Seeking Guidance on Working Through Tank Problem with Multiple Tanks

In summary: Sorry yeah I used little p for density.So putting D back in for p$${A_2\,d \,h_2\over dt}={A_1\,d\, h_1\over dt}-{({h_1*D*g})-({h_2*D*g})\over R_f}$$Then D can be canceled out still to make :$${A_2\,d \,h_2\over dt}={A_1\,d\, h_1\over dt}-{({h_1*g})-({h_2*g})\over R
  • #36
BvU said:
That is not the idea for this exercise, no. You now have $$ {d\,(h_1-h_2)\over dt} =
-{D\,g\over R_f} {A_1+A_2\over A_1 A_2} \,(h_1-h_2)$$Which is of the form ##\ y'= -k\,y\ ##, so easily solved if the initial ##h_{1,0}## and ##h_{2,0}## are given .

When I re-read your post #3 I understand where your interest in ##h_2## comes from. Well, at least now you have the answer for part (b) :wink: .

For part (a): Do you see a path to express ##h_2## when ##h_{1,0}## and ##h_{2,0}## are given and ##h_1-h_2## is known as a function of time ?

##\ ##

For Part (b) I can basically follow the other thread as we are at a point now that I can bring the equation to meet the equation here although I don't really see how they did the step shown by the red arrow
1611078081464.png


For part (a) could I integrate the left hand side between limits of h1-h2 and h1,0-h2,0 to give:
$$(h_1-h_2)-(h_{1,0}-h_{2,0}) = -{D\,g\over R_f} {A_1+A_2\over A_1 A_2} \,(h_1-h_2)t$$
$$-h_2 = -{D\,g\over R_f} {A_1+A_2\over A_1 A_2} \,(h_1-h_2)t+(h_{1,0}-h_{2,0})-h_1$$

Along the right lines here?
 

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  • #37
Jason-Li said:
I don't really see how they did the step shown by the red arrow
$$\Bigl .\ln x\,\Bigr |^x_{x_0} = \ln x -\ln x_0$$

Jason-Li said:
Along the right lines here?
I don't think so !

Jason-Li said:
For part (a) could I integrate the left hand side between limits of h1-h2 and h1,0-h2,0 to give:
What left hand side ? The ##\displaystyle{dx\over x}\ ## integral was done and gave ## \ln x -\ln x_0## !

How about 'stealing' the answer from the other thread ? Or don't you like it :rolleyes: ?

##\ ##
 
  • #38
BvU said:
$$\Bigl .\ln x\,\Bigr |^x_{x_0} = \ln x -\ln x_0$$

I don't think so !

What left hand side ? The ##\displaystyle{dx\over x}\ ## integral was done and gave ## \ln x -\ln x_0## !

How about 'stealing' the answer from the other thread ? Or don't you like it :rolleyes: ?

##\ ##
Thanks for that explanation.

So their final answer of
1611160223059.png

is suitable for part (a)?

Is their method of identifying the time constant of 63.2% also suitable?
 
  • #39
Jason-Li said:
So their final answer of ... is suitable for part (a)?

Is their method of identifying the time constant of 63.2% also suitable?
What do you think? Can you find fault?
 
  • #40
BvU said:
What do you think? Can you find fault?

I don't think I can see a mistake with the formula however going back to post #35 where you said to find h2, the attached image in #38 isn't expressly looking for h2 rather dh2/dt but this is likely still correct. Think I have a lot of doubts at the moment, apologies.

In terms of part b if my "time is over tau then the time constant should be 63.2% in line with first order systems
 
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  • #41
Jason-Li said:
the attached image in #38 isn't expressly looking for h2 rather dh2/dt but this is likely still correct
I see what you mean. But I suppose an extra integration is unavoidable. One way to look at it: ##h_2-h_1## (and thereby ##Q##) follows from solving the differential equation & initial conditions. Subsequently ##h_2## follows from ##\ \displaystyle{ \int {dQ\over dt}}##.
Jason-Li said:
Think I have a lot of doubts at the moment, apologies.
It's always good to be critical :smile: . And check things.
 
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