Pumping Water from a deep mine

TFM

1. The problem statement, all variables and given/known data

You are required to pump water out of a very deep mine. Where should you put the pump? Explain your answer fully.

2. Relevant equations

Not sure

3. The attempt at a solution

See a similar question I did regarding beer in a can, I explained, but I needed formulas apparently, and didn't get marks for my 'waffle' (which is strange because in A level, waffle = good)

So for this question, i feel that equations are neede, but I am not sure what equations will be useful. This question I feel is a fluid dynamics question. So I first thought of Benouilles Equation, but I don't think this is useful, since we don't know the speed the water is travelling.

I also feel I need to use F = mg somewhere, but this doesn't help me.

Could anyone recommend some equations or laws that might be useful?

TFM

dlgoff

Maybe you should read this first:
How It Works: Water Well Pump

TFM

Okay, reading from that implies that the pump should be at the bottom of the well, thus pushing the water up, not having to pull it up.

Thanks for the link.

So now, how do I prove this? What formulas/Laws are useful?

TFM

TFM

Is this a fluid dynamics question, since I have a book on formula and I have looked through the Fluid Dynamics Section several times and can't seem to find any related fomrula?

TFM

Vuldoraq

You can show that the water will only be sucked to a certain height using the formula for pressure.

How does a suction pump work?

By the way this question doesn't ask for you to prove anything (unlike the beer can one), so I think concise and true 'waffle' should suffice, although equations and math arguments are always better (in general-less chance of ambiguity).

TFM

It sucks (Sorry, couldn't help my self. )

Seriously, though.

- Wikipedia

So basically, a pump creates a vacuum at the top, and the wtare is sucked up to fill the vacuum

The formula for pressure is:

[tex] P = \frac{F}{A} [/tex]


And Boyles KLaw:

pV = Constant, could be useful

(I am sure there is another pressure law with rho in it, but cannot currently find it.)

TFM

Vuldoraq

True, but more importantly how does it suck up the water up? In space there is a perfect vacuum, but if you put a straw in a glass of water it wouldn't magically move up the straw (it would probably float off and freeze, but you see my point?).

So the pump creates a vacuum but something else pushes the water up the pipe. What is it?

You can get rho into your pressure expression by saying F=mg with m=???

TFM

well F = ma

m = rho * Volume

also, a = g?

thus:

[tex] P = \frac{\rho v g}{A} [/tex]

would it be the decreas in pressure, because the water is in equilibrium at 1 atms, so if a pump reduces the pressure, the water will increase in volume to make up for it, as in Boyles Law:

pV = Constant

???

TFM

Vuldoraq

Exactly. You can take this further by saying [tex]V=A*L[/tex]
where V is he volume, A is the cross sectional area and L is the height of fluid.

Yes, in other words atmospheric pressure is pushing the water up the pipe and will do so until the pressure at the bottom of the pipe is equal to atmospheric pressure.

Will the pump be able to continue sucking the water when this equilibrium is acheived? Why?

TFM

okay so:

[tex] P = \frac{\rho A*l g}{A} [/tex]

this will cancel down to:

[tex] P = \rho l g [/tex]

Well surely when equilibrium is achieved, the volume of vacuum will be 0, and this the water is at the top?

TFM

Vuldoraq

Yup.

Sadly not. When equilibrium is achieved the water will be at a height at which the pressure at the bottom of the pipe equals atmospheric pressure. There may still be a vacuum above it, depending upon how long the pipe is.

What this means is you have an upper limit on the pressure at the bottom of the pipe and can thus find the maximum l in your equation.

Vuldoraq

This site has some very good explanations about pump limitations;

http://www.newton.dep.anl.gov/askasc...9/eng99574.htm

TFM

Okay so:

[tex] P = \rho l g [/tex]

P = 1 atm = 101 325

[tex] \frac{101325}{9.8 \rho} = l [/tex]

Density of water: 1000 kg/m^3

this gives l to be:

l= 10.34m

So how will this help with were to put the pump?

TFM

Vuldoraq

Well it means that a suction pump can only raise the water to 10m. Since a suction pump is the only type of pump where it is entirley out of the water (except the pipe obviously) having such a type of pump further tan 10 m from the water would render it useless.

So you know where not to put the pump, which should help you choose where to put it.

The rest is down to arguments. You will have to have a pump somewhere that pushes rather than sucks the water. If this pump is under the water then it can exclusively push. If it's above the water then it will have to suck then push etc.

TFM

So the above is useful for showing why surface based pumps are not useful.

If the pump is below the water, it won't have to suck at all, just push the water upwards.

TFM