# Thread: Calculus tank being filled with fuel (volume and partial pressures?)

1. ## Calculus tank being filled with fuel (volume and partial pressures?)

The greatest risk of unintended discharge is when the tank is being filled or emptied through the valve (the diameter of which is approximately 9"), as through unintended contact or mechanical failure the valve could fail and allow unimpeded discharge through the pipe to which the valve is connected. In this case, it is known that the rate at which the height of the fuel in the tank will change is proportional to the ratio of the squares of the diameters of the valve and the tank and the square root of the height of the liquid in the tank, with constant of proportionality k=(2g)1/2 (where g is the acceleration due to gravity).

2. ## Re: Calculus tank being filled with fuel (volume and partial pressures?)

I understand ratios between two values, but not ratios among three. I am not sure what it is saying. Let $\displaystyle h$ be the height of the fuel in the tank. Let $\displaystyle v$ be the diameter of the valve (approximately 9"). Let $\displaystyle r$ be the diameter of the tank. Now, you say $\displaystyle k=(2g)1/2$. To me, that reads $\displaystyle 2g\dfrac{1}{2} = g$. But, why not just write $\displaystyle g$? What are the units for $\displaystyle k$? Do you know the diameter of the tank? Now, it might be the following:

$\displaystyle \dfrac{dh}{dt} = k\dfrac{v^2r^2}{\sqrt{h}}$

If that is the case, then since $\displaystyle v$ is fixed, and I assume $\displaystyle r$ is fixed (the tank has a fixed diameter at all heights), $\displaystyle \dfrac{2}{3}h^{3/2} = kv^2r^2$. But, I have no clue if my assumption for $\displaystyle \dfrac{dh}{dt}$ is correct, as I still don't understand the ratio among three values (and just guessed that the first two should be in the numerator).

3. ## Re: Calculus tank being filled with fuel (volume and partial pressures?)

Originally Posted by SlipEternal
If that is the case, then since $\displaystyle v$ is fixed, and I assume $\displaystyle r$ is fixed (the tank has a fixed diameter at all heights), $\displaystyle \dfrac{2}{3}h^{3/2} = kv^2r^2$. But, I have no clue if my assumption for $\displaystyle \dfrac{dh}{dt}$ is correct, as I still don't understand the ratio among three values (and just guessed that the first two should be in the numerator).
That formula lost the $\displaystyle t$ for time. It should be $\displaystyle \dfrac{2}{3}h^{3/2} = kv^2r^2t$.

4. ## Re: Calculus tank being filled with fuel (volume and partial pressures?)

Without details of the downstream piping system nothing can be calculated

5. ## Re: Calculus tank being filled with fuel (volume and partial pressures?)

Originally Posted by Lddenis
$\displaystyle \frac{dh}{dt}= k r_1r_2\sqrt{h}$ where k is the rate of proportionality. What you wrote was k= (2g)1/2= g but I suspect you mean (2g)^(1/2) (or $\displaystyle \sqrt{2g}$.