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P3= (187-2NBI) psia

2

4

T

s

P1=( 5461+2 NBI) psia

P2= (1161+2 NBI) psia

1

3

5

6

T1 = (1461+2 NBI)°F

78

MATAKULIAH : SISTEM PEMBANGKIT UAP DAN

 TURBINMODEL

 TUGAS/QUIZ

: ETS

MATERI : Sistem Air UmpanAAL TUGAS :

AKHIR TUGAS :NAMA : SIGIT EKA DUI TAMA

Ni!ai : NBI/KLAS : "#$#%"$&'/A

Di(eta)*i :

Ditan+a :La,* a!iran massa -an parameter +an. er(aitan -en.an si(!*s :000000000000001

 2a3a :P$ 4 555& psia6 T$ 4 $55&78 (ea-aan s*per)eat0

9T+pe tet; Pa.e $

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)$ 4 $##' $=>55

555&) $"$<

>%%%) $

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Ment*(an ) -an s pa-a te(anan 555& psia -en.an temperat*re $>%% 78

Ment*(an ) -an s pa-a te(anan 555& psia -en.an temperat*re $55& 78

9T+pe tet; Pa.e

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s# 4 s$ 4 $=>#'$ Bt*/!m0K= pa-a P# 4 $#5& psia0 !i)at ni!ai = ,i(a $ ma(a

prCses )# ter,a-i pa-a s*per)eate-0h2=1496,3 Btu/ lbm

  Bt*/

?!m0R@P ?psia@

S S.

$#%% %=

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s2=sf +( x . s fg)

 x=s2−sf sfg

 x=1,6281−0,7788

0,5824

 x=1,46

Karena $ ma(a prCses )# pa-a s*per)eate- P# 4 $#5& psia6 s# 4 s$ 4 $=>#'$

Bt*/!m0K0

9T+pe tet; Pa.e 5

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  T?78@P ?psia@

&%% $%%%

$%%%) $""'=$ $5%5=&s $=>$## $=>5#

$#5&) $" $"&>= $"&$

#%%% ) $"% $"

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Ment*(an ) pa-a te(anan $#5& psia -en.an temperat*re $=>"%% Bt*/!m

s 4 s$ 4 $=>#'$ Bt*/!m0K= pa-a P 4 '& psia0 !i)at ni!ai = ,i(a $ ma(a

prCses )# ter,a-i pa-a (ea-aan s*per)eate-0h3=1198,9Btu /lbm

P ?psia@ s s.

'% %="55 $=$>'$'& %=">#& $=$5%>

9T+pe tet; Pa.e '

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$%% %="

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 T?78@P ?psia@

%% "%%

'%

) #>&=

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Ment*(an ) -an s pa-a te(anan '& psia -en.an temperat*re "%% 78

Ment*(an ) pa-a te(anan '& psia -en.an temperat*re $=>#'$ Bt*/!m

9T+pe tet; Pa.e $$

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)" (ea-aan sat*rate- *i- =P" 4 P 4 '& psia0 )" 4 ) ?P '&@ 4 #'&=> Bt*/!m6

F" 4 F ?P '&@ 4 %=%$5 t/!m

P ?psia@F )  

'% %=%$

9T+pe tet; Pa.e $#

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)5 (ea-aan s*CC!0 P54 P# 4$#5& psia0h5=293,4Btu / lbm

W  p=h5−h4

 P

v4(¿¿5− P4)W  p=¿

 P

v4(¿¿5− P4)h5−h

4=¿

 P

h4+v4(¿¿5− P4)h5=¿

h5=289,6

 Btu

lbm+(0,01765   ft 

3

lbm. (1259−89)

 lbf 

i n2 × 144¿2

1 f t 2

  ×  1Btu

778,16lbf . ft )h5=293,4Btu / lbm

)> (ea-aan sat*rate- *i-0 P> 4 P5 4 P# 4 $#5& psia0 ma(a )> 4 ) ?P $#5&@ 4

5 Bt*/!m 6 F> 4 F ?P $#5&@ 4 %=%##5" t/!m

 T?C8@

P ?psia@

F )  

$#%% %=%### 5

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$#5& %=%##5" 5

$"%% %=%#%< 5&'=>

)< (ea-aan s*CC!0 P< 4 P$4 555& psia0h7=597,5Btu/ lbm

W  p=h7−h6

 P

v6(¿ ¿7− P6)

W  p=¿ P

v6(¿ ¿7− P6)h7−h

6=¿

 P

h6+v

6(¿¿7− P6)

h7=¿

h7=579,6 Btu

lbm +(0,02254

  ft 3

lbm × (5559−1259 )

 lbf 

i n2 ×

 144¿2

1 f t 2   ×

  1Btu

778,16 lbf . ft 

)9T+pe tet; Pa.e $"

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h7=597,5Btu/ lbm

La,* a!iran massa

ḿ2 (h2−h6 )=(1− ḿ2 ) (h6−h5 )

ḿ2 (1496,3−579,6 )=(1− ḿ2 ) (579,6−293,4 )

ḿ2(916,7 )=(1− ḿ2 ) (286,2)

916,7 ḿ2=286,2−286,2 ḿ

2

916,7 ḿ2+286,2. ḿ

2=286,2

1202,9 . ḿ2=286,2

ḿ2=0,238 lbm

Ker,a t*rin

w T =( h1−h2 )+(1− ḿ2 )( h2−h3 )

w T =(1771,9−1496,3 )+ (1−0,238) (1496,3−1198,9 )

wT =502,2Btu / lbm

Ker,a pCmpa

|∑ w p|=(1− ḿ2 ) (h5−h4 )+(h7−h6 )

|∑ w p

|= (1−0,238) (293,4−289,6 )+(597,5−579,6 )|∑ w p|=20,8Btu /lbm

Ker,a si(!*s

∆wnet =wT −|∑ w p|

∆wnet =502,2−20,8

∆wnet =481,4 Btu/ lbm

mas*( pa-a prCses pemanasan -i Ci!erq A=h1−h7

q A=1771,9−597,5Btu /lbm

q A=1174,4Btu / lbm

(e!*ar pa-a prCses pen-in.inan -i (Cn-ensCr

|q R|=(1− ḿ2 ) (h3−h4 )

|q R|=(1−0,238 ) (1198,9−289,6 )

|q R|=692,9Btu / lbm

9T+pe tet; Pa.e $5

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EJsiensi si(!*s

ηth=∆wnet 

q A

ηth= 481,4

1174,4

ηth=0,41

ηth=41

RasiC (er,a

WR=∆ wnet 

wT 

WR=481,4

502,2

WR=0,989

Kesimp*!an :Dari )asi! per)it*n.an= .raJ( si(!*s +an. enar men,a-i0

9T+pe tet; Pa.e $>