[themonk]

I have a single Feed at X lbm/hr of 46% B and 54%T which exits through feeds, R1 and R2. with R1 being 98% B and R2 being 95% T (all in moles percents). I need to compute the flow rates of R1 and R2.

I am pretty sure that the flow rates should add up to Feed X's flow rate, so R1 + R2 = X, but I do not know how to go from there. An explanation or some hints would be greatly appreciated.

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This is what I did.
I assumed 100 lbm/hr to make it simple, so 46 lbm/hr B and 54 lbm/hr T.
R1 is .98*46lbm/hr = 45.08 B and .02*54=1.08 T ==> 46.16
R2 is .95*54lbm/hr = 51.30 T and .05*46=2.3 B ==> 53.6
Total is 99.76, so it seems something I did is wrong.

( by lbm I mean lbmol )

[Lukec]

You have mass flow and mole percentage.. you need to assume in mol/hr than with molecular masses compute X in lbs/h

[typhoon2028]

Assume 100 lbm/hr

Feed = 46 lbm/hr B + 54 lbm/hr T

In = Out

For B

R1 B + R2 B = Feed B

I believe you are incorrect when you perform: " R1 is .98*46lbm/hr = 45.08 B and .02*54=1.08 T ==> 46.16 "


I setup an excel file to "goal seek" the solution.

A hint:
The total feeds of R1 & R2 need to be found.

R1 total + R2 Total = Feed Total

[Lukec]

yes .. but all percentages are in mol percents ... not mass... you need to calculate in moles per hour .. and than conver to mass....

[typhoon2028]

yes .. but all percentages are in mol percents ... not mass... you need to calculate in moles per hour .. and than conver to mass....

when assuming 100 lbm/hr, Feed = 46 lbm/hr of B & 54 lbm/hr T.

R1: B lbm/hr + R2: B lbm/hr = 46 lbm/hr

[DrCMS]

I do wish you'd go metric.

I have a single Feed at X lbm/hr of 46% B and 54%T which exits through feeds, R1 and R2. with R1 being 98% B and R2 being 95% T (all in moles percents). I need to compute the flow rates of R1 and R2.

I am pretty sure that the flow rates should add up to Feed X's flow rate, so R1 + R2 = X, but I do not know how to go from there. An explanation or some hints would be greatly appreciated.

You are correct that the amount of B in and amount of B out (similarly for T) should be equal but you have not done the correct calculations for that

R1 is .98*46lbm/hr = 45.08 B and .02*54=1.08 T ==> 46.16
R2 is .95*54lbm/hr = 51.30 T and .05*46=2.3 B ==> 53.6
Total is 99.76, so it seems something I did is wrong.

The correct calculations as typhoon2028 has pointed out relates the amount of B in and out:
R1x0.98 + R2x0.05 = 46

and the same for T
R1x0.02 + R2x0.95 = 54

you then need to solve these to get R1 and R2 which added together should give the same as the total input.

[themonk]

Thank you Lukec, typhoon2028, and DrCMS.

I think I worked it out correctly with R1 being 296.21 lbmol/hr and R2 being 323.76 for an initial 620 lbmol/hr (a later part of the question asked for 620 rather than 100).

I do wish you'd go metric.

You don't know how much I wish this.

[typhoon2028]

For an initial feed of 620 lbm/hr

I calculate
R1 = 273.33 lbm/hr
R2 = 346.67 lbm/hr

ignoring sig figs

[DrCMS]

I agree with typhoon2028's figures.

[themonk]

Ah, slight calculation mistake.

F=R1+R2=620 so 620-R1=R2
Basically did B*F=R1*B1+R2*B2 and plugged in above. and all other known ie all but R1.

Cheers again.