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Pumping Water from a deep mine 
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#1
Nov2808, 04:11 PM

P: 1,031

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 


#2
Nov2808, 04:35 PM

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#3
Nov2808, 04:44 PM

P: 1,031

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 


#4
Nov2908, 04:58 PM

P: 1,031

Pumping Water from a deep mine
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 


#5
Nov3008, 09:52 AM

P: 273

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 generalless chance of ambiguity). 


#6
Nov3008, 10:02 AM

P: 1,031

Seriously, though. 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 


#7
Nov3008, 10:15 AM

P: 273

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=??? 


#8
Nov3008, 10:21 AM

P: 1,031

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 


#9
Nov3008, 10:29 AM

P: 273

where V is he volume, A is the cross sectional area and L is the height of fluid. Will the pump be able to continue sucking the water when this equilibrium is acheived? Why? 


#10
Nov3008, 10:33 AM

P: 1,031

okay so:
[tex] P = \frac{\rho A*l g}{A} [/tex] this will cancel down to: [tex] P = \rho l g [/tex] TFM 


#11
Nov3008, 10:43 AM

P: 273

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. 


#12
Nov3008, 10:47 AM

P: 273

This site has some very good explanations about pump limitations;
http://www.newton.dep.anl.gov/askasc...9/eng99574.htm 


#13
Nov3008, 10:53 AM

P: 1,031

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 


#14
Nov3008, 11:04 AM

P: 273

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. 


#15
Nov3008, 11:28 AM

P: 1,031

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 


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