ADECH02 SOLUTION problem 8 5
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HANOI UNIVERSITY OF MINING AND GEOLOGY
ADVANCED PROGRAM IN CHEMICAL ENGINEERING
ADECH02-5/2017
Petroleum Refining
HOMEWORK HELP- SOLUTION PROB 8.5
Name : Dai Duong Van K4
The feed of 100,000 BPD with API = 24 (distillation column).
TBP(°R) = {( . ln
−𝐿𝑉%
API = -0.0004 𝑉% + .
T°C =
T°R =
° −
° +
SG = 𝐴𝑃𝐼+
.
.
+ }
𝑉% − .
(8.0)
𝑉% +
(8.1)
(8.2)
.
(8.3)
(8.4)
.
From the CDU unit, we get CDU-Gasoline cut as well as AGO and VGO cuts. FCC- Gasoline cut is also
produced from the FCC as well as LCO, which is then fed to a hydrocracking unit to produce
Hydrocracking-Gasoline.
We must calculate the total amount of gasoline (lb/h) that is produced from all units.
Consider the crude distillation unit :
190-380°F cut
IBP = 190 °F = 87.78°C = 649.67°R
1
EBP= 380°F= 193.33°C= 839.67°R
Substituting in eq ( 8.0) to get LV% at IBP and EBP :
Cut vol % = 44.71-35.95=8.76%
Cut mid volume = 40.33
Substituting the cut mid volume in eq ( 8.1) to get API= 30.29
From the Eq (8.4) ,we get SG = 0.875
Cut volume = 0.0876(100,000)= 8760 BPD
Cut amount = 8760(SG)(14.60)=8760(0.875)(14.60)=111,909 lb/h
Calculate of sulphur content :
M= 42.965[exp(2.097×
−
𝑏
− .
𝑏
M= 111.50 lb/lb mol
I=2.266×
n=
+ 𝐼
−𝐼
/
−
exp .
= 1.495
×
−
+
=
𝑏
𝑖
.
+ .
𝑖
m= M(n – 1.475) = 2.23 since M< 200
sulphur wt% = 177.448 – 170.946
+ .
+
.
×
−
+
− .
×
−
= .
=
sulphur amount = 0.0072(111,909) = 805.74 lb/h
−
+ .
650- 850°F cut ( AGO)
IBP = 650°F = 343.33°C = 1109.67°R
EBP= 850°F= 454.44°C= 1309.67°R
Substituting in eq ( 8.0) to get LV% at IBP and EBP :
Cut vol % = 55.54-51.87=3.67%
2
=
−
]
𝑏
𝑏
.
.
.
.
𝑏
𝑏
= .
𝑤%
− .
= .
Cut mid volume = 53.705
Substituting the cut mid volume in eq ( 8.1) to get API= 25.36
From the Eq (8.4) ,we get SG = 0.902
Cut volume = 0.0367(100,000)= 3670 BPD
Cut amount = 3670(SG)(14.60)=3670(0.902)(14.60)=48,330.964 lb/h
Calculate of sulphur content :
M= 42.965[exp(2.097×
−
𝑏
− .
𝑏
M= 340.35 lb/lb mol
I=2.266×
n=
+ 𝐼
−𝐼
/
−
exp .
= 1.516
×
−
=
𝑏
+ .
+
.
+ .
−
=
−
+
− .
𝑖
m= M(n – 1.475) = 13.95
×
×
−
=
−
𝑖
− .
+
𝑏
= .
.
sulphur amount = 0.02827(48,330.964) = 1366.32 lb/h
= .
850-1050°F cut ( VGO)
IBP = 850°F = 454.44°C = 1309.67°R
EBP= 1050°F= 565.56°C= 1509.67°R
Substituting in eq ( 8.0) to get LV% at IBP and EBP :
Cut vol % = 58.46-55.54=2.92%
Cut mid volume = 57.0
Substituting the cut mid volume in eq ( 8.1) to get API= 23.57
3
𝑏
.
.
.
since M>200
sulphur wt% = -58.02+ 38.463
]
𝑏
𝑤%
𝑏
.
− .
= .
From the Eq (8.4) ,we get SG = 0.912
Cut volume = 0.0292(100,000)= 2920 BPD
Cut amount = 2920(SG)(14.60)=2920(0.912)(14.60)= 38,880.384 lb/h
Total feed to FCC = 48,330.964 lb/h + 38,880.384 lb/h = 87,211.348 lb/h
SG for AGO = 0.902 and SG for VGO = 0.912
Then SG for mixed feed =
.
+
So API = 24.68
+
.
+
= .
Consider the FCC unit :
Conversion = 72%
Coke wt% = 0.05356×CONV – 0.18598×API + 5.966975 = 5.2333
𝑉 −
LCO LV%= 0.0047×
.
×
Gases wt% = 0.0552×CONV + 0.597 = 4.5714
𝑉+
.
=
.
Gasoline LV% = 0.7754 ×CONV – 0.7778 = 55.051
i
n
𝑉% = .
×
𝑉% = .
𝑉 + .
×
=
𝑉 + .
=
𝑉% = .
Gasoline API = - 0.19028 ×
LCO API = − .
×
×
𝑉% = .
𝑉% = .
×
𝑉+ .
𝑉+ .
𝑉+ .
×
×
𝑉 +
×
.
𝑎
= .
= .
𝑉− .
𝑉− .
𝑖
×
= .
= .
𝑉+ .
𝑉% +
×CONV + 1.725715 × ( Feed API ) = 17.63
Gasoline amount = 0.55051×(6590)BPD×0.7715×14.60 = 40,863.86 lb/h
Feed to hydrocracking ( LCO) = 0.1628×6590 BPD = 1072.852 BPD
Considering the gas oil hydrotreater unit :
Middle distillate HT
SCFB
=
.
𝑓
+
.
% −
4
.
= .
=
.
100% severity of the hydrotreater
SCFB
=
.
.
+
.
Since the feed is 1072.852 BPD:
Δ
𝑝
= .
− .
−
=
(
𝑝
=
.
𝑓)
.
+ .
+ .
= .
Gasoline amount =1072.852 × 0.919×14.6= 14,394.88 lb/h
=
.
The total amount of Gasoline = CDU-Gasoline+FCC-Gasoline+Hydrocracking-Gasoline =
111,909 lb/h +40,863.86 lb/h +14,394.88 lb/h = 167,168 lb/h
5
