# How do you work out the energy requirement for a submerged pump?

I was wondering if you were to place a pump in a submerged container and then were to open a hatch allowing water to flood the previously vacant cavity how much energy would be required to discharge the water at the entry rate. What I would like to know also whether or not the depth would have any influence on the pump's output and efficiency. I understand submersible vehicles rely on the flooding and flushing of water from their hulls to dive and ascend but am curious if pumps must do more work when at depth. Please and thank you.

• you would have to know how much water was going in to the container at what rate to be able to answer this question, when you get a submersible pump it will tell you its capability on the manufacturers spec and also how much power would be required, this is as far as my experience with pond pumps takes me anyway – Daz Hawley Dec 20 '14 at 11:18
• The deeper the submersible, the greater the pressure trying to force water into the buoyancy tank, and hence the more work needed to force water out of the tank. – Hot Licks Dec 21 '14 at 2:22

Suppose we have a pump submerged in a liquid of density $\rho$ at a depth $d$ which we want to use to raise the liquid to a height $h$. The pump has to do work to overcome the potential energy difference between the two heights of liquid. That is:

$$W(t)=\rho Qg (h-d)t$$

Where $g$ is the acceleration due to gravity (taken as $9.81m/s^2$). $Q$ is the volumetric flow rate (in $m^3/sec$) of liquid being pumped and $t$ is the time (in sec).

This would give us the minimum energy required. In practice, pumps have losses which are represented as 'percentage efficiency', $\eta$ so the actual mechanical energy required is:

$$E_M=W/ \eta$$

For example, the efficiency of a centrifugal pump might be $\eta=0.65$ (65%) due to losses in the pump such as turbulence. The actual efficiency of a pump will vary with flow rate $q$ and can be obtained from a manufacturers data sheet.

• Just looking for a clarification if possible, does the above apply when you are discharging the fluid perpendicular to the head (pump horizontal and discharge parallel to the axis of the outlet), instead of, for example pushing the fluid to a higher elevation? In my mind it would seem to be the same effect but im not certain. Thank you all again. – slowadult Dec 21 '14 at 13:51