Mixing Problem
HTML-код
- Опубликовано: 23 окт 2024
- In this video, I explained how to set up the differential equation to solve a mixing problem. It is assumed that there is uniform mixing of the solution in the tank at any time.
Get your t-shirt here:
shorturl.at/HNUX1
The Quality of this shirt is impressive. You will also be helping me a lot!
Colors: Black, Navy Blue, Gray, Red
This video takes me back 60 years to my undergrad days. Solutions to real-world problems are always worth learning. Your teaching technique is excellent.
I love mixture problems because they always have a solution (budum tss) thanks to you Prime Newtons! A fantastic video sir 🔥🔥
This guy is the best mathematician I have ever seen Keep going like this professor!
Problem is interesting but cracking a differential equation at 15 minutes is critical part.
Problem of everyday application for chemical engineers. Your excellence solving this problem
A golden video for chemical engineering students.
A Chemical Engineer must take in account that, when you mix alcohol and water there will be an exothermal contraction of the volume of the solution. That contraction is dependent on the diference of concentration on the inflow conc and the instant concentration inside the tank. So you should consider that the outflow concentration is slitely higher in volume. This problem wold be insanely hard this way. The problem should be put in terms of massflow instead of volume. By the way, I´m a huge fan of this channel.
@@eduardolacusta Ok, I’ll revise my statement to beginner chemical engineering students.
Never boring but your amazing presentation with flawless articulation on the subject has really impacted me..... certainly it is mesmerizing....
A good old mixture problem. Cant wait to get off work and solve this one.
" It is assumed that there is uniform mixing of the solution in the tank at any time."---damn big and unwarranted assumption
Yes. Given the flow rate is in minutes rather than seconds, I would have thought adding the sentence: "Assume the tank is stirred continuously for the duration." or some such would address that aspect. If the quoted rates were for hours rather than minutes I would have thought you could just rely on simple diffusion to do the job, as long as you can ignore evaporation. Also, as others have pointed out, it should also say "Assume no volume change on mixing." Highlights the differences between how a physicist and a mathematician thinks.
The only assumption not stated is that the function y(t) = amount of alcohol/volume of solution would properly describe an average concentration throughout the volume. The concentration at exit may not be equal to y(t) unless the solution is considered 'fully mixed' such that the concentration at the exit is always equal to the average concentration of the volume for the entire time interval considered.
The problem would be more difficult to resolve with the assumption that the solution in the container is "fully mixed" before exit. Your comment is much appreciated 👍
That was excellent… you went through every step slowly and explained everything you were doing
Excellent solution to an apparently very difficult math problem. Thank you for sharing.
When I was doing physics calculations I made a point of periodically stopping to assure myself that my units were correct. Example - in {D = 1/2 a*t^2}. acceleration has units [m per sec^2] . time^2 has units [sec^2}. IN the calculation a*t^2, the two appearances of m/sec^2 cancel out and leaves just m. Dimensionally, (1/2at^2) returns the same units as (D) Units on both sides of an equation must match. Any elements of a calculation that add or subtract must match. (you cant add 5 m and 2 miles and get 7 meter miles)
These is a trivial example, but when making more complex calculations about observable, measurable things, watching the dimension of your calculations can keep you from charging off in strange directions and lets you know you being consistent.
I agree; this is so important and useful, but often not emphasised; it has saved me many hours of wasted time. I used to say to my students "if the units are correct, it doesn't guarantee your equation is right, but if they're not correct, it is definitely wrong!"
Genius exercise. My favourite till now
Good illustration of why differential equations are an essential tool for physics
This is very good. Keep up the style and diversity of problems - it is very inspiring and no doubt has helped many who visit your channel. I just want to point out that within your Rate Out formulation, there is the implicit assumption of instantaneous homogenous mixing within the main tank before outflow. This is reasonable given we are looking at a model scenario of alcohol and water, but in practice may not apply with less miscible fluids. Not a big point of contention, but I imagine we could develop an equivalent question for say oil and vineagar where it is more complicated to work out the effective boundary mixing before one fluid is expelled from the tank. Thank you!
Very nice video. How did You find the integrating factor ?
Never stop watching Prime Newton's video, those who stop watching they stop learning.
These can be great fun. IF the outflow were equal to the inflow (volume of in tank is constant), they come down to simple exponential equations. N1 = N0 + (Nf-N0)e^(-lambda*t) where lambda is 'rate/volume' and t is time. (initial concentration 10%, final concentration 50%, lambda 4gpm/50g = 0.08 / m) N1 = 10% + (50%-10%)e^(-.08*t).
The equal inflow versus outflow is a special case that occurs quite often. Ventilating smoke or contaminant from a room is one example. Know the fan flow rate and size of room and you know how long it will take. (assuming perfect mixing so that outflow concentration equals room concentration)
Wow, that's the next level of difficulty. At what level is it studied? I believe we studied that in the 3rd year of University on physics department, the Differential Equations Class, we call it also MathematicalPhysics) and stil don't remember such problems, were mixture were involved (maybe I forgot already, it was quite while LOL)ю When I saw the mixture problem I thought of those problems to study perentage, which are statical, were some fixed amount is added at a time and no in/outflow happening. Wow that gives me real flashbacks of those times in the University. I'm impressed how broad is your scope is, sir Prime Newtons))
2nd year, 1st year is calculus. Ordinary differential equations is during the second year, usually after calculus III.
It was nice seeing a difficult hard math way to rhe way engineers and physicists solve the problem. I enjoyed seeing a rates priblem again without the way I remember from after differentiations and Integrations Calculus.
Since a presumed uniform 2 gallons/min of alcohol will fill the tank and (2.5 - (t/(t -1)2.5) leaves tge tank on a volume change of d(volume of mixture)/dt = volume change in liquid = something like (50 + (4 - 5)t) times t/(t - 1) sort of thing! I'll have to remember how I did the same problem awhile ago. It is still a Y' = KY linear 1st Order differential equation instead of jumping from exp() and then exp(ln()) type solution your method.
As RUclips grammar and spell check fail me: It looks as if we can solve the problem as Volume in the tank = 50 gallons at time t=0 and V = 0.1A + 0.9W. Where 0.1A is volume of alcohol at time t=0 equalling 5 gallons and 0.9W is volume of water at time t=0 equalling 45 gallons if water:
Volume(t) = [5 + (0.5(4t) - 5t] + [45 + (0.5(4t) - 5t]. At time t=10 Volume of mixed liquid is 90% of original t=0 volume V(0) = 50 gallons or 40 gallons by losing 1 gallon/min of in (4 - 5 out)! That is how I solved problems in Calculus like this by analyzing Volume change per time by the tank liquid and its volume changes in time.
Thank you ❤
Prime Newtons always mixes pizzazz with excellent teaching. Watching these videos is the solution! 🎉😊
What is the concentration of alcohol in the last drop of the mixture leaving the tank just before it is empty after 50 min ?
That's an interesting one. We could guess. Let's assume the last gal.
@@PrimeNewtons
I didn't assume but I calculated: the concentration of alcohol in the tank
rises whereas the volume decreases to zero after 50 min. The (maximum) alcohol- concentration is then 50 % , isn't it ?
This one was to hard for me to figure out myself. I did understand the explanation.
❤❤❤thank you sir...
شكراً لك
The total volume decreases when mixing ethanol with water. One gallon ethanol mixed with one gallon water gives less than two gallons of mixed liquid.
That means the tank drains faster than expected from an assumption of no volume change from mixing.
While an interesting fact to take into consideration, the problem doesn't say that the alcohol was ethanol or that it was dissolved in water.
@@jumpman8282 For liquids to be able to dissolve into each other, attractive forces between respective molecules are needed that usually reduce the volume after mixing.
Ethanol is usually what a non-chemist means when mentioning "alcohol".
Water is usually the implied solvent for ethanol if not explicitly stated.
No one cares about chemistry here
@@jacoboribilik3253 How do you know that?
what if the tank is one size but the initial liquid is another size
Why is the final answer in gallons? y describes the amount in a volume so i think you should have said 13.4%
Remember that we integrated dy/dt with respect to t. So, y is no longer gal per minute but just gal.
Genial
Chem Engineering problems let's gooooo
It seemed fairly simple until I realised the rate of accumulation is a variable 😅
I think you made a mistake in interating 2(50-t)^-5 it must be -1/2 (50-t)^-4 since you divide 2 by the new power which is -4
Yes, but we also divide by the derivative of (50 − 𝑡), which is (−1), so the integral does indeed become (1 ∕ 2)(50 − 𝑡)⁻⁴.
15:41 why minus there?
(t-50) = -(50-t)
@@hr5492 where do you see t-50?
When you integrate a function where the numerator is the derivative of the denominator, it turns into ln. However in this case, this is not true (d/dt[50-t]≠1, d/dt[50-t]=-1). He didn’t show this step but he rewrote 50-t as -(t-50) and he took the minus outside of the integral so now the numerator is the derivative of the denominator with the minus out the front.
I don’t see t-50 in this video, it’s 50-t the whole time