Compartment Modeling

H

HankSpank

New member
Joined
Sep 11, 2014
Messages
4
(1) Consider a pair of tanks in which tank 1 drains into tank 2 at a
rate of 5 liters/minute. Tank 1 intially contains 100% ethanol,
but is being refilled with pure water at a rate of 5 liters/minute.
Tank 2 initially has pure water in it and drains at a rate of 5
liters/minute. Tank 1 has a volume of 100 liters, and tank 2
has a volume of 10 liters.
(a) Write down ODEs for the amount of ethanol in each tank
and solve them. (Note that the amount in tank 1 does not
depend on the amount in tank 2.)
(b) Find the maximum amount of ethanol present in tank 2
(first find when this happens).


Anyways, I think I have the rate of change for concentration of the first tank. x1=(100/e)e-5t/100​ however further than that I am absolutely stuck. Any help is much appreciated.

Edit: My bad, accidentally left the t out of the function for x1​ that I found.
 
Last edited:
I'm not sure where you got the expression for the rate of change for concentration of the first tank. It is not something I would get. The following offers a hint on how to proceed:

Let Wj be the amount of water and Ej be the amount of ethanol in tank j, i.e. W1 is the amount of water in tank 1, etc. For tank 1 we are told
W1' = 5 (pure water is going into tank 1 at the rate of 5 liters/min)
and
E1' = -5 (ethanol is going out of tank 1 at the rate of 5 liters/min)
where the ' means derivative wrt the time t.

You are also given initial conditions, for example
W2(0) = 10 (the amount of pure water in tank two at time t=0 is 10 liters)
and
E2(0) = 0 (the amount of ethanol in tank two at time t=0 is 0).

From those equations, the other corresponding 4 equations, and the indicated range of the functions, you should be able to compute the amount of ethanol and water in either tank at any time. Hint, the contents in tank 2 change from water to ethanol and back to water.
 
Ishuda said:
I'm not sure where you got the expression for the rate of change for concentration of the first tank. It is not something I would get. The following offers a hint on how to proceed:

Let Wj be the amount of water and Ej be the amount of ethanol in tank j, i.e. W1 is the amount of water in tank 1, etc. For tank 1 we are told
W1' = 5 (pure water is going into tank 1 at the rate of 5 liters/min)
and
E1' = -5 (ethanol is going out of tank 1 at the rate of 5 liters/min)
where the ' means derivative wrt the time t.

You are also given initial conditions, for example
W2(0) = 10 (the amount of pure water in tank two at time t=0 is 10 liters)
and
E2(0) = 0 (the amount of ethanol in tank two at time t=0 is 0).

From those equations, the other corresponding 4 equations, and the indicated range of the functions, you should be able to compute the amount of ethanol and water in either tank at any time. Hint, the contents in tank 2 change from water to ethanol and back to water.
Click to expand...

I should have specified; the mixtures in the tanks are assumed to mix instantaneously. While it is true that there is pure ethanol coming out of tank one at t=0, the amount of ethanol that comes out of tank 1 is dependent on the concentration of the mixture for all time t>0.
 
HankSpank said:
(1) Consider a pair of tanks in which tank 1 drains into tank 2 at a
rate of 5 liters/minute. Tank 1 intially contains 100% ethanol,
but is being refilled with pure water at a rate of 5 liters/minute.
Tank 2 initially has pure water in it and drains at a rate of 5
liters/minute. Tank 1 has a volume of 100 liters, and tank 2
has a volume of 10 liters.
(a) Write down ODEs for the amount of ethanol in each tank
and solve them. (Note that the amount in tank 1 does not
depend on the amount in tank 2.)
(b) Find the maximum amount of ethanol present in tank 2
(first find when this happens).


Anyways, I think I have the rate of change for concentration of the first tank.

x1=(100/e)e-5/100​ ← This is incorrect because it is not a function of time


however further than that I am absolutely stuck. Any help is much appreciated.
Click to expand...

.
 
HankSpank said:
I should have specified; the mixtures in the tanks are assumed to mix instantaneously. While it is true that there is pure ethanol coming out of tank one at t=0, the amount of ethanol that comes out of tank 1 is dependent on the concentration of the mixture for all time t>0.
Click to expand...

And for tank 2?
 
Subhotosh Khan said:
​(1) Consider a pair of tanks in which tank 1 drains into tank 2 at a
rate of 5 liters/minute. Tank 1 intially contains 100% ethanol,
but is being refilled with pure water at a rate of 5 liters/minute.
Tank 2 initially has pure water in it and drains at a rate of 5
liters/minute. Tank 1 has a volume of 100 liters, and tank 2
has a volume of 10 liters.
(a) Write down ODEs for the amount of ethanol in each tank
and solve them. (Note that the amount in tank 1 does not
depend on the amount in tank 2.)
(b) Find the maximum amount of ethanol present in tank 2
(first find when this happens).


Anyways, I think I have the rate of change for concentration of the first tank.

x1=(100/e)e-5/100​ ← This is incorrect because it is not a function of time


however further than that I am absolutely stuck. Any help is much appreciated..
Click to expand...
Sorry, I accidentally left out the t. The correct function is as follows:

x1=(100/e)e-5t/100​
 
HankSpank said:
Same for tank 2: what comes out of both tanks is based on the contents. If a tank is 55% ethanol at some given time, then the flow out at that time is 55% ethanol.
Click to expand...

Do a Google search to get an example for three tanks of brine and apply the process to two tanks, i.e. search for
Math 2250 Final Exam Take-Home Portion Solutions
 
You must log in or register to reply here.
Top