Chemical Engineering

[Axioms]

I am busy studying for an exam and I am stuck on a particular question. I am not sure how to relate the flow rates to each other without getting more than one variable in the final equation.

When I get one variable it solves incorrectely because you using four equations that become dependent when you relate more than two variables. Picture is down below. I may be over complicating something but I am stumped at this point in time. Any help would be much appreciated.

[Axioms]

I'll post my working for the previous question later if that will help. I do have another question that I am struggling with.

 

They give me a table with water density at certain temperatures: Where the density of water at 20C is 0.9988kg/L

 

The question is: The SG of ethyl alcohol at 20C is 0.7893. What is the density of ethyl alcohol at 20C in lb/ft^3.

 

Ok so this is what I did. 0.7893(kg/m^3 ethyl)/(kg/m^3 water) * (0.9982kg/L water) * (1000L/m^3) * ((0.3048m/ft)^3) * (lb/0.4536kg) = 49.18 lb/ft^3

 

It is apparently wrong. The way the question is phrased is that the SG is at 20C (and not just the ethyl), I took into account the density change of the water. The answer is 49.3lb/ft^3 but they take the density of water to be 1kg/L (I think because I worked it out with that density and it gave me an answer of 49.27lb/ft^3). Anyone have an explination as to why they use the one density over the other? Or does anyone think that the question is phrased incorrectely?

 

My only thought: the SG is already corrected so that you can just multiply by 1kg/L water to find the density of the ethyl alcohol.

If so, can anyone explain why?

 

Any information would be appreciated.

[studiot]

Surely all the propane and above goes to D and all the isobutane and below goes to B

From then on is it not a question of proportion?

[Axioms]

Yes that is true but the isobutane and propane splits into B and D. They give percentages of Isobutane in B (its mol % in B not F) and propane in D. Then the mols in the other two streams will be unknown so we can call them x and y.

 

A total mol balance:

F = B + D

Therefore 100 mol= B + D

 

Mol balance of isobutane and propane:

 

25mol iC4H10 = 0.05(D) + xB and 15mol C3H8= yD + 0.008B (where y is unknown mol of propane and x is unknown mol of isobutane)

 

My answer at a point that I cannot solve: 100= (15-y)(39.2-100x/(0.042+y-x))/0.008 + (39.2-100x/(0.042+y-x)). (F = B + D, I solved for B and D and subbed into F = D + B)

[studiot]

Yes you are right there is overlap between the isobutane and the pentane.

 

However you are given two pieces of additional information which allows you to form two simultaneous equations in 2 unknowns.

 

I have attached a start, can you take it from there?

Edited July 1, 2012 by studiot

[Axioms]

Ooo that was a good way to look at it. I never noticed that you could compare the % like that. Thanks .

[studiot]

specific gravity definition sg (ethyl alcohol) = Density EA / Density of water at same temp.

 

Density EA = 0.9988 * 0.7893 kg/L

 

= 0.9988 * 0.7893 * 2.2046226 lbs/L

 

= 0.9988 * 0.7893 * 2.2046226 / 0.035314 lbs/cuft

 

= 49.216189 lbs/cuft

 

I note you have water density as 0.9988 in one place and 0.9982 in another in your post

 

go well in your exam.

Edited July 1, 2012 by studiot

[Axioms]

Thanks, answer that was given was 49.3 lb/cuft though. Not sure what to make of it.

[studiot]

49.216 * 1/0.9988 = 49.23

 

So they are correct if they used 1.000 kg/L