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, 0.220 gasoline, 0.130 kerosene, 0.130 diesel, 0.120 lubricating oil 0.120 heating oil, and the rest solids. The crude oil is fed into the first column at 1.190 x 103 kg/hr producing a stream of gas, a 5:1 gasoline to kerosene stream, a stream containing kerosene and diesel, and 672 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 oil, and the bottom stream contains the solids. If you want to produce 193 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

All the gas entering the system is removed from the top of 1st distillation column,

Mass flow rate of gas entering = 1.190*1000*0.04=47.6 kg/hr of gas is leaving as m2.

Composition of solids, x7= 1-0.04-0.22-0.130-0.130-0.130-0.120-0.120 = 0.24

Mass flow rate of solids entering the system= 1.190*1000*0.24 kg/hr =285.6 kg/hr. This is leaving as stream m6.

Writing overall balance across the 2nd distillation column

672 (m5)= 285.6(m8)+m6+193(m7)

Hence m6= 672-285.6-193= 193.4 kg/hr

All the lubricating oil entering the system is leaving vide m3. Hence writing lubricating oil balance

1.190*1000*0.120= 193*x19, x19= fraction of lubricating oil in m7.

Hence x19 = 0.74, hence mass fraction of heating oil in m7= x18= 1-0.74=0.26

Total of heating oil entering is leaving vide m6 and m7

Heating oil entering = heating oil in m7 + heating oil in m6

Hence writing overall heating oil balance, 1.190*1000*0.12= 193*0.26+ 193.4*x17, x17=0.479

Hence x16=1-0.479= 0.521 mass fraction of diesel

Writing balance of diesel across 2nd distillation column

672*x12= 193.4*0.521, x12= 0.149, All the heating oil entering the process enters the 2nd distillation column. Hence flow rate of heating oil= 1.190*1000*0.12= 142.8 kg/hr

Hence 672*x13= 142.8, x13= 142.8/672= 0.2125

Writing solids balance across the 2nd distillation column, 672*x15= 285.6, x15= 0.425

Hence x14= 1-x12-x13-x15= 1-0.149-0.2125-0.425= 0.2135 , mass fraction of lubricating oil

Writing overall mass balance

1.190*1000 ( entering the process)= 672 ( leaving vide m6, m7 and m8)+47.6 (m2)+m3+m4

Hence m3+m4= 1190-672-47.6= 470.4 kg/hr

Diesel entering the process is removed from 672 kg/hr stream from 2nd distillation column and through m4. Hence diesel in m4= 1.190*1000*0.130-672*0.149

Diesel in m4= 54.572 kg/hr . m4*x11= 54.572, x11= 54.572/m4, x10= 1-54.572/m4

All the gasoline entering the system is leaving through m3. Hence gasoline in m3= 1.190*1000*0.22= 261.8 kg/hr

Hence m3*x8= 261.8 kg/hr, x8= 261.8/m3, x9= 1-261.8/m3

Hence x8= 261.8/m3, given m3*x8/m3*x9= 5/1, x8= 5x9

Hence x8+x9=1,6x9=1, x9=1/6=0.167, x8= 1-0.167=0.833

Hence m3= 261.8/0.833 =314.3 kg/hr, m4= 470.4-314.3= 156.1 kg/hr

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