# Work done by pump under water

1. Sep 18, 2014

### kihel

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

• 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

2. Relevant equations

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

3. 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

2. Sep 18, 2014

### olivermsun

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

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).

3. Sep 18, 2014

### Staff: Mentor

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