Time required to fill a vacuum tank with water?

[siddharth23]

I have a 6500 litre or 6.5 m^3 tank.

A vacuum pump (40000 litres per minute) creates a 90 % vacuum inside the tank.

The vacuum is used to suck in water. The climb the water has to make is 6m.

How much time will it take to fill the tank, given that the pump is kept on all the while?

Thanks in advance!
Siddharth

 

[Simon Bridge]

If the pump did not keep operating, what would happen?

 

[mfb]

How does water enter your tank? That will influence the filling rate significantly (imagine a 1mm pipe compared to a 10cm pipe...)

 

[siddharth23]

It's a 4 inch pipe. I used Bernoulli's equation. That would be applicable here, right?

 

[siddharth23]

If the pump did not keep operating, what would happen?

I guess as the water would fill up the tank, the air would get less place leading to compression and the vacuum force would go on decreasing.

 

[mfb]

I guess as the water would fill up the tank, the air would get less place leading to compression and the vacuum force would go on decreasing.

That's right, but you have to compare the inflow of water with the pump capability.

As a first approximation, the velocity of the water will be of the order of 5m/s or less, this gives ~2500l/min. Your pump is way more powerful than the pipe leading to the tank.

0.67m^3/s... looks really over-sized for such a small tank.

 

[Simon Bridge]

If the pump did not keep operating, what would happen?

I guess as the water would fill up the tank, the air would get less place leading to compression and the vacuum force would go on decreasing.

... yes - under what conditions would the flow come to a stop?

I'm not sure it is helping you to think of the tank as containing a vacuum or the vacuum exerting a force - think of it as containing air at very low pressure. It's like the experiment you did as a kid where you light a candle in a dish of water and put a jar over it.

That's right, but you have to compare the inflow of water with the pump capability.

... i.e. since there is a pump: what difference does that make?
... will the air pressure at the top of the inrushing water remain the same, increase, or decrease, as a result of the action of the pump?

mfb has discussed this a bit:

As a first approximation, the velocity of the water will be of the order of 5m/s or less, this gives ~2500l/min. Your pump is way more powerful than the pipe leading to the tank.

0.67m^3/s... looks really over-sized for such a small tank.

... I suspect the person setting the problem is trying to make sure the math is simple-ish.

 

[siddharth23]

... I suspect the person setting the problem is trying to make sure the math is simple-ish.[/QUOTE]

This is being actually used by the company where I'm interning. They've been doing it by trial and error till date and are doing a decent job.

Simon - Nah I didn't literally mean vacuum force. I got what you meant. What I meant was as the vacuum would decrease, the negetive pressure head will decrease and the depth from which water can be sucked up will be affected.


Just tell me, Bernoulli's equation can be applied here, right?

 

[Simon Bridge]

http://www.princeton.edu/~asmits/Bicycle_web/Bernoulli.html
Sounds good to me - provided the pressure difference is maintained: should put you in the right ballpark at least. Check it by using a known result.

Yep - the water would be drawn into the tank until the pressure difference balances the weight of water.
The pump stops this from happening - if you didn't shut it off, it looks like you'd be pumping water instead of air at some point.

 

[siddharth23]

The vacuum pump is actually a compressor, so there's no worry of water being pumped out.

 

[Simon Bridge]

Well it's besides the point.

 

[siddharth23]

I mean it won't pump out water.

Anyways, thanks everyone :)

 

[Simon Bridge]

No worries - have fun :)