For m6 composition, It does not state that the balance is hexanol as water could be there as well. The water steam leaving the extractor is not necessarily all the water that went inside the system
Can it also be m1 be the basis instead? Or is it one of the right choice if we also use m1 as the basis of 100 g just like what you've done with m2 on a video? What makes the m2 be the best basis instead of m1?
I am struggling to find complex questions and solutions for mass balancing. The questions in Felder's Elementary principles of chemical processes has very easy questions in comparison to our courses questions. Does anyone know where I can find more questions and solutions for mass balancing?
since the composition of m3 was unknown, how'd you know that water was not part of that composition(even if a minute amount)? the composition of m6 was also not completely specified so it is possible that there was water in that portion as well right?
The problem statement says, "Assume that water is completely insoluble in hexanol." Thus, you know that they will be in separate streams. Then there will not be any water in m6 because no water went into the distillation column.
Once you have solved for m5, you can then use the other equations to find m1 and m4. Using the W balance, you see that m1 = 1.213m4, so you can substitute that into either the A balance or the total balance. I'll use the A balance here: 0.18(1.213m4) = 0.005m4 + 120 + 2.742. This simplifies to 0.21334m4 = 122.742, then m4 = 575. I hope that helps.
+Syvmana Thank you for your question. In this case, the fact that the H and A add to 1 is already accounted for in that we only used one unknown for both of those, the variable x. If we had used an unknown for H and an unknown for A, then the relationship of H+A=1 could be used. However, we would then have 6 unknowns in the LLE and 5 unknowns in the DIST, so the resulting number of degrees of freedom would be the same. I hope that helps.
Could you include an equation relating the hexanol in the distillate to the hexanol input to the liquid-liquid extraction? i.e., could you include .05*m2 = .04*m5 as another equation?
Thanks for the question. While you could use that equation, it would not be an independent equation from 0.95*m2=0.972*m6. Thus, you would need to use one or the other of those equations, but not both.
Andrew Tice Yes, you are correct. Remember that when the degrees of freedom were calculated, only the independent number of balances were used. At around 11 minutes, all of the species balances and the overall balance were written out simply to see which would be the easiest ones to solve. They aren't all needed. In this case, the instructor didn't use the overall balance.
Yes, the column he is referring to is the liquid-liquid extraction column. It could be more clear in the problem statement, but it is stated clearly at 2:10.
you chose the basis at m2 to be 100, its dificult to understand why this assumption is allowed,and if i'm to make this assumption in a similar DOF, is 100 moles like a constant to always use?
Thanks for your question. We are allowed to set a basis in this problem because no flow rates are given, and we are only solving for a percent recovery. You can actually choose any flowrate for your basis, but it is usually best to use a basis that is a multiple of 10 - like 10, 100, or 1000 moles - because it makes converting mole percents to flowrates easier. However, you could choose any flowrate for your basis and arrive at the same answer. Hope this helps!
The stream m3 is not in the overall equation because it is not a stream that enters or exits the entire system. It is an output if you are looking at just the extractor or an input if you are looking at just the distillation column. I hope that answers your question.
Your videos are helping me a lot.
Thank you very much.
Very useful video. Explained it very well...
How come you never solved for M3 ? or used it as part of the Total feed calculation?
Sir very very thank you for uploading this Leacture
Very well explained!
For m6 composition, It does not state that the balance is hexanol as water could be there as well. The water steam leaving the extractor is not necessarily all the water that went inside the system
Julio Cubillas Flores hexanol And water are assumed to be insoluble according to the problem.
That's exactly what I was looking for, thank you so much!
You're very welcome!
Thank u for this videos 🥰
Can it also be m1 be the basis instead? Or is it one of the right choice if we also use m1 as the basis of 100 g just like what you've done with m2 on a video? What makes the m2 be the best basis instead of m1?
as by taking m2 as basis wecan directly find m6
I am struggling to find complex questions and solutions for mass balancing. The questions in Felder's Elementary principles of chemical processes has very easy questions in comparison to our courses questions. Does anyone know where I can find more questions and solutions for mass balancing?
i mean, u can make ur own questions if u find everything easy, just pick a question from folder and make it harder
since the composition of m3 was unknown, how'd you know that water was not part of that composition(even if a minute amount)? the composition of m6 was also not completely specified so it is possible that there was water in that portion as well right?
or are we to assume that the extraction did its job and all water was separated as you did?
The problem statement says, "Assume that water is completely insoluble in hexanol." Thus, you know that they will be in separate streams. Then there will not be any water in m6 because no water went into the distillation column.
Gotcha
Very helpful_
How calculate m4 and m1
Once you have solved for m5, you can then use the other equations to find m1 and m4. Using the W balance, you see that m1 = 1.213m4, so you can substitute that into either the A balance or the total balance. I'll use the A balance here: 0.18(1.213m4) = 0.005m4 + 120 + 2.742. This simplifies to 0.21334m4 = 122.742, then m4 = 575. I hope that helps.
For 7:13 and before that, why didn't you include the relationship for the species in m3; that H and A equal 100?
+Syvmana Thank you for your question. In this case, the fact that the H and A add to 1 is already accounted for in that we only used one unknown for both of those, the variable x. If we had used an unknown for H and an unknown for A, then the relationship of H+A=1 could be used. However, we would then have 6 unknowns in the LLE and 5 unknowns in the DIST, so the resulting number of degrees of freedom would be the same. I hope that helps.
this is very helpfull...
Could you include an equation relating the hexanol in the distillate to the hexanol input to the liquid-liquid extraction? i.e., could you include .05*m2 = .04*m5 as another equation?
Thanks for the question. While you could use that equation, it would not be an independent equation from 0.95*m2=0.972*m6. Thus, you would need to use one or the other of those equations, but not both.
Hm, very interesting. Thank you for answering my question!
How come you are able to use all the species balances in addition to the overall balance? Wouldn't one of the balances not be independent?
Andrew Tice Yes, you are correct. Remember that when the degrees of freedom were calculated, only the independent number of balances were used. At around 11 minutes, all of the species balances and the overall balance were written out simply to see which would be the easiest ones to solve. They aren't all needed. In this case, the instructor didn't use the overall balance.
LearnChemE can you include what book did u get this problem from?
don't the question say that the pure hexanol is fed to the column (distillation column) not the vessel? I guess that should be the vessel!
Yes, the column he is referring to is the liquid-liquid extraction column. It could be more clear in the problem statement, but it is stated clearly at 2:10.
I got the value of m1= 7.02697 & m4= 579.10
Is it correct or not?
The answers are written out starting at 11:30. They do not match yours.
you
made a mistake on ( 11:25 ) the m5 u didnt multiply by the percent of hexanol in the stream...
He doesn't need to. He's just solving for m5... which is ( 100 - 0.972 (97.7) ) / 0.04 = 125.89 Should be 126 really rounding up.
If there is 97.7 hexanol in m6 and 100 in m2, then why isn't the recovery percentage 97.7% instead of 95%
Nevermind. 97.7 is the mase of hexanol acetic acid. I got it.
When I solved H and W simultaneously I got 837 for m1, could you please check again
Nevermind I've seen my mistake
Where m3 gone
you chose the basis at m2 to be 100, its dificult to understand why this assumption is allowed,and if i'm to make this assumption in a similar DOF, is 100 moles like a constant to always use?
Thanks for your question. We are allowed to set a basis in this problem because no flow rates are given, and we are only solving for a percent recovery. You can actually choose any flowrate for your basis, but it is usually best to use a basis that is a multiple of 10 - like 10, 100, or 1000 moles - because it makes converting mole percents to flowrates easier. However, you could choose any flowrate for your basis and arrive at the same answer. Hope this helps!
Where does the M3 goes???
The stream m3 is not in the overall equation because it is not a stream that enters or exits the entire system. It is an output if you are looking at just the extractor or an input if you are looking at just the distillation column. I hope that answers your question.
More or less good but, not clear!