calculus 2 mixing problem, CSTR, differential equation application
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- Опубліковано 3 жов 2017
- The mixing problem is an application in separable differential equation. This is also known as continuous stirred tank reactor (CSTR).
@bprpcalculusbasics
Why do I like these videos so much
Recently, you've been uploading the videos I needed in the moment I needed them!
I have diffeq and linear algebra midterm tomorrow. Perfect timing!
The best tutorial of all!! Please keep up the good work. Everyting is explained, nothing without explanation. That one example was like 10 examples. Thank you so much. Please keep doing tutorials sir..
I was studying this for hours, then I watched your video and it all made sense, thank you!
Those problems are amazing, please do more!!
Your videos are very interesting, I learn so much from them :) Keep up the great work :D
ooohhhh its the best channel on youtube
Thank you so much. This is the only way I have understood how to set up the equations.
Excellent explanation. You put my previous mathematics professors to shame.
I'm in calculus II right now and I though he was gonna do some shenanigans to solve it. I got really giddy when I figured out what he was doing was actually pretty straightforward.
I finally understood this concept! Thank you!
You're welcome!
nice english, nice presentation, nice explation ................. BE BLESSED BROTHER
Im so glad I watched this video. The C trick is really clever and I will definetely use it
So inspirering Thanks very much :) I love your videos.
Nice! More physics related videos please!
Thanks for doing my homework, muhaha
You could skip the constant term manipulation by setting the limits of integration from 0 to 10 on the time side and 9 to A|t=10 on the amount side.
You can avoid the undefined integral with the constant by doing a difined integral from t equals to 0 to t equals yo 10 and a equals to a(0) to the new A
Thankyou so much for this! This is exactly what we are learning in Specialists Maths (Australia) in year 12 right now and is perfect for revision for our upcoming exams. Are you able to do a D.E. where the tank is changing volume but pure water is being added to the mixture instead. Thankyou! c:
Caleb Jones
I will have to find time...
VCE represent
Link to a vid like this?
Why did you get rid of the +/- in front of the constant?
Is it normal to find setting up the differential equations so difficult?
Hi, what colour pens do you use? Thanks.
More Oreo!!! The rest of the video was pretty good, too ;)
When I first learned solving differential equations it is so hard for me
pretty good. thank's for the video!
Awesome video!
Why cant we put limits on integral instead of solving for c
so good
This video is worth watching.
amazing, thank you
Love your videos man!!!
Could you talk about the limits definition of logarithm?
Hi blackpenredpen, it's worthwhile to mention that the final concentration of tank is same as incoming solution.
Yes I should have.
That makes the absolute value around 30-A irrelevant right off, right?
Pleeease show us how to model more complicated systems (2nd order), rocket fuel / springs / chaotic motion. I know that treads on physics territory but would still be really cool.
Enjoying your videos! Would have liked this one better if you had kept the units throughout the calculation. Seeing the units at intermediate stages of integration, differentiation, algebra, etc. can be really useful!
On a pedantic note, it doesn't ask for the amount of alcohol, it asks for the percentage of alcohol, or rather, the concentration of alcohol. So the final answer should be roughly 32%.
Doesn't density slightly changes due to partial miscibility?
PackSciences yes
No water and alcohol are 100% miscible but due to volumecontraction the density changes slightly.
This is a Differential Equations problem so if you are smart to know that, which you are, know that in these type of math problems you assume the change in density to be negligible.
thank u sir,what is your name? you explian brifly and vividly.
A water tank of diameter 6m and depth of 4m. The inflow rate of water is 434.60 lps (liters per second) and rate of outflow is 467.8 lps (Liters per second). How much time will the the water outflow at the rate of 467.8lps. how much time will the tank get empty.? please help me with this problem or refer me some books for this.
I loved the approach in this solving problem , i always have a hard time interpreting the question , your house analogy was sooooooooo good.
there's one more method of solving these types of differential equations of the form dy/dx + Py=Q which uses the integrating factor method , Could you please please solve this using that method too?
ooooooooooooooooooooh i just found out you have videos on those too. Differential equation marathonn, Mad Men step aside
Yea I had a vid on that already. Yay
Thanks
In practice you'd never feed from the top and drain from the bottom. It's far easier to feed from the bottom and let it overflow the top. That way you always have equal mass out for the mass in.
Signed an old chemical engineer.
Holy molly. You make me like math again.
UnOrigionalOne I am glad to hear!!!!
講得很清楚!!
謝謝!
It is really interesting how you solve differential equations because that's not how we learn to solve them at all in France.
Really, don't we?
I know I didn't. Sometimes we used this method in physics class, but I don't know, playing around with dx and dy, I always found it kinda sloppy. Just my 2 cents ;)
sir i think got mistake for the it should be A = 30 -C5^-T/15 instead of add please notify me if i am wrong
thank you
Actually, alcohol-water is a bad example. When you mix them, the volume of the solution is less than the volumes put in. This " volume of mixing" effect varies with concentration, making the true solution very complicated.
Great point, examples like this a great mathematically but can be so far removed from the real world that they mislead students.
Yeah it gives a positive deviation
@@richardaversa7128 yeah but it doesn't matter though I like rigorous maths more than applied
This is a standard problem in chemical engineering, called continuously stirred tank reactor or CSTR. Google it or just check Wikipedia if you want to learn more.
14:50 best part :D
As I suspected, natural growth!
This... Amazing!!
Thanks!
blackpenredpen Going through this topic right now and it always makes me wonder how to make up the equations. This is a real help :)
you should come and teach in our Uni @Austria
Thanks sir
why is the incoming concentration 0.5, and not 1, as in one gallon of alcohol for every gallon of water? half and half is the same as a 1:1 ratio right? 1gal CH3OH/1gal H2O=1.
1:1 ratio represents 0.5
he didn't write out the units. its .5 gallons of alcohol per 1 gallon of liquid. 1/1 would be pure alcohol
can you show how to differentiate x^x by writing using the exponential function e^xlnx
Hasn't he done this already?
+tsujimasen yeah but seeing all possible methods would be cool
I did that too in that vid
"when helping your little brother or sister or child or whatever and they ask 'Whats 5+8?' Well... lets just say it's a constant!"
no wonder chinese r good at math
Thankyou
I used additional buffer tank to get easier mathematics :)
What kind of liters are those gallons?? :trollface:
:fuuu:
hello, i have a question. How do i know when to use separable and not linear?
Consider y as a function of t, where y is dependent, and t is independent.
As differential equations terms:
Linear means that the DiffEQ only uses y and its derivatives, directly. No squaring of y, no cubing of y, no exponentiating of y, no trigonometry or logs of y. Or of any of y's derivatives. The instances of y, and instances of derivatives of y, are either just multiplied by constants, or just multiplied by "pure functions" of t, and then added together.
Separable means that it is possible to treat the Leibnitz notation as a fraction, and completely separate y and dy on one side, from t and dt on the other side, and then integrate both sides to solve it.
Aren't you supposed to account for the fact that the concentration in the tank is constantly changing, so the amount of alcohol flowing out is also constantly changing and can't be regarded as a constant?
he is. A/60 is a changing term which represents changing conc of alcohol
@@JensenPlaysMCi see.
thanks
Suppose that m and n are integers such that both the quadratic equations x^2 + mx − n = 0
and x^2− mx + n = 0 have integer roots. Prove that n is divisible by 6.
challenge for you..
this war was supposed to be between black pen and red pen...not between 'Bob Tenwith' and 'Iamyou iamyou'.
I challenge you! Integral sin^2 (lnx)I know as a fact it can be solved
Should be easy if you use the exponential definitions of sin(x) = (1/2i)*(e^(ix) - e^(-ix))
Given:
integral sin(ln(x))^2 dx
Let u = ln(x), thus du = 1/x dx. Solve for dx:
dx = x du
and since u = ln(x), this means x = e^u
Thus, in the u-world, the integral becomes:
integral sin(u)*e^u du
This is a standard looper for integration by parts. Let sin(u) be differentiated, and e^u be integrated.
S _ _ _ D _ _ _ _ _ I
+ _ _ _ sin(u) _ _ e^u
- _ _ _ cos(u) _ _ e^u
+ _ _ _ -sin(u) _ _ e^u
Spot the original integral, call it I. Construct result in terms of I:
I = sin(u)*e^u - cos(u)*e^u - I
2*I = [sin(u) - cos(u)]*e^u
I = 1/2*[sin(u) - cos(u)]*e^u
Translate back to the x-world and add +C, and we're done:
1/2*x*[sin(ln(x)) - cos(ln(x))] + C
Extra part: how long will it take to get back to the original concentration if you pour in 100% alcohol
I will remember to tell my future kids that when they ask for math help 😂
i tried this problemm in real life and now im drunkkk
I solved it without calculus . My answer came 19.4 litres .
14:51 best
The only thing I learned is that 23 + 19 = C
That works!
WOW!!!!!
So much good alcohol and it is wasted in mixing and throwing out .. :)
Technically this is wrong....considering the alcohol (at lower density) is being poured into the top of the tank while drained at the bottom, you would have to take into account the rates of mixture, solubility speed, and membrane movement. Taking into account even the fastest soluble alcohol at the 4gal/min rate, you would reach 50% concentration in under 5 minutes.....which gives you the answer of closer to 30 gallons after 10 minutes.
But that's a physics problem instead of a calculus one. :)
There's also the fact that the volume of any alcohol/water mixture is actually less than the sum of the volumes of alcohol and water.
it's written that we assume that it's stirred constantly.
as you mentioned the physics and fluid dynamics would not be a question for students of calculus 2 only, applying actual physics to the questions always makes it harder and at the end this question would probably only be solveable for a physics student and even he would have an issue without being prepared for such a "nice" question.
TL:DR you are right, but physics only makes it harder, not a question you can ask on a mostly math channel.
It'd be a calculus problem, just a mutivariable one.
Better avoid to keep the focus on the mathematical problem - but stay being distracted by physical side effects.
This way you will never understand how to set a differential equation - is this really your goal ?
Your solution to diff equation is OK.
Your solution to the original solution problem is not OK. Volume/concentration ratio of water/alcohol solution is not linear. If you mix one gallon of water to one gallon water, you of course get 2 gallons. The same goes with one gallon of alcohol mixed with one gallon alcohol. BUT: If you mix one gallon of ethanol with one gallon of water, you only get 1.9 gallons of solution.
Solution 1: Do not use water/alcohol solutions in math problems so no wiseguy can come and make remarks.
Solution 2: Do not mix ethanol with water. You have much more to drink.
isn't this due to hydrogen bonding , so the change in volume is not zero and change in enthalpy is negative? After some googling the weak hydrogen bond interactions seem to play a larger role in the ethanol-water structure and the maximum hydrogen bonding is in 20% (v/v).
Quite estreme differences between 15 vol% and 50 vol% alcohol. Thx to witchcraft 4 gallons go in and 4 gallons go out but still the level in the reactor ... I think it decreases, but how much is that actually? I really like nonlinear equations if they can be solved analytically, though it seems quite hopeless in this case. I think you'll need to calculate V(t) which is the volume in the reactor and with the mass of the alcohol you find the rest.
Also I remmeber myself mixing alcohol with water ..... it doesnt blend so well, if you dont give it a very good stirring. Thus might be that some guests get much more alcohol than others :-D.
Olli Turunen You are completely right, great analysis of the real situation
Although, I think he is just trying to model a solution for ideal circunstances defining a physical property that is not true in real life, ist est, volumes are additive for every non reactive mixture
It's quite interesting your remark, I hope he reads it and tries to solve it for a real case
Greetings from Colombia, excuse me for my broken English
If u have so much problem then dont watch his video.This is what I haye most ,they just point someones for those things that are not even wrong.Go get a life.Nobody likes you.
Well I doubt that this would be possible without using numerical integration. Anyway there is some really great lectures by MIT that deals with solving systems of linear constant coefficient ODEs
ua-cam.com/video/XDhJ8lVGbl8/v-deo.html
Take a bow
: )
guess how many friends I have...
hint its a constant
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just kidding its null
aaaaahhh I'm so alone
First, but no one cares ^^
I care ^_^
jeromesnail I care too
I care too
Hey blackpenrodpen , I solved it using simple maths and using concept of percantage.
My answer came 17.4 litre