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### Topic: flowrate  (Read 3741 times)

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#### forum12345

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##### flowrate
« on: September 30, 2009, 09:55:28 PM »
A solution composed of 50% ethanol (EtOH), 10% methanol (MeOH), and 40% water (H2O) is fed at the rate of 100 kg/hr into a separator that produces one stream at the rate of 60 kg/hr with the composition of 80% EtOH, 15% MeOH, and 5% H2O, and a second stream of unknown composition. Calculate the composition (in %) of the three compounds in the unknown stream and its flowrate in kg/hr.
MASS BALANCE!!!  assuming no reaction in=out
so starting with the flowrate

100 kg/hr IN = 60 kg/hr + x kg/hr OUT
the unknown stream is 40 kg/hr

now we do this individually  for each component
for ethanol
we have (50%)(100 kg/ hr)= 50 kg/hr IN
therefore we must have 50 kg/hr coming OUT
so we sum the ethanol coming out (i.e. we add up the amount of ethanol in stream 1 coming up and the amount of ethanol in stream 2 coming out)
(80%)(60 kg/hr) + a(40 kg/hr) = 48 + 40a = 50
from that we can figure out a=(50-48)/40=2/40=5%

repeat for the other components!
Is thi correct

#### typhoon2028

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##### Re: flowrate
« Reply #1 on: October 01, 2009, 08:21:58 AM »
First, I would draw a diagram.
Second, Label the diagram Flow in, Split 1, & Split 2.
Third, I would convert the the Flow In components from % to kg/hr.
Fourth, I would calculate the Split 1 components in terms of kg/hr
Mass flow is conserved, so Split 2 can be calculated from simple subtraction.  Split (a) = Flow In (a) - Split 1 (a)
You know the total flow of Split 2, so percentages can be calculated.

Flow In
total   100   kg/hr

Et   50%   50
Me   10%   10
H20   40%   40

Split 1            Split 2
total   60   kg/hr         total   40   kg/hr

Et   80%   48   kg/hr      Et   2   5%
Me   15%   9   kg/hr      Me   1   3%
H20   5%   3   kg/hr      H20   37   93%