Steam Power Board Problems

 

SOLVED SOL VED MECHANICAL ENGINEERING PROBLEMS STEAM POWER PLANT Problem 1 A reheat steam has 13,850 kPa thrtt!e "ress#re at the t#r$%&e %&!et a&' a (800 kPa reheat "ress#re) *he thrtt thrtt!e !e a&' reheat reheat tem"er tem"erat# at#re re + the steam %s 50C, -&'e&s -&'e&ser er "ress# "ress#re re %s 3) kPa, kPa, e&. e&.%&e %&e e++%-%e&-/ + h%.h "ress#re a&' ! "ress#re %s 52) %&' the -/-!e therma! e++%-%e&-/) e++%-%e&-/) A) 3)42 C) 34)442

B) 35)542

D) 3)42

Solution: th



()8 MPa 50C

Reheat

t  " 6A

* 6A

1

B%!er 

13)85 kPa1 50C

3

m

(

H)P)

3

L)P)*#r$%&e) 

0)003 MPa 4

(

C&'e&ser 

m 5

4 5

P#m"

rm Steam *a$!e7 At 13)85 MPa a&' 50C7 h1 33)1 k9:k. ;%&ter"!ate'< At ()8 MPa a&' 50C7 h3 358)5 k9:k. h( at s(  s1  4)534 k9:k. 7 h(  (=)= k9:k. h at s  s3  ) )38 3810 10 k9: k9:k. 7 h  ((0)5 k9:k. h5  h+  at  at 0)003 MPa  10=)8 k9:k. 3 >5  > f at 0)003 MPa  0)001003( m  :k. S!>%&. +r h47 h4 ? h5  >5 ;P4 ? P5< h4 ? 10=)8  0)001003( ;13,850 ? 3)< h4  1(3)3 k9:k. t  @;33)1 ? (=)=<  ;358)5 ? ((0)5+  at  at 0)10 MPa< h4 ? h5  >5 ;P4 ? P5<

h4 ? 1)4  0)00103( ;3000 ? 100< h4  (0)8 k9:k. h  1008)( k9:k. ;h + at 3 MPa< *h#s, m ; (=(8)15 ? 1008)( <  ;1 ? m< ;1008)( ? (0)8 < m  #.2"$   Answer 

Problem $ A 'r#m -&ta%&%&. steam %th ()5 m %& '%ameter %s )5 m !&.) O+ the tta! >!#me, 1:3 -&ta%&s sat#rate' steam at 800kPa a&' the ther (:3 -&ta%&s sat#rate' ater) I+ th%s ta&k sh#!' e"!'e, h m#-h ater #!' e>a"rateF C&s%'er the "r-ess t $e -&sta&t e&tha!"/)

A) (,=8)11 k. B) (,()4( k.

C) (,451)( k. D) (,1(3)4 k.

Solution: V!#me + sat#rate' steam a&' ater7 1    (  3 Vs    ()5   )5    1()( m 3 

V 

(   

 

(

3    ()5   )5    ()5 m

3

rm Steam *a$!e (7 At 800 kPa ;0)80 MPa< 7

h+  (1)11 k9:k.

> +  0)001118 m3:k. > .  0)(0 m3:k.

h+.  (08)0 k9:k. h.  (4=)1 k9:k.

I&%t%a! mass + steam, mS7 mS 

1()( 0)(,0,



51)0, k.

I&%t%a! mass + ater, m7 m 

()5 0)001118

 ((,01()=( k.

I&%t%a! #a!%t/, 17    1

51)0

 0)00(31

51)0  ((,01( ,01()=( )=( h  h+   h+.  (1)11  0)00(31;(08< h  (5)8 k9:k.

 

A+ter e"!s%&, +%&a! -&'%t%& %!! $e atms"her%At 100C a&' 101)3 kPa h+  1=)0 k9:k. 3 h+.  ((5)0 k9:k. > +  0)001035 m :k. h.  (44)1 k9:k. > .  1)4(= m3:k. 6#a!%t/ + the +%&a! state7 h(  h1 h1  h+  ( h+. (5)8  1=)0   ( ;((5< (  0)135=3 Mass + steam at atms"her%- "ress#re7 mS  0)135=3 ;51)0  ((,01()=(< mS  (,===)15 k. *he am#&t + >a"r e>a"rate'7 mV  (,===)15 ? 51)0  2%9$&.11 k! 

 Answer 

Problem ' A $a-k "ress#re steam t#r$%&e + 100,000 k ser>es as a "r%me m>er %& a -.e&erat%& s/stem) *he $%!er a'm%ts the ret#r& ater at a tem"erat#re + 44 C a&' "r'#-es the steam at 4)5 MPa a&' 55C) Steam the& e&ters a $a-k "ress#re t#r$%&e a&' e"a&'s t the "ress#re + the "r-ess, h%-h %s 0)5( MPa) Ass#m%&. a $%!er e++%-%e&-/ + 802 a&' &e.!e-t%&. the e++e-t + "#m"%&. a&' the "ress#re 'r"s at >ar%#s !-at%&s, hat %s the %&-reme&ta! heat rate +r e!e-tr%-F *he +!!%&. e&tha!"%es ha>e $ee& +#&'7 t#r$%&e e&tra&-e  3304)8 k9:k., e%t  (00)8 k9:k., B%!er e&tra&-e  (4)(3 k9:k., e%t  3304)8 k9:k. A) ((,50 k9:khr C) (5,(00 k9:khr   B) (,500 k9:khr D) 30,504 k9:khr   Solution:

t  m ;h1 ? h(<  m ; 3304)8 ? (00)8 <  404 m k 3400  m  h1  h (  3400  m  3304)8  (00)8    13,43,545m k9:hr  6A   $ 0)80 Heat Rate 

13, 43, 545m 404m

 ((,50

k9

   Answer  kGhr  

Problem 6

A -a!+%re' "er "!a&t has a t#r$%&e.e&eratr rate' at 1000 M .rss) *he "!a&t re#%re' a$#t =2 + th%s "er +r %ts %&ter&a! "erat%&s) It #ses =800 t&s + -a! "er 'a/) *he -a! has a heat%&. >a!#e + 4,388)= k-a!:k. a&' the steam .e&eratr e++%-%e&-/ %s 842) hat %s the &et stat%& e++%-%e&-/ + the "!a&t %& "er-e&tF A) 33)02 C) 3)052 B) 35)02 D) ()052

 

Solution:

 Net O#t"#t  1000 ;1 ? 0)0=<  =10 M  =10,000 k

Heat I&"#t  m+  6  6h 

=800  =0  (  3400 

 Net Stat%& E++%-%e&-/  

388)=  )18   (, 5 5(, 00 001    4, 38

 Net O#t"#t Heat I&"#t =10,000

 0)330 1002  33)02    Answer 

(,5(,001 Problem 7 A s#"erheat steam Ra&k%&e -/-!e has t#r$%&e %&!et -&'%t%&s + 1)5 M"a a&' 530 C e"a&'s %& a t#r$%&e t 0)00 M"a) *he t#r$%&e a&' "#m" "!/tr"%- e++%-%e&-%es are 0)= a&' 0) res"e-t%>e!/) Press#re !sses $etee& "#m" a&' t#r$%&e %&!et are 1)5 M"a) hat sh#!' $e the "#m" rk %& k9:k.F A) 1)3 k9:k. C) 3)3 k9:k. B) ()3 k9:k. D) )3 k9:k. Solution:

 " 

>3  P  P3   "

here7 >3 

1

3  0)001 m k. 1000

P  1)5  1)5  1= MPa  1=, 000 kPa P3  0)00 MPa   kPa  "

 " 

 0)0

0)001 1=,000    0)0

 ()1 k9 k.  

 Answer 

Problem & A steam "!a&t "erates %th %&%t%a! "ress#re + 1)0 MPa a&' 30 C tem"erat#re a&' eha#st t a heat%&. s/stem at 0)1 MPa) *he -&'e&sate +rm the heat%&. s/stem %s ret#r&e' t the $%!er at 45)5 C a&' the heat%&. s/stem #t%!%es +rm %ts %&te&'e' "#r"se =02 + the e&er./ tra&s+erre' +rm the steam %t re-e%>es) re-e% >es) *he t#r$%&e t#r$%&e e++%-%e&e++%-%e&-/ /, t %s 02) I+ the $%!er e++%-%e&-/ %s 802, hat %s the -.e&erat%& e++%-%e&-/ + the s/stem %& "er-e&tF Ne.!e-t "#m" rk) Steam Pr"ert%es7 At 1) MPa a&' 30C7 h  318)1 k9:k. s  )1081 k9:k.   At 0)10 MPa7 h+  83)(0 k9:k. s+  1)5( k9:k.     h+.  ((14)0 k9:k. s+.  5)04( k9:k.   At 45)5C7 h+  ()1 k9:k.

A) 82 B) 10()102

C) =1)(2 D) 4=2

 

Solution:

h1  318)1 k9:k. S!>%&. +r h( 7 s1 s(  ;s+   s+.< ( )1081  1)5(   ( ;5)04(< (  0)=81 h(  ;h+   ( h+.<    83)(0  0)=81 ;((14)0<  (40)4 k9:k. h3  h   ()1 k9:k. t  ;h1 ? h(< t  ;318)1 ? (40)4< ;0)0<  341)55 k9:k. 6R     0)=0 ;h( ? h3<  0)=0 ;(40)4 ? ()1<  (154)81 k9:k. h1  h  318)1  ()1  341)( k9:k. 6A   $ 0)80 C.e&erat%& E++%-%e&-/ 

t  6 R 



6A

341)55  (154)81 341)(

 0)4=14  69.16(   Answer 

Problem 9 I& a Ra&k%&e -/-!e, steam e&ters the t#r$%&e at ()5 M"a a&' -&'e&ser "ress#re + 50 kPa) hat %s the therma! e++%-%e&-/ + the -/-!eF

Steam Pr"ert%es7 At ()5 M"a, h .  (803)1 k9:k., S.  4)(55 k9:k. At 50 kPa, S+  1)0=10 k9:k., S+.  4)50(= k9:k., h+     30)= k9:k., h+.  (305) k9:k., >+  0)0010300 m3:k. Solution:  NE* eth





6A

*  P



 h1  h (    h   h 3 

6A

h1  h 

here7 h . at ()5 MPa  h1  (803)1 k9:k.   h + at 50 kPa  h 3  30)= k9:k. S!>%&. +r h ( a&' h  7 S  S+   xS+. 4)(55  1)0=10   x  4)50(=   x

 0)=5

h (  h +   xh +.  30)=  0) 05  (305)   (1()13 k9:k. h   h +  >+  P(  P1   30)=  0)00103  (500  50   3()=8 k9:k. th#sJ  NE*  (803)1  (1()11   3()=8  30)=   6A (803)1  3()=8

eth



eth

 (5)552