Chemical Forums
Specialty Chemistry Forums => Chemical Engineering Forum => Topic started by: forum12345 on September 30, 2009, 09:55:28 PM
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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
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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%