My Math Forum water remaining in tank

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October 28th, 2013, 12:58 AM   #1
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water remaining in tank

a recangular tank consist of water at height h, width,w and length,l and is allow to flow through a rectangular hole of width, 0.1w and height 0.1h
what is the water remain in the tank after s sec?

I am stuck somewhere around there.
Attached Images
 rate flow of water.jpg (30.6 KB, 348 views)

 October 28th, 2013, 09:37 AM #2 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: water remaining in tank You need to know how fast the water is flowing out of the tank. That's not given in what you've posted.
October 28th, 2013, 06:39 PM   #3
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Re: water remaining in tank

Hi chuackl,

Quote:
 a recangular tank consist of water at height h, width,w and length,l and is allow to flow through a rectangular hole of width, 0.1w and height 0.1h
Do you mean:

1) a rectangular hole of width 0.1w and height 0.1l (note the l, not h) ?

2) a rectangular hole of width 0.1w and height 0.1h, with $\ 0.1h=0.1h_0, \ h_0 \$ being the height at t=0 ?

3) a rectangular hole of width 0.1w and height 0.1h, with 0.1h changing as h changes ?

In case you mean 3), I have worked it out as follows:

$v= \text{ velocity} = \sqrt{2gh}$

$V= \text{Volume} = l \cdot w \cdot h$

$\frac{dV}{dt}=\ -\text{area of rectanglar hole } \cdot \text{ velocity } = -0.1w \cdot 0.1h \cdot \sqrt{2gh}=-0.01w \cdot h^{\frac{3}{2}}\sqrt{2g}$

$\frac{dh}{dt}=\frac{dV}{dt} \cdot \frac{dh}{dV} \$ by the chain rule, and

Since $\ h=\frac{V}{l \cdot w}, \ \frac{dh}{dV}=\frac{1}{l \cdot w}$

$\frac{dh}{dt}=\frac{dV}{dt} \cdot \frac{dh}{dV}=-\frac{0.01\cancel{w} \cdot h^{\frac{3}{2}}\sqrt{2g}}{l \cdot \cancel{w}}$

$\frac{dh}{dt}=-\frac{0.01\sqrt{2g}}{l} \cdot h^{\frac{3}{2}}$

$h^{-\frac{3}{2}} \ dh= -\frac{0.01\sqrt{2g}}{l} \ dt \$ and integrating, we get

$-2h^{-\frac{1}{2}}= -\frac{0.01\sqrt{2g}}{l}\cdot t + C_1$

$\frac{1}{\sqrt{h}}= \frac{0.01 \cdot sqrt{2g}}{2l}\cdot t + C_2=\frac{0.01}{l}\cdot \sqrt{\frac{g}{2}} \cdot t+C_2$

$\text{At } t=0, \ h=h_0 \text{ so } C_2=\frac{1}{\sqrt{h_0}}$

$\frac{1}{\sqrt{h}}= \frac{0.01}{l}\cdot \sqrt{\frac{g}{2}} \cdot t + \frac{1}{\sqrt{h_0}}$

$h=\frac{1}{\left(\frac{0.01}{l}\cdot \sqrt{\frac{g}{2}} \cdot t + \frac{1}{\sqrt{h_0}}\right)^2}$

$V(t)=\frac{l \cdot w}{\left(\frac{0.01}{l}\cdot \sqrt{\frac{g}{2}} \cdot t + \frac{1}{\sqrt{h_0}}\right)^2}$

The simulation in the attachment appears to verify this result. l, w, and h were arbitrarily chosen and the volume in the tank is shown for t=0 to t=100.
Attached Images
 tank_2013_10_28.jpg (78.8 KB, 324 views)

 October 29th, 2013, 06:26 AM #4 Member   Joined: Sep 2013 Posts: 43 Thanks: 0 Re: water remaining in tank Thank for your effort in showing the step! This is a different consideration from my original question. Which consider the hole area reduce with h and this will be the result. So it make sense that the water will never drain out from the tank since hole become smaller and smaller. I am sorry as forgot to mentioned where is the hole situated. I believe this model is consider the hole at the base of the rectangular?(the distance h of the hole to the surface of the water at time t is the same for the hole, since consider the hole can be made out of infinitely small partition) what happen when the hole located at the side of the rectangular tank.(the height of water from each small compartment are different)
October 29th, 2013, 08:17 PM   #5
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Re: water remaining in tank

Hello again,

Quote:
 I believe this model is consider the hole at the base of the rectangular?
Yes, that is correct.

Quote:
 what happen when the hole located at the side of the rectangular tank.(the height of water from each small compartment are different)
It will make a little difference, at least for a hole size of 0.1h for one of the dimensions.

Ignoring any effects of the water moving at slightly different speeds, I think that we can still find $\ \frac{dV}{dt} \$ as a function of h by integrating:

$\frac{dV}{dt}=\int_{0.1h}\,^0 \ 0.1w\sqrt{2g(h-\Lambda)} \ d\Lambda=\int_{0.1h}\,^0 \ 0.1w\sqrt{2g}(h-\Lambda)^{\frac{1}{2}} \ d\Lambda$

$=\left. -0.1w\sqrt{2g}\left(\frac{2}{3}\right)\left(h-\Lambda \right)^{\frac{3}{2}} \ \right|_{0.1h}^0$

$=-0.1w\sqrt{2g}\left(\frac{2}{3}\right)\left(h^{\fra c{3}{2}}-(0.9h)^{\frac{3}{2}}\right)$

$= -0.1w\sqrt{2g}\left(\frac{2}{3}\right) h^{\frac{3}{2}} \left(1-(0.9)^{\frac{3}{2}}\right)$

$\frac{dV}{dt}=-0.00974567 \cdot w \cdot h^{\frac{3}{2}} \cdot \sqrt{2g}$

Following the procedure in my previous post results in:

$V(t)=\frac{l \cdot w}{\left(\frac{0.00974567}{l}\cdot \sqrt{\frac{g}{2}} \cdot t + \frac{1}{\sqrt{h_0}}\right)^2}$

I have run simulations to verify the results and the equation matches the simulation closely. Using the same numbers as in the simulation given previously, with the hole in the bottom, after 20 seconds the volume is 213.6 cubic units. With the hole in the side, after 20 seconds the volume is 216.0 cubic units.

For a hole height of $\ h_h \$ (and width 0.1w) I get a more general formula of:

$V(t)=\frac{l \cdot w}{\left(\frac{0.1 \cdot \frac{2}{3} \cdot \left(1-(1-h_h)^{\frac{3}{2}}\right)}{l}\cdot \sqrt{\frac{g}{2}}\cdot t+\frac{1}{sqrt{h_0}}\right)^2}$

I have simulated various hole heights (mainly 0.1h and 0.5h) and I have found excellent agreement between the formula and the simulations' computed values.

October 31st, 2013, 01:12 AM   #6
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Re: water remaining in tank

Thank again for all the work shown.
My consideration appear to be same as urs.
I using double integration over the surface of the hole.
Now i been trying to find the volume remain for the case the height of the hole at the side be a constant H.
I have 3 question in this model.
a)if that the case, I would get a model as in attached picture?
b)For the 1st equation, i am not sure how to integrate it. Would be greatly appreciate if you can show the way again thank
c)when the water reach height H, 1st equation would no longer be valid is it since a negative number of power 1.5 will not be define. After time t=t' at h=H, it appear it will follow the 2nd equation?
Attached Images
 hole with constant height,H.jpg (33.0 KB, 255 views)

October 31st, 2013, 07:36 PM   #7
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Re: water remaining in tank

Quote:
 a)if that the case, I would get a model as in attached picture?
Yes, but:

$\frac{dV}{dt}=-\int_h\,^{h-H} \sqrt{2gh}(0.1l) \$ is incomplete because you need to denote the variable of integration. Since I believe that it is improper to have integration limits that are the same as, or functions of, the variable of integration, we need to introduce another variable, say $\ \Lambda \$, to represent the slices of the slot that we are integrating over, keeping h constant.

This allows us to integrate over the area of the slot, ending up with $\ \frac{dV}{dt} \$ as a function of $\ h \$ as desired. So using $\ \Lambda \$ the integral would be:

$\frac{dV}{dt}=-\int_{h-H}\,^h \sqrt{2g\Lambda}(0.1l) \ d \Lambda \$ Note: Did you mean $\ w \$ instead of $\ l \$?

But I ended up with the same answer that you did except the terms in the parenthesis are swapped because the result must be negative. (I also used w instead of l):

$\frac{dV}{dt}=-0.1w\cdot \frac{2}{3} \cdot \sqrt{2g}\left(h^{\frac{3}{2}}-(h-H)^{\frac{3}{2}}\right)$

Quote:
 b)For the 1st equation, i am not sure how to integrate it. Would be greatly appreciate if you can show the way again thank
I think that we both end up with a function like:

$\int \frac{dh}{h^{\frac{3}{2}}-(h-H)^{\frac{3}{2}}}$

I do not know how to do this integral and apparently W|A cannot do it in a reasonable amount of time either. A (very) quick search shows calculators that just use a constant velocity throughout the hole. Maybe another forum member knows how to do this integral, or perhaps numerical methods could be used.

Quote:
 c) ... After time t=t' at h=H, it appear it will follow the 2nd equation?
Correct, but again I would use $\ \Lambda \$:

$\frac{dV}{dt}=-\int_0\,^{h} \sqrt{2g\Lambda}(0.1w) \ d \Lambda$

which as you know can be worked fairly easily.

 November 2nd, 2013, 06:43 AM #8 Member   Joined: Sep 2013 Posts: 43 Thanks: 0 Re: water remaining in tank Yes i do agree for all the correction u made to the equation. For the equation may be i can use excel to approximate the result. The link below is just an example the way of approximation a function and i did applied it before to check the adiabatic gas equation on diatomic gas by considering P and V be 1. http://postimg.org/image/7wcq2wk5p/15822e45/ Maybe it is possible to apply the same thing for this question.(approximate the curve to multiple straight line) I am sorry for not able to share the result now as quite busy this week and hopefully can get it by next week
 November 2nd, 2013, 09:25 PM #9 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications Re: water remaining in tank I am not an Excel expert by any means, but I think that this equation will be harder to do since for each straight line approximation you will have to do an integration (Edit: Wrong, see my next post). You will also need an IF statement for the case when h
 November 3rd, 2013, 10:54 AM #10 Member   Joined: Sep 2013 Posts: 43 Thanks: 0 Re: water remaining in tank Sorry i don't know any programming language. Just noticed that the software u use t_step=0.0001. In order to calculate the h at time t=100second i need to repeat the sequence 100/0.0001=1000k times. I couldnt done the approximation in excel as well. anyway the approximation sequence should look something like this correct? since dv/dt=f(h),dv=f(h)*dt lwh-f(h)dt=lw(h2) lw(h2)-f(h2)dt=lw(h3) and so on

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