A stream of 5.00 wt% oleic acid in cottonseed oil enters an extraction unit at a

Question

A stream of 5.00 wt% oleic acid in cottonseed oil enters an extraction unit at a rate of 150.0 kg/h. The unit operates as a single equilibrium stage (the streams leaving the unit are in equilbrium) at 85.0°C. At this temperature, propane and cottonseed oil are essentially immiscible, and the distribution coefficient (oleic acid mass fraction in propane/oleic acid mass fraction in cottonseed oil) is 0.150.

The vapor pressure of propane can be estimated from

A = 4.65678
B = 1149.36 K
C = 24.906 K

log10 (P) = A – (B/(T+C))

where P is pressure in bar and T is temperature in K.

a) Calculate the rate at which liquid propane must be fed to the unit to extract 85.0% of the oleic acid.

b)Suppose that the actual operating temperature were 55.0°C. Estimate the minimum operating pressure (in bar) of the extraction unit (i.e. the pressure required to keep the propane liquid at 55.0°C).

Explanation / Answer

Flow rate of oleic acid = 150*5/100 = 7.5 kg/hr

Flow rate of cotton seed oil =150-7.5= 142.5 kg/hr

85% of oleic acid needs to be extracted. Hence oleic acid in propane= 7.5*0.85=6.375 kg/hr

Oleic acid in cotton seed oil= 7.5-6.375=1.125 kg/hr

Let x= mass flow rate of of propane used

Mass fraction of oleic acid in propane = 6.375/(6.375+x)

Mass fraction of oleic acid in cotton seed oil = 1.125/(142.5+1.125)=0.007833

Oleic acid mass fraction in propane = 0.15* oleic acid mass fraction in cotton seed oil

6.375/(6.375+x)= 0.15*0.007833 = .001175

6.375 = 0.001175*(6.375+x)

6.375-6.375*0.001175 = 0.001175x

Hence x = 5419 kg/hr

2.

A = 4.65678
B = 1149.36 K
C = 24.906 K

At 55 deg.c (55+273= 328K), propane has to be in the form of liquid. So vapor pressure of log10 (P) = A – (B/(T+C)) = 4.65678- 1149.36/(24.906+328)
P= 3.76 bar. This is the minimum pressure that needs to be maintained

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