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

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SUMMARY

The discussion focuses on calculating the power required to pump air underwater, specifically using the formula P = p · V, where P is power in watts, p is pumping pressure in pascals, and V is the volumetric flow rate in cubic meters per second. Participants Jeff and Fred highlight that initial estimates for pumping air to a depth of 5 meters yield approximately 6.7 watts, which aligns with practical measurements from mini air compressors. The conversation also explores the feasibility of using exhaled air to assist in pumping new air, suggesting that a diver could potentially operate without an external power source if efficiency is optimized.

PREREQUISITES
  • Understanding of fluid dynamics principles
  • Familiarity with basic physics formulas related to pressure and power
  • Knowledge of volumetric flow rates and their significance in pumping systems
  • Experience with energy efficiency concepts in mechanical systems
NEXT STEPS
  • Research the effects of isothermal compression on air pumping efficiency
  • Explore advanced fluid dynamics simulations for underwater air pumping
  • Investigate energy reclamation techniques for diving systems
  • Learn about the design and efficiency of double-action pumps
USEFUL FOR

This discussion is beneficial for engineers, underwater divers, and researchers interested in the mechanics of air pumping systems and energy efficiency in aquatic environments.

Rayner
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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.
 
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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?
 

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