How Does a Pump Calculate Work When Transitioning from Water to Air Underwater?

[kihel]

Homework Statement

• We have a container/volume of 1m3 100 m below mean sea level,
• A pump connected to it, pumping from inside the volume with outlet in the surrounding water at same height.
• And there is an pipeline to air, with a one-way valve allowing air to be sucked down to the volume and preventing anything from moving up.
  State 1 the volume is filled with water
  State 2 the volume is filled with air

What are the forces involved and how to calculate work done by a pump going from State 1 to state 2 - pumping water out of the closed volume and into the surrounding water at same height, hence creating a vacuum which sucks air down through the pipe.

Assumptions:

Ignore efficiency of the pump
Water as incompressible fluid

Homework Equations

Energy = Pressure*Volume
Pressure = Density * Gravity * Height
Force = Pressure * Area
Pressure = Force/Area

The Attempt at a Solution

My initial thought is that the work done that need to be done by the pump must equal the potential energy of State 2.

This energy is: E=PV, P=DGH

Pressure = ~1000 * 9,81 * 100m = ca. 10 bar = 10 000 N/m2

Energy= 1 m3 * 10 000 N/m2 = 10 000 Nm

Delta E = Heat transfer + Work done

Assuming no heat transfer and energy in state 1 is zero:

W= 1000 N/m



This however seems to me like a derived answer, I am looking for a different method, more direct calculation of actually moving the water.

Best regards Kihel

 

[olivermsun]

Homework Statement

• We have a container/volume of 1m3 100 m below mean sea level,
• A pump connected to it, pumping from inside the volume with outlet in the surrounding water at same height.
• And there is an pipeline to air, with a one-way valve allowing air to be sucked down to the volume and preventing anything from moving up.
  State 1 the volume is filled with water
  State 2 the volume is filled with air

...
This energy is: E=PV, P=DGH

Pressure = ~1000 * 9,81 * 100m = ca. 10 bar = 10 000 N/m2

I think 1 bar = 100 000 Pa = 100 000 N/m².

This however seems to me like a derived answer, I am looking for a different method, more direct calculation of actually moving the water.

Well, it amounts to the same thing as what you've done, but you could think of it this way:

Your container is basically a cylinder with height 1 m and cross-sectional area 1 m².
Your pump needs to push the water out against the ambient pressure, which would be just like pushing a piston into the cylinder to push out the water.

The force is therefore F = P * area of piston, and the total work would be F*(height) = P(area)(height).

 

[Chestermiller]

To understand what is happening mechanistically, note that, as soon as you pump the slightest amount of water out of the volume, the pressure within the volume will drop to 1 atm. This is because the volume is connected by a column of air (with negligible static head) directly to the air at the surface. So the water inside the volume is being pumped from a pressure of 1 atm (1 bar) to a pressure of 11 bars (static head + surface pressure) at depth.

Chet