Power Required to Pump Air Underwater: Calculate & Analyze Metric Results

  • Thread starter Thread starter Rayner
  • Start date Start date
  • Tags Tags
    Air Underwater
AI Thread Summary
The discussion focuses on calculating the theoretical power required to pump air underwater to a specified depth, emphasizing the need for a formula that incorporates pressure and flow rate. The key formula presented is P = p · V, where P is power in watts, p is pumping pressure in pascals, and V is volumetric flow rate in cubic meters per second. Initial estimates suggest that the power needed to pump air, even at depths like 5 meters, is significantly lower than expected, with calculations yielding around 6.7 watts. The conversation also explores the possibility of using exhaled air to assist in pumping new air, potentially allowing divers to operate without external power sources. Overall, the calculations indicate that with a system efficiency of 25%, it may be feasible for a diver to sustain air supply using minimal energy.
Rayner
Messages
3
Reaction score
0
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.
 
Engineering news on Phys.org
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.
 
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.
 
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?
 
Hi all, i have some questions about the tesla turbine: is a tesla turbine more efficient than a steam engine or a stirling engine ? about the discs of the tesla turbine warping because of the high speed rotations; does running the engine on a lower speed solve that or will the discs warp anyway after time ? what is the difference in efficiency between the tesla turbine running at high speed and running it at a lower speed ( as fast as possible but low enough to not warp de discs) and: i...
Thread 'Where is my curb stop?'
My water meter is submerged under water for about 95% of the year. Today I took a photograph of the inside of my water meter box because today is one of the rare days that my water meter is not submerged in water. Here is the photograph that I took of my water meter with the cover on: Here is a photograph I took of my water meter with the cover off: I edited the photograph to draw a red circle around a knob on my water meter. Is that knob that I drew a red circle around my meter...
Back
Top