Crude oil is fractioned by two separating columns. The incoming crude oil is com

Question

Crude oil is fractioned by two separating columns. The incoming crude oil is composed of the following weight fractions: 0.0400 gas, 0.210 gasoline, 0.120 kerosene, 0.110 diesel, 0.100 lubricating oil, 0.110 heating oil, and the rest solids. The crude oil is fed into the first column at 1.250 x 103 kg/hr producing a stream of gas, a 5:1 gasoline to kerosene stream, a stream containing kerosene and diesel, and 683 kg/hr stream of the bottoms. The 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 o and the bottom stream contains the solids. If you want to produce 169 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.

Explanation / Answer

Mass fraction of solids in feed= 1-0.04-0.210-0.120-0.110-0.100-0.110 =0.31

Flow rate of curde oil= 1250 kg/hr. All the solid are removed from stream-8, mass flow rate of solids entering = mass flow rate of solids in stream-8

Hence m8= 1.250*1000*0.31= m8, m8= 387.5 kg/hr.

Writing balance of lubricating oil from 2nd distillation column, 683*x14= 169*0.814

Hence x14= 0.20

Writing overall balance across the 2nd distillation column

683= 387.5(solids)+169(m7)+m6, m6= 683- 387.5-169= 126.5 kg/hr

Writing overall heating oil balance of the entire plant

1.250*1000*0.1= 169*0.186+x17*126.5, x17=0.74, x16=1-0.74=0.26

Writing heating oil balance across the 2nd distillation column

683*x13= 125 , x13= 125/683= 0.183, x15= m8/m5= 387.5/683= 0.567

Hence x12= 1-x14-x13-x15= 1-0.2-0.183-0.567= 0.05

Writing overall gas balance, gas entering = 1.250*1000*0.04= 50 kg/hr

This is leaving as m2. Hence flow rate, m2= 50 kg/hr

Wiring overall plant balance, 1250= m2+m5+m3+m4

Hence m3+m4= 1250-50-683= 517

Writing gasoline balance across the 1st distillation column( Gasoline is leaving from 1st distillation column)

1250*0.120= m3*x8 =150

Given m3*x8/m3*x9= 5/1 hence x8/x9= 5, x8= 5x9

But x8+x9= 1, 5x9+x9=1, x9=0.167, x8=1-0.167=0.833

Hence m3=150/x8= 150/0.833= 180 kg/hr

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.