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Chapter 9

GAS POWER CYCLES

Actual and Ideal Cycles, Carnot cycle, Air-Standard Assuptions

9-1C The Carnot cycle is not suitable as an ideal cycle for all power producing devices because it cannotbe approximated using the hardware of actual power producing devices.

9-2C It is less than the thermal efficiency of a Carnot cycle.

9-3C It represents the net work on both diagrams.

9-4C The cold air standard assumptions involves the additional assumption that air can be treated as anideal gas with constant specific heats at room temperature.

9-5C Under the air standard assumptions, the combustion process is modeled as a heat addition process,and the exhaust process as a heat rejection process.

9-6C The air standard assumptions are !"# the working fluid is air which behaves as an ideal gas, !$# allthe processes are internally reversible, !%# the combustion process is replaced by the heat addition process,and ! the exhaust process is replaced by the heat rejection process which returns the working fluid to itsoriginal state.

9-7C The clearance volume is the minimum volume formed in the cylinder whereas the displacementvolume is the volume displaced by the piston as the piston moves between the top dead center and the

bottom dead center.

9-8C It is the ratio of the maximum to minimum volumes in the cylinder.

9-9C The '() is the fictitious pressure which, if acted on the piston during the entire power stroke,would produce the same amount of net work as that produced during the actual cycle.

9-10C *es.

9-11C +ssuming no accumulation of carbon deposits on the piston face, the compression ratio willremain the same !otherwise it will increase#. The mean effective pressure, on the other hand, willdecrease as a car gets older as a result of wear and tear.

9-12C The I and CI engines differ from each other in the way combustion is initiated- by a spark in Iengines, and by compressing the air above the selfignition temperature of the fuel in CI engines.

PROPRIETARY MATERIAL. / $001 The 'c2raw3ill Companies, Inc. 4imited distribution permitted only to teachers andeducators for course preparation. If you are a student using this 'anual, you are using it without permission.

5"

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9-13C troke is the distance between the T6C and the 76C, bore is the diameter of the cylinder, T6C isthe position of the piston when it forms the smallest volume in the cylinder, and clearance volume is theminimum volume formed in the cylinder.

9-14 The temperatures of the energy reservoirs of an ideal gas power cycle are given. It is to bedetermined if this cycle can have a thermal efficiency greater than 88 percent.

Analysis The maximum efficiency any engine using the specified reservoirs can have is

!"#$%=++

==9$:%#!;$:

9$:%#!":""Carnotth,

H

L

T

T

Therefore, an efficiency of 88 percent is possible.


5$

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9-15 The four processes of an airstandard cycle are described. The cycle is to be shown onP-vand T-sdiagrams, and the net work output and the thermal efficiency are to be determined.

Assumptions1The airstandard assumptions are applicable. 29inetic and potential energy changes arenegligible. 3+ir is an ideal gas with variable specific heats.

PropertiesThe properties of air are given in Table +":.

Analysis!b# The properties of air at various states are

( )

( )

( ) k

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T3[K] t in!tota[k)%k*]

;net[k)%k*]

1 1384 (2'22300

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.3!! .$!! .9!! /.!! /0!! /3!!3!"3

3.

3."3

3/

3/"3

30

30"3

31

&0 &+

th

.3!! .$!! .9!! /.!! /0!! /3!!%!!

.!!!

./!!

.1!!

.#!!

.%!!

&0, &+,

4in,total

&'()'*

.3!! .$!! .9!! /.!! /0!! /3!!1!!

3!!

#!!

$!!

%!!

9!!

.!!!

&0, &+,

5

net

&'()'*


5;

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9-17 The four processes of an airstandard cycle are described. The cycle is to be shown onP-vand T-sdiagrams, and the maximum temperature in the cycle and the thermal efficiency are to be determined.

Assumptions1The airstandard assumptions are applicable. 29inetic and potential energy changes arenegligible. 3+ir is an ideal gas with constant specific heats.

PropertiesThe properties of air at room temperature are cp @ ".008 krom the ideal gas isentropic relations and energy balance,

( )

( ) 98:5.$k)a"00

k)a"0009%00

0.&=".&="

"

$"$ =

=

=

kk

P

PTT

( )

( )( ) K3360===

==

%max%

$%$%in

8:5.$9k

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9-18E The four processes of an airstandard cycle are described. The cycle is to be shown onP-vand T-sdiagrams, and the total heat input and the thermal efficiency are to be determined.


PropertiesThe properties of air are given in Table +":(.


7tu=lbm"$5.0;7tu=lbm,5$.0&B8&0 """ === huT

( )

( ) 7tu=lbm85%.$$%":.0"$&$psia8:.;

psia"&.:

"$&$

7tu=lbm&1.1&5B%$00

psia8:.;psia"&.:B8&0

B$"";

7tu=lbm".8%:,B$"";

7tu=lbm0&.%5$%000&.5$

&%

&

%

%

""

$$

"

""

$

$$

$$

in,"$"$

"$in,"$

%&

%

====

==

=

====

==

=+=+==

hPP

PP

P

hT

PT

TP

T

P

T

P

hT

quuuuq

rr

r

vv

>rom energy balance,

7tu=lbm&;&.";0;."$5$$.85%

%1.%"$%00

7tu=lbm%"$.%1".8%:&1.1&5

"&out

in$%,in"$,in

$%in$%,

===

=+=+=

===

hhq

qqq

hhq

!"/#$612.38

!c# Then the thermal efficiency becomes

24.2%===7tu=lbm;"$.%1

7tu=lbm&;&.";""

in

outth

q

q


51

v

P

%

&

$

"

?"$

q$%

qout

s

T

%

&

$

" q

out

q$%

q"$

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9-19E The four processes of an airstandard cycle are described. The cycle is to be shown onP-vand T-sdiagrams, and the total heat input and the thermal efficiency are to be determined.


PropertiesThe properties of air at room temperature are cp @ 0.$&0 7tu=lbm.B, cv @ 0.":" 7tu=lbm.B,

and k@ ".& !Table +$(#.Analysis!b#

( )

( ) ( )

( )

( ) ( )( ) 7tu=lbm$":.&B$$5&%$00B7tu=lbm0.$&

psia;$.&;psia"&.:B8&0

B$$5&

B$$5&

B8&07tu=lbm.B0.":"7tu=lbm%00

$%$%in,$%

""

$$

"

""

$

$$

$

$

"$"$in,"$

====

====

==

==

TTchhq

PT

TP

T

P

T

P

T

T

TTcuuq

P

vv

v

)rocess %& is isentropic

( )( )

( ) ( )( ) 7tu=lbm%:1.88&0$"":7tu=lbm.B0.$&0

&.$":%00

B$"":psia;$.&;

psia"&.:B%$00

"&"&out

in,$%in,"$in

0.&=".&="

%

&%&

====

=+=+=

=

=

=

TTchhq

qqq

P

PTT

p

kk

!"/#$517.4

!c# 26.8%===7tu=lbm8":.&

7tu=lbm%:1.8""

in

outth

q

q


55

v

P

%

&

$

"

q"$

q$%

qout

s

T

%

&

$

" q

out

q$%

q"$

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9-20 The three processes of an airstandard cycle are described. The cycle is to be shown onP-vand T-sdiagrams, and the heat rejected and the thermal efficiency are to be determined.


PropertiesThe properties of air at room temperature are cp @ ".008 k

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9-21 The three processes of an airstandard cycle are described. The cycle is to be shown onP-vand T-sdiagrams, and the net work per cycle and the thermal efficiency are to be determined.




( )

( )

( ) ( )( )

( ) ( )( )

kJ0.422;8"."0:%.$

k

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9-22 The three processes of an airstandard cycle are described. The cycle is to be shown onP-vand T-sdiagrams, and the net work per cycle and the thermal efficiency are to be determined.


PropertiesThe properties of air at room temperature are cp @ ".008 krom the isentropic relations and energy balance,

( )

( )

( )

( ) ( )

( )( )( )

( ) ( )

( )( )( )

kJ0.39&1."1:."

k

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9-23 + Carnot cycle with the specified temperature limits is considered. The net work output per cycle isto be determined.

Assumptions+ir is an ideal gas with constant specific heats.


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9-24+ Carnot cycle executed in a closed system with air as the working fluid is considered. The minimumpressure in the cycle, the heat rejection from the cycle, the thermal efficiency of the cycle, and the secondlaw efficiency of an actual cycle operating between the same temperature limits are to be determined.


PropertiesThe properties of air at room temperatures areR@ 0.$1: k

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9-25 +n ideal gas Carnot cycle with air as the working fluid is considered. The maximum temperature ofthe lowtemperature energy reservoir, the cycleEs thermal efficiency, and the amount of heat that must besupplied per cycle are to be determined.



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Otto Cycle

9-27C The four processes that make up the Ftto cycle are !"# isentropic compression, !$# v@ constantheat addition, !%# isentropic expansion, and ! v@ constant heat rejection.

9-28C The ideal Ftto cycle involves external irreversibilities, and thus it has a lower thermal efficiency.

9-29C >or actual fourstroke engines, the rpm is twice the number of thermodynamic cycles- for twostroke engines, it is e?ual to the number of thermodynamic cycles.

9-30C They are analyGed as closed system processes because no mass crosses the system boundariesduring any of the processes.

9-31C It increases with both of them.

9-32C 7ecause high compression ratios cause engine knock.

9-33C The thermal efficiency will be the highest for argon because it has the highest specific heat ratio,k@ ".;;:.

9-34C The fuel is injected into the cylinder in both engines, but it is ignited with a spark plug in gasolineengines.

9-35 +n ideal Ftto cycle is considered. The thermal efficiency and the rate of heat input are to bedetermined.



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9-36+n ideal Ftto cycle is considered. The thermal efficiency and the rate of heat input are to bedetermined.



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9-37E+n Ftto cycle with nonisentropic compression and expansion processes is considered. The thermalefficiency, the heat addition, and the mean effective pressure are to be determined.


PropertiesThe properties of air at room temperature are R@ 0.%:0& psiaAft%=lbm.B !Table +"(#, cp @

0.$&0 7tu=lbmAB, cv @ 0.":" 7tu=lbmAB, and k@ ".& !Table +$(a#.

Analysis De begin by determining the temperatures of the cycle statesusing the process e?uations and component efficiencies. The idealtemperature at the end of the compression is then

( ) B""581B#8$0! "&."""

"

$

""$ ===

=

k

k

s rTTTv

v

Dith the isentropic compression efficiency, the actual temperature atthe end of the compression is

B"%"&0.18

B8$0#!""58B#8$0!"$"$

"$

"$ =

+=

+=

=

TT

TTTT

TT ss

imilarly for the expansion,

B"$0"1

"B#&;0$%00!

""&.""

%

"

&

%%& =

+=

=

=

kk

sr

TTTv

v

B"$:5B#"$0"!$:;0#58.0!B#$:;0!#! &%%&&%

&% ===

= ss

TTTTTT

TT

The specific heat addition is that of process $%,

6tu)l7/1$"0=== B#"%"&$:;0#!B7tu=lbm":".0!#! $%in TTcq v

The net work production is the difference between the work produced by the expansion and that used bythe compression,

7tu=lbm8."":

B#8$0"%"!B7tu=lbm":".0!B#"$:5$:;0#!B7tu=lbm":".0!

#!#! "$&%net

==

= TTcTTcw vv

The thermal efficiency of this cycle is then

!"1$3===7tu=lbm$&:.%

7tu=lbm"":.8

in

netth

q

w

+t the beginning of compression, the maximum specific volume of this cycle is

=lbmft1$."&psia"%

B#8$0#!B=lbmftpsia%:0&.0! %%

"

"" =

==

P

RTv

while the minimum specific volume of the cycle occurs at the end of the compression

=lbmft18$."1

=lbmft1$."& %%

"$ ===

rvv

The engineHs mean effective pressure is then

psia19"!=

=

=

7tu"

ftpsia&0&.8

=lbmft#18$."1$."&!

7tu=lbm8."":'()

%

%$"

net

vv

w


5"1

v

P

&

"

%

$ q

out

qin

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9-38 +n ideal Ftto cycle with air as the working fluid has a compression ratio of 5.8. The highestpressure and temperature in the cycle, the amount of heat transferred, the thermal efficiency, and the meaneffective pressure are to be determined.



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9-39 +n Ftto cycle with air as the working fluid has a compression ratio of 5.8. The highest pressure andtemperature in the cycle, the amount of heat transferred, the thermal efficiency, and the mean effective

pressure are to be determined.Assumptions1The airstandard assumptions are applicable.29inetic and potential energy changes are negligible. 3+ir isan ideal gas with constant specific heats.PropertiesThe properties of air at room temperature are cp@

".008 k

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9-40E +n ideal Ftto cycle with air as the working fluid has a compression ratio of 1. The amount of heattransferred to the air during the heat addition process, the thermal efficiency, and the thermal efficiency ofa Carnot cycle operating between the same temperature limits are to be determined.

Assumptions1The airstandard assumptions areapplicable. 29inetic and potential energy changes arenegligible. 3+ir is an ideal gas with variable specificheats.


Analysis!a# )rocess "$ isentropic compression.

%$."&&

7tu=lbm5$.0&B8&0

"

"

" ==

=r

uT

v

( ) 7tu=lbm"".$1$0&."1%$."&&1

""$

"

$

$$$===== u

r rrr vv

v

vv

)rocess $% v@ constant heat addition.

6tu)l7/1."1/====

=

=$1.$"":0.&8$

&"5.$

7tu=lbm&8$.:0

B$&00

$%

%

%%

uuq

u

T

in

rv

!b# )rocess %& isentropic expansion.

( ) ( ) 7tu=lbm$08.8&%8."5&"5.$1 &%

&

%%&===== ur rrr vv

v

vv

)rocess &" v@ constant heat rejection.

7tu=lbm80.""%0&.5$8&.$08"&out === uuq

30"!8===

7tu=lbm$&".&$

7tu=lbm""%.80""

in

outth

q

q

!c# $$"38===B$&00

B8&0""Cth,

H

L

T

T


5$"

v

P

&

"

%

$

qin

qout

$&00 B

8&0 B

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9-41E +n ideal Ftto cycle with argon as the working fluid has a compression ratio of 1. The amount ofheat transferred to the argon during the heat addition process, the thermal efficiency, and the thermalefficiency of a Carnot cycle operating between the same temperature limits are to be determined.

Assumptions1The airstandard assumptions are applicable with argon as the working fluid. 29ineticand potential energy changes are negligible. 3+rgon is an ideal gas with constant specific heats.

Properties The properties of argon are cp @ 0."$8% 7tu=lbm.B,cv @ 0.0:8; 7tu=lbm.B, and k@ ".;;: !Table +$(#.


( )( ) B$";"1B8&0 0.;;:"

$

""$ ==

=

k

TTv

v

)rocess $% v@ constant heat addition.

( )

( ) ( )

!"/#$18.07==

==B$";"$&007tu=lbm.B0.0:8;

$%$%in TTcuuq v

!b# )rocess %& isentropic expansion.

( ) B;001

"B$&00

0.;;:"

&

%%& =

=

=

k

TTv

v


( ) ( )( ) 7tu=lbm&.8%;B8&0;007tu=lbm.B0.0:8;"&"&out ==== TTcuuq v

$1"98===7tu=lbm"1.0:

7tu=lbm&.8%;""

in

outth

q

q

!c# $$"38===B$&00

B8&0""Cth,

H

L

T

T


5$$

v

P

&

"

%

$

qin

qout

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9-42+ gasoline engine operates on an Ftto cycle. The compression and expansion processes are modeledas polytropic. The temperature at the end of expansion process, the net work output, the thermalefficiency, the mean effective pressure, the engine speed for a given net power, and the specific fuelconsumption are to be determined.


PropertiesThe properties of air at 180 9 are cp @ ".""0 k

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!d# The clearance volume and the total volume of the engine at the beginning of compression process!state "# are

%%

m000$&&&.0m00$$.0

"0 =+

=+

= cc

c

c

dcr V

V

V

V

VV

%

"m00$&&&.000$$.0000$&&&.0

=+=+= dc VVV

The total mass contained in the cylinder is

( )( )kg0.00$881

9%%%9=kgmk)a0.$1:

#m&&&k)a#=0.00$"00!%

%

"

"" =

==RT

VPmt

The engine speed for a net power output of :0 kD is

re2)in1!!.=

==

min"

s;0

cycle#k =

=

== ff

f

f

ft

f

am

m

m

m

mm

m

m

>inally, the specific fuel consumption is

*)'Wh/3%"!=

==

kDh"

k

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9-43EThe properties at various states of an ideal Ftto cycle are given. The mean effective pressure is tobe determined.



0.$&0 7tu=lbmAB, cv

@ 0.":" 7tu=lbmAB, and k@ ".& !Table +$(a#.Analysis +t the end of the compression, the temperature is

( ) B"$8$5B#8$0! "&."""

"

$

""$ ===

=

k

k

rTTTv

v

while the air temperature at the end of the expansion is

B5.1"%5

"B#"5;0!

""&.""

%

"

&

%%& =

=

=

=

kk

rTTT

v

v

+pplication of the first law to the compression and expansion processes gives

7tu=lbm1".:0

B#8$0"$8$#!B7tu=lbm":".0!B#5.1"%"5;0#!B7tu=lbm":".0!

#!#!"$&%net

== =

TTcTTcw vv

+t the beginning of the compression, the specific volume is

=lbmft:;."%psia"&

B#8$0#!B=lbmftpsia%:0&.0! %%

"

"" =

==

P

RTv

while the specific volume at the end of the compression is

=lbmft8$5."5

=lbmft:;."% %%

"$ ===

r

vv

The engineHs mean effective pressure is then

psia0."0=

=

=

7tu"

ftpsia&0&.8

=lbmft#8$5.":;."%!

7tu=lbm1".:0'()

%

%$"

net

vv

w


5$8

v

P

&

"

%

$

qin

qout

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9-44EThe power produced by an ideal Ftto cycle is given. The rate of heat addition and rejection are tobe determined.



0.$&0 7tu=lbmAB, cv

@ 0.":" 7tu=lbmAB, and k@ ".& !Table +$(a#.Analysis The thermal efficiency of the cycle is

81&1.05

""

""

"".&"th ===

kr

+ccording to the definition of the thermal efficiency, the rate ofheat addition to this cycle is

6tu)h#!9,.!!=

==

hp"

7tu=h$8&&.8

0.81&1

hp"&0

th

netin

WQ

The rate of heat rejection is then

6tu)h/3/,9!!=== 7tu=h#8.$8&&"&0!"00,;05netinout WQQ

9-45The expressions for the maximum gas temperature and pressure of an ideal Ftto cycle are to bedetermined when the compression ratio is doubled.


Analysis The temperature at the end of the compression varies with the compression ratio as

""

"

$

""$

=

= k

k

rTTTv

v

since T"is fixed. The temperature rise during the combustionremains constant since the amount of heat addition is fixed.Then, the maximum cycle temperature is given by

""in$in%

+=+= krTqTqT

The smallest gas specific volume during the cycle is

r

"%

vv =

Dhen this is combined with the maximum temperature, the maximum pressure is given by

( )""in"%

%%

+== krTqRrRT

Pvv


5$;

v

P

&

"

%

$ qout

qin

v

P

&

"

%

$ qout

qin

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9-46It is to be determined if the polytropic exponent to be used in an Ftto cycle model will be greaterthan or less than the isentropic exponent.


Analysis 6uring a polytropic process,

constant

constant

=#"! =

= nn

n

TP

Pv

and for an isentropic process,

constant

constant

=#"! =

= kk

k

TP

Pv

If heat is lost during the expansion of the gas,

sTT &&>

where T&sis the temperature that would occur if the expansion were reversible and adiabatic !n@k#. Thiscan only occur when

kn


5$:

v

P

&

"

%

$ qout

qin

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9-47+n ideal Ftto cycle is considered. The heat rejection, the net work production, the thermal efficiency,and the mean effective pressure are to be determined.


PropertiesThe properties of air at room temperature are R@ 0.$1: k

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9-48The power produced by an ideal Ftto cycle is given. The rate of heat addition is to be determined.


PropertiesThe properties of air at room temperature are R@ 0.$1: k)aAm%=kg.9, cp @ ".008 k

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iesel Cycle

9-49C + diesel engine differs from the gasoline engine in the way combustion is initiated. In dieselengines combustion is initiated by compressing the air above the selfignition temperature of the fuelwhereas it is initiated by a spark plug in a gasoline engine.

9-50C The 6iesel cycle differs from the Ftto cycle in the heat addition process only- it takes place atconstant volume in the Ftto cycle, but at constant pressure in the 6iesel cycle.

9-51C The gasoline engine.

9-52C 6iesel engines operate at high compression ratios because the diesel engines do not have theengine knock problem.

9-53CCutoff ratio is the ratio of the cylinder volumes after and before the combustion process. +s the

cutoff ratio decreases, the efficiency of the diesel cycle increases.

9-54+n expression for cutoff ratio of an ideal diesel cycle is to be developed.


Analysis(mploying the isentropic process e?uations,

""$

= krTT

while the ideal gas law gives

""

$% TrrrTT k

cc==

Dhen the first law and the closed system work integral isapplied to the constant pressure heat addition, the result is

#!#! ""

""

$%in TrTrrcTTcq kk

cpp ==

Dhen this is solved for cutoff ratio, the result is

""

in"Trrc

qr

kcp

c +=


5%0

v

P

&

"

$ %q

in

qout

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9-55+n ideal diesel cycle has a compression ratio of "1 and a cutoff ratio of ".8. The maximumtemperature of the air and the rate of heat addition are to be determined.


PropertiesThe properties of air at room temperature are R@ 0.$1: k)aAm%=kg9, cp @ ".008 k

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9-56+ 6iesel cycle with nonisentropic compression and expansion processes is considered. Themaximum temperature of the air and the rate of heat addition are to be determined.


PropertiesThe properties of air at room temperature are R@ 0.$1: k)aAm%=kg9, cp @ ".008 k

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9-57+n ideal diesel cycle has a a cutoff ratio of ".$. The power produced is to be determined.



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9-58E+n ideal dual cycle has a compression ratio of $0 and cutoff ratio of ".%. The thermal efficiency,amount of heat added, and the maximum gas pressure and temperature are to be determined.



0.$&0 7tu=lbmAB, cv

@ 0.":" 7tu=lbmAB, and k@ ".& !Table +$(a#.AnalysisDorking around the cycle, the germane propertiesat the various states are

( ) B":8:$0B#8%0! "&."""

"

$

""$ ===

=

k

k

rTTTv

v

( ) psia5$1$0psia#"&! &.""$

""$ ===

= k

k

rPPPv

v

psia...1==== psia#5$1#!$."!$% PrPP p$

B$"05psia5$1psia"""&B#":8:!

$$ =

=

=PPTT $$

R/$1/===

= B#!".%#$"05!%% c$

$

$ rTTTv

v

B1.5"1$0

".%B#$:&$!

"&.""

%

"

&

%%& =

=

=

=

kc

k

r

rTTT

v

v

+pplying the first law and work expression to the heat addition processes gives

6tu)l7/./".= +=

+=

B#$"05$:&$#!B7tu=lbm$&0.0!B#":8:$"05#!B7tu=lbm":".0!

#!#! %$in $p$ TTcTTcq v

The heat rejected is

7tu=lbm&1.;;B#8%01.5"1#!B7tu=lbm":".0!#! "&out === TTcq v

Then,

!"#%$===7tu=lbm$"$."

7tu=lbm;;.&1""

in

outth

q

q


5%&

v

P

&

"

$

%

qout

$

qin

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9-59E+n ideal dual cycle has a compression ratio of "$ and cutoff ratio of ".%. The thermal efficiency,amount of heat added, and the maximum gas pressure and temperature are to be determined.



0.$&0 7tu=lbmAB, cv

@ 0.":" 7tu=lbmAB, and k@ ".& !Table +$(a#.AnalysisDorking around the cycle, the germane propertiesat the various states are

( ) B"&%$"$B#8%0! "&."""

"

$

""$ ===

=

k

k

rTTTv

v

( ) psia5.&8%"$psia#"&! &.""$

""$ ===

= k

k

rPPPv

v

psia311"$==== psia#5.&8%#!$."!$% PrPP p$

B":"1psia&8%.5psia8&&.:B#"&%$!

$$ =

=

=PPTT $$

R//00===

= B#!".%#":"1!%% c$

$

$ rTTTv

v

B5.5":"$

".%B#$$%%!

"&.""

%

"

&

%%& =

=

=

=

kc

k

r

rTTT

v

v

+pplying the first law and work expression to the heat addition processes gives

6tu)l7.$/"3= +=

+=

B#":"1$$%%#!B7tu=lbm$&0.0!B#"&%$":"1#!B7tu=lbm":".0!

#!#! %$in $p$ TTcTTcq v

The heat rejected is

7tu=lbm%%.;;B#8%05.5":#!B7tu=lbm":".0!#! "&out === TTcq v

Then,

!"#.3===7tu=lbm":$.8

7tu=lbm;;.%%""

in

outth

q

q


5%8

v

P

&

"

$

%

qout

$

qin

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9-60E+n airstandard 6iesel cycle with a compression ratio of "1.$ is considered. The cutoff ratio, theheat rejection per unit mass, and the thermal efficiency are to be determined.




%$."&&

7tu=lbm&0.5$B8&0

"

"

" ==

=r

uT

v

( )7tu=lbm&0$.08

B";$%.;5%.:%$."&&

$."1

""

$

$

"

$

""$ ==

====h

T

r rrr vv

v

vv

)rocess $%P@ constant heat addition.

."%1%====B";$%.;

B%000

$

%

$

%

$

$$

%

%%

T

T

T

P

T

P

v

vvv

!b#

7tu=lbm%11.;%08.&0$;1.:50

"10."7tu=lbm:50.;1B%000

$%in

%%

%

===

===

hhq

hTrv

)rocess %& isentropic expansion.

( ) 7tu=lbm5".$80;$"."""10."1&1."

$."1

1&1."1&1."&

$

&

%

&

%%%&====== u

rrrrr vv

v

vv

v

vv


!c#

59.1%

!"/#$158.87

===

===

7tu=lbm%11.;%

7tu=lbm"81.1:

""

0&.5$5".$80

in

out

th

"&out

q

q

uuq


5%;

v

P

&

"

$ %q

qout

%000 B

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9-61E +n airstandard 6iesel cycle with a compression ratio of "1.$ is considered. The cutoff ratio, theheat rejection per unit mass, and the thermal efficiency are to be determined.


PropertiesThe properties of air at room temperature are cp @ 0.$&0 7tu=lbm.B, cv @ 0.":" 7tu=lbm.B,

and k@ ".& !Table +$(#.Analysis!a# )rocess "$ isentropic compression.

( ) ( ) B":$&"1.$B8&0 0.&"

$

""$ ==

=

k

TTv

v

)rocess $% ) @ constant heat addition.

."$1.====B":$&

B%000

$

%

$

%

$

$$

%

%%

T

T

T

P

T

P

v

vvv

!b# ( ) ( )( ) 7tu=lbm%0;B":$&%0007tu=lbm.B0.$&0$%$%in ==== TTchhq p


( ) B"":%"1.$

".:&"B%000

:&"."0.&"

&

$%

"

&

%%& =

=

=

=

kk

vTTT

v

v

v


!c#

( )

( ) ( )

#1"#8

6tu)l7.!%

===

====

7tu=lbm%0;

7tu=lbm"01""

B08&"":%7tu=lbm.B0.":"

in

outth

"&"&out

q

q

TTcuuq

v


5%:

v

P

&

"

$ %q

in

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9-62+n ideal diesel engine with air as the working fluid has a compression ratio of $0. The thermalefficiency and the mean effective pressure are to be determined.



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9-63+ diesel engine with air as the working fluid has a compression ratio of $0. The thermal efficiencyand the mean effective pressure are to be determined.


PropertiesThe properties of air at room temperature are cp @

".008 k

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9-64 EES )roblem 5;% is reconsidered. The effect of the compression ratio on the net work output, mean

effective pressure, and thermal efficiency is to be investigated. +lso, TsandPvdiagrams for the cycleare to be plotted.

AnalysisUsing ((, the problem is solved as follows

Proce5ure ?Tota.12!.23!.34!.41@ .in.tota!.out.tota.in.tota = 0.out.tota = 0IA .12 B 0 TC .in.tota = .12 E .out.tota = - .12I, .23 B 0 ten .in.tota = .in.tota : .23 ese .out.tota = .out.tota - .23I, .34 B 0 ten .in.tota = .in.tota : .34 ese .out.tota = .out.tota - .34I, .41 B 0 ten .in.tota = .in.tota : .41 ese .out.tota = .out.tota - .41D

"Input Data"T[1]=23 [K]P[1]=< [kPa]

T[3] = 2200 [K]n=1'3 (0'8 (('1> 82'820 24 '

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.!-/ .!-. .!! .!. .!/.!.

.!/

.!0

.!1

.!.

.!/

.!0

.!1

2 &0)'*

P&'Pa

293 K

1049 K

2200 K

5.69

6.74kJ/kg-K

Air

.1 .# .% /! // /1

$9!

%!!

%.!

%/!

%0!

%1!

%3!

rcop

5net

&'()'*

.1 .# .% /! // /11$

19

3.

30

33

3$

rcop

th


5&$

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.1 .# .% /! // /19$!

9$3

9%!

9%3

99!

993

.!!!

rcop

:EP

&'Pa


5&%

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9-65 + fourcylinder ideal diesel engine with air as the working fluid has a compression ratio of ": and acutoff ratio of $.$. The power the engine will deliver at "800 rpm is to be determined.

Assumptions1The cold airstandard assumptions are applicable. 29inetic and potential energy changesare negligible. 3+ir is an ideal gas with constant specific heats.


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9-66 + fourcylinder ideal diesel engine with nitrogen as the working fluid has a compression ratio of ":and a cutoff ratio of $.$. The power the engine will deliver at "800 rpm is to be determined.

Assumptions1The airstandard assumptions are applicable with nitrogen as the working fluid. 29ineticand potential energy changes are negligible. 3itrogen is an ideal gas with constant specific heats.

PropertiesThe properties of nitrogen at room temperature are cp @ ".0%5 k

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9-67+n ideal dual cycle has a compression ratio of "1 and cutoff ratio of ".". The power produced by thecycle is to be determined.



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9-68 + dual cycle with nonisentropic compression and expansion processes is considered. The powerproduced by the cycle is to be determined.



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9-69E+n ideal dual cycle has a compression ratio of "8 and cutoff ratio of ".&. The net work, heataddition, and the thermal efficiency are to be determined.



0.$&0 7tu=lbmAB, cv

@ 0.":" 7tu=lbmAB, and k@ ".& !Table +$(a#.AnalysisDorking around the cycle, the germane properties at the various states are

( ) B"810"8B#8%8! "&."""

"

$

""$ ===

=

k

k

rTTTv

v

( ) psia$.;$5"8psia#$."&! &.""$

""$ ===

= k

k

rPPPv

v

psia".;5$psia#$.;$5#!"."!$% ==== PrPP p$

B":%1

psia;$5.$

psia;5$."B#"810!

$$ =

=

=

P

PTT $

$

B$&%%B#!".":%1!%

% ===

= c$

$

$ rTTTv

v

B$.5&$"8

".&B#$&%%!

"&.""

%

"

&

%%& =

=

=

=

kc

k

r

rTTT

v

v

+pplying the first law to each of the processes gives

7tu=lbm:.":1B#8%8"810#!B7tu=lbm":".0!#! "$$" === TTcw v

7tu=lbm0$.$:B#"810":%1#!B7tu=lbm":".0!#! $$ === TTcq $$ v

7tu=lbm1.";;B#":%1$&%%#!B7tu=lbm$&0.0!#! %% === $p$ TTcq

7tu=lbm5;.&:B#":%1$&%%#!B7tu=lbm":".0!7tu=lbm1.";;#! %%% === $$$ TTcqw v

7tu=lbm5.$8&B#$.5&$$&%%#!B7tu=lbm":".0!#! &%&% === TTcw v

The net work of the cycle is

6tu)l7./1"/=+=+= :.":15;.&:5.$8&$"%&%net wwww $

and the net heat addition is

6tu)l7.90"%=+=+= 1.";;0$.$:%$in $$ qqq

3ence, the thermal efficiency is

!"#1.===7tu=lbm"5%.1

7tu=lbm"$&.$

in

netth

q

w


5&1

v

P

&

"

$

%

qout

$

qin

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9-70+n expression for the thermal efficiency of a dual cycle is to be developed and the thermal efficiencyfor a given case is to be calculated.


PropertiesThe properties of air at room temperature are cp @ ".008 kor the case r@ $0 and rp@ $,

!"##!=

+

=

+

"$#$$!&."

"$

$0

""

$

"&."

"&."th


5&5

v

P

&

"

$

%

qout

$

qin

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9-71+n expression regarding the thermal efficiency of a dual cycle for a special case is to be obtained.


AnalysisThe thermal efficiency of a dual cycle may be expressed as

#!#!

#!

""%$

"&

in

out

th$p$ TTcTTc

TTc

q

q

+

== vv

7y applying the isentropic process relations for ideal gases with constant specific heats to the processes "$ and %&, as well as the ideal gas e?uation of state, the temperatures may be eliminated from the thermalefficiency expression. This yields the result

+

=

"#"!

"""

"thpcp

kcp

k rrkr

rr

r

where

$P

Pr

$p = and

$cr

v

v%=

Dhen rc@ rp, we obtain

+

=

+

"#!

"""

$

"

"th

ppp

kp

krrrk

r

r

Bearrangement of this result gives

"th$

"

#"!"#!

" +

=+

k

ppp

kp

rrrrk

r


580

v

P

&

"

$

%

qout

$

qin

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9-72+ sixcylinder compression ignition engine operates on the ideal 6iesel cycle. The maximumtemperature in the cycle, the cutoff ratio, the net work output per cycle, the thermal efficiency, the meaneffective pressure, the net power output, and the specific fuel consumption are to be determined.


PropertiesThe properties of air at 180 9 are cp @ ".""0 k

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( )

( ) k)a$.%;%m0.00&:1"

m0.000:;

k)a&%&"

9"$8&m0.00&:1"

m0.000:;9$%1%

".%&5

%

%

&

%

%&

"".%&5

%

%"

&

%%&

=

=

=

=

=

=

k

k

PP

TT

V

V

V

V

)rocess &" constant voume heat rejection.

( ) ( )( ) k

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Stirlin* and Ericsson Cycles

9-73C The efficiencies of the Carnot and the tirling cycles would be the same, the efficiency of the Fttocycle would be less.

9-74C The efficiencies of the Carnot and the (ricsson cycles would be the same, the efficiency of the6iesel cycle would be less.

9-75C The tirling cycle.

9-76C The two isentropic processes of the Carnot cycle are replaced by two constant pressureregeneration processes in the (ricsson cycle.

9-77+n ideal steadyflow (ricsson engine with air as the working fluid is considered. The maximumpressure in the cycle, the net work output, and the thermal efficiency of the cycle are to be determined.

Assumptions+ir is an ideal gas.

PropertiesThe gas constant of air is R@ 0.$1: k

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9-78+n ideal tirling engine with air as the working fluid operates between the specified temperature andpressure limits. The net work produced per cycle and the thermal efficiency of the cycle are to bedetermined.



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9-79+n ideal tirling engine with air as the working fluid operates between the specified temperature andpressure limits. The power produced and the rate of heat input are to be determined.



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9-80E+n ideal tirling engine with hydrogen as the working fluid operates between the specifiedtemperature limits. The amount of external heat addition, external heat rejection, and heat transfer

between the working fluid and regenerator per cycle are to be determined.

Assumptions3ydrogen is an ideal gas with constant specific heats.

PropertiesThe properties of hydrogen at room temperature are R@ 8.%$$& psiaAft%=lbm.B @ 0.518"

7tu=lbmAB, cp @ %.&% 7tu=lbmAB, cv

@ $.&& 7tu=lbmAB, and k@ ".&0& !Table +$(a#.AnalysisThe mass of the air contained in this engine is

lbm00;1%$.0B#""00#!B=lbmftpsia%$$&.8!

#ftpsia#!0."&00!%

%

"

"" =

==RT

Pm

V

+t the end of the compression, the pressure will be

psia&0ft"

ft0."psia#&00!

%

%

$

""$ =

==

V

VPP

The entropy change is

B7tu=lbm$;1.$psia&00

psia&0ln#B7tu=lbm0.518"!0

lnln"

$

0

"

$&%"$

=

=

== PPR

TTcssss

v

ince the processes are reversible,

6tu.$"!=== #B7tu=lbm$;1.$#!B00""!lbm#0.00;1%$!#! "$"in ssmTQ

6tu$"$3=== #B7tu=lbm$;1.$#!B008!lbm#0.00;1%$!#! %&%out ssmTQ

+pplying the first law to the process where the gas passes through the regenerator gives

6tu.!"!=== B#800""00#!B7tu=lbm&&.$!lbm#0.00;1%$!#! &"regen TTmcQ v


58;

s

T

%

$q

in

qout

&

"""00B

800 B

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9-81E+n ideal tirling engine with air as the working fluid operates between specified pressure limits.The heat added to and rejected by this cycle, and the net work produced by the cycle are to be determined.


PropertiesThe properties of air at room temperature are R@ 0.%:0& psiaAft%=lbm.B @ 0.0;188 7tu=lbmAB,cp @ 0.$&0 7tu=lbmAB, cv @ 0.":" 7tu=lbmAB, and k@ ".& !Table +$(a#.

Analysis+pplying the ideal gas e?uation to the isothermal process %& gives

psia"00psia#!"0#"0!&

%%& ===v

vPP

ince process &" is a constant volume process,,

B%%;0psia"00

psia;00B#8;0!

&

"&" =

=

=

P

PTT

+ccording to first law and work integral,

6tu)l730!"0==== B#ln!"0#%%;0#!B7tu=lbm0.0;188!ln"

$"$"in

v

vRTwq

and

6tu)l7%%"1=

=== "0

"B#ln8;0#!B7tu=lbm0.0;188!ln

%

&%&%out

v

vRTwq

The net work is then

6tu)l711."9=== &.11%.8%0outinnet qqw

9-82E+n ideal tirling engine with air as the working fluid operates between specified pressure limits.The heat transfer in the regenerator is to be determined.


PropertiesThe properties of air at room temperature are R@ 0.%:0& psiaAft%=lbm.B, cp @ 0.$&0

7tu=lbmAB, cv @ 0.":" 7tu=lbmAB, and k@ ".& !Table +$(a#.

Analysis+pplying the ideal gas e?uation to the isothermal process "$ gives

psia;0"0

"psia#;00!

$

""$ =

==

v

vPP

ince process $% is a constantvolume process,

B%%;0psia"0

psia;0B#8;0!

%

$%$ =

=

=P

PTT

+pplication of the first law to process $% gives

6tu)l71$%"%=== B#8;0%%;0#!B7tu=lbm":".0!#! %$regen TTcq v


58:

s

T

%

$q

in

qout

&

"

8;0 B

s

T

%

$q

in

qout

&

"

8;0 B

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9-83 +n ideal (ricsson cycle operates between the specified temperature limits. The rate of heat additionis to be determined.

AnalysisThe thermal efficiency of this totally reversiblecycle is determined from

0.;115

9500

9$10""th ===

H

L

T

T

+ccording to the general definition of the thermal efficiency,the rate of heat addition is

'W$/#===0.;115

kD800

th

netin

WQ

9-84+n ideal (ricsson cycle operates between the specified temperature limits. The power produced by

the cycle is to be determined.

AnalysisThe power output is 800 kD when the cycle isrepeated $000 times per minute. Then the work per cycle is

k

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Ideal and Actual Gas-ur7ine ;6rayton< Cycles

9-85C In gas turbine engines a gas is compressed, and thus the compression work re?uirements are verylarge since the steadyflow work is proportional to the specific volume.

9-86C They are !"# isentropic compression !in a compressor#, !$#P@ constant heat addition, !%#isentropic expansion !in a turbine#, and !P@ constant heat rejection.

9-87C >or fixed maximum and minimum temperatures, !a# the thermal efficiency increases with pressureratio, !b# the net work first increases with pressure ratio, reaches a maximum, and then decreases.

9-88C 7ack work ratio is the ratio of the compressor !or pump# work input to the turbine work output. Itis usually between 0.&0 and 0.; for gas turbine engines.

9-89C +s a result of turbine and compressor inefficiencies, !a# the back work ratio increases, and !b# the

thermal efficiency decreases.


585

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9-90E + simple ideal 7rayton cycle with air as the working fluid has a pressure ratio of "0. The airtemperature at the compressor exit, the back work ratio, and the thermal efficiency are to be determined.

Assumptions1teady operating conditions exist. 2The airstandard assumptions are applicable. 39inetic and potential energy changes are negligible. 4+ir is an ideal gas with variable specific heats.


Analysis!a# oting that process "$ is isentropic,

Th

Pr

1

1

112147

= =

=520 R

124.27 Btu / lbm

.

( )( )7tu=lbm$&0.""

"&:."$$"&:.""0

$

$

"

$

"$ ==

===h

TP

P

PP rr

R996.5

!b# )rocess %& is isentropic, and thus

( )

7tu=lbm%1.11$1%.$;8:".80&

7tu=lbm""8.1&$:."$&"".$&0

7tu=lbm$;8.1%&.":0.":&"0

"

0.":&

7tu=lbm80&.:"B$000

&%outT,

"$inC,

&%

&

%

%

%&

%

===

===

====

==

=

hhw

hhw

hPP

PP

P

hT

rr

r

Then the backwork ratio becomes

48.5%===7tu=lbm$%1.11

7tu=lbm""8.1&

outT,

inC,

bww

wr

!c#

46.5%===

===

===

7tu=lbm$;&.;0

7tu=lbm"$%.0&

7tu=lbm"$%.0&1&.""811.$%1

7tu=lbm$;&.;0"".$&0:".80&

in

outnet,

th

inC,outT,outnet,

$%in

q

w

www

hhq


5;0

s

T

"

$

&

%q

in

qout

$000

8$0

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9-91 (%'s( s('ved by EE! (n enc'(sed )*)+ simple 7rayton cycle with air as the working fluid has apressure ratio of 1. The air temperature at the turbine exit, the net work output, and the thermal efficiencyare to be determined.

Assumptions1teady operating conditions exist. 2Theairstandard assumptions are applicable. 39inetic and

potential energy changes are negligible. 4+ir is an

ideal gas with variable specific heats.


Analysis!a# oting that process "$s is isentropic,

Th

Pr

1

1

115546

= =

=310 K

310.24 kJ / kg

.

( )( )

( )

( )

( ) ( )

k

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9-92 EES )roblem 55" is reconsidered. The mass flow rate, pressure ratio, turbine inlet temperature, andthe isentropic efficiencies of the turbine and compressor are to be varied and a general solution for the

problem by taking advantage of the diagram window method for supplying data to (( is to be developed.


"Input 5ata - ,rom 5ia*ram /in5o/"6P.ratio = 876T[1] = 310 [K]P[1]= 100 [kPa]

T[3] = 11>0 [K]m.5ot = 20 [k*%s]ta.c = (

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T[4]=temperatureOairO!=[4]s[2]=entropyOairO!T=T[2]!P=P[2]s[4]=entropyOairO!T=T[4]!P=P[4]

N/r Pratio ;c[k;] ;net[k;] ;t[k;] ?in[k;]0'30< 0'1>44 4 4033 23>4 >3> 143(30'(038 0'1814 > 20'(>11 0'180> 8 >(23 2110 8833 11>820'8088 0'1(02 10 ((0< 1822 4 (800'(>( 0'03>(< 20 111>< 2>>'1 11431 (241

5.0 5.5 6.0 6.5 7.0 7.5

0

500

1000

1500

s &'()'*-+

-&+

100 kPa

800 kPa

.

/s

/

0

1

1s

Air Standard 6rayton CyclePressure ratio = % and a>= ..#!+


5;%

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2 4 6 8 10 12 14 16 18 20

0.00

0.05

0.10

0.15

0.20

0.25

0

500

1000

1500

2000

2500

Pratio

Cyclee??icien

cy,

Wnet&'W

Wnet

a>=..#! +

@ote Pratio?or a>iu 5or' and

c= !"$3

t = !"%/


5;&

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9-93 + simple 7rayton cycle with air as the working fluid has a pressure ratio of 1. The air temperature atthe turbine exit, the net work output, and the thermal efficiency are to be determined.

Assumptions1teady operating conditions exist. 2The airstandard assumptions are applicable. 39inetic and potential energy changes are negligible. 4+ir is an ideal gas with constant specific heats.


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9-94+ simple ideal 7rayton cycle with air as the working fluid operates between the specified temperatureand pressure limits. The net work and the thermal efficiency are to be determined.



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9-95+ simple 7rayton cycle with air as the working fluid operates between the specified temperature andpressure limits. The net work and the thermal efficiency are to be determined.



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PropertiesThe properties of air at room temperature are

cp @ ".008 kor the expansion process,

95.&$&k)a$000

k)a"009#"000!

0.&=".&=#"!

%

&%& =

=

=

kk

sP

PTT

9&.&1$

#5.&$&"000#!50.0!"000

#!#!

#!&%%&

&%

&%

&%

&%

==

=

=

= sTsp

p

sT TTTT

TTc

TTc

hh

hh

+pplying the first law to the constantpressure heat addition process $% produces

k

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PropertiesThe properties of air at room temperature are

cp @ ".008 kor the expansion process,

90.&$1k)a"580

k)a"009#"000!

0.&=".&=#"!

%

&%& =

=

=

kk

sP

PTT

9$.&18

#0.&$1"000#!50.0!"000

#!#!

#!&%%&

&%

&%

&%

&%

==

=

=

= sTsp

p

sT TTTT

TTc

TTc

hh

hh

+pplying the first law to the constantpressure heat addition process $% produces

k

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9-98 + gas turbine power plant that operates on the simple 7rayton cycle with air as the working fluidhas a specified pressure ratio. The re?uired mass flow rate of air is to be determined for two cases.


PropertiesThe properties of air at room temperature

are cp @ ".008 k

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9-99 + stationary gasturbine power plant operates on a simple ideal 7rayton cycle with air as theworking fluid. The power delivered by this plant is to be determined assuming constant and variablespecific heats.

Assumptions1teady operating conditions exist. 2The airstandard assumptions are applicable. 39inetic and potential energy changes are negligible. 4+ir is an ideal gas.

Analysis!a# +ssuming constant specific heats,

( )

( )( )

( )

( )

( )

( )

( ) ( ) k'15+680===

=

=

=

==

=

=

=

==

=

kD%8,0000.&&1

&&1.0%.8$8""00

$50$.;0:""""

9;0:.$1

"9""00

98$8.%19$50

inthoutnet,

$%

"&

$%

"&

in

outth

0.&=".&="

%

&%&

0.&=".&

="

"

$"$

QW

TT

TT

TTc

TTc

q

q

P

PTT

P

PTT

p

p

kk

s

kk

s

!b# +ssuming variable specific heats !Table +":#,

( )( )

( )

( ) ( ) k'15+085===

=

=

===

==

==

==

=

====

==

=

kD%8,0000.&%"

&%".0"".8$;0:."";"

";.$50%:.;8""""

k

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9-100 +n actual gasturbine power plant operates at specified conditions. The fraction of the turbine workoutput used to drive the compressor and the thermal efficiency are to be determined.

Assumptions1teady operating conditions exist. 2The airstandard assumptions are applicable. 39inetic and potential energy changes are negligible. 4+ir is an ideal gas with variable specific heats.


Analysis!a# Using the isentropic relations,

T h

T h

1 1

2 2

= =

= =

300 K 300.19 kJ / kg

580 K 586.04 kJ / kg

( )

( ) ( )( ) k

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9-101 + gasturbine power plant operates at specified conditions. The fraction of the turbine work outputused to drive the compressor and the thermal efficiency are to be determined.

Assumptions1teady operating conditions exist. 2The airstandard assumptions are applicable. 39inetic and potentialenergy changes are negligible. 4+ir is an ideal gas withconstant specific heats.

PropertiesThe properties of air at room temperature arecp @ ".008 k

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9-102E+ simple ideal 7rayton cycle with argon as the working fluid operates between the specifiedtemperature and pressure limits. The rate of heat addition, the power produced, and the thermal efficiencyare to be determined.

Assumptions1teady operating conditions exist. 29inetic and potential energy changes are negligible. 3+rgon is an ideal gas with constant specific heats.

PropertiesThe properties of argon at room temperature areR@ 0.$;1; psiaft%

=lbmAB !Table +"(#,cp@0."$8% 7tu=lbmAB and k@ ".;;: !Table +$(a#.

Analysis +t the compressor inlet,

=lbmft;:0.5psia"8

B#8&0#!B=lbmftpsia$;1;.0! %%

"

"" =

==

P

RTv

lbm=s08.;$=lbmft;:0.5

ft=s##!$00ft%!%

$

"

"" ===v

V%m

+ccording to the isentropic process expressions for an ideal gas,

B"%8:psia"8

psia"80B#8&0!

:0.;;:=".;;=#"!

"

$

"$ =

=

=

kk

P

PTT

B:.;;0psia"80

psia"8B#";;0!

:0.;;:=".;;=#"!

%

&%& =

=

=

kk

P

PTT

+pplying the first law to the constantpressure heat addition process $% gives

6tu)s/03#=== B#"%8:";;0#!B7tu=lbm"$8%.0!lbm=s#08.;$!#! $%in TTcmQ p

The net power output is

6tu)s.1.$=

+=

+=

B#"%8:8&0:.;;0";;0#!B7tu=lbm"$8%.0!lbm=s#08.;$!

#! $"&%net TTTTcmW p

The thermal efficiency of this cycle is then

!"#!.===7tu=s$%8;

7tu=s"&":

in

netth

Q

W


5:&

s

T

"

$

&

%

qin

qout

";;0 B

8&0 B

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9-103+n aircraft engine operates as a simple ideal 7rayton cycle with air as the working fluid. Thepressure ratio and the rate of heat input are given. The net power and the thermal efficiency are to bedetermined.



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9-104+n aircraft engine operates as a simple ideal 7rayton cycle with air as the working fluid. Thepressure ratio and the rate of heat input are given. The net power and the thermal efficiency are to bedetermined.



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9-105+ gasturbine plant operates on the simple 7rayton cycle. The net power output, the back workratio, and the thermal efficiency are to be determined.


PropertiesThe gas constant of air isR@ 0.$1: k

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6rayton Cycle 5ith Re*eneration

9-106C Begeneration increases the thermal efficiency of a 7rayton cycle by capturing some of the wasteheat from the exhaust gases and preheating the air before it enters the combustion chamber.

9-107C *es. +t very high compression ratios, the gas temperature at the turbine exit may be lower thanthe temperature at the compressor exit. Therefore, if these two streams are brought into thermal contactin a regenerator, heat will flow to the exhaust gases instead of from the exhaust gases. +s a result, thethermal efficiency will decrease.

9-108C The extent to which a regenerator approaches an ideal regenerator is called the effectiveness ,

and is defined as + qregen, act,qregen, max.

9-109C !b# turbine exit.

9-110C The steam injected increases the mass flow rate through the turbine and thus the power output.This, in turn, increases the thermal efficiency since in= QW= and W increases while Qinremainsconstant. team can be obtained by utiliGing the hot exhaust gases.


5:1

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9-111+ 7rayton cycle with regeneration produces "80 kD power. The rates of heat addition and rejectionare to be determined.

Assumptions1 The air standard assumptions are applicable. 2 +ir is an ideal gas with constant specificheats at room temperature. 3 9inetic and potential energy changes are negligible.

PropertiesThe properties of air at room temperature are cp@ ".008 k

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9-112+ 7rayton cycle with regeneration produces "80 kD power. The rates of heat addition and rejectionare to be determined.



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9-113+ 7rayton cycle with regeneration is considered. The thermal efficiencies of the cycle for parallelflow and counterflow arrangements of the regenerator are to be compared.



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9-114E+n ideal 7rayton cycle with regeneration has a pressure ratio of 1. The thermal efficiency of thecycle is to be determined with and without regenerator cases.


PropertiesThe properties of air at room temperature are cp@ 0.$& 7tu=lbmB and k @ ".& !Table +$(a#.

Analysis+ccording to the isentropic process expressions for an ideal gas,

B1.5$%B#!1#8"0!0.&=".&=#"!

"$ === kkprTT

B"01$1

"B#"5;0!

"0.&=".&=#"!

&8 =

=

=

kk

prTT

The regenerator is ideal !i.e., the effectiveness is "00# and thus,

B1.5$%

B"01$

$;

8%

====

TT

TT

The thermal efficiency of the cycle is then

!"3/9=

=

=="01$"5;0

8"05$%.1"""

%&

";

in

outth

TT

TT

q

q

The solution without a regenerator is as follows

B1.5$%B#!1#8"0!0.&=".&=#"!

"$ === kkprTT

B"01$1

"B#"5;0!

"0.&=".&=#"!

%& =

=

=

kk

prTT

!"11%=

=

==1.5$%"5;0

8"0"01$"""

$%

"&

in

outth

TT

TT

q

q


51$

s

T

"

$8

&qin"5;0 B

8"0 B

%

;q

out

s

T

"

$

&

%q

in

qout

"5;0 B

8"0 B

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9-115+n expression for the thermal efficiency of an ideal 7rayton cycle with an ideal regenerator is to bedeveloped.


AnalysisThe expressions for the isentropic compression and expansion processes are

kkprTT

=#"!"$

=

kk

prTT

=#"!

%&

"

=

>or an ideal regenerator,

$;

&8

TT

TT

==

The thermal efficiency of the cycle is

kkp

kkp

kkp

rT

T

r

r

T

T

TT

TT

T

T

TT

TT

T

T

TT

TT

q

q

=#"!

%

"

=#"!

=#"!

%

"

%&

"$

%

"

%8

";

%

"

8%

";

in

out

th

"

"

""

#=!"

"#=!"

#=!"

"#=!

"""

=

=

=

=

==


51%

s

T

"

$&

%qin

8

;q

out

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9-116E + car is powered by a gas turbine with a pressure ratio of &. The thermal efficiency of the car andthe mass flow rate of air for a net power output of 58 hp are to be determined.

Assumptions1 teady operating conditions exist. 2+ir is an ideal gas with variable specific heats. 3 Theambient air is 8&0 B and "&.8 psia. 4 The effectiveness of the regenerator is 0.5, and the isentropicefficiencies for both the compressor and the turbine are 10. 5 The combustion gases can be treated asair.

PropertiesThe properties of air at the compressor and turbine inlet temperatures can be obtained fromTable +":(.

Analysis The gas turbine cycle with regeneration can be analyGed as follows

( ) ( )

( ) 7tu=lbm%:$.$8%.8:"$.$%0&"

"$.$%0

7tu=lbm8&5.%8B$";0

7tu=lbm0."5$8&&.8%1;."&

%1;."

7tu=lbm"$5.0;B08&

&%

&

%

%

$"

$

"

"

%&

%

"$

"

==

==

==

=

====

==

=

srr

r

srr

r

hPPPP

P

hT

hPP

PP

P

hT

and

7tu=lbm;%.&0:$.%:$%8.8&5

%8.8&50.10

7tu=lbm:&.$0:0;."$5

0;."$50."5$0.10

&&

&%

&%turb

$$"$

"$comp

=

=

=

=

=

=

hh

hh

hh

hhhh

hh

s

s

Then the thermal efficiency of the gas turbine cycle becomes

7tu=lbm":5.5#:&.$0:;%.&0:!5.0#! $&regen === hhq

7tu=lbm;%.0#0;."$5:&.$0:!#;%.&0:%8.8&5!#!#!

7tu=lbm";".:@5.":5#:&.$0:%8.8&5!#!

"$&%inC,outT,outnet,

regen$%in

======

hhhhwww

qhhq

0.%57tu=lbm";".:

7tu=lbm;%.0

in

outnet,

th %39====q

w

>inally, the mass flow rate of air through the turbine becomes

#$/*1.07=

==

hp"

7tu=s:0;1.0

7tu=lbm;%.0

hp58

net

netair

w

Wm


51&

s

T

"

$s&s

%qin$";0 B

8&0 B

8&

$

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9-117(%'s( s('ved by EE! (n enc'(sed )*)The thermal efficiency and power output of an actual gasturbine are given. The isentropic efficiency of the turbine and of the compressor, and the thermalefficiency of the gas turbine modified with a regenerator are to be determined.

Assumptions1+ir is an ideal gas with variable specific heats. 2 9inetic and potential energy changes arenegligible. 3 The mass flow rates of air and of the combustion gases are the same, and the properties ofcombustion gases are the same as those of air.


AnalysisThe properties at various states are

( )( )

( ) k

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9-118 EES )roblem 5"": is reconsidered. + solution that allows different isentropic efficiencies for thecompressor and turbine is to be developed and the effect of the isentropic efficiencies on net work doneand the heat supplied to the cycle is to be studied. +lso, the Tsdiagram for the cycle is to be plotted.


"Input 5ata"T[3] = 1288 [+]Pratio = 14'(

T[1] = 20 [+]P[1]= 100 [kPa]6T[4]=

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ta.t = ;.5ot.turL %;.5ot.turLisen "turLine a5iaLatic e,,iciency! ;.5ot.turLisen B;.5ot.turL"

"EEEA Airst a/ ,or te isentropic turLine! assumin*@ a5iaLatic! ke=pe=0.5ot.in -.5ot.out = DT.5ot.c$ = 0 ,or stea5y-,o/"m.5ot#[3] = ;.5ot.turLisen : m.5ot#.s[4]

.s[4]=TCPir!T=T.s[4]"ctua TurLine anaysis@"m.5ot#[3] = ;.5ot.turL : m.5ot#[4][4]=TCPir!T=T[4]s[4]=T&JPir!T=T[4]! P=P[4]

"+yce anaysis""sin* te 5e,inition o, te net cyce /ork an5 1 H; = 1000 k;@";.5ot.net#1000=;.5ot.turL-;.5ot.comp "k)%s"ta.t.nore*=;.5ot.net#1000%?.5ot.in.nore*"+yce terma e,,iciency"N/r=;.5ot.comp%;.5ot.turL"Nack /ork ratio"

";it te re*enerator te eat a55e5 in te eterna eat ecan*er is"

m.5ot#[]=P[4]

"+yce terma e,,iciency /it re*enerator"ta.t./itre*=;.5ot.net#1000%?.5ot.in./itre*

"Te ,oo/in* 5ata is use5 to compete te rray TaLe ,or pottin* purposes'"s.s[1]=s[1]

T.s[1]=T[1]s.s[3]=s[3]

T.s[3]=T[3]s.s[

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t c t!nore* t!/itre* ?innore*[k;]

?in/itre*[k;]

;net[k;]

0'( 0'82 0'230 0'340< 4420>3 2(>> 102'10'(< 0'82 0'2(3> 0'3841 4420>3 3148>3 120'

0'8 0'82 0'31>3 0'423( 4420>3 32>0 13'80'8< 0'82 0'33 3>01

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!"$ !"$3 !"% !"%3 !"9 !"93 ./$3!!!

0.!!!!

013!!!

0%!!!!

1.3!!!

13!!!!

t

Adot,in

no re*eneration

5ith re*eneration

!"$ !"$3 !"% !"%3 !"9 !"93 .!"/

!"/3

!"0

!"03

!"1

!"13

!"3

!"33

!"#

t

Etath

5ith re*eneration

no re*eneration


515

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9-119+ 7rayton cycle with regeneration using air as the working fluid is considered. The air temperatureat the turbine exit, the net work output, and the thermal efficiency are to be determined.

Assumptions1 The air standard assumptions areapplicable. 2 +ir is an ideal gas with variable specificheats. 3 9inetic and potential energy changes arenegligible.


Analysis!a# The properties of air at various states are

( ) ( )

( ) ( ) ( )

( )

( ) ( )( ) k

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9-120+ stationary gasturbine power plant operating on an ideal regenerative 7rayton cycle with air asthe working fluid is considered. The power delivered by this plant is to be determined for two cases.

Assumptions1 The air standard assumptions are applicable. 2 +ir is an ideal gas. 3 9inetic and potentialenergy changes are negligible.

PropertiesDhen assuming constant specific heats, the properties of air at room temperature are cp@

".008 k

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9-121+ regenerative gasturbine engine using air as the working fluid is considered. The amount of heattransfer in the regenerator and the thermal efficiency are to be determined.

Assumptions1 The air standard assumptions are applicable. 2 +ir is an ideal gas with variable specificheats. 3 9inetic and potential energy changes are negligible.


Analysis!a# The properties at various states are

( )

( )

( ) ( )

( ) ( )( ) kJ/kg152.50&.81;11.:5::$.0

k

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Assumptions1 The air standard assumptions are applicable. 2 +ir is an ideal gas with constant specificheats. 3 9inetic and potential energy changes are negligible.

PropertiesThe properties of air at room temperature

are cp@ ".008 k

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Assumptions1 The air standard assumptions are applicable. 2 +ir is an ideal gas with variable specificheats. 3 9inetic and potential energy changes are negligible.


Analysis!a# The properties of air at various states are

( )

( ) ( )( )

( ) ( )( ) kJ/kg148.30&.81;11.:5::0.0

k

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6rayton Cycle 5ith Intercoolin*, Reheatin*, and Re*eneration

9-124C +s the number of compression and expansion stages are increased and regeneration is employed,the ideal 7rayton cycle will approach the (ricsson cycle.

9-125C !a# decrease, !b# decrease, and !c# decrease.

9-126C !a# increase, !b# decrease, and !c# decrease.

9-127C

7173597-Chap-9

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