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
  • #1
Jason-Li
119
14
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
1610545186726.png


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:
1610545510046.png

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?
1610546231787.png


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

Thanks in advance.
Jason
 

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  • #2
Jason-Li said:
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...

Jason-Li said:
There is another thread
what thread ? Link ?

Jason-Li said:
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) ## ?
 
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  • #3
BvU said:
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/
 
  • #4
Ok, let's forget the other thread. ( but it did contain the problem specification :wink: )
I edited my post #2
 
  • #5
BvU said:
Ok, let's forget the other thread. ( but it did contain the problem specification :wink: )
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
 
  • #6
BvU said:
What is the dimension
Do you see a [length]3/[time] and doesn't that remind you of the dimension of ##q## ?
 
  • #7
BvU said:
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

$${A_2\,d \,h_2\over dt}={A_1\,d\, h_1\over dt}-Q$$
$${A_2\,d \,h_2\over dt}={A_1\,d\, h_1\over dt}-{P_in-P_out\over R_f}$$
 
  • Like
Likes BvU
  • #8
Jason-Li said:
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?
 
  • #9
Yes. Next step ?
(Hint: we need another relevant equation that wasn't in the list thus far :wink: )
 
  • #10
BvU said:
Yes. Next step ?
(Hint: we need another relevant equation that wasn't in the list thus far :wink: )

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?
 
  • #11
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 ?
 
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  • #12
BvU said:
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.
 
  • #13
Jason-Li said:
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
Jason-Li said:
D can be canceled out
You can not simply divide by ##D## in one term on side of an equation :mad: ! 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 ?
 
  • #14
@BvU

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

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?
 
  • #15
$${-A_2*h_2}={A_1*h_1}$$ this is the result of an integration, but without the initial conditions !
$$-A_2*(h_2-h_{2,0})=A_1*(h_1-h_{1,0})$$

Jason-Li said:
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 ?
 
  • #16
@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)
 
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  • #17
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
Jason-Li said:
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 ?
 
  • #18
@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:
$${A_2\Delta h_2 = A_1 \Delta h_1}-{({\Delta h_1*D*g})-({\Delta h_2*D*g})\over R_f}$$
 
  • #19
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 ?
 
  • #20
BvU said:
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.
 
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  • #21
Jason-Li said:
So by changing the pressures to the other formula it would still be a flow due to the Rf 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
 
  • #22
BvU said:
$$Q={P_{\text {in}}-P_{\text {out}}\over R_f}= {({h_1*D*g})-({h_2*D*g})\over R_f}$$
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?
 
  • #23
Jason-Li said:
if I change Equations 1 and 2 to:
You can't just change an equation :mad: ! The dimensions left and right become different !
All three equations are in [##Q##] = m3/s
[##Ah##] is m3
 
  • #24
BvU said:
You can't just change an equation :mad: ! The dimensions left and right become different !
All three equations are in [##Q##] = m3/s
[##Ah##] is m3

:oops::oops::oops: Apologies, I'll look at this again.
 
  • #25
@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
$${dh \over dt} = {Q_i-Q_o \over A}$$
 
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  • #26
Jason-Li said:
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})##?

##\ ##
 
  • #27
BvU said:
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
 
  • #28
Jason-Li said:
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.
 
  • #29
BvU said:
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?
 
  • #30
Jason-Li said:
I don't think so
Think again.

Jason-Li said:
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.

Jason-Li said:
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 ?
 
  • #31
Jason-Li said:
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}\;## :smile:

##\ ##
 
  • #32
BvU said:
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?

BvU said:
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.
BvU said:
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.
BvU said:
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:
1610997947589.png

But we are already at this point in post 21
 
  • #33
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)##.

Jason-Li said:
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 }}\ \ ## o_O .

Note: the ... part is constant and will go into the ##k## and that means
Jason-Li said:
so k would be DG/rf
Is not completely correct.
---- also: try to be consistent in your notation: Dg/Rf
So far step 1.
 
  • #34
@BvU
Apologies I seem to be losing my mind... the only explanation for not seeing thiso_O 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:
$${dh_2\over dt}=...$$
 
  • #35
Jason-Li said:
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) :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 ?

##\ ##
 
<h2>1. What is a tank problem with multiple tanks?</h2><p>A tank problem with multiple tanks refers to a mathematical problem that involves calculating the amount of liquid in a series of interconnected tanks over a period of time. This type of problem is commonly encountered in fields such as engineering, physics, and chemistry.</p><h2>2. How do I approach solving a tank problem with multiple tanks?</h2><p>The first step in solving a tank problem with multiple tanks is to carefully analyze the problem and identify all the given information. Then, you can use mathematical equations and principles such as conservation of mass and volume to set up a system of equations and solve for the unknown variables.</p><h2>3. What are some common challenges when working through a tank problem with multiple tanks?</h2><p>Some common challenges when working through a tank problem with multiple tanks include keeping track of the different variables and units, understanding the relationships between the tanks, and setting up the correct equations. It is important to approach the problem systematically and double-check your calculations to avoid errors.</p><h2>4. Are there any tips or tricks for solving tank problems with multiple tanks?</h2><p>Some tips for solving tank problems with multiple tanks include drawing diagrams to visualize the problem, labeling all the tanks and variables, and breaking down the problem into smaller, more manageable parts. It can also be helpful to practice with simpler tank problems before attempting more complex ones.</p><h2>5. How can I check if my solution to a tank problem with multiple tanks is correct?</h2><p>To check if your solution to a tank problem with multiple tanks is correct, you can plug your values back into the original equations and see if they satisfy all the given conditions. You can also try using a different method or approach to solving the problem to see if you get the same result. Additionally, it is always a good idea to have someone else review your work to catch any potential errors.</p>

1. What is a tank problem with multiple tanks?

A tank problem with multiple tanks refers to a mathematical problem that involves calculating the amount of liquid in a series of interconnected tanks over a period of time. This type of problem is commonly encountered in fields such as engineering, physics, and chemistry.

2. How do I approach solving a tank problem with multiple tanks?

The first step in solving a tank problem with multiple tanks is to carefully analyze the problem and identify all the given information. Then, you can use mathematical equations and principles such as conservation of mass and volume to set up a system of equations and solve for the unknown variables.

3. What are some common challenges when working through a tank problem with multiple tanks?

Some common challenges when working through a tank problem with multiple tanks include keeping track of the different variables and units, understanding the relationships between the tanks, and setting up the correct equations. It is important to approach the problem systematically and double-check your calculations to avoid errors.

4. Are there any tips or tricks for solving tank problems with multiple tanks?

Some tips for solving tank problems with multiple tanks include drawing diagrams to visualize the problem, labeling all the tanks and variables, and breaking down the problem into smaller, more manageable parts. It can also be helpful to practice with simpler tank problems before attempting more complex ones.

5. How can I check if my solution to a tank problem with multiple tanks is correct?

To check if your solution to a tank problem with multiple tanks is correct, you can plug your values back into the original equations and see if they satisfy all the given conditions. You can also try using a different method or approach to solving the problem to see if you get the same result. Additionally, it is always a good idea to have someone else review your work to catch any potential errors.

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