Forcing water from a pipe underwater

  • #1

Main Question or Discussion Point

Hi

I have a problem beyond my scope. I am not a lerned physics chap so please, any answers would be great if kept simple :)

I have a pipe containing 300 litres of water. The pipe is 346mm diameter and 1 meter long. The top edge of the pipe is 1200mm under the sea. The pipe lies horizontally (i,e, on its side)

I need to push the water out of the pipe using either compressed air or a ram powered by electricity or hydraulic force. The pipe will fill with air (normal atmospheric pressure) as the water is pushed out.

How much energy is required to do this. If the answer can be given in watts, that would be marvellous.

Many thanks and no, this sin't a homework question,, at 47, I am let off that peril :))
 

Answers and Replies

  • #2
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Energy is in Joules, Watts is power. The energy, neglecting losses, is equal to the pressure times the volume. The pressure is up to 1546 mmH2O and the volume is 300 L.
 
  • #3
Thank you DaleSpam,

That helps a bit. 'The pressure is up to 1546 mm H2O and the volume is 300 L' so if I take the middle of the of the points of pressure as 1373mm,

-1200mm at the top edge and 1546mm at the bottom edge, middle = 1373mm,

and multiply that by 300 litres, which = 411,900 (joules). Taking that joules equate to about 1 watt per second, heck, I'm going to need a bigger power station.

I'm obviously not doing somethng quite right. Could someone please explain. Thx.
 
  • #4
Andy Resnick
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I'm having a hard time picturing this- the pipe is submerged in water (~ 1 meter below sea level, or buried ~ 1m under the ground at the bottom of the sea?), and you are merely displacing the interior water somehow, leaving the interior full or air at atmospheric proessure? But the water will tend to rush back in, since it's at greater than atmospheric pressure.

If the pipe is open at the exit, and initially there is no flow, then the water inside the pipe is at hydrostatic equilibrium with the ocean, so any applied pressure is sufficient to displace the water- the requirement for atmospheric pressure in the back-filled air is then a requirement on the stiffness of the ram pushing the water out.

Like I said, I am having trouble picturing what the situation is.
 
  • #5
Hi Andy,

Thank you for your response. You are more or less on the right lines in your las paragraph.

Imagine a hydraulic ram (as per off an excavator or such machine) lying on its side. One end is sealed, the end where the ram pushes from. Thus, yes, as the ram pushes, atmospheric air pressure will be allowed in behind the ram.

The other end is open to the surrounding water.

The pipe (ram) is 1373 mm deep in water (sea water as it happens which is a tad heavier but of no great import here).

So it really comes down to how much pressure is required to push the water out of the ram, into the open water around it. I appreciate that it probably comes down to the amount of pressure from the water pressure that is surround the ram at that depth, but when I tried working it out, the force required seemd rather high.
 
  • #6
Mapes
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Let's assume 100% efficient machines for simplicity.

The required power (in W) is the required energy divided by the time you want this to take.

The required energy (in J) is the work you need to do, which is the force multiplied by the distance.

The force (in N) is the pressure difference (in Pa) multiplied by the pipe area, ignoring friction and assuming a supply of air at atmospheric pressure.

Does this help?
 
  • #7
28,561
4,867
Thank you DaleSpam,

That helps a bit. 'The pressure is up to 1546 mm H2O and the volume is 300 L' so if I take the middle of the of the points of pressure as 1373mm,

-1200mm at the top edge and 1546mm at the bottom edge, middle = 1373mm,

and multiply that by 300 litres, which = 411,900 (joules). Taking that joules equate to about 1 watt per second, heck, I'm going to need a bigger power station.

I'm obviously not doing somethng quite right. Could someone please explain. Thx.
Your units are off. You need to convert mmH2O into Pa and L into m^3 in order to get energy in J.

1373 mm = 54 in
54 inH2O = 13.4 kPa
300 L = 0.3 m^3

13.4 kPa 0.3 m^3 = 4031 J

Also, you said "joules equate to 1 watt per second". That is backwards. The correct relationship is 1 W = 1 J/s. I don't know what your power supply is or how long you have to clear it. If you have a 100 W supply you will be able to clear it in 40 s neglecting inefficiencies.
 

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