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: 0.0300 gas, 0.240 gasoline, 0.100 kerosene, 0.120 diesel, 0.110 heating oil, 0.100 lubricating oil and the rest solids. The crude oil is fed into the first column at 1415 kg/hr producing a stream of gas, a 5:1 asoline to kerosene stream, a stream containing kerosene and diesel, and 822 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 204 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. 3 m2 kg/hr gas m 1415 kg/hr Crude Oil gas Column #1 diesel heating oil m4 k kerosene diesel heating oil lubricating oil solids umn # ms 822 kg/hr Bottoms -204 k heating oil lubricating oil lubricating oil mg kg/hr solids

Explanation / Answer

ANS 1

The mass of crude oil= 1415kg/hr

The mass fraction of gasoline =0.24

Hence m_{gasolene} = 1415 X 0.24 Kg/hr =339.6 Kg/ hr ---(1a)

As m3 contains gasolene & kerosene in 5:1 ratio; hence

m3 = (6/5) X 339.6 Kg/hr = 407.52 Kg/hr --(1)

Ans 2

To calculate the composition of lubricating oil and heating oil:

let us 1st calculate m2 = mass of crude oil X gas fraction = (1415 X 0.03) Kg/hr =42.45 Kg/hr --(2)

Hence from (1) & (2):

m4 = m1 - m2- m3 = 143.03 Kg/ hr ---(3)

Kerosene fraction= 1415 X 0.1 = 141.5 kg/hr --(4)

Now in m3 , kerosene is 1/5(m_gasolene) = 67.92 from (1a)

So in m4, mass of kerosene = 141.5 -67.92 =73.58 kg/hr

Hence, in m4, the mass of diesel = 143.03 - 73.58 =69.45 kg/hr --(5)

Now total mass of diesel = m1 X diesel fraction = 1415 X 0.12 = 169.8 Kg/hr

using equ(5): diesel in the bottom = (169.8 - 69.45) =100.35 Kg/hr ---(6)

Now the solid fraction in m1 = 1 - sum of all fractions = 0.23

Hence mass of solid = 1415 X 0.23 = 325.45 Kg/hr --(7)

So m6 = m5 -m7-m8 = 822-325.45-204=292.55 Kg/hr ---(8)

m_heating oil in m6 = m6-m_diesel = 292.55- 100.35=192.2 Kg/hr --(9)

Now total heating oil = m1 X 0.1 = 141.5 Kg/hr

The required heating oil = 204-141.5 = 62.5 Kg/hr

Ans 3

As we have already shown in (6) & (9)

mass of diesel =100.4 kg/hr

mass of heating oil=192.2 kg/hr

When expressed in fractions: the numerical values of the results will be kg/kg.

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