That’s one of the material balances. You can either write a mass balance for A, B and C, or one overall mass balance (which is 100 = m2 + m3) and two individual mass balances. If you use all four (3 balances and the overall), they are no longer independent. Adding the three mass balances will give us the overall balance.
In example 1can we write the fraction of C as y=1-x and take it as an equation ? in this case we have 4 unknowns, 3 independent variables and 1 equation. So DOF=0
I'd like to know. For instance, on the product of a stream if we have the mass flow of the solution and the mass fraction of component A and the mass fraction of component B, the formula (Xa+Xb=1) working that out should be understood to fall under “other equations”?
sorry I am confused. why didn't u say : y = 1 - x - 0.2. like you did with z, to become z-1 for B??? i am looking for your help and explanation regarding this question thx q
You absolutely can do that. In this case, that was considered an other equation. If you got rid of the y in the beginning, you would only have 4 unknowns and wouldn't need the other equation. Both ways work out exactly the same.
Thanks for the question. At first, we do not know the mole fractions of the components of stream 3. So we are going to solve for those. We use x as the mole fraction of A in that stream (which has units of kg of A per kg of that stream). We know that there are only 2 components of stream 3 and that their mole fractions must add to 1. Therefore, the mole fraction of C will be (1-x). Does that make more sense? [It's the same as saying you know that if your drink is made of only coffee and milk, and you know that it is 20% milk, you automatically know that it is 80% coffee.]
Great question. This just depends on how you want to do your accounting. If you claim m3 as an unknown, then you use the equation m3 = 0.1m1 as one of the "other equations". However, if you calculate m3 first, then you can no longer use that equation as one of the "other equations". Either way, you'll get the same answer. Just be careful not to count anything twice.
Thank you for the question. Here 1-z isn't an equation. It is simply the mole fraction of B in stream 2. We could write it another way and say that r is the mole fraction of B in stream 2. Then we would use the equation z+r=1, but we would have added another unknown, r. The way it was done in the screencast is a quick simplification. (Here the equation would be z+(1-z)=1, which is not an independent equation.) Does that help?
LearnChemE but then how is y=1-.02-x and independent equation? You literally said in the video it was the same thing you did on the right side but you didn’t count that one at and indp. eqn?
The degree of freedom will depend on what other parameters you know vs. how many unknowns you have. That may not be a satisfactory answer, so please be more specific if you still have a question.
This can be handled in different ways. In this case, as shown at 7:26, another equation is used. y = 1 - 0.2 - x . You could also just write that instead of y and have 4 unknowns. Both ways work.
I have an exam in my chemical engineering class tomorrow and now this concept is beginning to make sense. I thank you great sir.
you got this brethren
This is a bit late but i aced it, 94%, thank you for this helpful video.
I just wanna say that I am so happy that I came across this channel, Thank You for all of this!!
You guys are saving a life.Thank you so much.
Wish you guys were here when I was taking the chemical engineering introductory course ☹️
superb.Well explained
Really really good explanations
✨ Finally !
Thank you so much!
very good explanation (:
can 100= m2 + m3 be considered an equation?
That’s one of the material balances. You can either write a mass balance for A, B and C, or one overall mass balance (which is 100 = m2 + m3) and two individual mass balances. If you use all four (3 balances and the overall), they are no longer independent. Adding the three mass balances will give us the overall balance.
In example 1can we write the fraction of C as y=1-x and take it as an equation ? in this case we have 4 unknowns, 3 independent variables and 1 equation. So DOF=0
+LearnChemE very good explanation. I am a fan of yours now ;-)
if we use y in terms of x for the second equation, wouldnt that make one less unknownvariable?
Yes, this is discussed at 7:26.
very well explained
Well explain
Owk, understand now. Thanks
Awesome!
For example2 ,there's no m1 on the chart and I wonder if this makes the example unsolvable ?
In example 2 m1 = 100kg/h. It is an input value that is already known.
Mostafizor Rahman Thank you for explaining :D
I'd like to know. For instance, on the product of a stream if we have the mass flow of the solution and the mass fraction of component A and the mass fraction of component B, the formula (Xa+Xb=1) working that out should be understood to fall under “other equations”?
yes
Why not 1 = -0.2-x-y? On the second example?
You have to put one variable in terms of the other.
Timothy Christopher thanks
sorry I am confused. why didn't u say : y = 1 - x - 0.2. like you did with z, to become z-1 for B???
i am looking for your help and explanation regarding this question thx q
You absolutely can do that. In this case, that was considered an other equation. If you got rid of the y in the beginning, you would only have 4 unknowns and wouldn't need the other equation. Both ways work out exactly the same.
sorry I am confused. why didn't u say : y = 1 - x - 0.2. like you did with z, to become z-1 for B????
Hahahahahahahaha, U did so, thanks. God bless u
can we get a word problem solved, cause the part on stream 3 were (1-x) is the mass of c/kg confuses me.
Thanks for the question. At first, we do not know the mole fractions of the components of stream 3. So we are going to solve for those. We use x as the mole fraction of A in that stream (which has units of kg of A per kg of that stream). We know that there are only 2 components of stream 3 and that their mole fractions must add to 1. Therefore, the mole fraction of C will be (1-x). Does that make more sense? [It's the same as saying you know that if your drink is made of only coffee and milk, and you know that it is 20% milk, you automatically know that it is 80% coffee.]
How is m3 an unknown here? it is given that it's 0.1m1 and m1 is known so how is m3 an unknown?
Great question. This just depends on how you want to do your accounting. If you claim m3 as an unknown, then you use the equation m3 = 0.1m1 as one of the "other equations". However, if you calculate m3 first, then you can no longer use that equation as one of the "other equations". Either way, you'll get the same answer. Just be careful not to count anything twice.
@@LearnChemE So the dof sud b (-1) r8?
hi, i'm a bit confused why you haven't taken 1-z as one of the equations
Thank you for the question. Here 1-z isn't an equation. It is simply the mole fraction of B in stream 2. We could write it another way and say that r is the mole fraction of B in stream 2. Then we would use the equation z+r=1, but we would have added another unknown, r. The way it was done in the screencast is a quick simplification. (Here the equation would be z+(1-z)=1, which is not an independent equation.) Does that help?
LearnChemE but then how is y=1-.02-x and independent equation? You literally said in the video it was the same thing you did on the right side but you didn’t count that one at and indp. eqn?
What if I have pressure and temperature as parameters
The degree of freedom will depend on what other parameters you know vs. how many unknowns you have. That may not be a satisfactory answer, so please be more specific if you still have a question.
@@LearnChemE When to use degrees of freedom to solve a question!!! I do not know when to use it, I wish a reply
6:36
y . Is not unknown
This can be handled in different ways. In this case, as shown at 7:26, another equation is used. y = 1 - 0.2 - x . You could also just write that instead of y and have 4 unknowns. Both ways work.
LearnChemE thanks mr❤️
The lessons are not beginner friendly