work ratio - Lecture 05 Problems

−

  =



 



= 

p + p +

/ + 2

=

⁵⁷⁹.²

k k

p p + +

4 3

−

  

 



= 

( )¹¹⁴⁴¹

1650 10

. . −

=

/ + 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

−

=

∆

( )( 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

+ 1  / 

&&.+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

&-.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 2

−

 



 

= 

( )¹¹⁴⁴¹

2 7 5

298

. .

. ⁻

=

/ + 2

=

⁵³⁰.⁰

(

+ 3 + 2

)

mc ,_A

=

−

( )( ¹⁰ ¹³⁷³ ⁵³⁰)

500

=

^m ^.

−

hr kg s

=

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 + +

/ 5 /

−

  

 

= 

+ ^/^.^,&

5

5&1 ¹^.⁵

1 5 . 1

 =



 



= 

−

;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

B. 57$-2 C. 43,250 D. /57&, Solution(

( )

( ) ^L

t m. . ,

h h m

= − + − −

1000 2

2 1 2 2 1

( )

[

( ) ( )

]

( ¹⁰⁰⁰) ¹⁵⁰⁰

50 200

500 40 1600 40

2 2

− − +

−

=

% _t

=

⁴³7²⁵⁰

-.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. 1&,

B. 1$,

C. 1325 D. 1-/2 Solution(

p34 p1  

+ 1  2,  2-/  2$  + 2  11  2-/  1/-/ 

k k

p p + +

1 2

−

 



 

= 

( )¹¹⁴⁴¹

2 9

298 ^.

. −

=

/ + 2

=

⁵⁵⁸.³

(

+ 3 + 2

)

mc ,_A

=

−

( ) ¹⁰ ( ¹³⁷³ ⁵⁵⁸ ³)

300

=

m .

−

hr kg s

=

⁰^.³⁶⁸²

=

¹³²⁵^.⁵

S+EA! ;O"ER ;LA=+

-1.+he ener0% input to the tur9ine is 2$ *J4*0 and the ener0% at the exit is 15 *J4*07 e@#ien#% of the en0ine is -,6 and power output at full load is & *". "hat is the tur9ine Now rate at full load in *04*"3hr8

A. ,.-B. 5.//

C. 3.43 D. .2/

Solution(

% t

rate m 5ow

+urbine

=

³⁶⁰⁰

( h h ) _t

m rate m 5ow +urbine

2 1

3600

= −

(

h h

)

rate 5ow +urbine

2 1

3600

= −

( ²⁸⁰⁰ ¹⁴⁰⁰)( )⁰⁷⁵

3600

−

=

rate 5ow +urbine

k%h kg

rate 5ow

+urbine

=

³.⁴³

-2.A water3tu9e #ondenser has a total of  tu9es. If these t%pe of #ondenser are two passes7 then #ompute the num9er of tu9es per pass.

A. 45 &'($

B. / tu9es '. & tu9es D. 1$ tu9es Solution(

<or two pass. +he num9er of tu9es per pass   tu9es 4 2 pass  5, tu9es

-/.A steam #ondenser re#eies 1 *0s per se#ond of steam with an enthalp% of 2,- *J4*0.

Steam #ondensers into a liuid and leaes with an enthalp% of 1& *J4*0. 'oolin0 water passes throu0h the #ondenser with temperature in#reases from 1/ de0ree ' to 25 de0ree '.

'al#ulate the #oolin0 water Now rate in *0s per se#ond.

A. ,//.2 B. ,1$.5 '. ,2$.

D. 523.2 Solution(

(

h h

)

m_wc_pw t _w m² ²

−

= ∆

(²⁵⁷⁰ ¹⁶⁰) (⁴¹⁸⁷ )(²⁴ ¹³)

− =

m_w ^.

−

s kg m_w

=

⁵²³^.³

-5.In a steam Ran*ine #%#le7 saturated liuid at 15.- psia Cf  .1&-2 #u. ft. per l97 is pump to a hi0h pressure liuid at 1 psia. ?ow mu#h wor* is reuired to pump one l9m of water8 A. 11$ Btu

B. 1$., Btu '. .12 Btu D. 0.264 B&

Solution(

i f f

p mv p p

= −

( ) ( ) ⁽ ⁾ ( )

Btu lb ft

ft in psia

lb ft

% _p lb

−

= −

778

144 7

14 100 01672

1 . ³ . ² ²

Btu

% _p

=

⁰^.²⁶⁴

-,.A steam power plant has a tur9ine wor* of /2 *J4*0 and the pump wor* is 2 *J4*0. If the mass Now rate is 1 *0 per se#ond7 #ompute the power produ#ed 9% this plant8

A. /2 *w B. /22 *w C. 3180 kw D. 2$ *w Solution(

p t %

= −

( ) k%

=

¹⁰ ³²⁰

−

=

³¹⁸⁰

-&.A tur9ine has an e@#ien#% of -,6 and has a steam Now rate of 227 *04hr. If the aaila9le ener0% is 12 *J4*07 determine the *ilowatt output at full load.

A. 5500 B. /,

'. 5,

D. 5

Solution(

( ) _t

t mh h

=

−

2η

( )( ) k%

% _t ¹²⁰⁰ ⁰ ⁷⁵ ⁵⁵⁰⁰

3600 000

 =



 



= 

⁷ ^.

--.+he aaila9le ener0% of a tur9ine is 15, *J4*07 e@#ien#% of the en0ine is -6 and the power output at full load is ,., !". "hat is the en0ine Now rate at full load in *04*"3hr8

A. 2., B. 3.5 '. /.

D. 2.-, Solution(

% t

rate m 5ow

!ngine

=

³⁶⁰⁰

(

h h

)

m rate m 5ow

!ngine

2 1

3600

= −

( h h ) _t

rate 5ow

!ngine

2 1

3600

= −

( ¹⁴⁵⁰)( )⁰⁷⁰

3600

=

. rate 5ow

!ngine

k%h kg

rate 5ow

!ngine

=

³^.⁵⁴

-$.A steam plant operates with an initial pressure of 1.- !;a and /- ' temperature and to a

#ondenser at 1- *;a. If the tur9ine e@#ien#% and 9oiler e@#ien#% are -,6 and $,67 respe#tiel%7 determine the #%#le thermal e@#ien#%.

Steam properties(

At 1.- !;a  /- '(

h  /1$-.1 *J4*0 s  -.1$1 *J4*0 At 1- *;a(

hf  5$/.2

hfg  221&.

sf  1.5-,2 sfg  ,.-&2 A. 2,.&6 B. ,2.&6 C. 12.1%

D. /,.&6 Solution(

kg k-h1

=

³¹⁸⁷.¹

/ kg k-s

=

⁷.¹⁰⁸¹

−

f xs

s s2

= +

( ⁵⁷⁰⁶²)

4752 1 1081

=

+

x .

987 0.

=

f xh

h h2

= +

( ²²¹⁶ ⁰)

987 0 2

=

483.

+

. .

kg k-h2

=

²⁶⁷⁰.⁶

( ) _t

t h h

=

−

2 η

(³¹⁸⁷^.¹

−

²⁶⁷⁰^.⁶)(⁰^.⁷⁵ )

=

k-% _t

=

³⁸⁷.³⁶

(

4 3

)

3 p p

% _p

= −

Assume water densit% at 1 *04m^/. kg

k-% _p ¹⁵³

1000 170 1700

− =

=

4 h

% _p

= −

2 483 53

=

h₄

−

/-kg k-h4

=

⁴⁸⁴.⁷³

b A

h , h

4 1

−

=

85 0

73 484 1 3187

. . .

−

=

kg k-,_A

=

³¹⁷⁹^.²⁵

A p t

% e %

−

=

25 3179

53 1 36 387

. . .

−

=

6 . .¹²¹ ¹²¹

=

-.=a0a priate power plant is usin0 water as the wor*in0 Nuid that operates on an ideal Ran*ine #%#le. Superheated apor enters the tur9ine at 17 *;a and ,2 ' and is exhausted into the #ondenser at $ *;a. +he net output of the Ran*ine #%#le is 17 *".

Determine the mass Now rate of steam in *0 per se#ond.

Steam properties(

At 17 *;a C1 9ar and ,2 '(

h  /52,.17 s  &.&22 At $ *;a(

hf  1-/.$$7 sf  .,2&

hg  2,--7 sg  $.22$-v f  .1$5

A. &- *04se#

B. 2&- *04se#

C. 74.4 kg/$(c D. 2.1 *04se#

Solution(

kg k-h1

=

³⁴²⁵.¹

/ kg k-s

=

⁶.⁶²²

−

f xs

s s2

= +

( ⁸²²⁸⁷ ⁰⁵⁹²⁶)

5926 0 622

=

+

x .

−

7896 0.

=

f xh

h h2

= +

( ²⁵⁷⁷ ¹⁷³⁸⁸)

7896 0 88

=

173.

+

−

kg k-h2

=

²⁰⁷¹.³⁸

1 h

m h

% _t

−

=

kg m

k-% _t

72 1353 38

2071 1

3425.

−

=

( 4 3)

3 p p m v

% _p

−

=

( )( ) k- kg

% _p

08 10 8 000 10 0010084

0. 7

− =

=

% m

=

−

08 10 72 000 1353

1007 . .

−

=

se . kg m

=

⁷⁴ ⁴

$.A 1,7 *" steam tur9ine30enerator power plant has full load steam rate of -., *04*"3hr. =o load steam #onsumption is around 16 of full load steam #onsumption. 'al#ulate the hourl%

steam #onsumption at three3fourths load in *04hr.

A. 1127, *04hr B. 1,7, *04hr

C. 87,187.50 kg/"r D. &7-, *04hr

Solution(

b kx m_s

= +

At full load

( ) lbhr

m_s

=

⁷.⁵¹⁵7⁰⁰⁰

=

¹¹²7⁵⁰⁰

At no load

( ) lbhr

=

⁰.¹⁰¹¹²7⁵⁰⁰

=

¹¹7²⁵⁰

+hen7

000 157

=

x b kx m_s

= +

( ¹⁵⁰⁰⁰) ¹¹²⁵⁰

500

1127

=

^k 7

+

75 6.

=

/

At P load7

( )( ) k%

=

¹⁵7⁰⁰⁰ ⁰.⁷⁵

=

¹¹7²⁵⁰

( ¹¹²⁵⁰) ¹¹²⁵⁰

6. 7

+

=

hr kg m_s

=

⁸⁷⁷¹⁸⁷^.⁵

$1.A t%pi#al steam 0enerator with an e@#ien#% of $,6 is produ#in0 1$ *04se# of steam. +he enthalp% of the enterin0 water is 15 *J4*0 and is superheated to an enthalp% of //.,

*J4*0. +he fuel used has a heatin0 alue of 557, *J4*0. +he A4< ratio 9% wei0ht is 2.

Determine the amount of air needed in *04hr.

A. 1&$7/1-B. 1-$7/2$

'. &712/

D. 108,288 Solution(

( )

.

m h h m

f s b

1 2

−

=

( )( )

( ⁴⁴⁵⁰⁰)

140 5 3300 3600

85 18

0 7

. .

= −

hr kg m_f

=

⁵⁴¹⁴^.⁴²

( ) kghr

m r

m_a

=

_fa _f

=

²⁰ ⁵⁴¹⁴^.⁴²

=

¹⁰⁸⁷²⁸⁸

$2.A 127 *" steam tur9ine30enerator power plant has full load steam rate of ,., *04*"3hr. =o load steam #onsumption is around 16 of full load steam #onsumption. 'al#ulate the hourl%

steam #onsumption at two3third of load in *04hr.

A. 572,

B. /&7$/

C. 46,200 D. /,7$2

Solution(

b kx m_s

= +

At full load

( ) lbhr

m_s

=

⁵^.⁵¹²⁷⁰⁰⁰

=

⁶⁶⁷⁰⁰⁰

At no load

( ) lbhr

=

⁰^.¹⁰ ⁶⁶⁷⁰⁰⁰

=

⁶⁷⁶⁰⁰

+hen7

000 127

=

x b kx m_s

= +

( ⁶⁶⁰⁰) ¹² ⁰⁰⁰

000

667

=

+

95 4.

=

At P load7

( )( ) k%

=

¹²7⁰⁰⁰ ² ³

=

⁸7⁰⁰⁰

5

( ⁸⁰⁰⁰) ⁶ ⁶⁰⁰

4. 7

+

=

hr kg m_s

=

⁴⁶⁷²⁰⁰

$/.A #ondenser 0au0e reads 2,., in ?0 a#uum when the 9arometer stands at 1./2 m of ?2O.

Determine the a9solute pressure in the #ondenser in mm ?0.

A. -$. mm ?0 B. 2/,.& mm ?0 '. 1,-.& mm ?0 D. 111.6 mm )g Solution(

  

 



− 

=

ing

g

g mm in p_g

. . .

92 29 5 760

g mm p_g

= −

⁶⁴⁷^.⁷³

(

1032m2#

) (

981k&m³

)

p_atm

=

. . kPa p_atm

=

¹⁰¹^.²⁴

  

 



= 

kPa

g kPa mm

p_atm

325 101 24 760

101. .

  

 



= 

kPa

g kPa mm

p_atm

325 101 24 760

101. .

atm g

abs p p

= +

4 759 73

647.

+

−

abs

=

g mm p_abs

=

¹¹¹.⁶⁷

$5.Steam at a pressure of 1 !;a Chf  -&$.$1 *J4*07 hfg  21,./ *J4*0 in the main steam line is passed throu0h a throttlin0 #alorimeter with an a9solute pressure of 11 *;a and a temperature of 1, ' Ch  2&$,.1, *J4*0. 'al#ulate the steam ualit%.

A. 95.39%

B. /.,6 '. .,/6 D. 1./,6 Solution(

6 . .

. 9509

3 2015

81 768 15

2685

− =

=

fg f

h h x h

$,.A steam tur9ine is powered 9% a set of >xed ori>#e noles. Ea#h has an isentropi# e@#ien#%

of 6. 1-&.&- ' sat. steam Chg  2--/.1 *J*0 at $,.5 m4s enters the noles. +he steam expands adia9ati#all% to ,//., m4s. "hat is the enthalp% of the steam as it leaes the noles8

A. 2&1 *J4*0 B. 2&5/ *J4*0 C. 2634 k*/kg D. 2&1 *J4*0

Solution(

( ) ¹⁰⁰⁰ ²( )¹⁰⁰⁰

2 2 2

2 1 1

h v h

+

= +

( )

( ¹⁰⁰⁰)

2 5 533 1000

2 4 10 85

2773

2 .

+

=

+

kg k-h2

=

²⁶³⁴

$&.An industrial plant operates a , *" tur9ine hain0 an en0ine e@#ien#% of -/6. +he initial steam #onditions are at /2.5 *J4*0 enthalp% and &.52 *J4*03 entrop%. +he 9a#* pressure is /.5 *;a Csf  ./$ *J4*037 sg  $.,25& *J4*037 hf  1./2 *J4*07 hg  2,5&.$2 *J4*0. +he tur9ine steam rate in *04*w3hr is7

A. /./2 B. 4.54 '. /.2/

D. 5.5, Solution(

/ kg k-s1

=

⁶⁴²⁰⁰.

−

/ kg k-s

=

⁶.⁴²⁰⁰

−

7415 0 3818 0 5246 8

3818 0 4200

2 6

. . .

=

−

= −

−

= −

f g

s s

s x s

( ²⁵⁴⁶ ⁸² ¹⁰⁹³²)

7415 0 32

=

^h_f

+

^x^h_g

−

^h_f

=

109.

+

. .

−

kg k-h2

=

¹⁹¹⁶.⁷³

kg k-h1

=

³⁰⁰².⁴

( )

(

k- kg

)

^kg^k% ^hr

hr rate s

steam

= −

⁴ ⁵⁴

73 1916 4

3002 73 0

3600 .

. .

$-.A water tu9e 9oiler has a heatin0 surfa#e area of , m². <or a deeloped 9oiler hp of $2,.

Determine the per#ent ratin0 of the 9oiler.

A. 1&2.5,6 B. 15$.$6 C. 153.37%

D. 152.1$6 Solution(

, m²  ,/$/ ft² De. Bo. ?p.  $2,

Rated Bo. ?p.  ,/$/ 4 1  ,/$./

;er#ent ratin0  De. Bo. ?p. 4 Rated Bo. ?p.  C$2, 4 ,/$./ C16  1,/.2&6

$$.1 liters per min of hot water at $2 ' Cv f 1./, x 1^3/ m^/4*0 is produ#ed in a Now s%stem 9% inFe#tion and #ondensation of low pressure steam at 1/$.&-, *;a0 and $6 ualit% into

#old water at 1& '. 'al#ulate the steam Now rate in *04min. Q 1/$.&-, *;a07 hf  ,2.&,

*J4*07 h0  2-1,. *J4*0.

A. /.&

B. 15.-, C. 2.80

D. 1.5, de#reased to .1 !;aa. If its dr%ness fra#tion is 67 what is the >nal olume of the steam in L8 Q 2., !;aa  5 '7 v  12,.2 x 1^3/ m^/4*0. Q .1 !;aa7 v f  1.12 x 1^3/ m^/4*07 v g  157&-5 x 1^3/ m^/4*0.

A. 264 B. 1&5 '. /&5 D. 5&5 Solution(

.+he euialent eaporation of a 9oiler7 from and at 1 ' is 1, *0 steam per *0 fuel and the

#alori># alue of the fuel 9urned is 51. !J4*0. "hat is the 9oiler e@#ien#%8 Q 2., !;aa  5 ': hf0  22,- *J4*0.

1.In a steam power plant7 the #apa#it% is 2 !" and ,6 is for auxiliaries. If the ener0% #har0ed to the station is 1 !" and the 9oiler e@#ien#% is -,67 the 0ross station heat rate in *J4*"3 hr is .

A. 15721

B. 1/7,

'. 1$7,

D. 18,000 Solution(

( )( )

hr k%

k-hr k%

k-rate $%

heat station

gross

− = −

=

¹⁸⁰⁰⁰

20 3600 100

2.+he power deeloped 9% the steam tur9ine is 2$ *" while the power reuired 9% the feedwater pump is 12 *". If the heat supplied to the 9oiler is /, *J4*0 and the heat reFe#ted from the #ondenser is 22 *J4*07 >nd the mass Now rate of the steam in *04s.

A. 2.1 B. 1.-'. 2., D. /.1 Solution(

s kg

m ² ¹⁴⁵

2200 3500

12 2800

=

−

= −

/.Determine the wor* of a pump in *J reuired in raisin0 the pressure of 1 *0 of water from 2

*;aa to / !;aa7 assumin0 that the spe#i># olume of saturated water at 2 *;aa is 1.1- x 1^3/ m^/4*0.

A. 30.3 B. 2.&, '. /./

D. 2&., Solution(

( )

(

m kg

) (

kPa

)

k- p mv

= ∆ =

11.1 -

×

1⁻^/ ^/ /

−

2

=

/./

5.)ien the steam pressure of  l94ft²7 temperature of / <7 spe#i># olume of ,.$ ft^/4l9. If the spe#i># enthalp% is , ft3l94l9. "hat is the internal ener0% per pound of the s%stem8 A. 55 ft3l9

B. / ft3l9 C. 4280 +!' D. /5 ft3l9 Solution(

( ) ( ) ft lb lb pv

= − =

⁹⁵⁰⁰

−

⁹⁰⁰ ⁵^.⁸

=

⁴²⁸⁰

−

,.Determine the heat transferred to the #oolin0 Nuid in a #ondenser operatin0 under stead%

Now #onditions with steam enterin0 with an enthalp% of 2/ *J4*0 and a elo#it% of /, m4s.

+he #ondensate leaes with an enthalp% of 1& *J4*0 and elo#it% of - m4s.

A. 312 *J4*0 B. 21 *J4*0 C. !2199 k*/kg D. 21 *J4*0 Solution(

( ¹⁰⁰⁰)

2 1 2 2 1 2

. h .

= − + −

( ) ( ) ( ¹⁰⁰⁰)

350 2300 70

160

2 2

−

+

−

=

( ) ( ) ( ¹⁰⁰⁰)

350 2300 70

160

2 2

−

+

−

=

kg k-,

=

²¹⁹⁹

−

&.Steam at the rate of , *04hr is produ#ed 9% a 9oiler from 5 ' feedwater. If the enthalp% of the steam is 2,12 *J4*07 >nd the hourl% rate of heat reuired.

A. 171,17/-

B. 171/572-

C. 1,172,264 D. 1712172-5 Solution(

( ) ( t )( ) k-kg

h_f

=

_pw

∆ =

⁴.¹⁸⁷ ⁴⁰

−

⁰

=

¹⁶⁷.⁴⁸ kg

k-h1

=

²⁵¹²

hr kg m^ _s

=

⁵⁰⁰

(

)

A m h h

,

=

 1

−

( )( ) k- hr

,^_A

=

⁵⁰⁰ ²⁵¹²

−

¹⁶⁷^.⁴⁸

=

¹⁷¹⁷²⁷²⁶⁰

-.A simple Ran*ine #%#le operates 9etween superheated steam at & !;a. & ' enterin0 the tur9ine7 and 1 *;a enterin0 the pump. Determine the maximum possi9le #%#le thermal e@#ien#%. Steam properties( at & !;a7 & '7 h  /&,$7 s  -.1&$,: at 1 *;a7 hf  127 hfg  2//7 sf  .&57 sfg  -.,27 f  .1 m^/4*0.

A. /,.

B. /-., C. 40.0 D. 52., Solution(

=

3658

( ⁷⁵⁰²)

649 0 1685

7 ₂

2 s . . x .

= = = +

869

=

0. x

( ²³⁹³) ²²⁷¹ ⁵²

869 0

2 192

=

+

x hfg

= +

=

192

=

^h ^at10^kPa

=

h _f

( 4 3) ¹⁹² ⁰⁰⁰¹( ⁶⁰⁰⁰ ¹⁰) ¹⁹⁷ ⁹⁹

3 3

=

+

v p

−

= +

− =

(

4 3

)

⁰⁰⁰¹( ⁶⁰⁰⁰ ¹⁰) ⁵⁹³

− =

=

^v^p ^p

% _p

6 . . .

. 039894 39894 99

197 3658

93 5 52 2271 3658

4 1

= =

−

= −

−

= −

h h

% h

e h ^p

$.An open feedwater heater utilies saturated steam at 1, ' extra#ted from the steam tur9ine. <eedwater to 9e heated enters at & '. If the mixture leaes the heater at the rate of 1$ *04hr7 #al#ulate for the uantit% of steam extra#ted from the tur9ine7 in *04hr. Steam properties at 1, '7 hf  &/2.27 hfg  2115./: at & '7 hf  2,1.1/.

A. 2-5 B. 2749 '. 52-

D. 5-2 Solution(

hr kg m_s

=

¹⁸⁷⁰⁰⁰

Ener0% 9alan#e

( ) 5 6

2 m m h mh

m_e

+

−

=

At 1, '

5 2746 3

2114 2

=

^h_f

+

^h_fg

=

632.

+

=

. h

=

632. h

At & '

=

251. h

( )

hr h kg

h h h

m_e m^s ²⁷⁴⁸ ⁷⁹⁵

13 251 5 2746

13 251 2 632 000 18

5 2

6 .

. .

=

−

= −

−

= −

.A steam 0eneratin0 unit produ#es steam at the rate of 1 *04s at , !;a7 5, ' whi#h is

#ontinuousl% 9lown down at .2, *04s. <eedwater enters the e#onomier at 1 '. 'oal with a heatin0 alue of 527 *J4*0 as >red is 9urned at the rate of 1 *04s. "hat is the oer3all e@#ien#% of the steam 9oiler8 Steam properties at , !;a7 5, '7 h  //1&.2: at , !;a7 1 '7 h  522.-2: at , !;a7 hf  11,5.2/7 hg 2-5./.

A. 69.33 B. &/./

'. &&./&

D. -2./

Solution(

23 1154

=

^h ^at ^$Pa

h_bo _f

Oer3all 9oiler e@#ien#%7 eo. ( . )

h h m h

h e m

fw bo bo fw

s s o

− +

= −

( ) ( )

( )( ⁷ ) ⁶

. .

. 100

000 42 1

72 422 23 1154 25 0 72 422 2 3316

− + − ×

=

6 .³³

=

C#.

1. A horiontal return tu9ular 9oiler with a heatin0 surfa#e of 2 m² 0enerates steam at the rate of 5., *04s. <eedwater enters at , !;a7 12 ' and steam leaes at , !;a7 /, '.

'al#ulate for the per#ent ratin0 deeloped if #oal with a heatin0 alue of 27 *J4*0 is 9urned at the rate of  !tons4hr. Steam properties at , !;a7 /, '7 h  /&$.5: at , !;a7 12

'7 h  ,-..

A. &2.5&

B. &&.-$

C. 64.62 D. &$.5 Solution(

<or >re tu9e 9oiler

18 1818 1

1 2000

2 2

=

m k

surface heating

total hp

bo rated

Deeloped 9oiler hp(

( )( )

91 1174 316

09 507 4 3068 3600

5 4 316

35 .

. .

− = − =

=

k-hr

h h hp m

dev ^s ^s ^fw

;er#ent ratin0

6 . .

6 . 100 6462

18 1818

100

=

1174

× =

×

=

ratedbohp hp bo rating dev

11. A ,3!" steam tur9ine 0enerator power plant has a full3load steam rate of &. *04*"3hr.

Assumin0 that no3load steam #onsumption as 16 of full3load steam #onsumption7 #ompute for the hourl% steam #onsumption at &6 load7 in *04he.

A. 127

B. 2,7,

C. 19,200 D. /17$

Solution(

At full load

Steam #onsumption7 ms3

ms3  C, *wC&. *04*"3hr  /7 *04hr At no3load

Steam #onsumption7 ms1

ms1  C.1C/7 *04hr  /7 *04hr Gsin0 two3point form(

(

)

1 2

1 x x

x x

y y y

   −

 





−

= −

−

"here(

;1Cx17%1  ;1C7/

;2Cx27%2  ;2C,7/

;Cx7%  ;Cms7L

+hus

ms  ,.5L  /

At & 6 load7 L  .&C,  /

ms  ,.5C/  /  172 *04hr

12. +he a#uum in the surfa#e #ondenser of a small #ondensin0 steam power plant is &5 mm

?0 as referred to a -& mm ?0 9arometer. If the temperature in the #ondenser is /, ' C#orrespondin0 pressure of ,.&2$ *;a7 the a#uum e@#ien#% is approximatel%(

A. $1.&

B. $&.1

C. 89.16 D. /./

Solution(

5-sat atm

cond atm

vac p p

p p

−

= −

patm  -& mm ?0

pcond  &5 mm ?0 a#  12 mm ?0

psat  ,.&2$ *;a C-& mm ?0 4 11./2, *;a  52.21 mm ?0

6 .21 ¹⁰⁰ ⁸⁹.¹⁶

42 760

120

760

× =

−

= −

η vac

1/. A 9oiler feedpump re#eies water at 2 ' Centhalp%7 h  $, *J4*0 from a surfa#e

#ondenser at the rate of 1 L4s. It operates a0ainst a total head of $, meters. Determine the enthalp% leain0 the pump7 in *J4*0.

A. $-$., B. $.2 C. 857.8 D. 1.

Solution(

;ump wor*7 "p

% p  mCh3 H h1  mgh41

h3 H $,  C.$1C$41

h3  $,-.$ *J4*0

15. A 9inar% mer#ur%3steam #%#le produ#es a net power output of / !". Steam enters the tur9ine at 5 !;a7 5 ' where it is 9ein0 exhausted to a #ondenser pressure of 1 *;a.

Assumin0 an oerall thermal e@#ien#% of $6 for the #%#le and 16 heat transfer e@#ien#%

in the heat ex#han0er7 determine the enthalp% diMeren#e in the 9oiler per *ilo0ram of mer#ur% per se#ond. Steam properties at 5 !;a7 5 '7 h  /21/.&7 s  &.-&: at 1 *;a7 hf  11.$/7 hfg 2/2.$7 sf  .&/57 sfg  -.,.

A. 37,500 B. /7

'. /27,

D. /,7

Solution(

A net

b ,

=

e k%

, %

b net

A 37500

80 0

000

30 7

=

(

_a _d

)

g

A m h h

= −

( )

k- kg

m h , h

g A d

a 37500

1 500

377 7

=

−

1,. In a #o30eneration steam power plant7 steam enters the tur9ine at , !;a7 5 '. A uarter of the steam is extra#ted from the tur9ine at 1- *;a while the remainin0 steam is allowed to expand to 1 *;a. +he extra#ted steam is then #ondensed and mixed with feedwater at

#onstant pressure. +he mixture is then pumped to a 9oiler pressure of , !;a. !ass Nowrate of steam throu0h the 9oiler is 2, *04s. Assumin0 no pressure drops and heat losses in the pipin0 s%stem7 #al#ulate for the pro#ess heat reuired in *". Steam properties at , !;a7 5 '7 h  /1,.-7 s  &.&5,: at .1- !;a7 hf  5$/.27 hfg  221&7 sf  1.5-,27 sfg  ,.-&2: at .1

!;a7 hf 11.$/7 hfg  2/2.$7 sf  .&5/7 sfg  -.,.

A. 12,551

In document Lecture 05 Problems (Page 32-49)