4iii7 Heat #eJe ti$ i $ %e se# i W. 4i 7 Ra i e " *e e++i ie "4 7 A t)a* the#ma* e++i ie " $+ the 0*a t i+ the !$i*e# e++i ie " is =6

S$*)ti$ K>efer fig .-i e

p% 3 '7 bar T sup 3 77 C,m 3 kg;s, p b 3 7.% bar

4i7 Net 0$(e# $)t0)tK0rom steam Table 5 at

p% 3'7 bar,Tsup 3 77 C, we get Fi3@ .h 3 7'E.7 kA;kg ( 39.


109/224

0rom steam Table5' at p b 3 7.% bar, h %3h f% 3h f@ 3 %D%.? kA;kghfg@ 3 ' D'.? kA;kg ( f@ 3 7.9@D kA;kg = (fg@ 3


110/224

4!a#7 Tem0. C h + h +3 h 3 S+ S+3 S37. 9D.%' '?D. ' 9.% '9'E.@ 7.D@@ 9.?'E


111/224

>ankine efficiency,T p

*i

W W !!!1.2 2.$$"0.2 3$ or 2 .3$-

Q 3010 .!$ = = =

UNIT III

AIR COMPRESSORS

PART A

1. What is a +) ti$ $+ $m0#ess$#s' C*assi+" the a#i$)s t"0es.

Compressors are used to compress air, gas or vapour. Compressed air has wide applications in industry aswell as in commercial application.

T"0es%. >eciprocating compressors'. >otary compressors

,. What a#e the a% a ta3es $+ m)*ti sta3e $m0#essi$ (ith i te# $$*i 3 $ e# si 3*e sta3e$m0#essi$ +$# the same 0#ess)#e #ati$ ' 4A.U.7

%. The work done on the air is reduced.'. It increases the efficiency of the compressor.

. De+i e +#ee ai# %e*i e#e% i ai# $m0#ess$#.

The free air delivered in the actual volume delivered at the stated pressure, reduced to a standard pressureand temperature and e!pressed in cubic meter per minute.

2. De+i e $*)met#i e++i ie " $+ the $m0#ess$# 4A.U.7

The ratio of actual free air delivered by the compressor to the displacement of the compressor is known asvolumetric efficiency of the compressor.

8. Dis )ss the e++e t $+ *ea#a e )0$ the 0e#+$#ma e $+ a ai# $m0#ess$#.

The clearance volume does not affect the work of compression ; kg of air. It reduces the volumetricefficiency of the compressor.9. De+i e the e++e ti e ess 4 7 $+ i te# $$*e# i ai# $m0#ess$#.

$ctual drop in temperature of air


%%%


112/224

3 5555555555555555555555555555555555555555555555555555555555 &a!imum possible drop in temperature of air

PART B

1. Dete#mi e the si e $+ the "*i %e# $+ %$)!*e a ti 3 ai# $m0#ess$# $+ , ( IP i (hi h ai# is %#a(i at 1!a# a % $m0#esse% t$ 19 !a# a $#%i 3 t$ the *a( P: 1.,8 C RPM 66@ Pist$ S0ee% 1


113/224

3


114/224

: p ' >#&I# 3 '#% 4- s - ) 5 % !

n5% p% 97 !%77

(ubstituting the values

- s 3 7.7EE%' m .

#iston (peed 3 '1 ! >#& 3 '77'77

1 3 3 7.? m 3 ? . cm ' ! %'7

-s 3 ' 1 @

@ 7.7EE%'' 3 ! 3 7.'D m.

7.?

the volume efficiency is given by - c # '

-3%5 5 % - s # % - c

7.D 3 %5 %7 5 % - s

- c 3 7.% - s

3 7.7 ''' 3 '.'' 4 %7 5 %)

2. E/0*ai the a% a ta3e a % *imitati$ $+ #$ta#" a % #e i0#$ ati 3 $m0#ess$#s 4A.U.7

>eciprocating compressors are suitable for low discharge at very high discharge pressure. The speedof reciprocating compressor is very low compared with rotary compressors due to inertia of reciprocating

parts.


%%@

-( )

1

n

-( ) 1

1.35

1

1.35


115/224

>otary compressors are suitable for delivering of large volumes at low discharge pressure. Thema!imum free air discharge can be as high as 777 m ;min and ma!imum delivery pressure is about %7 bar.The speed of the rotary compressor can be as high as E7,777 rpm.

$ir supply is continuous in rotary compressors and therefore no receiver is re+uired. Hut receiver isan essential re+uirement of a reciprocating compressor.

>otary compressors are small in si e for the same discharge compared with reciprocatingcompressors.

There is no balancing problem in rotary compressor whereas it is one of the maGor problem inreciprocating compressor.

The air delivered by rotary compressor is more clean as it does not come in contact with lubricatingoil.

1ubricating system is more complicated in reciprocating compressor where as it is very simple inrotary compressor.

8. The +#ee ai# %e*i e#e% $+ a si 3*e "*i %e# si 3*e sta3e #e i0#$ ati 3 ai# $m0#ess$# is ,.8 m mi .the am!ie t ai# is at STP $ %iti$ *ea#a e. $*)me is 8 $+ the st#$ e $*)me a % *a( $+

$m0#essi$ a % e/0a si$ is P: 1.,8 C. I+ L 1.,D a % the $m0#ess$# #) s at 186 #0m@ %ete#mi ethe 0$(e# #e )i#e% a % me0 a % si e $+ the "*i %e#. De*i e#" 0#ess)#e is ; !a#.

S$*)ti$ K #-

The mass of free an delivered per second is given by ma 3>T

Where > 3 '?< G ; kgi

%.7% ! %7E ! '.E %ma 3 ! 3 7.7E@ kg;s

'?< ! '< 97

n # '4 i ) Work done, w 3 m a >T, 5 % n 5% #%

%.'E 4T ' 5T%) n5%

3 9 .@D kw - @ p 3- p@

- @ 3 7.' - s.- % B - @ 3 -% B 7.' -s

3 %.7E -s B 7.' -s3 7.?%< -s

p- p % 4- % 5- @ )m 3 3

>T >T %. T p %

Ie 0$ ; cycle - 3 4- % 5- @) .T% p

- 3 7.sta3e $m0#ess$# %e*i e#s ,m +#ee ai# 0e# mi )te. The tem0e#at)#e a % 0#ess)#e $+ ai# atthe s) ti$ a#e ,; $C a % 1 !a#. The 0#ess)#e at the %e*i e#" is 86 !a#. The *ea#a e is 8 $+ thest#$ e i L.P "*i %e# as (e** as i H.P. C"*i %e#. Ass)mi 3 0e#+e t i te#> $$*i 3 !et(ee the t($sta3es@ +i % the mi im)m 0$(e# #e )i#e% t$ #) the $m0#ess$#.

I+ the $m0#ess$# is t$ #) t$ ,66 #.0.m +i % the %iamete#s a % st#$ es ass)mi 3 the st#$ es $+!$th "*i %e#s a#e e )a* t$ the %iamete# $+ LP "*i %e#. What is the #ati$ $+ "*i %e# $*)mes' La( $+

$m0#essi$ a % #e>e/0a si$ i !$th "*i %e# is 0. . 1. 8 CA*s$ ass)me that the am!ie t ai# $ %iti$ is same as s) ti$ $ %iti$ .

S$* K

#ower of : stages compressor is given by 4as clearance does not affect the work ; kg) # p: 6 % %

I.# 3 . # %- % : 5 % ! kW.n 5% p% %777

$s it is two stage compressor " :3'.

p: 6 % 3 p 3 E7, p % 3%


%'7

( ) 1n

-( ) n 1n


121/224

%. E ! ' % ! %7E ! '

I# 3 . 4E7) 5 % 3 %9.D? kW7. E %777

p ' 3 p% p 3 % ! E7 3


122/224

1,. A %$)!*e a ti 3 t($ sta3e $m0#ess$# %e*i e#s ai# at ,8 !a#. The 0#ess)#e a % tem0e#at)#e $+ theai# at the !e3i i 3 $+ $m0#essi$ i L.P. "*i %e# a#e 1 !a# a % ,6 $C. The tem0e#at)#e $+ ai#

$mi 3 $)t +#$m the i te#> $$*e# !et(ee the t($ sta3es is 26 $C a % the 0#ess)#e is ; !a#. The%iamete# a % st#$ e $+ L.P. "*i %e# a#e 96 m a %


123/224

4a7 The 0#$ ess a#e sh$( 0> a % T>s %ia3#am.

p% p % ! '7# ' ' 3 3 p' 3 @.E?? bar

C 7.DE

p ' @.E?? p '73 3 @.E?? 3 3 @. E?D

p% % p' @.E??

- 1p 3 %5 7.7@ [email protected]?? ) 5% 3 7.D%7D

- hp 3 %5 7.7@ 4@. E?D) 5% 3 7.D%ED

-olume of air taken in the 1# cylinder per second3 7.D%7D !;@ ! 7.@' ! 7.E ! '77 ; 97 m;s 3 7.%D7? m ; s.&ass of air ;s 3 4% ! %7 E ! 7.%D7?) ; 4'?< ! 77) 3 7.''%9 kg;s

%. '?< ! 7.''%9 ! 77Work in 1#cyl inder;s 3 ! [email protected]?? )5%QkA;s3 @.? kW

7. %777Work done in F# cylinder

%. '?< ! 7.''%9 ! 773 ! [email protected]?? ) 5%QkA;s3 .@E kW 7. %777

#ower re+uired 3 9?.'? kwFeat reGected in the intercooler per minute

mC# PT ' 5 TEQ but TE 3 T %T ' 3 77 ! @.E?? 3 @'9. D =

2 3 mC 4T' B TE)

2 3 7.''%9 ! 97 ! %.77E P @'9. D B 77 Q 3 %9?D kA;min.12. A th#ee sta3e ai#> $m0#ess$# %e*i e#s 8., m $+ ai# 0e# mi )te. The s) ti$ 0#ess)#e a %tem0e#at)#e a#e 1 !a# a % 6 $C. The 0#ess)#e a % tem0e#at)#e a#e 1.6 !a# a % ,6 $Cat the +#ee ai#

$ %iti$ . The ai# is $$*e% t$ 6 $C a+te# ea h sta3e $+ $m0#essi$ . The %e*i e#" 0#ess)#e $+ the$m0#ess$# is 186 !a#. The R.P.M $+ the $m0#ess$# is 66. The *ea#a e $+ L.P.@ I.P.@ a % H.P.@"*i %e#s a#e 8 $+ the #es0e ti e st#$ es. The i %e/ $+ $m0#essi$ a % #e>e/0a si$ i a** sta3es is

1. 8. Ne3*e ti 3 0#ess)#e *$sses@ +i % the B.P. $+ the m$t$# #e )i#e% t$ #) the $m0#ess$# i+ theme ha i a* e++i ie " is


124/224

T% 3 7 6 '< 3 7 =.$s the perfect inter5cooling is given.

p ' p p@3 3 3 3C p% p ' p

p@ %E7Where C 3 3 3 E. %

p% %

The mass of air delivered per minute is given by

# o- o %.7 ! %7E ! E.'m 3 3 3 9. < kg;min. >T o '?< ! 'D

I.# of the compressor is given by

n %I.#. 3 m>T % : 4c ) 5% !

n B % %777

%. E 9. < %3 ! ! '?< ! 7 ! 4E. %) 5% ! 3 E


125/224

T% p%3 3

T p

T% @' p% 3 p 3 7.DE 3 .%? bar

T ' '7T% 3 T 3 T E 3 7 6 '< 3 7 =.

$fter the first stage, perfect inter5cooling is used in subse+uent stages, therefore the intermediate pressurefor minimum work are to be found.

p ' p p@ 3 3 3C

p% p ' p

p@Where C 3 and : 3

p%

'77 C 3 3 .D


126/224

300"0 100

3 '%.E9 kW.

19. A t($ sta3e a#e $m0#ess$# (ith 0e#+e t i te# $$*i 3 ta es i ai# at 1 !a# a % ,; $C. The$m0#essi$ 0#$ ess is 0$*"t#$0i $+ i %e/ 1. . The $m0#esse% ai# is %e*i e#e% at = !a#. Ca* )*ate 0e#3 $+ ai#@ the mi im)m ($# #e )i#e% a % heat #eJe te% t$ the i te# $$*e#. 4AU De ,66


127/224

? The lubricating system is complicated. The lubricating system is simple.

D The air delivered is less clean, as it comes in contactwith the lubricating oil.The air delivered is more clean, as it does not comes incontact with the lubricating oil.

%7 Isothermal fficiency is used for all sorts ofcalculation. Isentropic fficiency is used for all sorts of calculation.

1


128/224

I:7 A IAL FLOW COMPRESSOR

1=. C$m0a#e e t#i+)3a* a % a/ia* +*$( $m0#ess$#s.

COMPARISON OF CENTRIFU-AL AND A IAL FLOW AIR COMPRESSORS

S. N$ Re i0#$ ati 3 Ai# C$m0#ess$# R$ta#" Ai# C$m0#ess$#s

% The flow of air is perpendicular to the a!is ofcompressor The flow of air is parallel to the a!is of compressor

' It has low manufacturing and running It has high manufacturing and running cost.

It re+uires low starting tor+ue It re+uires high starting tor+ue.


%'?


129/224

@ It is suitable for multi5staging It is suitable for multi5staging

E It re+uires large frontal area for a given rateof flow.It re+uire less frontal area for a given rate of flow. Itmakes the compressor suitable for air crafts.

UNIT I:

REFRI-ERATION & AIR CONDITIONIN-

PART A

1. De+i e t$ $+ #e+#i3e#ati$ a % COP $+ #e+#i3e#at$#. ?AU APR ,662

$ ton of refrigeration is defined as the amount of refrigeration effect that can free e % tonne of waterat 7[C with ice at 7[C in '@ hours.

% tonne of refrigeration 3 .E >G;( :et refrigerating effect

4CO#) > 3Work input

,. List %$( the a00*i ati$ s $+ #"$3e i +*)i%s. ? AU APR ,662


%'D


130/224

%. (toring food, medicine etc'. Cryo engines

. Cryo treatment of metals.

. List the im0$#ta t i %)st#ia* a % $mme# ia* a00*i ati$ s $+ #e+#i3e#ati$ .

%. Ice making'. 0ood #reservation.

. &ilk preservation@. Industrial air conditioningE. Comfort air conditioning.9. Chemical and related industries

2. What a#e the 0#$0e#ties $+ a i%ea* #e+#i3e#a t'

%. 1ow boiling point.'. Figh critical temperature.

. Figh latent heat of vapourisation.@. 1ow specific heat of li+uid.E. :on5flammable and non e!plosive.9. 1ow cost %%, > %', > '' , $mmonia, CO ' etc.

9. E/0*ai the me#its a % %eme#its $+ ai# #e+#i3e#ati$ s"stem.

Me#its K$s air is non flammable, there is no risk of fire as that of a machine using :F as the refrigerant.

It is cheaper and easily available than other refrigerant.

$s compared to the other refrigeration system the T> is +uite low, because of this reason this system isemployed in air crafts.


131/224

=. De+i e Ps" h#$met#".

#sychrometry is a study of properties of moist air.

16. De+i e (et !)*! tem0e#at)#e.

It is the temperature of air measured by a thermometer when its bulb is covered with wet cloth and ise!posed to a current of rapidly moving air. It is denoted by t w.

11. De+i e %#" !)*! tem0e#at)#e.

The temperature of moist air as indicated by an ordinary thermometer is called as HT. It is denoted by t d.

1,. De+i e %e3#ee $+ sat)#ati$ .

egree of saturation is the ratio of actual specific humidity to the saturated specific humidity both atthe same temperature of moisture.

1 . De+i e De( 0$i t tem0e#at)#e.

The temperature at which the water vapour present in air begins to condense when the air is cooled,is known as dew point temperature. It is denoted by t dp.

12. De+i e Re*ati e h)mi%it".

>F is the ratio of the mass of water vapour in a certain volume of moist air at a given temperature tothe mass of water vapour in the same volume of saturated air at the same temperature.

18. De+i e S0e i+i h)mi%it".

(pecific humidity is the ratio of mass of water vapour to the mass of associated dry air in a givenvolume of moist air.

19. H$( %$es the h)mi%it" a++e t h)ma $m+$#t'

If the humidity is above a certain level, water vapour from human body, cannot be absorbed by theatmospheric air.

1;. What a#e the a#i$)s s$)# es $+ heat 3ai $+ a ai# $ %iti$ e% s0a e'


% %


132/224

i. (olar gain through glass panelsii. (olar gain through roof and walls

iii. Feat gain from occupantsiv. uct leakagev. Infiltration

1emoval of heat of occupants

. >emoval of moisture of occupants@. Mood air distribution

1=. What is H)mi%i+i ati$ '

The addition of water vapour to the air is known as Fumidification.

PART B

1. E/0*ai the +) ti$ $+ a#i$)s $m0$ e ts $+ a :a0$)# $m0#essi$ #e+#i3e#ati$ s"stem. ? AUAPR ,662


% '


133/224

-apour compression refrigerator is an improved type of air refrigerator with different workingsubstance called S >efregerantJ. The refrigerant condenses and evaporates alternatively at temperatures closeto atmosphere. The system works as closed system and hence the refrigerant does not leave the system.

The schematic diagram of vapour compression refrigeration system is shown in fig.E.% It consist ofi) compressor ii) condenser iii) e!pansion valve and iv) evaporator.

The compressor draws the low pressure and temperature vapour from the evaporator through thefunction valve and compresses it to high pressure and temperature and delivery value H. In the condenser,the compressed vapour is cooled by the circulating water to form high pressure and low temperature li+uid.The evaporator consists of coils which provides a heat transfer surface through which heat from the brinesolution is e!tracted. The cold bourine solution is circulated through coils around the cold chamber fromwhich it e!tracts heat and become hot refrigerant at low pressure and low temperature. The low pressure lowtemperature vapour is converted to low pressure high temperature vapour by absorption of latent heatevaporation from the brine.

,. Dis )ss the 0#i i0*e $+ $0e#ati$ $+ a a0$)# a!s$#0ti$ #e+#i3e#ati$ s"stem. 4AU AP ,6627.


%


134/224

In this type of refrigeration system. The refrigerant vapour is absorbed in a absorbing mediuminstead of compressing it in the compressor. The absorbing medium is the mi!ture of ammonia and water.The following figure shows the schematic arrangement of the essential elements of a vapour absorptionrefrigeration system.

S hemati %ia3#am $+ NH H , O 0a#ti *e a0$)# a!s$#0ti$ s"stem

In this ammonia is the refrigerant and water is the refrigerant and water is the absorbent. This isknown as a+ua ammonia vapour leaving the evaporator enters the absorber and it is absorbed in the weakammonia solution. This process is taking place at a temperature slightly above, that of the surroundings. Inthis process some heat is reGected to the surroundings.

Then the strong ammonia solution is pumped through heat e!changer and is heated in the generatorto higher pressure and temperature. ue to reduced solubility of ammonia in water at the higher pressure andtemperature the vapour is removed from the solution. The vapour then passes to the condenser where itcondensed in the same way as in the vapour compression system. Then the condensed high pressure lowtemperature refrigerant is passed through the e!pansion value where it is brought to low pressure and lowtemperature.


% @


135/224

. ,< t$ es $+ i e +#$m a % 6 $ is 0#$%) e% 0e# %a" i a amm$ ia #e+#i3e#at$#. The tem0e#at)#e#a 3e i the $m0#ess$# is +#$m ,8 $C t$ >18 $C. The a0$)# is %#" a % sat)#ate% at the e % $+

$m0#essi$ a % a e/0a si$ a*)e is )se%. Ass)mi 3 $+ C$>e++i ie t $+ 0e#+$#ma e $+ 9, $+ thethe$#eti a*@ a* )*ate the 0$(e# #e )i#e% t$ %#i e the $m0#ess$#. The 0#$0e#t" $+ amm$ ia a#e 3i ei the +$**$(i 3 ta!*e. 4AU APR ,662 7.

Tem0e#at)#e$C

E tha*0" G3E t#$0" $+ Li )i% G3G E t#$0" $+ :a0$)#G3G 1i+uid -apour

'E %77.@ % %D.'' 7. @< @.@?E'5%E [email protected] % [email protected] 5'.% ? E.7E?E

S$* K h ' B h%

Theoretical CO# 3h B h'

h 3 % %D.'' kA;kgh% 3 h @ 3 %77.@ kA;kg( g 3 ( f' 6 ! ' (f g'! ' 3 7.D'

h ' 3 h f' 6 ! ' hf g' 3 %%D9.' kA;kg

CO# Theoretical 3 ?.D%

$ctual refrigeration effect ; kg

3 CO# 4actual) ! work done3 E.E' ! 4h B h' )3 9


136/224

i sat)#ate% *i )i% a % $m0#essi$ is ise t#$0i . The a0$)# *ea i 3 the e a0$#at$# is %#" sat)#ate%.Ass)me that e tha*0" at the e % $+ ise t#$0i $m0#essi$ ,16 3. Dete#mi e.4i7 The #e+#i3e#ati$ +*$( #ate.4ii7 The 0$(e# #e )i#e% t$ #) the $m0#ess$#.

The heat #eJe te% i the 0*a t.COP $+ the s"stem.

The properly of r%' are listed below

Tem0 P#ess)#e h+ h3 S+ S34$C7 !a# GJ G3 GJ G3 GJ G3G GJ G3G

E7 %'.%DD ?@.?9? '79.'D? 7. 7 @ 7.9


137/224

D CO# Figh 1ow%7. &aintenance &ore 1ess

9. A amm$ ia #e+#i3e#at$# 0#$%) es 18 t$ s $+ i e +#$m a % at 6 $C i a %a". The tem0e#at)#e #a 3e$+ the ($# i 3 " *e is ,8 $C a % >19$C. The amm$ ia a0$)# is %#" a % sat)#ate% at the e % $+

$m0#essi$ . Ass)me a t)a* C$0 is 88 $+ the the$#eti a* a*)e. Ca* )*ate the 0$(e# #e )i#e% t$%#i e the $m0#ess$# a % mass +*$( #ate i 3 mi . Ta e *ate t heat $+ i e 8 3 a % C 0 4(ate#

2., 3 $C.

S$*)ti$ K0orm the table h ' 3 % %D.'% kA;kg.h 3 h @ 3 ?7.efrigeration effect re+uired 3 4mass of ice ; sec ) ! 1atent heat

%E ! %7773 ! E 3 E?.%9 kA;s 3 E?.%9 kW. '@ ! 977

>efrigeration effect ; kg of refrigerant 3 h % B h@

$s the process %5' is isentropic compression, we can write

(f %6 ! % (fg % 3 (g '7.'% @ 6 !@ 6 ! % ! 4E.7E?E B 457.'% @) 3 @.@?D@Z % 3 7.?D'

h% 3 [email protected] 67.?D' P4% [email protected] B 45E@.?9)Q 3 %%E?.' kA;kg.

>efrigeration effect 3 %%E?.' B ?7.


138/224

$ctual CO# 3 @.? ! 7.EE 3 '.99.

>e+uired refrigeration effect#ower re+uired 3

$ctual CO# E?.%93 3 '%.?9 kw. '.99

;. Re+#i3e#ati$ 0*a t )si 3 C$ , as #e+#i3e#a t ($# s !et(ee ,8 $C a % >8$C. The %#" ess $+ C$ , is 6.9at the e t#" $+ the $m0#ess$#. Fi % the i e +$#me% %a" i+ the i e is +$#me% at 6 $C a % +#$m the (ate#at 116 $C )a tit" $+ C$ , i# )*ate% 16 3 mi . Ta e #e*ati e e++i ie " 6.9.C04(ate#7 2., 3@ Late t heat OF i e 8 GJ G3.

Tem0$C

Li )i% heatG3

Late t heat J 3 E t#$0" $+ *i )i% J 3

'E ?%.'E %'%.9 7.'E%5E 5


139/224

m 3 3 7.%9< kg;s 97h% 3 hf % 6 ! % hf g% 3 5 efrigeration effect ; day 3 E?< ! 97 ! '@ kA;day

E?< ! 97 ! '@ Ice formed ; day 3 3 ''@' kg


140/224

#&) @


146/224

%%.D'@ 7 7.7%D9? ' 9.9D @% .D %.%'E@ %.


147/224

Harometric pressure, # b 3 % bar T$ +i %K

:a0$)# 0#ess)#e4P : 7s0e i+i h)mi%it" 4W7sat)#ati$ #ati$ 4 7#e*ati e h)mi%it" 4 7%e( 0$i t tem0e#at)#e 4t %07

a0$)# %e sit" 4 7e tha*0"4h7

S$*)ti$ K%. -apour pressurewe know that,

# v 3 # sw 3( ) ( ); ? d ?

?

P P t tf

1$2 . 1.3t

Pfrom e+uation 4% )Q

Where# b 3 Harometric pressure 4or) atmospheric pressureTd 3 ry bulb temperatureTw 3 wet bulb temperature# sw 3 (aturation pressure corresponding to wet bulb temperature0rom steam table, we find that for %E C wet bulb temperature corresponding saturation pressure # sw is7.7%


148/224

# s B (aturation pressure corresponding to dry bulb temperature0rom steam table, we find that for that for 'E C dry bulb temperature, corresponding saturation pressure is7.7 %99 bar i.e. # s 3 7.7 %99bar

0.010$ 1 0.031""0.031"" 1 0.010$

0.32

= =

2. Re*ati e h)mi%it"K

( )#P

....... 10P

0.010$ 5

0.031"" 5 0.33 5 33-

=

8. De( 0$i t tem0e#at)#e 4t %07K0rom steam table, we find that for partial pressure 4# v) 3 7.77E bar, corresponding specific volume t dp

is ? C T dp3 ? C9. :a0$)# %e sit" 4 7K0rom steam table, we find that for 'E C dry bulb temperature, corresponding specific volume, 4- g) is@ .@7m' ;kg- g3 @ .@7m' ;kg-apour density, v 3 %;vg 3 %;@ .@7 3 7.7' 7v 3 7.7' 7kg;m of saturated steamFence vapour density at 3 37.' 7 ! v 3 7.77


149/224

Res)*tsK1. -apour pressure, v 3 7.7%7Ebar '. specific humidity, W 3 9.9 ! %7 5 kg;kg of dry air 3. saturation ratio, 3 7. '@

. relative humidity, 3 7.$. dew point temp, 4t dp) 3 ? C". vapour density, v 3 7.77


150/224

0rom steam table, we find that for '7 C dry bulb temperature corresponding saturation pressure is 7.7' < bar i.e. # s 3 7.7' 37.'?< =A;kgk Ta 3 t d 6 '< 3 '7 6 '< 3 'Da 3 7.D?%@bar 3 7.D?%@ ! %

Documents

Aero Engineering Thermo Dynamics