Seeking Guidance on Working Through Tank Problem with Multiple Tanks

[Jason-Li]

Homework Statement

FIGURE 2 shows two cylindrical tanks interconnected with a pipe which has a valve that creates a constant resistance to flow of Rf when fully open. The height of liquid (of density D) in the first tank is Hin and the second tank Hout. The cross-sectional area of the first tank is Ain m2 and the second tank Aout m2.

Relevant Equations

Q = ( Pin - Pout) / Rf

Hello all I come to again seeking guidance... There is another thread but he seems to have pulled formulas from somewhere other than our learning materials to reach the conclusions.

Looking at this question and I can't figure out where to start really, the examples in out learning materials show the use of just one tank so its more easily done. For example with density being P:

But as qvi = qvo then the right hand side would be effectively 0 as density would also be constant so can't make use of that one..

Was thinking of making an like however unsure of how to work the Q equation into this?

If someone could assist in the way to go with this then I can work through it.

Thanks in advance.
Jason

 

[BvU]

Looking at this question

What is being asked ?

Note: look at the dimension of $Q$ [edit] never mind-- volume flow. I misinterpreted $h$ on first reading...

There is another thread

what thread ? Link ?

unsure of how to work the Q equation into this?

Assume density $D$ (is given) is a constant. What is the dimension of your $${A_1\,d\, \rho h_1\over dt}= {A_2\,d \,\rho h_2\over dt}$$ when left and right are divided by $\rho\ (\equiv D)$ ?

 

[Jason-Li]

What is being asked ?

Note: look at the dimension of $Q$ [edit] never mind-- volume flow. I misinterpreted $h$ on first reading...

what thread ? Link ?

Apologies, the question is as follows:
(a) Produce a mathematical model of the process to determine the change in height of fluid in the second tank when the valve is open.
(b) Determine the time constant for the system.

This is the thread, can't make sense of what is going on.
https://www.physicsforums.com/threads/tank-change-in-flow-process-modeling.966659/

 

[BvU]

Ok, let's forget the other thread. ( but it did contain the problem specification )
I edited my post #2

 

[Jason-Li]

Ok, let's forget the other thread. ( but it did contain the problem specification )
I edited my post #2

Yeah that will cancel out the densities on both side so that

$${A_1\,d\, h_1\over dt}= {A_2\,d \,h_2\over dt}$$

Not sure how to work in the Q figure unless I somehow equate Hin, Ain & Pin

 

[BvU]

What is the dimension

Do you see a [length]³/[time] and doesn't that remind you of the dimension of $q$ ?

 

[Jason-Li]

Do you see a [length]³/[time] and doesn't that remind you of the dimension of $q$ ?

Ah so I effectively have Volume / Time, hence volumetric flow rate hence Q = Volume / Time, So I could make the formula:

$$Q={A_1\,d\, h_1\over dt}= {A_2\,d \,h_2\over dt}$$

This look correct?

If so I could make

$${A_2\,d \,h_2\over dt}={A_1\,d\, h_1\over dt}-Q$$
$${A_2\,d \,h_2\over dt}={A_1\,d\, h_1\over dt}-{P_in-P_out\over R_f}$$

 

[Jason-Li]

Ah so I effectively have Volume / Time, hence volumetric flow rate hence Q = Volume / Time, So I could make the formula:

$$Q={A_1\,d\, h_1\over dt}= {A_2\,d \,h_2\over dt}$$

This look correct?

If so I could make

$${A_2\,d \,h_2\over dt}={A_1\,d\, h_1\over dt}-Q$$
$${A_2\,d \,h_2\over dt}={A_1\,d\, h_1\over dt}-{P_in-P_out\over R_f}$$

@BvU think I need to simplify this down further however for the question?

 

[BvU]

Yes. Next step ?
(Hint: we need another relevant equation that wasn't in the list thus far )

 

[Jason-Li]

Yes. Next step ?
(Hint: we need another relevant equation that wasn't in the list thus far )

So if I look at
$${A_2\,d \,h_2\over dt}={A_1\,d\, h_1\over dt}-{P_in-P_out\over R_f}$$
then if
$${P}={m\,g\over a}$$
So
$${P}={p\,g\,a\,h\over a}$$
So
$${P}={h*p*g}$$
So would it be productive to make:
$${A_2\,d \,h_2\over dt}={A_1\,d\, h_1\over dt}-{({h_1*p*g})-({h_2*p*g})\over R_f}$$
Then as pressure is constant
$${A_2\,d \,h_2\over dt}={A_1\,d\, h_1\over dt}-{({h_1*g})-({h_2*g})\over R_f}$$

This look correct? If so I will continue to simplify it down?

 

[BvU]

I lost the distinction between $P$ and $p$. Do you mean pressure and density (which is $D$ in the exercise, but usually designated $\rho$) ?

Coming back to $${A_2\,d \,h_2\over dt}={A_1\,d\, h_1\over dt}$$with the dimension of volumeflow. Check it: is it likely $h_i$ and $h_o$ move in the same direction ?

 

[Jason-Li]

I lost the distinction between $P$ and $p$. Do you mean pressure and density (which is $D$ in the exercise, but usually designated $\rho$) ?

Coming back to $${A_2\,d \,h_2\over dt}={A_1\,d\, h_1\over dt}$$with the dimension of volumeflow. Check it: is it likely $h_i$ and $h_o$ move in the dame dierction ?

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 :
$${A_2\,d \,h_2\over dt}={A_1\,d\, h_1\over dt}-{({h_1*g})-({h_2*g})\over R_f}$$

No in regards to the system as H1 decreases, H2 would increase.

 

[BvU]

No in regards to the system as H1 decreases, H2 would increase.

Correct. So $$\begin {align*} Q &= - {A_1\,d\, h_1\over dt} \\ \ \\ Q & = \phantom{-} {A_2\,d \,h_2\over dt} \end{align*}$$ (you want to be a bit more consistent with upper/lower case and subscripts)
and I strongly object against

D can be canceled out

You can not simply divide by $D$ in one term on side of an equation ! Check the dimensions !

Instead, you have $$ Q = {\text {see above}}$$

Now, do you see an opportunity to further simplify this ? Does the number of equations match the number of unknowns ?

 

[Jason-Li]

@BvU

Oh yeah sorry I will work to correct this issue in future. Was a bit lax in my workings there.

I could equate them
$${-A_2\,d \,h_2\over dt}={A_1\,d\, h_1\over dt}$$

I was thinking of doing simultaneous equations but I only have two equations for 4 variables?
I could make
$${-A_2*h_2}={A_1*h_1}$$

Again if i rearrange this and substitute back in it doesn't seem to get me any further?

 

[BvU]

$${-A_2*h_2}={A_1*h_1}$$ this is the result of an integration, but without the initial conditions !
$$-A_2*(h_2-h_{2,0})=A_1*(h_1-h_{1,0})$$

I only have two equations for 4 variables

I count 3 equations for 3 unknowns ($h_1, h_2, Q$) !? What is your 4th unknown ?

 

[Jason-Li]

@BvU Apologies, I was counting A_1 & A_2 as unknowns but they are still constant.

For your integration I could rearrange it to be

$${-A_2\over A_1} = {h_1-h_{1,0}\over h_2-h_{2,0}}$$

Can't see where to go from here (also you subscripts 0, is the t=0 or 0s)

 

[BvU]

It's not my integration ! you went from $${-A_2\,d \,h_2\over dt}={A_1\,d\, h_1\over dt}$$ to $${-A_2*h_2}={A_1*h_1}$$ in post #14 and I corrected. However, for the dynamic model this volume balance is not all that interesting. With $A_1\Delta h_1 = - A_2 \Delta h_2$ we can link $h_1$ and $h_2$. The interesting variable is the driving force

So can you now correct

So putting D back in for p
$${A_2\,d \,h_2\over dt}={A_1\,d\, h_1\over dt}-{({h_1*D*g})-({h_2*D*g})\over R_f} ? $$

and work it around to one equation with one unknown ?

 

[Jason-Li]

@BvU

Okay so I could change

$${A_2\,d \,h_2\over dt}={A_1\,d\, h_1\over dt}-{({h_1*D*g})-({h_2*D*g})\over R_f}$$

to

$${A_2\Delta h_2 = A_1 \Delta h_1}-{({h_1*D*g})-({h_2*D*g})\over R_f}$$

But the flow rate will also change with the pressure or height so:
$${A_2\Delta h_2 = A_1 \Delta h_1}-{({\Delta h_1*D*g})-({\Delta h_2*D*g})\over R_f}$$

 

[BvU]

This does not make sense. First two terms are volumes, third term is a volume flow ! Did you check the dimensions ?

Can you do the exercise without the tank on the right ?

 

[Jason-Li]

This does not make sense. First two terms are volumes, third term is a volume flow ! Did you check the dimensions ?

Can you do the exercise without the tank on the right ?

Ah thought I could because the flow rate was given by

$$Q={P_in-P_out\over R_f}$$
So by changing the pressures to the other formula it would still be a flow due to the Rf figure.

Not sure how to do this without the tank on the right as the flowrate is determined by the differential in pressure of the two tanks?

If i go back to:
$${A_2\Delta h_2 = A_1 \Delta h_1}-{({h_1*D*g})-({h_2*D*g})\over R_f}$$
Which I can equate to
$$\Delta h_2 = {A_1*\Delta h_1*R_f+D*g*(h_2-h_1)\over (A_2*Rf)}$$
And change that to
$$\Delta h_2 = {A_1*\Delta h_1\over A_2}+{D*g*(h_2-h_1)\over (A_2*Rf)}$$

Not sure if I'm going down the correct path here.

 

[BvU]

So by changing the pressures to the other formula it would still be a flow due to the R_f figure.

$$Q={P_{\text {in}}-P_{\text {out}}\over R_f}= {({h_1*D*g})-({h_2*D*g})\over R_f}$$
This was the third equation I was referring to in post #13. I now regret that at the time I didn't post it explicitly with the other two. Since then we are running around in circles and maze-like sidetracks ! So this time I'll post the complete the set here:
$$\begin {align*}
Q &= - A_1\,{d\, h_1\over dt} \tag{1}\\ \ \\
Q & = \phantom{-} A_2\,{d \,h_2\over dt} \tag{2}\\ \ \\
Q &= \phantom{-} {D\,g\over R_f}\,(h_1-h_2)\tag{3}
\end{align*}$$
no more, except initial conditions $\;h_{1,0}\;$ and $\;h_{2,0}\;$.

I count eight variables, of which 5 are known. There are three unknowns and three equations. So modelling-wise we are ok. We could give this to an equation solver like Aspen Custom Modeller and be happy with the outcome. However, this is an exercise and the system can be simplified:

Since there is only one driving force $(h_1-h_2)$ , we now try to work around to one equation in ${h_1-h_2}$ as unknown with an initial condition $\;h_{1,0}-h_{2,0}\;$. Namely by eliminating $Q$.
If you can do that, that would be nice. Otherwise, you can shop around in the other thread, or re-do the single tank exercise

 

[Jason-Li]

$$Q={P_{\text {in}}-P_{\text {out}}\over R_f}= {({h_1*D*g})-({h_2*D*g})\over R_f}$$
This was the third equation I was referring to in post #13. I now regret that at the time I didn't post it explicitly with the other two. Since then we are running around in circles and maze-like sidetracks ! So this time I'll post the complete the set here:
$$\begin {align*}
Q &= - A_1\,{d\, h_1\over dt} \tag{1}\\ \ \\
Q & = \phantom{-} A_2\,{d \,h_2\over dt} \tag{2}\\ \ \\
Q &= \phantom{-} {D\,g\over R_f}\,(h_1-h_2)\tag{3}
\end{align*}$$
no more, except initial conditions $\;h_{1,0}\;$ and $\;h_{2,0}\;$.

I count eight variables, of which 5 are known. There are three unknowns and three equations. So modelling-wise we are ok. We could give this to an equation solver like Aspen Custom Modeller and be happy with the outcome. However, this is an exercise and the system can be simplified:

Since there is only one driving force $(h_1-h_2)$ , we now try to work around to one equation in ${h_1-h_2}$ as unknown with an initial condition $\;h_{1,0}-h_{2,0}\;$. Namely by eliminating $Q$.
If you can do that, that would be nice. Otherwise, you can shop around in the other thread, or re-do the single tank exercise

Thanks for all the help, I feel I may be starting to understand this, thanks for the continued insight!

Okay looking at this if I change Equations 1 and 2 to:
$$Q = -A_1(h_1-h_{1,0})$$
$$Q = A_2(h_2-h_{2,0})$$
Then subtract so
$$0 = A_2(h_2-h_{2,0}) +A_1(h_1-h_{1,0})$$
Then also combine Equations 2 & 3:
$$0 = {D\,g\over R_f}\,(h_1-h_2)-A_2(h_2-h_{2,0})$$
Then equate
$$A_2(h_2-h_{2,0}) +A_1(h_1-h_{1,0}) = {D\,g\over R_f}\,(h_1-h_2)-A_2(h_2-h_{2,0})$$
Which can be made to
$$H_2=(-R_f*A_1(H_1-H_{1,0})-A_2*H_1*R_f+A_2*H_{2,0}*R_f+D*g*H_1) /(D*g)$$
I've eliminated Q but still not sure if I'm pursuing the correct path here?

 

[BvU]

if I change Equations 1 and 2 to:

You can't just change an equation ! The dimensions left and right become different !
All three equations are in [$Q$] = m^3/s
[$Ah$] is m³

 

[Jason-Li]

You can't just change an equation ! The dimensions left and right become different !
All three equations are in [$Q$] = m^3/s
[$Ah$] is m³

Apologies, I'll look at this again.

 

[Jason-Li]

@BvU
So would it work if I were to
$$0=A_2{dh_2\over dt}+A_1{dh_1\over dt}$$
and
$$0= {D\,g\over R_f}\,(h_1-h_2)-A_2{dh_2\over dt}$$
and
$$A_2{dh_2\over dt}+A_1{dh_1\over dt}={D\,g\over R_f}\,(h_1-h_2)-A_2{dh_2\over dt}$$
and then rearrange for $${dh_2\over dt}$$
Am I pursuing the wrong path here? I am trying to eliminate Q but having a hard time knowing how if I'm honest. Looking to try and make sense of it all and the learning materials are limited to say the least.

The single tank exercise I found quite a bit easier and made
$${dh \over dt} = {Q_i-Q_o \over A}$$

 

[BvU]

The single tank exercise I found quite a bit easier and made

And with $Q_i=0$ and $Q_o = {p\over R_f}$ you get the one-tank version of our exercise that I was aiming at:$$\begin{align*}A{dh\over dt} &= -{Dg\over R_f}\,h \tag{a}\end{align*}$$
The physics: the driving force to reduce $h$ is proportional to $h$ itself. Occurs in radioactivty, height of beer foam, discharging a capacitor, etc.

For the two-tank exercise: what is the driving force and how do you maniplulate $(1)$, $(2)$ and $(3)$ to get something of the form $y'= -ky$, like $(\text{a})$?

$\$

 

[Jason-Li]

And with $Q_i=0$ and $Q_o = {p\over R_f}$ you get the one-tank version of our exercise that I was aiming at:$$\begin{align*}A{dh\over dt} &= -{Dg\over R_f}\,h \tag{a}\end{align*}$$
The physics: the driving force to reduce $h$ is proportional to $h$ itself. Occurs in radioactivty, height of beer foam, discharging a capacitor, etc.

For the two-tank exercise: what is the driving force and how do you maniplulate $(1)$, $(2)$ and $(3)$ to get something of the form $y'= -ky$, like $(\text{a})$?

$\$

Well the driving force is the differential of h1 & h2.
Looking at equations 2 & 3 I could make (with the issue being "k" being positive in the form $y'= -ky$
$$A_2{dh_2\over dt}={Dg\over R_f}(h_1-h_2)$$
I feel like I need a time aspect on the Right hand side of this formula?

Thanks for sticking with me for this I am learning... slowly

 

[BvU]

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

Do you have that in $y'= -ky$ ??

You want something else on the left side ! Something that goes to zero by the time $h_1=h_2$ and you have reached equilibrium.

 

[Jason-Li]

Do you have that in $y'= -ky$ ??

You want something else on the left side ! Something that goes to zero by the time $h_1=h_2$ and you have reached equilibrium.

I don't think so due to the -ky side being positive in my example? also there is a h1 in the k value...

So on the left side, the flow would be zero at equilibrium? Also the difference between the two tanks so I could either include flow or equation 1?

 

[BvU]

I don't think so

Think again.

also there is a h1 in the k value...

You don't want that. Just like in the 1-tank case you want a constant factor there.

Well the driving force is the differential 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 ?