Compartmental analysis-Two joined tanks

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Mildred

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Can somebody help me with this problem?

There are two tanks, Tank1 and Tank2 with volume of water V1 and V2 respectively.The two tanks are joined by a pipe so that liquid can flow between the tanks with a flow rate of Q1. Tank1 also leaks fluid with a flow rate of Q2. If D grams of salt is poured in Tank1 at time=0, what will the ratio of amount of salt in tank1 to the amount of salt in Tank2 be as time approaches infinity.

I have written the differential equations as:

A1' = (Q1/V2)*A2 - (Q1/V1)*A1 - (Q2/V1)*A1
A2' = (Q1/V1)*A1 - (Q1/V2)*A2

with A1 And A2 the amount of salt in Tank1 and Tank2 respectively.

Initial conditions are given by A1(0) = D A2(0) = 0
 
Mildred said:
Can somebody help me with this problem?

There are two tanks, Tank1 and Tank2 with volume of water V1 and V2 respectively.The two tanks are joined by a pipe so that liquid can flow between the tanks with a flow rate of Q1. Tank1 also leaks fluid with a flow rate of Q2. If D grams of salt is poured in Tank1 at time=0, what will the ratio of amount of salt in tank1 to the amount of salt in Tank2 be as time approaches infinity.

I have written the differential equations as:

A1' = (Q1/V2)*A2 - (Q1/V1)*A1 - (Q2/V1)*A1
A2' = (Q1/V1)*A1 - (Q1/V2)*A2

with A1 And A2 the amount of salt in Tank1 and Tank2 respectively.

Initial conditions are given by A1(0) = D A2(0) = 0
Click to expand...
The first equation can be written \(\displaystyle A1'= (Q1/V1)A2- (Q1/V1- Q2/V1)A1\).
Differentiate again to get the second order equation \(\displaystyle A1''= (Q1/V1)A2'- (Q1/V1- Q2/V1)A1'\)
Now replace A2' in that with the second equation and then solve the first equation (algebraically) for A2 to replace A2 in that equation.
 
A little more help

HallsofIvy said:
The first equation can be written \(\displaystyle A1'= (Q1/V1)A2- (Q1/V1- Q2/V1)A1\).
Differentiate again to get the second order equation \(\displaystyle A1''= (Q1/V1)A2'- (Q1/V1- Q2/V1)A1'\)
Now replace A2' in that with the second equation and then solve the first equation (algebraically) for A2 to replace A2 in that equation.
Click to expand...

I did as you sugested:

Then I have A1'' = (Q1/V2)((Q1/V1)A1 - (Q1/V2)A2) -(Q1/V1 -Q2/V1)A1'

which simplifies to (V2/V1)A1 - ((V2/Q1)^2)A1'' - (Q2/V1)(V2/Q1)A1' = A2

Now replace A2 in second equation: A2' = (Q1/V1)A1 -(Q1/V2) [(V2/V1)A1 - (V2/Q1)(V2/Q1)A1'' - (Q2/V1)(V2/Q1)A1']

which simplifies to A2' = (V2/Q1)A1'' + (Q2/V1)A1'

Now I am not sure what to do?

I tried substituting A1''= (Q1/V2)A2' - (Q1/V1 - Q2/V1)A1' back into the above equation

which simplifies to A1[(Q2/V1) - (V2/V1) - (V2/Q1)(Q2/V1)] = 0

I don't think this helps me


Can you help me please?

Thank you
 
Mistake in first equation

HallsofIvy said:
The first equation can be written \(\displaystyle A1'= (Q1/V1)A2- (Q1/V1- Q2/V1)A1\).
Differentiate again to get the second order equation \(\displaystyle A1''= (Q1/V1)A2'- (Q1/V1- Q2/V1)A1'\)
Now replace A2' in that with the second equation and then solve the first equation (algebraically) for A2 to replace A2 in that equation.
Click to expand...


Should the first equation not be written as: A1' =(Q1/V1)A2 -(Q1/V1 + Q2/V1)A1 ?
 
Sorry if I'm being obtuse but I think the answer is not solvable as written without some further assumptions which are 'not standard'. For example, assume that the connection with flow rate Q1 and the leak with flow rate Q2 are 'at the bottom'. Then, eventually (and as time approaches infinity), all the water (and all the salt, assuming ...) will leak out.

If we assume, just as an extreme assumption, the cross sectional area for both tanks are the same and the connection between the tanks is above the height of the smaller tank, no flow will occur but, depending on where the leak is. I know, but I said it was an extreme assumption (just to make a point).

Of course, if this were not a 'test/practice problem', one could make assumptions about these other needed conditions and, hopefully, develop an answer.
 
Last edited:
HallsofIvy said:
?? That's exactly what I have.
Click to expand...

quote_icon.png
Originally Posted by HallsofIvy
The first equation can be written [FONT=MathJax_Math-italic]A[FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]′[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]Q[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Math-italic]V[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Math-italic]A[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]Q[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Math-italic]V[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]Q[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Math-italic]V[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Math-italic]A[/FONT][FONT=MathJax_Main]1[/FONT].
Differentiate again to get the second order equation [FONT=MathJax_Math-italic]A[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]′′[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]Q[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Math-italic]V[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Math-italic]A[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]′[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]Q[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Math-italic]V[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]Q[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Math-italic]V[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Math-italic]A[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]′[/FONT]
Now replace A2' in that with the second equation and then solve the first equation (algebraically) for A2 to replace A2 in that equation.

[/FONT]


Should the first equation not be written as: A1' =(Q1/V1)A2 -(Q1/V1 + Q2/V1)A1 ?

The sign in the second bracket is positive in my equation and negative in yours. I am sorry if I did not make this clear in my previous post, I am new to using forums.

Any help is much appreciated.

Mildred
 
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