# Fuel Tank Question

Could some one please help me with this question as my physics knowledge does not extend this far.

I have a large fuel tank with a capacity of 21000L. The tank currently is holding 2000L of fuel. On one end of the tank there are two four inch pipes where the fuel comes in at about 1000L a minute for each pipe. On the other end is a two inch pipe for breathing. Is the two inch pipe enough for the amount of fuel coming in? The only other numbers I have is the density of the fuel which is 0.7335.

Thanks in advance for any help.

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Is that a homework? – J. C. Leitão Nov 17 '11 at 9:59
No it was asked of me by another person that is trying to figure this out. – user174084 Nov 17 '11 at 10:05
""Is the two inch pipe enough for the amount of fuel coming in?"" Of course not, if the full capacity of those pipes coming in is concerned. But with respect to the gas phase going out, the two inch pipe might be sufficient. But: a major problem can be foaming whne the fuel splashes into the tank. Did You ever refill a diesel car? – Georg Nov 17 '11 at 10:26
@user174084, is the other person that asked you this question doing homework? It sure looks like a homework problem. I'm adding the homework tag till convinced otherwise... – FrankH Nov 17 '11 at 10:53
Regardless of whether it really is a homework problem or not, the homework tag applies to all homework-like problems. It definitely belongs here. (This also runs afoul of some of our guidelines, but since it's a day old and has already accumulated some decent answers I'm not going to bother closing it.) – David Z Nov 19 '11 at 1:31

In physics a sketch of the problem is often a nice step towards the answer to the problem. If you look at your tank and the connected pipes and imagine what is happening when you fill the tank it is clear that neither the fuel density, nor the capacity of the tank matter at all. As the tank is a closed system with only the incoming and outgoing pipes connecting it to the environment it is clear that $$V_{incoming} = V_{outgoing}$$ if we do this very slowly. So the answer of your question is yes, a small pipe is enough.

Now we can try to think about what happens when we fill up the tank faster. This is much harder to answer as it involves looking at the time dependence of the whole setup. There is not much else you can do but to state this in differential equations and solve the problem to get an exact answer. To do this you look at a small time interval and calculate what happens. For a tiny amount of oil $dV$, the same amount of the air volume $dV$ will be decreased. The air can only react in two ways here, it can lower it's volume by compression which can be approximated by the ideal gas law $P\ V = N\ k\ T$ and because the pressure inside will be higher than outside it will start to flow through the outgoing pipe. Over time the pressure inside will reach a constant value until you stop pumping more oil in.

This constant value will determine whether the outgoing pipe is enough here. To reach your constant flow rate the oil pump has to pump against this air pressure inside the tank. So the answer is depends on the maximum pressure the pump can create.

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Downvoter? What's wrong with this answer? – Peter Shor Nov 17 '11 at 14:38

If I understand you correctly, then this is definitely not a homework question, because it is too undefined.

So, first, what happens: The tank will fill up at $2 \mathrm{m^3/min}$. Consequently, $2 \mathrm{m^3}$ of air and fumes have to be expelled from the tank through the third tube. With the given diameter, by $V/(A t)$ that means an air velocity of about $16 \mathrm{m/s}$, if I calculated correctly.

Given the density of Nitrogen, by $dp = v^2 \rho / 2$ this leads to an increased pressure of $1.8\mathrm{hPa}$ in the tank, which is not very much (somebody should check these values).

It is definitely not a students job to know if a standard tank can take this, but I am quite confident.

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How do You calculate the delta p of the vent line without knowing its length? – Georg Nov 17 '11 at 14:04
I just ignored the resistance of the tube. This is just the exhaust velocity from the pressure. Same for the compression of the gas. If I calculated correctly, then both can be easily left out. – mcandril Nov 17 '11 at 14:08