example 7.25 karina
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7/21/2019 Example 7.25 Karina
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1. A process gas stream at 400oC is to be rapidly cooled to 2000C by direct
quenching with cold liquid benzene at 20oC. If the hot stream consist of
40 C!"!# $0 C!"%C"$. 10 C"4 and 20 "2 calculate the benzene
quench rate required for a gas feed rate of 1000 &gmol'h# assuming
quencing occurs adiabatically. ()ith *ref + 200oC,
-olution
/iasumsi&an + 0 arena process teradi secara adiabati&
) + 0 arena tida& ada perubahan 3olume yang teradi
-ehingga dQ
dt −
dW
dt =∑ outlet F [ H (Ti )− H (Tr ) ]−∑ inlet F [ H (Ti)− H (Tr ) ]
H 2 (200 oC )− ( H 2 (20oC ) ]} F 2¿
F 1 ( H 1 (400oC )− H 1 (200 oC ) ]+¿ H 3 (200oC )− H 3 (200oC )−{¿
0= F 3¿
a&a
F 2 ( H 2 (200oC )− H 2 (20oC ) ] F 1 ( H 1 (400oC )− H 1 (200 oC ) ]=¿
¿
F 1 ( H 1 (400oC )− H 1 (200 oC ) ]−→ dianggap sebagai equation1¿
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F 2 ( H 2 (200oC )− H 2 (20oC ) ]¿ Dianggap sebagai equation 2
enyelesai&an equation 1
F 1 ( H 1 (400oC )− H 1 (200 oC ) ]=∑ F 1 ∫473.2 K
673.2 K
CpdT
¿
∑ F 1 ∫473.2 K
673.2 K
Cp dT = (T 2−T 1 )∑ F 1a+T 2
2−T 12
2 ∑ F 1b+
T 23−T 1
3
3 ∑ F 1c+
T 24−T 1
4
4 ∑ F 1d+
T 25
ema&ai 5 ( sigma , di&arena&an ada 4 gas yang masu& pada aliran# ya&ni
benzene#metana# toluen dan hidrogen.
6A* A 7 8 101 C 8 102 / 8 10% 5 8 109 :low
C!"! 19.%9; <0.11;44 0.12;%1 <0.20;=9 0.10%$$ 400
C!"%C"$ $1.920 <0.1!1!% 0.1444; <0.229=% 0.11$%; $00
C"4 $9.$9; <0.;$!!4 0.02=0=9 <
0.02!$9%
0.00900!
9
100
"2 1;.!$= 0.!;00! <0.01$14=
0.010%99 <0.002=19 200
>ang&ah selanutnya# dicari nilai masing<masing 5(sigma,
∑ F 1a= F (c 6h6 ) a (c6h6 )+ F (C 6 H 6CH 3)a(C 6 H 6CH 3)+ F (CH 4)a(CH 4)+ F ( H 2)a( H 2)
400 x18.587
∑ F 1a=¿ , ? ($00 8 $1.920, ? (100 8 $9.$9;, ? (200 8 1;.!$=,
∑ F 1a=2.43473 x 104
/engan cara yang sama# dicari ∑ F 1b ,∑ F 1c ,∑ F 1d dan∑ F 1e
/an didapat&an hasilnya
∑ F 1b=−3.5123
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∑ F 1c=0.94625
∑ F 1d=−1.5240 x10−3
∑ F 1e=0.76420 x10−6
/an dimasu&&an &epersamaan
∑ F 1 ∫473.2 K
673.2 K
Cp dT = (T 2−T 1 )∑ F 1a+T 2
2−T 12
2 ∑ F 1b+
T 23−T 1
3
3 ∑ F 1c+
T 24−T 1
4
4 ∑ F 1d+
T 25
¿ (200 K )2.43473 x 104+
673.22−473.2
2
2
x−3.5123+673.2
3−473.23
3
0.94625+673.2
4−473.24
4
x−1.5240 x 1
∑ F 1 ∫473.2 K
673.2 K
Cp dT =2.5634 x107 Kj/h
a&a didapat&anlah equation 1 + 2.%!$4 8 10; ' h
emudian dicara 5quation 2
F 2 ( H 2(v )(200oC )− H 2(l) (20oC ) ] F 2 ( H 2 (200oC )− H 2 (20oC ) ]=¿
¿
/engan data @ormal 7oilling point of benzene is $%$.2! # dan ada benzen
yang berubah fasa dari gas<cair
F 2 ( H 2 ( v ) (200oC )− H 2 ( l ) (20oC ) ]= F 2( ∫353.3 K
473.2 K
Cp (V ) dT + ∫293.2 K
353.2 K
Cp ( L ) dT +Hvl (353.3 K ))
¿
Hvl (353.3 K )¿=30,763.4 Kj /!g"ol
∫353.2 K
473.2 K
Cp (V ) dT =(473.2−353.3)18.587+ 473.22−353.3
2
2 x−1.174 x 10
−2+ 473.2
3−353.33
31.2751 x 10
−3+47
-ehingga
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∫353.2 K
473.2 K
Cp (V ) dT =1.3808 x104
!j/!g"ol
∫293.2 K
353.3# K
Cp ( L ) dT =(353.3−293.2)−7.273+
353.32−293.2
2
2 7.705 x10
−1
+
353.33−293.2
3
3 x−1.6481 x 10
−3
+
35
∫293.2 K
353.3# K
Cp ( L ) dT =8.0979 x103
!j/!g"ol
a&a
F 2 ( H 2 ( v ) (200oC )− H 2 ( l ) (20oC ) ]= F 2(30,673.4 Kj
!g"ol
+1.3808 x 10
4!j
!g"ol
+ 8.0979 x 103 !j
!g"ol
)
¿
F 2 ( H 2 ( v ) (200oC )− H 2 ( l ) (20oC ) ]= F 2 (5.2669 x 104 ) !j/!g"ol
/i&etahui tadi bahwa equation 1+ equation 2
Atau
F 2 ( H 2 (200oC )− H 2 (20oC ) ]
F 1 ( H 1 (400oC )− H 1 (200 oC ) ]=¿¿
a&a
F 1 ( H 1 (400oC )− H 1 (200 oC ) ]¿¿
F 2=¿
F 2= 2.5634 x107 !j /h5.2669 x10
4!j /!g"ol
=486.7 !g"ol/h
2. A process gas stream at 400oC is to be rapidly cooled to 200 pC by direct
quenching with cold liquid benzene at 20oC. If the hot stream consist of
40 C!"!# $0 C!"%C"$. 10 C"4 and 20 "2 calculate the benzene
quench rate required for a gas feed rate of 1000 &gmol'h# assuming
quencing occurs adiabatically. ()ith *ref + 20oC,
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-olution
/iasumsi&an + 0 arena process teradi secara adiabati&
) + 0 arena tida& ada perubahan 3olume yang teradi
-ehingga
dQ
dt −
dW
dt =∑ outlet F [ H (Ti )− H (Tr ) ]−∑ inlet F [ H (Ti)− H (Tr ) ]
H 2 (20oC )− ( H 2 (20oC ) ]} F 2¿
F 1 ( H 1 (400oC )− H 1 (20 oC ) ]+¿ H 3 (200oC )− H 3 (20oC )−{¿
0= F 3¿
a&a
F 3 ( H 2 (200oC )− H 2 (20oC ) ] F 1 ( H 1 (400oC )− H 1 (20 oC ) ]=¿
¿
F 1 ( H 1 (400oC )− H 1 (20 oC ) ]−→ dianggap sebagai equation1¿
F 3 ( H 3 (200oC )− H 3 (20oC ) ]¿ Dianggap sebagai equation 2
enyelesai&an equation 1
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∫353.3 K
673.2 K
Cp (V ) dT + ∫293.2 K
353.2 K
Cp ( L )dT + Hvl (353.3 K )
F 1 ( H 1 (400oC )− H 1 (20oC ) ]=∑ F 1¿¿
∑ F 1 ∫293.2 K
673.2 K
Cp dT =(T 2−T 1 )∑ F 1a+T 2
2−T 12
2 ∑ F 1b+
T 23−T 1
3
3 ∑ F 1c+
T 24−T 1
4
4 ∑ F 1d+
T 25
ema&ai 5 ( sigma , di&arena&an ada 4 gas yang masu& pada aliran# ya&ni
benzene#metana# toluen dan hidrogen.
6A* A 7 8 101
C 8 102
/ 8 10%
5 8 109
:lowC!"! 19.%9; <0.11;44 0.12;%1 <0.20;=9 0.10%$$ 400
C!"%C"$ $1.920 <0.1!1!% 0.1444; <0.229=% 0.11$%; $00
C"4 $9.$9; <0.;$!!4 0.02=0=9 <
0.02!$9%
0.00900!
9
100
"2 1;.!$= 0.!;00! <
0.01$14=
0.010%99 <0.002=19 200
>ang&ah selanutnya# dicari nilai masing<masing 5(sigma,
∑ F 1a= F (c 6h6 )a (c6h6 )+ F (C 6 H 6CH 3)a(C 6 H 6CH 3)+ F (CH 4 )a(CH 4 )+ F ( H 2 )a( H 2)
400 x18.587
∑ F 1a=¿ , ? ($00 8 $1.920, ? (100 8 $9.$9;, ? (200 8 1;.!$=,
∑ F 1a=2.43473 x 104
/engan cara yang sama# dicari ∑ F 1b ,∑ F 1c ,∑ F 1d dan∑ F 1e
/an didapat&an hasilnya
∑ F 1b=−3.5123
∑ F 1c=0.94625
∑ F 1d=−1.5240 x10−3
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∑ F 1e=0.76420 x10−6
/an dimasu&&an &epersamaan
∑ F 1 ∫293.2 K
673.2 K
Cp dT (v)=(T 2−T 1 )∑ F 1a+ T 22
−T 12
2 ∑ F 1b+T 23
−T 13
3 ∑ F 1c+T 24
−T 14
4 ∑ F 1d+ T
¿ (673.2−293.2 )2.43473 x 104+
673.22−293.2
2
2 x−3.5123+
673.23−293.2
3
30.94625+
673.24−293.2
4
4 x−1.5
∑ F 1 ∫293.2 K
673.2 K
CpdT (v)=4.2256 x107 Kj/h
emudian dicara 5quation 2
F 3 ( H 3(v ) (200oC )− H 3(l) (20oC ) ] F 3 ( H 3 (200oC )− H 3 (20oC ) ]=¿
¿
/engan data @ormal 7oilling point of benzene is $%$.2! # dan ada benzen
yang berubah fasa dari gas<cair
F 3 ( H 3 ( v ) (200oC )− H 3 (l ) (20oC ) ]= F 3 ( ∫353.3 K
473.2 K
Cp (V ) dT + ∫293.2 K
353.2 K
Cp ( L )dT + Hvl (353.3 K ))
¿
Hvl (353.3 K )¿=30,763.4 Kj /!g"ol
∫353.2 K
473.2 K
Cp (V ) dT =(473.2−353.3)18.587+ 473.2
2−353.32
2 x−1.174 x 10
−2
+ 473.2
3−353.33
3 1.2751 x 10
−3
+47
-ehingga
∫353.2 K
473.2 K
Cp (V ) dT =1.3808 x104
!j/!g"ol
∫293.2 K
353.3# K
Cp ( L ) dT =(353.3−293.2)−7.273+353.3
2−293.22
2
7.705 x10−1+
353.33−293.2
3
3
x−1.6481 x 10−3+
35
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∫293.2 K
353.3# K
Cp ( L ) dT =8.0979 x103
!j/!g"ol
a&a
F 3 ( H 3 ( v ) (200oC )− H 3 (l ) (20oC ) ]= F 3 (30,673.4 Kj
!g"ol+1.3808
x104
!j
!g"ol +8.0979 x 10
3 !j
!g"ol)
¿
F 3 ( H 3 ( v ) (200oC )− H 3 (l ) (20oC ) ]= F 3 (5.2669 x104 ) !j/!g"ol
/i&etahui tadi bahwa equation 1+ equation 2
Atau F 3 ( H 3 (200oC )− H 3 (20oC ) ]
F 1 ( H 1 (400oC )− H 1 (200 oC ) ]=¿¿
a&a
F 1 ( H 1 (400oC )− H 1 (20 oC ) ]¿¿
F 3=¿
F 3= 4.2256 x10
7 Kj /h
5.2669 x104!j /!g"ol
=801.97 !g"ol/h