Why does one co-ordinate of the operating line pass through (xf, xf). One co-ordinate lie in the equilibrium line (for liquid and vapor leaving the column), and to find the other co-ordinate the composition of liquid is assumed to be same as feed (xb=xf). Why is it so? Is there any possible explanation for this without the use of the operating line equation?
I just want to make sure that I have the math right. At 6:09, If you're at the bubble point how does that work out in the terms of the equation to be infinity?
In 5:09, can I use the DePriester chart to find the operating temperature, as the operating pressure is given and also the K-value can be obtained from y = Kx ?
Thanks for the question. You are trying to draw the operating line and you have the equation yH = -1.5(xH) + 1.25. You find the point at xH = 0.5, which means you solve for yH and you get yH = 0.5, so that point is on the y=x line.
Okay thank you for your assistance, so we use the composition of the feed zH in this case to find the intersection of y=x and operating line drawing all the way to equilibrium curve?
Take a look at some of the distillation videos where bottoms operating lines are determined using boil up ratios. There is an operating equation that is derived from material balances on that section. Assuming no special cases (i.e. side streams, multiple feeds, etc) the bottoms operating line is y = [(Vb+1)/(Vb)]*x - [(1/Vb)*xb] where Vb is your boilup ratio and xb is your bottoms composition of the more volatile species.
in 4:08 you mention that the fraction of the feed vaporized is 0.4 (40%), and the feed composition is 0.5 (50%) therefore when dividing the two getting V/F = 0.4 I dont understand how you achieved 0.4 shouldnt it be 0.8? sorry just confused :s
@Nuria Manuel it is 40% vapour / total feed that is 100% and thats how you get 0.4=V/T=40/100. btw its been 7 years since youve asked the question and you definitely are in no need for the answer but im really curious where are you now
At 2:50 ish, why couldn't you jut leave it as Yh = (1-(F/V))X + (F/V)Z? Is that not in y = mx + b form? Why did you multiply by (V/F) is my question, I suppose.
Good question. Yes, that is already in y=mx+b form. The term V/F is one that is frequently given, so changing it to include V/F usually makes it easier to use.
Thank you! As a studying chemical engineer, some of these videos where actually assigned as study material a few semesters ago and they've been heavily referenced since!
nevermind was confusing myself completely forgot that V/F is the fraction of the feed vaporized....awesome video
Glad it helped.
Why does one co-ordinate of the operating line pass through (xf, xf). One co-ordinate lie in the equilibrium line (for liquid and vapor leaving the column), and to find the other co-ordinate the composition of liquid is assumed to be same as feed (xb=xf). Why is it so? Is there any possible explanation for this without the use of the operating line equation?
I just want to make sure that I have the math right. At 6:09, If you're at the bubble point how does that work out in the terms of the equation to be infinity?
In 5:09, can I use the DePriester chart to find the operating temperature, as the operating pressure is given and also the K-value can be obtained from y = Kx ?
At 4:26, how was the value of 0.5 for Xh chosen? Thank you.
That was randomly chosen to find a point on the graph. It happens to land on the y=x line.
@@LearnChemE thanks! i don't do engineering anymore now after graduating but ur vids saved me in my upper level classes. so much respect
Why did you chose 0.5 as the point of intersection y=X line n the operating line?
Thanks for the question. You are trying to draw the operating line and you have the equation yH = -1.5(xH) + 1.25. You find the point at xH = 0.5, which means you solve for yH and you get yH = 0.5, so that point is on the y=x line.
Okay thank you for your assistance, so we use the composition of the feed zH in this case to find the intersection of y=x and operating line drawing all the way to equilibrium curve?
Anybody know how to find the Bottom Operating Line if you're only given the boil up ratio on a column with a partial reboiler?
Take a look at some of the distillation videos where bottoms operating lines are determined using boil up ratios. There is an operating equation that is derived from material balances on that section. Assuming no special cases (i.e. side streams, multiple feeds, etc) the bottoms operating line is y = [(Vb+1)/(Vb)]*x - [(1/Vb)*xb] where Vb is your boilup ratio and xb is your bottoms composition of the more volatile species.
in 4:08 you mention that the fraction of the feed vaporized is 0.4 (40%), and the feed composition is 0.5 (50%) therefore when dividing the two getting V/F = 0.4 I dont understand how you achieved 0.4 shouldnt it be 0.8?
sorry just confused :s
@Nuria Manuel it is 40% vapour / total feed that is 100% and thats how you get 0.4=V/T=40/100.
btw its been 7 years since youve asked the question and you definitely are in no need for the answer but im really curious where are you now
At 2:50 ish, why couldn't you jut leave it as Yh = (1-(F/V))X + (F/V)Z? Is that not in y = mx + b form? Why did you multiply by (V/F) is my question, I suppose.
Good question. Yes, that is already in y=mx+b form. The term V/F is one that is frequently given, so changing it to include V/F usually makes it easier to use.
Thank you! As a studying chemical engineer, some of these videos where actually assigned as study material a few semesters ago and they've been heavily referenced since!
That's great! We love to hear about professors assigning these!
awesome, great job