Comp Sci Seeking Guidance on Working Through Tank Problem with Multiple Tanks

[BvU]

I feel like I need a time aspect on the Right hand side of this formula?

To avoid confusion: the expression time aspect does not mean anything to me. Can you explain what you mean ?

Going back to post #21: eight variables, 5 knowns, 3 unknowns and 3 equations. I see I have given away the recipe already but you started to cook up something different.

Let me be more precise:

1. Work around $(1)$ and $(2)$ to get $Q$ as a function of the driving force $h_1-h_2$. Don't use anything from $(3)$.

2. Substitute that $\ Q(h_1-h_2)\$ in $(3)$. We now have one first order differential equation (DE) in terms of one time-dependent variable $h_1-h_2$ and the form is like $\ y'= -ky\$ with $k>0$

$Q$ has now been eliminated and can be forgotten. If we need it, we can dig up $(3)$ or one of the other two.

3. Invent some initial condition $\;h_{1,0}, h_{2,0}\;$ and and some parameter values (the 5 constants) and solve the DE . Make some plots.

4. See if it all also works if $\;h_{1,0}< h_{2,0}\;$

$\$

 

[Jason-Li]

Think again.

For the form of y'=-ky k represents a constant, so k would be DG/rf and the y value would be the driving force of h1-h2?

What is that ? Or do I have to ask: what are those ? Can you spell it out for me ?When does it (they?) become zero ?

The difference in the hieght gives the driving force e.g. when h1>h2 the pressure in the first tank will cause flow to try and equalise the tanks. h1-h2 will be zero when the tanks are at equilibrium.

To avoid confusion: the expression time aspect does not mean anything to me. Can you explain what you mean ?

I thought before I would need a value of "t" on the right hand side I could hypothetically substitute in.

1. Work around (1) and (2) to get Q as a function of the driving force h1−h2. Don't use anything from (3).

Not going to lie I am unsure of how to do this without touching equation 3, I've done some reading and youtube video watching but can't quite find what I need, e.g. (this uses a state-space model).

I can't equate them as I will lose the Q variable I need to create \ Q(h_1-h_2)\ I could add them together to create
$$2Q = A_2{dh_2\over dt}-A_1{dh_1\over dt}$$
However even if i rearrange this I don't see how I can get to the $$\ Q(h_1-h_2)\$$ ?
Sorry to keep bothering you I feel like we are near a turning point though!

I have spoken to someone else and they advised this:

But we are already at this point in post 21

 

[BvU]

Task at hand:

1. Work around $(1)$ and $(2)$ to get $Q$ as a function of the driving force $h_1-h_2$. Don't use anything from $(3)$.

Not going to lie I am unsure of how to do this without touching equation 3, I've done some reading and youtube video watching but can't quite find what I need

I'm baffled that you go searching instead of doing some simple math. I can't provide more guidance without robbing you of the exercise, but that can't be helped. See below for what I meant. I am even more baffled that you go so far out in searching, instead of looking in the other thread that you yourself mentioned in post #3 !

Note: I don't mean to be critical or anything, the subject matter is obviously new to you and you need to get acquainted with it. So:What I meant is: $\ \$You have $$
\begin {align*}

Q &= - A_1\,{d\, h_1\over dt} \tag{1}\\ \ \\

Q & = \phantom{-} A_2\,{d \,h_2\over dt} \tag{2}\\ \ \\

\end{align*}$$ and you want something in terms of $h_1-h_2$.

Perhaps it is a good idea to divide left and right side of $(1)$ by $A_1$ and divide left and right side of $(2)$ by $A_2$. You then have expressions for $\ \displaystyle{dh_1\over dt}$ and for $\ \displaystyle {dh_2\over dt}$. What do you get for $\ \displaystyle{d(h_1-h_2)\over dt \ \ }$ ?

So $\ \displaystyle{Q=\ ...\ {d(h_1-h_2)\over dt }}\ \$ .

Note: the ... part is constant and will go into the $k$ and that means​

​

so k would be DG/rf​

​

Is not completely correct.​

---- also: try to be consistent in your notation: Dg/R_f​

​

So far step 1.

 

[Jason-Li]

@BvU
Apologies I seem to be losing my mind... the only explanation for not seeing this No offense taken!
So working (1) which you have kindly done.
\begin {align*}
{Q\over A_1}= -{d\, h_1\over dt} \tag{1}\\ \ \\
{Q\over A_2} ={d \,h_2\over dt} \tag{2}\\ \ \\
{Q\over A_1}+{Q\over A_2} =-{d\, h_1\over dt}+{d \,h_2\over dt} \tag{1&2}\\ \ \\
-{Q\over A_1}-{Q\over A_2} ={d\, h_1\over dt}-{d \,h_2\over dt} \tag{1&2 x-1}\\ \ \\
-{Q\over A_1}-{Q\over A_2} ={d\,(h_1-h_2)\over dt} \\ \ \\
Q(-{1\over A_1}-{1\over A_2})=-{d\,(h_1-h_2)\over dt} \\ \ \\
Q= {d\,(h_1-h_2)\over dt} / (-{1\over A_1}-{1\over A_2})\\ \ \\
Q= {d\,(h_1-h_2)\over dt} / ({-A_1-A_2\over A_1A_2})\\ \ \\
Q= -{A_1A_2\over A_1+A_2} * {d\,(h_1-h_2)\over dt} \\ \ \\
\end{align*}
Does this look correct now? with the k factor being:
If so for step 2:
$$-{A_1A_2\over A_1+A_2} * {d\,(h_1-h_2)\over dt} = {D\,g\over R_f}\,(h_1-h_2)$$
So would I work this through until I have:
$${dh_2\over dt}=...$$

 

[BvU]

So would I work this through until I have: $\ \displaystyle {dh_2\over dt}$

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) .

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 ?

$\$

 

[Jason-Li]

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) .

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

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?

 

[BvU]

I don't really see how they did the step shown by the red arrow

$$\Bigl .\ln x\,\Bigr |^x_{x_0} = \ln x -\ln x_0$$

Along the right lines here?

I don't think so !

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 ?

$\$

 

[Jason-Li]

$$\Bigl .\ln x\,\Bigr |^x_{x_0} = \ln x -\ln x_0$$

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 ?

$\$

Thanks for that explanation.

So their final answer of

is suitable for part (a)?

Is their method of identifying the time constant of 63.2% also suitable?

 

[BvU]

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?

 

[Jason-Li]

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

 

[BvU]

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 $\ \displaystyle{ \int {dQ\over dt}}$.

Think I have a lot of doubts at the moment, apologies.

It's always good to be critical . And check things.