Lecture 05 Problems

(1)

VARIABLE LOADS VARIABLE LOADS 1.

1. A A DiDiesesel powel power plaer plant has nt has a a mamaxiximum demmum demanand d of 12 !" of 12 !" wiwith a th a loload fa#ad fa#totor r of .$of .$  anandd #apa#it% of .&. Estimate the plant #apa#it%.

#apa#it% of .&. Estimate the plant #apa#it%. A A.. 11  !!"" B. 160 MW B. 160 MW ' '.. 11$$  !!"" D D.. 22  !!"" Solution( Solution( load load Peak Peak load load Average Average factor factor Load Load

==

120 120 80 80 0

0..

==

Average Average loadload kw kw load load Average Average

==

9696 capacit capacit Plant Plant load load Average Average factor factor Use Use

==

capacit capacit Plant Plant 96 96 60 60 0 0_..

==

kw kw capacity capacity Plant Plant

==

160160 2.

2. 'al'al#ul#ulate the !" ate the !" powpower #apaer #apa#it#it% % of a of a )eo)eotherthermal plamal plant with a nt with a loaload d fa#fa#tor of .$2 and 12tor of .$2 and 12 !" pea* load. +he operation is limited to $, hours

!" pea* load. +he operation is limited to $, hours a %ear with a use a %ear with a use fa#tor of .-.fa#tor of .-. A A.. 11--,, B B.. 11 ' '.. 22//// D. 145 D. 145 Solution( Solution( load load Peak Peak load load Average Average factor factor Load Load

==

120 120 82 82 0

0..

==

Average Average loadload kw kw load load Average Average

==

9898..44 capacit capacit Plant Plant load load Average Average factor factor Use Use

==

capacit capacit Plant Plant 4 4 98 98 70 70 0 0..

==

.. kw kw capacity capacity Plant Plant

==

144144..99 /.

/. A 0ien euipA 0ien euipment #onsument #onsumes , *w3hr4mes , *w3hr4month at 256 rated plamonth at 256 rated plant #apa#int #apa#it%t%. It operates at. It operates at 25 hours7 / da%s4month. "hat

25 hours7 / da%s4month. "hat is the rated #apa#it%8is the rated #apa#it%8 A

A.. 11--..,,  *w*w

1 1

(2)

B B.. 22//..  *w*w C. 28.90 kw C. 28.90 kw D. D. $$2.2.5 5 *w*w Solution( Solution(

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hrs hrs day day days days month month month month hr hr kw kw load load Average Average 24 24 1 1 30 30 1 1 5000 5000 kw kw load load Average Average

==

66..9494 equipmen equipmen of of Rating Rating load load Average Average factor factor Plant Plant

==

equipme equipme of of Rating Rating 94 94 6 6 24 24 0 0..

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.. kw kw equipment equipment of of Rating Rating

==

2828..9393 5.

5. A #entral staA #entral station is suppl%tion is suppl%in0 ener0in0 ener0% to a % to a #ommun#ommunit% throit% throu0h two su9stau0h two su9stations. One su9stions. One su9statiotationn feeds four distri9ution #ir#uits: the

feeds four distri9ution #ir#uits: the other7 six. +he other7 six. +he maximum dail% re#orded demandsare(maximum dail% re#orded demandsare( ; ;oowweer r ssttaattiioonn 112277  **ww S Suu99ssttaattiioon n AA &&77  **ww < <eeeeddeer r 11 11-- < <eeeeddeer r 22 11$$ < <eeeeddeer r // 22$$ <

<eeeeddeer r 55 && S

Suu99ssttaattiioon n BB 77  **ww <

<eeeeddeer r 11 &&22 < <eeeeddeer r 22 11,, < <eeeeddeer r // 11 < <eeeeddeer r 55 22 < <eeeeddeer r ,, 2222 <

<eeeeddeer r && //

'al#ulate the diersit% fa#tor amon0 feeders in su9station B. 'al#ulate the diersit% fa#tor amon0 feeders in su9station B. A A.. 11..,, B. 1.25 B. 1.25 ' '.. 11..11,, D D.. 11..//,, Solution( Solution(

Diersit% fa#tors amon0 feeders in Su9station B Diersit% fa#tors amon0 feeders in Su9station B

B B substation substation of of demand demand imum imum max max feede feede each each of of demand demand imum imum max max individual individual of of sum sum

==

9000 9000 3000 3000 2200 2200 2900 2900 1000 1000 1500 1500 600 600

++

==

247 247 1 1..

==

,.

,. +he annual pea* load on a +he annual pea* load on a 1,71,73*w powe3*w power plant is r plant is 17,17, *" *". +. +wo su9statwo su9stations are 9ein0ions are 9ein0 supplied 9% this plant. Annual ener0% dispat#hed throu0h su9station A is 2-7,7 *"3hr supplied 9% this plant. Annual ener0% dispat#hed throu0h su9station A is 2-7,7 *"3hr with a pea* at $7 *w7 while 1&7,7 *"3hr are 9ein0 sent throu0h su9station B with a with a pea* at $7 *w7 while 1&7,7 *"3hr are 9ein0 sent throu0h su9station B with a pea* at &7&, *w. =e0le#tin0 line losses7 >nd the

pea* at &7&, *w. =e0le#tin0 line losses7 >nd the #apa#it% fa#tor of the #apa#it% fa#tor of the power plant.power plant. A A.. ..//,,// B B.. ..,,//// 2 2

(3)

B B.. 22//..  *w*w C. 28.90 kw C. 28.90 kw D. D. $$2.2.5 5 *w*w Solution( Solution(

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==

hrs hrs day day days days month month month month hr hr kw kw load load Average Average 24 24 1 1 30 30 1 1 5000 5000 kw kw load load Average Average

==

66..9494 equipmen equipmen of of Rating Rating load load Average Average factor factor Plant Plant

==

equipme equipme of of Rating Rating 94 94 6 6 24 24 0 0..

==

.. kw kw equipment equipment of of Rating Rating

==

2828..9393 5.

5. A #entral staA #entral station is suppl%tion is suppl%in0 ener0in0 ener0% to a % to a #ommun#ommunit% throit% throu0h two su9stau0h two su9stations. One su9stions. One su9statiotationn feeds four distri9ution #ir#uits: the

feeds four distri9ution #ir#uits: the other7 six. +he other7 six. +he maximum dail% re#orded demandsare(maximum dail% re#orded demandsare( ; ;oowweer r ssttaattiioonn 112277  **ww S Suu99ssttaattiioon n AA &&77  **ww < <eeeeddeer r 11 11-- < <eeeeddeer r 22 11$$ < <eeeeddeer r // 22$$ <

<eeeeddeer r 55 && S

Suu99ssttaattiioon n BB 77  **ww <

<eeeeddeer r 11 &&22 < <eeeeddeer r 22 11,, < <eeeeddeer r // 11 < <eeeeddeer r 55 22 < <eeeeddeer r ,, 2222 <

<eeeeddeer r && //

'al#ulate the diersit% fa#tor amon0 feeders in su9station B. 'al#ulate the diersit% fa#tor amon0 feeders in su9station B. A A.. 11..,, B. 1.25 B. 1.25 ' '.. 11..11,, D D.. 11..//,, Solution( Solution(

Diersit% fa#tors amon0 feeders in Su9station B Diersit% fa#tors amon0 feeders in Su9station B

B B substation substation of of demand demand imum imum max max feede feede each each of of demand demand imum imum max max individual individual of of sum sum

==

9000 9000 3000 3000 2200 2200 2900 2900 1000 1000 1500 1500 600 600

++

==

247 247 1 1..

==

,.

,. +he annual pea* load on a +he annual pea* load on a 1,71,73*w powe3*w power plant is r plant is 17,17, *" *". +. +wo su9statwo su9stations are 9ein0ions are 9ein0 supplied 9% this plant. Annual ener0% dispat#hed throu0h su9station A is 2-7,7 *"3hr supplied 9% this plant. Annual ener0% dispat#hed throu0h su9station A is 2-7,7 *"3hr with a pea* at $7 *w7 while 1&7,7 *"3hr are 9ein0 sent throu0h su9station B with a with a pea* at $7 *w7 while 1&7,7 *"3hr are 9ein0 sent throu0h su9station B with a pea* at &7&, *w. =e0le#tin0 line losses7 >nd the

pea* at &7&, *w. =e0le#tin0 line losses7 >nd the #apa#it% fa#tor of the #apa#it% fa#tor of the power plant.power plant. A A.. ..//,,// B B.. ..,,//// 2 2

(4)

C. 0.335 C. 0.335 D D.. ..,,//,, Solution( Solution( peri perioo same same the the for for energy energy possib possibllee imum imum max max produ producedced energy energy actual actual factor factor capacity capacity Plant Plant

==

(

kwkw

))(

(

hr hr

))

hr hr kw kw factor factor capacity capacity Plant Plant 8760 8760 00 0000 15 15 000 000 500 500 16 16 000 000 50 5000 27 27 77 77 77 77 77

++

−−

==

33 3355 0 0..

==

factor factor capacity capacity Plant Plant &.

&. A power plA power plant is said to haant is said to hae had a use fa#toe had a use fa#tor of 5$.,6 and a #ar of 5$.,6 and a #apa#itpa#it% fa#tor of 52% fa#tor of 52.56. ?ow.56. ?ow man% hours did it operate durin0 the %ear8

man% hours did it operate durin0 the %ear8

A. A. 76607660 B B.. $$--&& ' '.. $$22 D D.. 117722 Solution( Solution( hour hour operating operating capacity capacity pl plant ant kw kw prod produceduced kwhr kwhr actual actual factor factor use use Plant Plant

××

==

hours hours capacity capacity pl plant ant kw kw produce producedd kwhr kwhr actual actual factor factor capacity capacity Plant Plant 8760 8760

××

==

+herefore +herefore hour hour operating operating hrs hrs factor factor capacity capacity Plant Plant factor factor use use Plant Plant 87608760

==

hour hour operating operating hrs hrs 8760 8760 4 4 42 42 5 5 48 48

==

.. .. hr hr hours hours operating operating

==

76607660 -.

-. A distri9A distri9ution transution transformeformer supplier supplies a s a 0ro0roup of 0eneral powup of 0eneral power #ustomerer #ustomers hain0 a #onne#ts hain0 a #onne#teded load of 1$& *w. If ener0% sells at /., #ents per *"3hr7 what will 9e the monthl% in#ome from load of 1$& *w. If ener0% sells at /., #ents per *"3hr7 what will 9e the monthl% in#ome from ener0% deliered throu0h this transformer hain0 an aera0e motor e@#ien#% of -,6.

ener0% deliered throu0h this transformer hain0 an aera0e motor e@#ien#% of -,6. )eneral data(

)eneral data( Demand

Demand fa#tor fa#tor .-,.-, Diersit% <a#tor 1., Diersit% <a#tor 1., L

Looaad d ffaa##ttoorr ..55,, A A.. ;;17175-5-.. B B.. ;;1717-5-5.. C. P1,407.00 C. P1,407.00 D. D. ;1;17575--.. Solution( Solution(

A#tual maximum demand  demand fa#tor x #onne#ted load A#tual maximum demand  demand fa#tor x #onne#ted load A#tual maximum demand  .-, x 1$&  1/., *w

A#tual maximum demand  .-, x 1$&  1/., *w Simultaneous maximum demand  sum of

Simultaneous maximum demand  sum of indiidual maximum demands 4 diersit% fa#torindiidual maximum demands 4 diersit% fa#tor Simultaneous maximum demand  1/., 4 1., 

Simultaneous maximum demand  1/., 4 1.,  / *w/ *w "ith -,6 motor e@#ien#%

"ith -,6 motor e@#ien#%

Simultaneous maximum demand  / 4

Simultaneous maximum demand  / 4 .-,  125 *w.-,  125 *w Aera

Aera0e load on 0e load on transformer  load fa#tor x transformer  load fa#tor x pea* loadpea* load /

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Aera

Aera0e load on transformer  .5, x 0e load on transformer  .5, x 125  ,,.$ *w125  ,,.$ *w w3hrs deliered monthl% 

w3hrs deliered monthl%  C,,.$ *wC/ da%s4mo.C25 hrs4da%C,,.$ *wC/ da%s4mo.C25 hrs4da% w3hrs deliered monthl%  571-& *w3hrs4mo.

w3hrs deliered monthl%  571-& *w3hrs4mo. !onthl% in#ome from ener0% deliered 

!onthl% in#ome from ener0% deliered  C./,4*w3hrC571-C./,4*w3hrC571-& *w3hr4mo.& *w3hr4mo. !onthl% in#ome from ener0% deliered 

!onthl% in#ome from ener0% deliered  ;175&.1&;175&.1& $.

$. A dail% loA dail% load #ure wad #ure whi#h exhhi#h exhi9iti9ited a 1,3mined a 1,3minute pea* of &7ute pea* of &7,2 *" is dra,2 *" is drawn to s#ales of 1 #m wn to s#ales of 1 #m  2 hrs and 1 #m

2 hrs and 1 #m  , *w , *w. +he area under the #ure . +he area under the #ure is measured 9% a planimeter and found tois measured 9% a planimeter and found to 9e 5$.-2 #m

9e 5$.-2 #m22_{. Determine the load fa#tor 9ased on the 1,3minute}_{. Determine the load fa#tor 9ased on the 1,3minute pea*.}_pea*.

A A.. ..221111 B B.. ..551111 C. 0.311 C. 0.311 D D.. ..,,1111 Solution( Solution( 1 #m 1 #m22_{ ,C2  1 *w3hrs}_{ ,C2  1 *w3hrs} Aera0e load  C5$.-2 #m

Aera0e load  C5$.-2 #m22_{C1 *w3hrs4#m}_{C1 *w3hrs4#m}22_{C 1 da%425 hours}_{C 1 da%425 hours}

Aera0e load  2/ *w Aera0e load  2/ *w

Load fa#tor  Aera0e load 4 ;ea* load Load fa#tor  Aera0e load 4 ;ea* load Load fa#tor  2/ 4 &,2

Load fa#tor  2/ 4 &,2 Load fa#tor  ./11

Load fa#tor  ./11 .

. A A ununit it is is rarateted d at at $$77  *"*". . It It opopereratates es fofor r &&, , hrhrs s duduririn0 n0 ththe e %e%ear ar anand d 0e0eneneraratetess /,77 *"3hr. "ith a pea* load of -,7, *"7 #al#ulate the #apa#it% fa#tor.

/,77 *"3hr. "ith a pea* load of -,7, *"7 #al#ulate the #apa#it% fa#tor.

A. A. 50%50% B B.. ,,//66 ' '.. --66 D D.. &&,,66 Solution( Solution( ( ) ( )( ( 77 )) .. 66 77 77 50 50 50 50 0 0 000 000 80 80 8760 8760 000 000 000 000 350 350

==

factor factor Capacity Capacity

1.A 2,7 *" plant has a utiliation fa#tor of -16 and a load fa#tor of /.&6. "hat is the 1.A 2,7 *" plant has a utiliation fa#tor of -16 and a load fa#tor of /.&6. "hat is the

aera0e load on the plant in aera0e load on the plant in *"8*"8 A A.. 225577$$,, B B.. 11//77$$&& C. 9,841 C. 9,841 D D.. 1177,,2211 Solution( Solution( Aera

Aera0e load  0e load  C./&C2,7C./&C2,7 *"   *"  7 *"7 *"

11.A steam #%#le produ#es 5 !" of power7 , !" of pro#ess heat and reFe#ts & !" of heat. 11.A steam #%#le produ#es 5 !" of power7 , !" of pro#ess heat and reFe#ts & !" of heat.

"hat is the utiliation fa#tor for this

"hat is the utiliation fa#tor for this #%#le8#%#le8 A A.. ,,66 B. 60% B. 60% ' '.. --66 D D.. $$66 Solution( Solution( Gtiliation <a#tor  Gtiliation <a#tor  % % 60 60 60 60 .. 0 0 60 60 50 50 40 40 50 50 40 40

==

++

==

12.<or a #ertain power plant7 the load fa#tor is 5$6. +he plant #apa#it% is 52 !". If the resere 12.<or a #ertain power plant7 the load fa#tor is 5$6. +he plant #apa#it% is 52 !". If the resere oer pea* is , !" and the num9er of hours the plant not in seri#e per %ear is 527 #al#ulate oer pea* is , !" and the num9er of hours the plant not in seri#e per %ear is 527 #al#ulate the plant #apa#it% fa#tor.

the plant #apa#it% fa#tor. A A.. ..//2255 B B.. ..//$$- -C. 0.423 C. 0.423 5 5

(6)

D. .5& Solution(

(

Plant capacity

)(

ours per year

)

year per produced !nergy "actor Capacity Plant

=

(

Average Load

)(

Hours per year

)

produced Energy

=

Load Peak Loa Average "actor Load

=

Pea #ver rve Rese Capacity Plant Load Peak

=

−

$% Load Peak

=

42

−

5

=

37

(

) ( )

$% Load Average

=

.5$ /-

=

1-.-&

(

Plant capacity

)(

ours per year

)

year per produced !nergy "actor Capacity Plant

=

(

)(

)

(

)(

)

. . 6 . 3 42 423 0 8760 42 8760 76 17

=

year hrs $% year hrs $% "actor Capacity Plant

1/.+he resere oer and a9oe the pea* load of a power station is 2 !". <or an annual fa#tors as follows( load7 -,6 and #apa#it%7 &6: determine the rated #apa#it% of the power plant. A. & !"

B. -, !"

C. 100 MW

D. 55.55 !" Solution(

CLoad <a#torC;ea* Load  C'apa#it% <a#torCRated 'apa#it%  Aera0e Load C.-,C;ea* Load  C.&CRated 'apa#it%

Rated 'apa#it% H ;ea* Load  2 !"  Resere

C.-,CRated 'apa#it% H 2  C.&CRated 'apa#it% Rated 'apa#it%  1 !"

15.A power plant is to hae had a use fa#tor of 5&.,6 and a #apa#it% fa#tor of 5.,6. Determine the num9er of hours did the plant operated durin0 the %ear.

A. -2& B. -&2

C. 7629

D. -&2 Solution(

(

Plant capacity

)(

noof hrs per year

)

year per produced !nergy "actor Capacity Plant .

=

( )( ) ( )( ) PC A!P PC A!P PC"

=

8760 8760

(

Plant capacity

)(

&oof hoursnot inoperation per year

)

year per produced !nergy "actor Use Plant .

=

,

(7)

( )( )

( )PC

(

oursnot inoperation per year

)

A!P PU"

=

$-& Euate A! ( ( )( ) ( )( )

(

)

8760 year per operation in not ours PC PU" PC PC"

=

( )( ) ( )( ) hrs PU" PC" year per operation in not ours 7629677 465 0 8760 405 0 8760 . . .

=

<GELS A=D 'O!BGS+IO=

1,.<ind the air fuel ratio7 in *0 air per *0 of fuel7 for a #om9ustion pro#ess to whi#h the fuel is '$?2 with 26 ex#ess air.

A. 1,.-B. 18.5 '. 1.-D. 2., Solution( ( ) 2 2 2 ( ) 2 2 20 8 a# a376 & bC# c# a3 76 & C

+

.

→

+

.

Solin0 for a7 b and c

( ) 2 2 2 ( ) 2 2 20 8 13# 133 76 & 8C# 10 # 133 76& C

+

.

→

+

.

( ) (

)( )

( )

12 20

( )

1 15 38 8 28 76 3 13 32 13 . .

=

+

=

ta % ( 1538) 18 456 20 1 20 1.

=

. .

=

.

=

ta aa % %

1&.A fuel 0as has the followin0 olumetri# anal%sis( '?5  &$6 '2?&  /26

Assume #omplete #om9ustion with /6 ex#ess air at 11./2, *;a7 21 ' wet 9ul9 and 2- ' dr% 9ul9. "hat is the partial pressure of the water apor in *;a8

A. $.&2 B. 1.2$ '. 1&.5 D. 14.24 Solution( +heoreti#al Air ( ) ( ) 2 2 2 2 2 6 2 4 76 3 76 3 32 0 68 0 & a # c bC# & a a#  C C . . . .

+

→

+

(

)

(

)

2 2 2 2 2 6 2 4 76 3 48 2 32 2 32 1 76 3 48 2 48 2 32 0 68 0 & #  C# & #  C C . . . . . . . . .

+

→

+

"ith /6 Ex#ess air

( ) ( )( ) ( ) 2 ( )( ) 2 2 2 2 2 6 2 4 76 3 3 1 48 2 48 2 3 0 32 2 32 1 76 3 3 1 48 2 3 1 48 2 32 0 68 0 & # #  C# & #  C C . . . . . . . . . . . . . . + + + → + + + &

(8)

+otal moles in produ#ts  1./2  2./2  ./C2.5$  1./C2.5$C/.-& +otal moles in produ#ts  1&.,&

;artial pressure  C2./2 4 1&.,&C11./2,  15.25 *;a

1-.A diesel ele#tri# plant supplies ener0% for !E'O. Durin0 a 25hr period7 the plant #onsumed / 0allons of fuel at 2$ ' and produ#ed 5 *w3hr. Industrial fuel used is /2 A;I and was pur#hased at ;/2. per liter at 1,.& '. "hat should the #ost of fuel to produ#e one *w3hr8 A. ;-.,

B. P7.35

'. ;-.& D. ;&. Solution(

'omputin0 for pri#e per *0 from per liter at 1,.& ' 5 131 5 141 6 15 . . .

+

=

AP' () 8654 0 5 131 32 5 141 6 15 . . . .

=

+

=

()

Densit% at 1,.& '  .$&,5 C1 *04L  .$&,5 *04L ;ri#e per *0  ;/2 4 .$&,5 ; /&.-- per *0

'onertin0 / 0al per *w3hr to *0 per *w3hr at 2$ '

( )

[

1 00007 156

]

6 15.

−

.

−

.

=

() t ()_t ( )

[

1 00007 28 156

]

8654 0 28

=

.

−

.

−

. () 8579 0 28

=

. () hr kw kg L kg gal L hr kw gal

−

=

×

−

01988 8579 0 7854 3 4900 300 . . .

'ost per *w3hr  .1$$ x ; /&.--  ;-./, per *w3hr

1$.'al#ulate the mass of the #om9ustion produ#ts of a diesel fuel oil C'1&?/ with /6 ex#ess air

if the mass of fuel is / *0. A. 1$.- *0 B. 1&.1 *0 C. 599.1 kg D. ,&.1 *0 Solution( ( ) 2 2 2 ( ) 2 2 30 16 a# a376 & bC# c# a376 & C

+

.

→

+

.

Solin0 for a7 b and c

( ) 2 2 2 ( ) 2

2 /

1& 2/.,# 2/.,/.-&& 1&C# 1,# 2/.,/.-&&

C

+

→

+

( )

( )( )

( )

12 30

( )

1 14 53 16 28 76 3 5 23 32 5 23 . . . .

=

+

=

ta % ( 1453) 18889 30 1 30 1.

=

. .

=

.

=

ta aa % % mf  / *0

!ass of produ#t of #om9ustion

(9)

- C1  % tamf  C11$.$$C/  ,&.&- *0

1.A mixture of 11 *;a and 2/ ' that is /6 9% wei0ht 'O 2 and -6 9% wei0ht =27 determine

the partial pressure of 'O2.

A. /.1$2 *;a B. 21.65 kPa '. 1 *;a D. 2, *;a Solution( + C# C# P n+ n p













=

2 2

<or 1 *0 of 0as mixture(

!ass !ol 'O2  / *0 /455  .&$2 mol =2  - *0 -42$  2., mol n+  .&$2  2.,  /.1$2 mol ( ) kPa p_C# 101 2165 182 3 682 0 2 _. . .

=













=

2.<ind the air fuel ratio7 in *0 air per *0 fuel7 for a #om9ustion pro#ess to whi#h the fuel is ' $?2

with /6 ex#ess air. A. /.-& B. 1 '. 1/ D. 20 Solution(

(

)

(

)

2 2 2 2 2 20 8 76 3 76 3 & a # c bC# & a a#  C . .

+

→

+

Solin0 for a7 b and c ( ) ( ) 2 2 2 2 2 20 8 76 3 13 10 8 76 3 13 13 & #  C# & #  C . .

+

→

+

( ) ( )( ) ( ) 12 20( )1 1538 8 28 76 3 13 32 13 . .

=

+

=

ta % ( 15 38) 20 30 1 30 1

=

. _ta . . aa % %

21.At 2$ A;I spe#i># 0rait%7 what is the heatin0 alue of oil in *J4*08 A. 5&7- B. 5-71/ C. 45,039 D. 5$7/ Solution( ( AP') ,_h

=

417130

+

139.6 ( )28 6 139 130 417

+

.

=

h , kg k-,_h

=

457038.80 $

(10)

22.Dode#ane C'12?2& is 9urned at #onstant pressure with 156 ex#ess air. Determine the

num9er of moles of ox%0en in the produ#ts of #om9ustion8 A. 22. B. 25.9 '. 55., D. ,,., Solution( +heoreti#al Air ( ) 2 2 2 ( ) 2 2 2&

12 a# a/.-&& bC# c# a/.-&&

C

+

→

+

(

)

2 2 2

(

)

2

2 2&

12 1$.,# 1$.,/.-&& 12C# 1/# 1$.,/.-&&

C

+

→

+

"ith 156 Ex#ess air

( ) 2 ( )( ) 2 2 2 ( ) 2 ( )( ) 2 2&

12 1$.,1 1.5# 1$.,1 1.5 /.-&& 12C# 1/# 1.51$.,# 1$.,1 1.5 /.-&&

C

+

→

+

!ols of O2  1.5 x 1$.,  2,. mols

2/.A mixture of 15.- psia and &$ < that is /6 9% wei0ht 'O2 and -6 9% wei0ht =2 has a partial

pressure of 'O2 in psia that is nearest to(

A. 2.15 B. $.$/ '. -.$$ D. 3.15 Solution( + C# C# P n+ n p













=

2 2

<or 1 *0 of 0as mixture(

!ass !ol 'O2  / l9 /455  .&$2 mol =2  - l9 -42$  2., mol n+  .&$2  2.,  /.1$2 mol

(

)

psi p_C# 147 315 182 3 682 0 2 _. . . .

=













=

25.)ien 0aseous fuel '$?1$ and the olumetri# produ#ts of #om9ustion(

'O2  1.167 'O  .$,67 =2  $,67 O2  156. Determine the air3fuel ratio.

A. 1$. B. 20.1 '. 1,. D. 25. Solution(

(

2 2

)

2 2 2 2 1$

$ b# /.-&& 1.1C# .$,C# .15# c# $,&

aC

+

→

+

'ar9on Balan#e( $a  1.1  .$,7 a  1./,-, =itro0en Balan#e( 2C9C/.-&  2C$,7 9  22.&& =o need to determine #. ( )

[

]

( ) ( )

[

]

kg kg % _aa 201 1 18 12 8 3575 1 28 76 3 32 606 22 . . . .

=

+

=

C9 

(11)

2,.A manufa#turin0 #ompan% in Bi#ol operates a Diesel Ele#tri# ;lant to suppl% its ele#tri# ener0% reuirements. Durin0 a 25 hr period7 the plant #onsumed 2, 0allons of fuel at $ < and produ#ed 2- *w3hrs. Industrial fuel used is / A;I and was pur#hased at ;/. per liter at & <. Determine the oer3all e@#ien#% of the plant.

A. 2,.&6 B. 2.-6 C. 26.08% D. /.16 Solution( input outpu e

=

h m, hr hr kw e 24 2700

−

=

Solin0 for h , (

(

AP'

)

( )

k-kg ,_h

=

5171/

+

1/.&

=

5171/

+

1/.&/

=

5,7/1$ Solin0 for m ( AP' ) ( ) ( _" _C 

+

=

5 131 5 141 6 15 60 . . . . . . _. 876 0 30 5 131 5 141 6 15 60 . . . . . . . _.

=

+

=

C " () ) (

(

)

[

1 00007 26 67 156

]

0869 876 0 6 26 80 . . . . . .) _"

=

() _. _C

=

−

=

( . m

=

ρ

(

)

(

)

(

)

_ 











 











=

s hr gal m hr gal m kg m /& 1 /-$, .  25 2, 1 $& .  / / s kg m

=

009520. +hen7

(

kgs

)(

k-kg

)

hr hr k% e 318 45 00952 0 24 2700 7 .

−

=

.2&$2&.$6 2&.A #ertain #oal has the followin0 ultimate anal%sis(

A  ,6 '  -6 ?2  2.,6

!  16 =2  56

S  ,6 K2  /.,6

<ind a#tual air3fuel for an ex#ess air of /6.

A. 11.67

B. .,5

(12)

'. 1,.5 D. 1/.-, Solution( S O H C F A 4.32 8 5 . 34 5 . 11 2 2



+











₋

+

=

(

)

4.32

(

0.05

)

8 035 . 0 025 . 0 5 . 34 70 . 0 5 . 11



+











₋

+

=

F A coal kg air kg F A

=

8.977

<or an ex#ess air  /6

( )

A F _actual

=

(

8.977

)( )

1.30

=

11.671kg air kg coal

2-.It is reuired to >nd the theoreti#al olume of air at 2 ' and 1 *;a a9solute pressure to 9urn one *ilo0ram #oal. +he ultimate anal%sis of #oal as3>red is as follows(

'  &,.&,6 O  1$.&16 S  1.,16 !oist.  /6 ?  ,.$-6 =  1./6 Ash  ,./&6 A. .5/ m/_air4*0#oal B. 7.43 m3_air/kgcoa '. $.5/ m/_air4*0#oal D. &.5/ m/_air4*0#oal Solution(

+he theoreti#al olume of air reuired:

p mR .

=

Solin0 for m (

(

A"

)

C  # ( m _t 5./ $ , . /5 , . 11

_

+









 −

+

=

( ) 5./(.1, ) $ 1$&1 .  ,$-.  , . /5 &,&, .  , . 11

_

+











₋

+

=

m coa kg air kg m

=

$.$5

(

)

(

)(

)

2 1 2-/ 2 2$-.  $5 . $ m k& / / air kg k-coal kg air kg .  

+

=

coa kg air m .

=

-.5/ /

2$.A steam 9oiler 9urns fuel oil with /6 ex#ess air and represented as '1&?/2. +he fuel 0as leaes the preheater at 22 *;a. <ind the a#tual3air fuel ratio.

A. 1,.& B. 19.2 '. 1&.-, D. 1$.1 Solution( 32 16 C  C_n _m

=

11

(13)

B% <ormula

(

)

m n m n % _ta

+

=

12 25 0 28 137. . ( ) 708 14 32 16 12 32 25 0 16 28 137 . . .

=

+

×

+

=

ta % ( ) 1

+

=

( )( 1.30 14.708)

=

19.12

=

ta aa e% %

2.In 9urnin0 a t%pi#al #oal7 the theoreti#al air3fuel ratio was found to 9e 1, *04*0. +he approximate heatin0 alue of the #oal in *J4*0 is7

A. 57&, B. 45,610 '. 5&7, D. 517,1 Solution( 000 10 65 7 7 . . % _ta

=

( )(

)

kg k-lb Btu lb Btu . 613 45 84 607 19 65 7 000 10 15 7 . 7 . 7

=

I=+ER=AL 'O!BGS+IO= E=)I=E

/.Determine the thermal e@#ien#% of an air standard Otto #%#le with 16 #learan#e. A. ,/.$6 B. 61.68% '. ,$.//6 D. -2.//6 Solution( 1 1 1 −

−

=

_k r η 2 1 . . n compressio of ratio r

=

2 1 2 . . . c

−

=

2 2 1 c. . c.

−

=

11 1 0 1 1 0 1 2 1

=

+

=

+

=

. . c c . . 6 . . . 06168 6168 11 1 1 1 4 1

=

−

=

₋ η

/1.A two3stro*e Diesel en0ine has a fri#tion power of 1/6 of the heat 0enerated. Determine the 9ra*e thermal e@#ien#% if the indi#ated thermal e@#ien#% of the en0ine is ,6.

A. //6

(14)

B. 37% '. /56 D. /&6 Solution( A f , %

=

0.13 A i , %

=

0.5 A A A f i b % % , , , %

=

−

=

0.5

−

0.13

=

0.37 37 0.

=

A b , % e or /-6

/2.Durin0 a !orse test on a four3#%linder automotie en0ine7 the speed was *ept #onstant at 25., rps 9% the 9ra*e and the followin0 9ra*e torue readin0s were o9tained(

All #%linders >rin0 1/.$ =3m =o. 1 #%linder #ut3out 1/.$ =3m =o. 2 #%linder #ut3out 1/.2 =3m =o. / #%linder #ut3out 12. =3m =o. 5 #%linder #ut3out 1/1.1 =3m A##ount for the fri#tional power7 in *". A. $.-2 B. 9.13 '. 1.1, D. 12.-Solution( ( )( ) k% +& BP 29 83 1000 5 24 8 193 2 2

=

. .

=

.

=

π π

=o. 1 #%linder #ut3out(

(

)( )

k% BP 20 14 1000 5 24 8 130 2 1 . . .

=

π k% 'P₁

=

29.83

−

20.14

=

9.69 =o. 2 #%linder #ut3out(

( )( ) k% BP 2004 1000 5 24 2 130 2 2 . . .

=

π k% 'P₂

=

29.83

−

20.04

=

9.79 =o. / #%linder #ut3out(

(

)( )

k% BP 2000 1000 5 24 9 129 2 3 . . .

=

π k% 'P₃

=

29.83

−

20.00

=

9.83 =o. 5 #%linder #ut3out(

( )( ) k% BP 20 18 1000 5 24 1 131 2 4 . . .

=

π 1/

(15)

k 'P₄

=

29.83

−

20.18

=

9.65 k% 'P 'P 'P 'P 'P

=

₁

+

₂

+

₃

+

₄

=

38.96 k% BP 'P "P

=

−

=

38.96

−

29.83

=

9.13

//.An en0ine 9urns a liter of fuel ea#h 12 min. +he fuel has a spe#i># 0rait% of .$ and a heatin0 alue of 5, !J4*0. +he thermal e@#ien#% of the en0ine is 2,6. "hat is the 9ra*e horsepower of the en0ine8

A. 11 B. 17 '. 1/ D. 1, Solution( m ,_A

=

(

)

( )

_ 











=

min . . 12 1_Liter )R (P m ρ _w

(

)

(

)

. mi min . . kgL Liter kg m 006667 12 1 0 1 8 0

=

 











=

(

kg

)(

k- kg

)

k% ,_A 50 60 1 000 45 06667 0

=

 











=

se# min 7 min .

(

k%

)

k% e, %

=

_A

=

0.2550

=

12.5 hp hp k% k% hp 17 746 0 5 12

=

. .

/5.A Diesel en0ine used 2- tons of fuel per da% when deelopin0 5& *" indi#ated power and 5& *" 9ra*e power. Of the total heat supplied to the en0ine7 /1.-6 is #arried awa% 9% #oolin0 water and radiation7 and /.$6 in the exhaust 0ases. 'al#ulate the indi#ated thermal e@#ien#%. A. 37.5% B. /,.-6 '. /1.2,6 D. /2.-,6 Solution( 6 . . .7 308 37 5 31 100

−

=

i e

/,.+he #om9ined me#hani#al ele#tri#al e@#ien#% of an en0ine30enerator set is $26. If the #om9ined spe#i># fuel #onsumption is .2, *04*"3hr7 the indi#ated spe#i># fuel #onsumption is7 A. .2/, *04*"3hr B. .2, *04*"3hr C. 0.168 kg/kW!"r D. .1-, *04*"3hr Solution(

(

combsf c

)

( )( )kgk% hr c f s ind. . . .

=

. . . .η _comb

=

.2,.$2

−

15

(16)

 .1&$ *04*"3hr

/&.+he thermal e@#ien#% of an en0ine is /6 and the fuel used hae a heatin0 alue of 57, *J4*0. +he en0ines spe#i># fuel #onsumption is7

A. /./-, *04*"3hr B. .2$, *04*"3hr C. 0.296 kg/kW!"r D. ./-, *04*"3hr Solution(

(

k- kg

)

kgk% hr hr k% k-c f s

=

⋅

=

0296

⋅

500 40 30 0 3600 . 7 . . .

/-.A #ar en0ine produ#es an aera0e of 2, *" of power in a period of 1 min. durin0 whi#h 1.2, *0 of 0asoline is #onsumed. If the heatin0 alue of 0asoline is 5/7 *J4*07 the e@#ien#% of this en0ine durin0 the period is7

A. 256 B. 28% '. /26 D. /&6 Solution(

(

k- kg

)

k% s kg ,_A 43000 8958 60 1 10 25 1 . 7 min min .

=

 











 











=

k% %

=

25

(

6

)

6 .58 100 28 89 25

=













=

A , % e

/$.An en0ine 9urns a liter of fuel ea#h 12 minutes. +he fuel has a relatie densit% .$ and a heatin0 alue of 5, !J4*0. +he en0ine has a 9ra*e thermal e@#ien#% of 2,6. "hat is the 9ra*e horsepower of the en0ine8

A. 16.76 B. 12.&$ '. 1/.,& D. 2$./5 Solution( 1 Liter  .1 m/ . m ,

=

_f

(

kgm

)

m

(

k-kg

)

, 45000 60 1 12 001 0 9000 80 0 3 3 7 se# min min . .

_ 











 











=

k% ,

=

50

(

)

hp k% hp k% bhp 16 76 746 0 1 50 25 0 . . .

_



=









=

/.A , *" Diesel en0ine deelops torue of 2 *=3m. +he en0ine dries an alternatin0 #urrent 0enerator with $ poles produ#in0 #urrent at & ?. "hat is the speed ratio of redu#tion 0ear8 A. /.&(1

B. 1.&(1

C. 2.6#1

(17)

D. 5.&(1 Solution( 60 2_+& P

=

( ) 60 2 2 500

=

& rpm &

=

2387.32 ( ) rpm p f &_s 900 8 60 120 120

=

+hus7 the speed ratio is

1 6 2 900 32 2387. .

=

s & & R

5.A 1, *" diesel en0ine operates at an altitude of 12 meters eleation. <ind the power deeloped 9% the en0ine at hi0her eleation.

A. 11 *w

B. 1285 kw

'. 11, *w D. 12, *w Solution(

'onsiderin0 the eMe#t of 9oth pressure and temperature #han0e(

Bra*e power at a hi0her eleation  Bra*e power at a lower eleation

L  L  + + p p

 











 p

 pressure at a hi0her eleation L

p

 pressure at sea leel Cstandard 

+

 temperature at a hi0her eleation L

+

 temperature at a lower eleation Cstandard <or eer% 1 feet a9oe sea leel:

De#rease in pressure is 1 in#h ?0 a9solute

De#rease in temperature is /.,&& de0rees Ran*ine

( m)( ft m) ing ft g in g in p . . . . . . 1200 328 25984 1000 1 92 29

=











−

=

( ) ( m)( ft m) R ft R R + _ 1200 328 515 964 1000 566 3 460 70 .

_ 

.

=

.











−

+

=

1&

(18)

(

)

k% BP_ 1285 460 70 964 515 92 29 984 25 1500

=

+













=

. . .

51.Determine the power output of a Diesel power plant if the en0ine and 0enerator e@#ien#% is $/6 and ,6 respe#tiel%. +he en0ine uses 2,o_{A;I fuel and has a fuel #onsumption of 22$}

*04hr. A. 2815 kw B. 2, *w '. 2&, *w D. 2-, *w Solution(

(

eat generatedfromfuel

)

_m _g #utput Plant

=

η η

(

mf .

)

m g #utput Plant

=

η η

(

AP'

)

( ) k- kg .

=

417130

+

139.6 

=

417130

+

139.6 25

=

447620 hr kg m_f

=

288 m η

 en0ine me#hani#al e@#ien#%  $/6 g η  0enerator e@#ien#%  ,6 ( ) (44620 )( )( )0 83 0 95 3600 1 288 7 k-kg . . s hr s kg #utput Plant

_ 











=

k% #utput Plant

=

2815

52.An en0ine deelopin0 / hp7 is transferrin0 1 hp more to the #oolin0 tower. If the top water and 9ottom water of the radiator are 2 < and 1 <7 respe#tiel%7 #al#ulate the reuired water Now rate in 0allons per minute.

A. 20.34 B. 1,.2, '. ,.$ D. 1.1-Solution(

(

hp

)(

Btuhp hr

)

Btuhr ,

=

30

+

10 2544

−

=

1017760 " +

=

200

−

190

=

10

∆

" lb Btu c_p

=

1.0

−

+ mc ,

=

_p

∆

( )( )10 10 760 1017

=

m . hr lb m

=

107176 In 0pm(

(19)

1-(

lb hr

)

hr gal 2034gpm 34 8 1 60 1 176 10 _. . min 7

_ 

=











 











5/.+he e@#ien#% of an Otto en0ine is ,,6 and the stro*e olume is /, liters. If the heat added at the 9e0innin0 of the #om9ustion is 12.& *J7 #al#ulate the mean eMe#tie pressure.

A. 15 *;a B. 1$$ *;a C. 198 kPa D. 1$5 *;a Solution( 0 net m . % p

=

"here: 3 035 0 35_liters _m . ₀

=

. Solin0 for net W ( A net , % e

=

6 12 55 0 . .

=

% net k-% _net

=

6.93 +hus7 kP p_m 198 035 0 93 6

=

. .

55.A test 9ed7 1&3#%linder7 V3t%pe7 tur9o3#har0ed Diesel en0ine deeloped 7&1 metri# horsepower at ,15 rpm. ?eat reFe#tion to lu9ri#atin0 oil is -& *'al per metri# hp3hr. Spe#i># 0rait% of lu9ri#atin0 oil used is . and spe#i># heat of oil is ., Btu4l93<. 'ompute the uantit% of lu9ri#atin0 oil #ir#ulated in m/_{4hr if the temperature of lu9ri#atin0 oil enterin0 the}

en0ine is ,, ' and &, ' leain0. A. 2,$./5 m/_4hr B. 12/.5, m/_4hr C. 168.23 m3_/"r D. /25.,& m/_4hr Solution( ρ m .

=

Solin0 for mass Now rate of lu9ri#atin0 oil(

,

 heat #arried awa% 9% lu9ri#atin0 oil  C&1C-&  -,-7/& *'al4hr

C kg kCal " lb Btu c_p

=

₀_.₅ ο

=

₀_.₅ ο 1$

(20)

t mc ,

=

_p

∆

( ) 05 ( 65 55) 036 7577

=

m .

−

hr kg m

=

1517407.2

(

)

3 900 1000 90 0

=

_kg_m

=

. ρ +hen: hr m m kg hr kg . 3 3 16823 900 2 407 151 . . 7

=

5,.In an ideal Otto #%#le7 the initial pressure and temperature of air are 1 *;a and 1$ de0ree '. Determine the maximum pressure in the #%#le if the maximum temperature in the #%#le is & de0ree '7 and the #ompression ratio is $.

A. 1$ *;a B. 225 *;a C. 2400 kPa D. 1, *;a Solution( p1  1 *;a + 1  1$  2-/  21  + 2  &  2-/  $-/  r k  . 14. 3  $ 2 2 3 3 + p + p

=

( )

1 1 2 −

=

k k r + +

( )

k k r p p2

=

1

( )

1 1 1 1 1 3 3 + r p r + r p + p _k k k k k

₌

=

₋

( )

291 8 100 873 3

=

p kP p₃

=

2400 1

(21)

5&.Air enters the #%linder of an internal #om9ustion en0ine at an initial pressure and temperature of , *;a and 2$ de0 '. A four #%linder7 sin0le3a#tin07 5 mm x ,, mm7 four stro*e en0ine runs at 5, rpm. +he mass Now rate of air per se#onds is .1 *0. "hat is the olumetri# e@#ien#% of the en0ine8

A. -26 B. $16 C. 87% D. -.-6 Solution( 1 1 p mR .

′

=

( )( )( ) s m . 1 000909 3 95 273 28 287 0 01 0 . . .

=

+

=

′

c p 0 0L&& . 2 4 π

=

rpm &_p 2250 2 4500

=

(

) (

)

( )

m s . ₀ 3 3 01037 0 4 60 2250 055 0 040 0 4 . .





=

.









=

π 6 . . . . 7 87 877 0 01037 0 00909 0 1

=

′

=

0 v . . η

5-.A 2 #m x /& #m7 twin #%linder7 two stro*e #%#le diesel en0ine runnin0 at &, rpm. +he en0ine rate is 112 *w. Determine the en0ine displa#ement in #u9i# meter4se# per 9ra*e horsepower. A. .1&/ B. .1&/ C. 0.00163 D. .&/1 Solution( c p 0 0L&& . 2 4 π

=

rpm &_p

=

650 BP & L& 0 BP . p c 0 2 4 π

=

( ) ( )

( )

746 0 112 2 60 650 36 0 20 0 4 2 . . .

_











=

π BP . ₀ bhp m BP . ₀

−

=

. 3 se# 001632 0 2

(22)

5$.A 1 *w3hr diesel 0eneratin0 set has a 0enerator e@#ien#% of $,6. If the mass of the fuel is 1, *07 #ompute the en0ine fuel rate.

A. .1, *04*w3hr B. .11-, *04*w3hr C. 0.127 kg/kw!"r D. 22.- *04*w3hr Solution( k%h kg BP m_f 1275 0 85 0 1000 150 . .

=

5.In an air standard diesel #%#le7 #ompression starts at 1 *;a and // . +he #ompression ratio is 2 to 1. +he maximum #%#le temperature is 22 . Determine the thermal e@#ien#%. A. 5&6 B. 64% '. &6 D. -6 Solution(

(

)



_









−

=

₋ 1 1 1 1 1 c k c k k k r r r e 20 2 1

=

. . r _k 2 3 2 3 + + . . r _c

=

( )( )14 1 1 2 1 1 2 303 20 − −

=













=

. k . . + + / + ₂

=

1004 / + ₃

=

2200 19 2 1004 2200 .

=

c r

(

)



_









−

=

₋ 1 1 1 1 1 c k c k k k r r r e 21

(23)

( ) ( ) ( )



_









−

=

₋ 1 19 2 4 1 1 19 2 20 1 1 4 1 1 4 1 . . . . . e 6 . .6384 6384 0

=

e

,.If the pressure at the 9e0innin0 and end of #ompression in an Otto #%#le en0ine are 1 *;a and , *;a7 respe#tiel%7 the maximum #%#le temperature is 1, . Determine the air standard e@#ien#%. A. /,.-6 B. /.,6 C. 47.5% D. ,-.-6 Solution( 1 1 1 −

−

=

_k k r e k k p p . . r 1 1 2 2 1

 











=

993 4 100 950 14 1 . .

=













=

k r 6 . . .993. 04744 4744 4 1 1 1 4 1

=

−

=

₋ e C#

,1.An Otto en0ine has a #learan#e olume of $6. It produ#es , *w power. 'ompression starts at 1 *;a and 2, '. "hat is the heat reFe#ted in *w8

A. 273 kw B. /-2 *w '. /2- *w D. -/2 *w Solution( 1 1 1 −

−

=

_k k r e 5 13 08 0 08 0 1 1 . . .

=

+

=

+

=

c c r _k 22

(24)

(

135

)

06469 1 1 1 4 1 . . .

=

−

=

₋ e A , % e

=

A , 500 6469 0.

=

k% ,_A

=

772.92 R A % , ,

=

+

R ,

+

=

500 92 772. k% ,_R

=

272.92

,2.+he followin0 data are the results on a test of a two #%linder7 four stro*e #%#le Otto en0ine( torue  2 =3m: indi#ated mean eMe#tie pressure  - *;a: fuel #onsumption  .& *04se#: fuel heatin0 alue  5/7 *J4*0: 9ore x stro*e  / #m x 5 xm: speed  & rpm. 'al#ulate the 9ra*e mean eMe#tie pressure.

A. /-, *;a B. /2, *;a C. 445 kPa D. 2$, *;a Solution(

(

)

k% +& BP 12566 60 600 0 2 2 2 .

_

=

.











=

π π c p 0 0L&& . 2 4 π

=

rpm &_p 300 2 600

=

( ) ( )

( )

m s . ₀ 3 3 2827 0 2 60 300 40 0 30 0 4 . .





=

.









=

π kPa . BP p 0 mb 444 49 2827 0 66 125 . . .

=

,/. A four3#%linder7 four3#%#le en0ine with 13#m diameter pistons and a 1&3#m stro*e operates at a speed of & rpm and %ields an indi#ator dia0ram. +he area under the #ure CpV dia0ram is eual to 12. #m2_{. +he len0th of the dia0ram is . #m7 and the sprin0 #onstant of}

the indi#ator sprin0 is & *;a4#m. Determine the mean eMe#tie pressure.

A. 800 kPa B. 5 *;a '. & *;a D. , *;a Solution( 2/

(25)

scal spring stroke curve under Area p_m

=

×

kPa p_m 600 800 9 12

=

×

=

,5.+he #ompression ratio of an air3standard 0asoline en0ine is -.,. +he heat added is 1$ *J4*0. If the initial pressure and temperature are 1 9ar and 2- ' respe#tiel%7 determine the temperature in ' at the end of the isentropi# expansion.

A. 1206 B. 2&$ '. 1-&, D. 1/5-Solution( + 1  2-  2-/  /  r k  -., , A4m  1$ *J4*0

(

)

/ r + + ₂

=

₁ _kk −1

=

3007.51.4−1

=

671.7

(

+ 3 + 2

)

c m , v A

₌

₋

(

671 7

)

716 0 1890

=

. + ₃

−

. / + ₃

=

3311.4 1 4 3 −

=

k k r + +

(

)

14 1 4 5 7 4 3311 ₋

=

. . . + / + ₄

=

1479 C t ₄

=

1479

−

273

=

1206

,,.'al#ulate the thermal e@#ien#% of an air3standard Diesel #%#le operatin0 with a #ompression ratio of 21 and #ut3oM ratio of /.

A. ,,.6

B. 61.4%

'. &&.6 D. -1.6

(26)

Solution(

(

)



_









−

=

₋ 1 1 1 1 1 c k c k k k r r r e ( ) ( ) ( )



_









−

=

₋ 1 3 4 1 1 3 21 1 1 4 1 1 4 1 . . . e 6 . .6137 61 37 0

=

e

)AS +GRBI=E ;O"ER ;LA=+

,&.A 0as tur9ine plant wor*in0 on a #losed air3standard Bra%ton #%#le has a pressure ratio of ,. +he heat reFe#ted in the heat sin* is 2 *". If one3>fth of the expansion wor* is supplied for

#ompression ideall%7 determine the theoreti#al power of the 0as tur9ine.

A. 1460 kW B. -5 *" '. 12& *" D. 1&5 *" Solution( 5

=

p r

( )

( ) 3686 0 5 1 1 1 1 4 1 1 4 1 1 . . .

=

−

=

−

=

₋ ₋ k k p r η A R A A net , , , , %

−

=

η k% k% , ,_A R 3168 3686 0 1 2000 1

−

=

−

=

. η

(

)( )

k , % _net

=

η _A

=

0.3686 3168

=

1168 5 / 5 / 5 / 2 1 5 / , 5 , 1 − − − − −

−

=

−

=

% % % % % % _net ( ) k% % 1168 1460 4 5 4 3−

=

,-.A 0as tur9ine power plant operatin0 in a Bra%ton #%#le7 the #ompressor power needed is 2, *" whi#h is drien 9% a 1 *" tur9ine. +he #ompressor e@#ien#% is 6 and that of the tur9ine is 6. Determine the 9a#* wor* ratio.

A. /6 B. /56

(27)

C. 25%

D. 56 Solution(

Ba#* wor* ratio  #ompressor power 4 tur9ine power  2, *w 4 1 *w  .2,  2,6

,$.A 0as tur9ine plant utiliin0 a #losed #%#le has the followin0 temperatures at ea#h state point(

+ 1   <7 + 3  , <7 + 2  155 <7 and + 4  ,/ <. Based on these alues7 #al#ulate the

thermal e@#ien#% of the #%#le.

A. 48% B. 5,6 '. 5&6 D. 5-6 Solution( 48 0 590 1440 90 530 1 1 2 3 1 4 .

=

−

=

−

=

+ + + + e or 5$6

,.In a simple 0as tur9ine plant wor*in0 on the ideal #onstant pressure #%#le7 air is ta*en into the #ompressor at 1 9ar7 1& ' and deliered at ,.5 9ar. If the temperature at tur9ine inlet is - '7 #al#ulate the ideal thermal e@#ien#%. +a*e *  1.5.

A. /,.2$6 B. 52.&16 C. 38.23% D. 5,.&&6 Solution( 2 3 1 4 1 + + + + e_t

−

=

Solin0 for 2 + 7 3 + 7 and 4 + ( k k p p + + 1 1 2 1 2 −













=

/ + ₁

=

16

+

273

=

289 4 1 1 4 1 2 1 4 5 289 . . . −













=

+ / + ₁

=

467.90 k k p p + + 1 3 4 3 4 −











=

/ + ₃

=

700

+

273

=

973 2&

(28)

4 1 1 4 1 4 1 4 5 973 . . . −













=

+ / + ₄

=

600.98 +hen: 6 . . . . 23 38 3823 0 9 467 973 289 98 600 1

=

−

=

t e

&.+he net power output of an air3standard Bra%ton #%#le is 2 *". Air enters the #ompressors at /2 ' and leaes the hi0h3temperature heat ex#han0er at $ '. "hat is the mass Now rate of air if it leaes the tur9ines at /, '.

A. .,- *04s B. .&- *04s C. 0.87 kg/$ D. .-- *04s Solution( c t n % % %

=

−

Solin0 for t % and c % . / + ₁

=

32

+

273

=

305 / + ₃

=

800

+

273

=

1073 / + ₄

=

350

+

273

=

623 4 3 1 2 + + + +

=

623 1073 305 2

=

+ / + ₂

=

525.30

(

+ 2 + 1

)

mc % _c

=

_p

−

(

)

(

)

m m % _c

=

1 525.30

−

305

=

220.3

(

+ 3 + 4

)

mc % _t

=

_p

−

( )

(

)

m m % _t

=

1 1073

−

623

=

450 +hen: m m 220 3 450 200

=

−

.

(29)

2-s kg m

=

0.87

&1.Air enters the #ompressor of a 0as tur9ine at 11 *;a and 2- ' with a olume Now rate of $., m/_{4se#. +he #ompressor pressure ratio is 1 and its isentropi# e@#ien#% is $26. At the inlet to}

the tur9ine7 the pressure is , *;a and the temperature is 15 . +he tur9ine has an isentropi# e@#ien#% of $,6 and the exit pressure is 1 *;a. On the 9asis of an air standard anal%sis7 what is the thermal e@#ien#% of the #%#le in per#ent8

A. 27.3 B. 22. '. 21.$ D. 2.2 Solution( p1  11 *;a7 + 1  2-  2-/  /  . 1  $., m/4se# p34 p1  17 η c  .$2 p2  , *;a7 p4  1 *;a7 η t  .$, + 2  15 













−













−

=

− 1 1 1 1 2 1 1 k k c p p k . kp % ( )( ) ( )









₋

−

=

10 − 1 1 4 1 5 8 110 4 1 4 1 1 4 1 . . . . . c % kw % _c

=

3045.7 kw kw % % c c c ₀₈₂ 3714 3 7 3045 . . .

=

′

η

























−

=

− k k t p p k . kp % 1 3 4 3 3 1 1

























−













−

=

− k k t p p + + k . kp % 1 3 4 1 3 1 1 1 1 ( )( )

























−













−

=

− 4 1 1 4 1 950 100 1 300 1400 1 4 1 5 8 110 4 1 . . . . . t % kw % _t

=

7245.0 ( )

(

kw

)

kw % % _t

′

=

η _t _t

=

0.85 7245.0

=

6158.3 kw % % % _net

′

=

_t

′

−

_c

′

=

6158 .3

−

3714.3

=

2444

(

+ 3 + 2

)

mc ,_A

=

_p

−

2$

(30)

(

+ 3 + 2

)

mc ,_A

=

_p

−

k k p p + + 1 1 2 1 2 −

  







=

( )

























−

=

− k k A p p + + + k . kp , 1 1 2 1 3 1 1 1 1 ( )( ) ( )( ) ( )









₋

−

=

14− 1 4 1 10 300 1400 300 1 4 1 5 8 110 4 1 . . . . . A , kw ,_A

=

8953.5

(

6

)

. 6 .5 100 2730 8953 2444

=

′

=

kw kw , % e A net

&2.Air enters a 0as tur9ine power plant at 1 9ar and / ' with a olume Now rate of , m /_{4s. +he}

#ompressor pressure is 1 and its isentropi# e@#ien#% is $6. +he tur9ine inlet pressure and temperature are 1 *;a and 1/ '7 respe#tiel%. +he tur9ine isentropi# e@#ien#% is $,6 and the exit pressure is 1 *;a. Determine the thermal e@#ien#% of the #%#le in per#ent8 A. ,2.6 B. /.,6 C. 29.5% D. 2.,6 Solution( p1  1 9ar  1 *;a + 1  /  2-/  //  . 1  , m/4s p34 p1  1 η c  .$ p2  1 *;a + 2  1/  2-/  1,-/  ht  .$, p4  1 *;a













−













−

=

− 1 1 1 1 2 1 1 k k c p p k . kp % ( )( ) ( )









₋

−

=

10 − 1 1 4 1 5 100 4 1 4 1 1 4 1 . . . . c % kw % _c

=

1628.7 kw kw % % c c c 2035 9 80 0 7 1628 . . .

=

′

η 2

(31)

























−

=

− k k t p p k . kp % 1 3 4 3 3 1 1

























−













−

=

− k k t p p + + k . kp % 1 3 4 1 3 1 1 1 1

( )( )

























−













−

=

− 4 1 1 4 1 1000 100 1 303 1573 1 4 1 5 100 4 1 . . . . t % kw % _t

=

4379.4 ( )

(

kw

)

kw % % _t

′

=

η _t _t

=

0.85 4379.4

=

3722.5 kw % % % _net

′

=

_t

′

−

_c

′

=

3722.5

−

2035.9

=

1686.6

(

+ 3 + 2

)

mc ,_A

=

_p

−

(

+ 3 + 2

)

mc ,_A

=

_p

−

k k p p + + 1 1 2 1 2 −

  







=

( )

























−

=

− k k A p p + + + k . kp , 1 1 2 1 3 1 1 1 1 ( )( ) ( )( ) ( )









₋

−

=

14− 1 4 1 10 303 1573 303 1 4 1 5 100 4 1 . . . . A , kw ,_A

=

5706.3

(

6

)

. 6 . . 56 29 100 3 5706 6 1686

=

′

=

kw kw , % e A net

&/.A -, !" stationar% 0as tur9ine power plant has air enters the #ompressor at 1 *;a7 /  temperature and pressure ratio of 1. +he tur9ine inlet temperature is 1&, . Gsin0 the standard Bra%ton #%#le7 determine the 9a#*3wor* ratio of the 0as tur9ine power plant.

A. .2,, B. .$- '. 1.&1 D. 0.351 Solution( p1  1 *;a + 1  /  + 2  1&,  p34 p1  p24 p4  1 /

(32)

( )

14 1 4 1 1 1 2 1 2 30010 . . − −

=

 











=

k k p p + + / + ₂

=

579.2 k k p p + + 1 4 3 4 3 −

 











=

( )

14 1 4 1 4 10 1650 . . −

=

+ / + ₄

=

854.6 4 3 1 2 + + + + ratio work Back

−

=

−

6 854 1650 300 2 579 . .

−

=

−

work ratio Back 351 0.

=

−

work ratio Back

&5.A 27 !" 0as tur9ine operatin0 in the simple open #%#le has an exhaust 0as Now of 1 *0 per se#ond. +he 0as enters a waste heat re#oer% 9oiler at 5, ' and leaes at 1$ '7 spe#i># heat of the 0as is .$, *J4*037 heat reuired to produ#ed 1 *0 of steam at 5., !;a and /2 ' from feedwater at 11 ' is ,., *J. 'al#ulate the uantit% of steam that #an 9e produ#e in *0 per se#ond.

A. 4173 kg/$

B. 1-2/ *04s '. 21-/ *04s D. /1-/ *04s Solution(

?eat reFe#ted  heat a9sor9ed

(

t t

)

m h mc_p 4

−

5

=

_s

∆

( )( 085 450 180) ( )5 5 100 .

−

=

m_s . se -2 . 51-2 kg m_s

=

&,.A simple Bra%ton #%#le whi#h uses helium as the wor*in0 Nuid7 has a maximum temperature of 11, 7 and a pressure ratio of 2.. At the start of the #ompression7 the helium pressure and temperature are  *;a and / . Based upon #old air3standard anal%sis assumptions7 determine the thermal e@#ien#% of the #%#le in per#ent8

A. 34.7%

B. 1-.,6 '. 2/.-6 D. 2$.-6 Solution(

k  1.&&- for helium

p1   *;a

(33)

+ 1  /  + 2  11,  p34 p1  p24 p4  2.

(

)

1667 1 667 1 1 1 2 1 2 300 29 . . . − −

=

 











=

k k p p + + / + ₂

=

459_.34 k k p p + + 1 4 3 4 3 −

 











=

(

)

1667 1 667 1 4 9 2 1150 . . . −

=

+ / + ₄

=

751_.07

(

+ 2 + 1

)

mc % _c

=

_p

−

(

+ 3 + 4

)

mc % _t

=

_p

−

c t net % % %

=

−

(

+ 3 + 2

)

mc ,_A

=

_p

−

(

) (

)

2 3 1 2 4 3 + + + + + + , % e A net

−

=

( ) ( ) ( 6) . 6 . . . 69 34 100 34 459 1150 300 34 459 07 751 1150

=

−

=

e

&&.+he mass Now rate of the 0as in a 0as tur9ine is / *04se#. +he spe#i># enthalp% and elo#it% in the inlet are 1/ *J4*0 and 2 m4se# respe#tiel% while in the outlet are / *J4*0 and , m4s7 respe#tiel%. 'al#ulate the power output in *w of the tur9ine if there is a heat loss of / !". A. /27&-$., *w B. 27,562.50 kw '. /-712,., *w D. 5,7,., *w Solution(

(

)

(

)

( ) L t , . . m h h m %

=

−

+

−

1000 2 2 1 2 2 1 2 ( )

[

( ) ( )

]

( 1000) 3000 2 50 200 30 300 1300 30 2 2

−

+

−

=

t % k% % _t

=

277562.5 /2

(34)

&-.An ideal 0as tur9ine operates with a pressure ratio of -., and temperature limits of 2, ' and 11 '. +he ener0% input in the hi0h temperature heat ex#han0er is , *w. Determine the air Now rate in *04hr.

A. ,./ B. 2135 '. 2/, D. /5, Solution( p34 p1  -., + 1  2,  2-/  2$  + 2  11  2-/  1/-/  k k p p + + 1 1 2 1 2 −











=

( ) 14 1 4 1 2 5 7 298 . . . −

=

+ / + ₂

=

530.0

(

+ 3 + 2

)

mc ,_A

=

_p

−

( )( 10 1373 530) 500

=

m .

−

hr kg s kg m

=

0.593

=

2135

&$.Air at 5 *;a and ,  is extra#ted from a Fet en0ine #ompressor to 9e used for the 0eneration of auxiliar% power for the #a9in. +he extra#ted air is #ooled in a #onstant pressure heat ex#han0er down to 5& . It then enters an isentropi# tur9ine and expands to 1 *;a 9efore 9ein0 reFe#ted into the #a9in. If the mass Now is 1, *04min7 determine the power deeloped 9% the tur9ine in *w.

A. &/.- *w B. -/.& *w C. 37.6 kw D. -&./ *w Solution( + 3  , 7 + 2  5&  p4  1 *;a7 p2  5 *;a m  1, *04min k k p p + + 1 / 5 / 5 −

  







=

/ + /.,& 5 1 5& 1.5 1 5 . 1 5



=











=

−

;ower deeloped 9% the tur9ine  mc pC+ 2H + 4

 C1,4&C1.C5& H /.,&  /-.&1 *"

&.+he mass Now rate of the 0as in 0as tur9ine is 5 *04se#. +he spe#i># enthalp% and elo#it% in the inlet are 1& *J4*0 and 2 m4se# respe#tiel% while the outlet are , *J4*0 and , m4se#7 respe#tiel%. 'al#ulate the power output in *w of the tur9ine if there is a heat loss of 1., !".

A. /-75&2