# Pumping air underwater

1. Sep 12, 2007

### Rayner

Would be grateful for a formula for the power required (theoretically) to move air down to a specified depth underwater. Metric if possible please.

My calculations seem an order of magnitude too low.

2. Sep 12, 2007

### rcgldr

The pressure of the air just needs to be higher than the pressure of the water at the desired depth. The power would be a combination of this pressure, and flow rate.

3. Sep 12, 2007

### FredGarvin

A rough first estimate can be had by using

$$P = p \dot{V}$$

where

$$P$$ is the power in watts
$$p$$ is the pumping pressure in Pa
$$\dot{V}$$ is the volumetric flow rate in $$\displaystyle{\frac{m^3}{s}}$$

This assumes things like an isothermal compression and neglects frictional effects and other losses. Like I said, this is a first approximation.

4. Sep 12, 2007

### Rayner

Many thanks Jeff and Fred

I had previously assumed that the power to pump air to a diver, even in shallow water, would be far beyond what the diver could provide using his or her own muscle power alone. But I thought that if one could somehow part-use the energy in the exhaled air to help pumping down new air it might just be possible to dive without an external power source, using surface air tubes and some kind of double-action pump strapped onto the diver.

On initial estimation (given that 12 liters at the surface would become 8 liters at 5 meters depth), it appeared that to provide 8 liters per minute would require 98 x 5 / 60 or only about 8 watts. Intuitively this seemed an order of magnitude too low. But your formula seems to give a fairly similar result; 50,000Pa x 8/1000m³/60 or 6.7W (and checks quite closely with a mini air compressor for which I found figures).

If a real system were only 25% efficient it should be possible, even without energy reclamation (30-40 W for air and the rest for finning around) given that 0.1hp is supposedly a rate of work that a healthy person can sustain for several hours. See you in the sea?

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