Chemical engineering question Crude oil is fractioned by two separating columns.
ID: 889527 • Letter: C
Question
Chemical engineering question
Crude oil is fractioned by two separating columns.
Crude oil is fractioned by two separating columns. The incoming crude oil is composed of the following weight fractions: 0.0300 gas, 0.240 gasoline, 0.100 kerosene, 0.130 diesel, 0.110 heating oil, 0.130 lubricating oil and the rest solids. The crude oil is fed into the first column at 1435 kg/hr producing a stream of gas, a 5:1 asoline to kerosene stream, a stream containing kerosene and diesel, and 828 kg/hr stream of the bottoms he bottoms, containing diesel, lubricating oil, heating oil, and solids, is fed into a second distillation column that separates the components into three streams. The top stream contains diesel and heating oil. The middle stream contains heating oil and lubricating oil, and the bottom stream contains the solids. If you want to produce 265 kg/hr of the stream containing heating and lubricating oil, find 1 - The flow rate, m3, of the gasoline and kerosene stream 2 - The composition of the heating oil and lubricating oil stream 3 - The composition of the diesel and heating oil stream m2 kg/hr gas m 1435 kg/hr Crude Oil gas asoline Separation kerosene ka/hr diesel heating oil Column #1 soline erosene diesel heating oil lubricating oil m4 kg/hr kerosene diesel arat Column #2 ms 828 kg/hr Bottoms ottomsCo ies heating ol lubricating oil solids 265 k heating oil lubricating oil mg kg/hr solidsExplanation / Answer
Input
0.0300x1435=
43.05 kg gas
0.240x x1435=
344.4 kg gasoline
0.100x1435=
143.5 kg kerosene
0.130x1435=
186.55 kg diesel
0.110x1435=
157.85 kg heating oil
0.130x1435=
186.55 kg lubricating oil
Rest 0.26 x 1435 =
373.1 kg solids
Input, per hour
Separation, 1 per hour
Separation 2, per hour
43.05 kg gas
m2 43.05 kg gas
-
344.4 kg gasoline
m3 = 344.4 kg gasoline + 344.4/5 kg kerosene = 413.28 kg
* m4 = 150.67 kg = 74.62 kg kerosene + 76.05 kg diesel
m5 = 828 kg
-
143.5 kg kerosene
-
186.55 kg diesel
**m6 = 189.9 kg = 110.5 kg diesel + 79.4 kg heating oil
***m7 = 265 kg = 186.55 kg lubricating oil + 78.45 kg heating oil
157.85 kg heating oil
186.55 kg lubricating oil
373.1 kg solids
m8 = 373.1 kg solids
* m4 = m1 –m2- m3- m5 = 150.67 kg
Rest of kerosene 143.5-68.88 = 74.62 kg
Diesel in m4 = 150.67 – 74.62 = 76.05 kg diesel
** m6 = m5 – m8 – m7 = 828 kg – 373.1 – 265 = 189.9 kg
Rest of diesel in m6 is 186.55 - 76.05 = 110.5 kg
so heating oil is 189.9 - 110.5 = 79.4 kg
*** heating oil in m7 is 265 – 186.55 = 78.45 kg
Verification: 79.4 kg + 78.45 = 157.85 heating oil
Answers:
1. m3 = 344.4 kg/h gasoline + 68.88 kg/h kerosene = 413.28 kg/h
2. m7 = 265 kg/h = 186.55 /h lubricating oil + 78.45 /h heating oil
3. m6 = 189.9 kg/h = 110.5 /h diesel + 79.4 kg/h heating oil
0.0300x1435=
43.05 kg gas
0.240x x1435=
344.4 kg gasoline
0.100x1435=
143.5 kg kerosene
0.130x1435=
186.55 kg diesel
0.110x1435=
157.85 kg heating oil
0.130x1435=
186.55 kg lubricating oil
Rest 0.26 x 1435 =
373.1 kg solids
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