Sizing a 30kW gross heat flow bubble pump air compressor

This design uses water and air and ignores any difficulties with water vapor and other than perfect slug flow in the tubes. Also, it aims for the given arbitrary target of 30kW gross heat flow into the expansion process. The available work will depend on the temperature difference, and on efficiencies. However, 30kW of heat flow ought to be enough to produce 1kW of work, which is probably enough for one family. Note that the amount of work you can extract from a given amount of compressed air depends on how much heat you put into it during expansion, and the starting temperature. Here, the work represents how much work went into compressing the air, not how much might come out of it.

We assume a pressure ratio of 2: the low pressure reservoir is at half the pressure of the high pressure reservoir. If the machine is 30 feet tall, the pressures are 13 psi and 26 psi. Higher pressures with a 30 foot machine would reduce the pressure ratio.

The required moles per second of air to handle 30kW of heat flow is 30kW = nRT ln(pressure ratio). Using gnu units, this gives

  • 30kW / (R 300K ln(26psi/13psi)) = 17.351613 mol / s
  • (29 g / mol) (30kW / (R 300K ln(26psi/13psi))) = 0.50319679 kg / s.
  • 30kW / (R 300K ln(100psi/87psi)) = 86.363948 mol / s

Substituting this formula for the number of moles (which assumes isothermal expansion) into V = nRT/P, the temperature and gas constant cancels out, and the volume per second at a given pressure is just

  • volume / second = power / (P ln(ratio)).
  • 30kW / (26 psi ln(26psi/13psi)) = 241.4368 liters / second
  • 30kW / (100 psi ln(100psi/87psi)) = 312.44202 liters / second

This is the required air flow volume into the expansion hyperbola at the high-pressure end. The gas expands throughout the hyperbola so the total volume will be larger. This value can be used as volume per unit height, as the expansion is accomodated by bending the tube to be more nearly horizontal. The space required for the slugs of liquid is accomodated in the same way. A reasonable minimum amount of liquid would provide equal volume slugs of gas and liquid at the low-pressure end.

The required volume of the expansion hyperbola depends on the expected rate of heat flow per unit area of hyperbola tubing surface, or on the maximum velocity of the fluid and gas, whichever gives the more stringent limit.

Considering 1 meter / second (2.2 mph) a reasonable velocity for the fluid, and 30 feet (9.14 meters) a reasonable height, we first need the hyperbola length.

Still working on this…


Leave a Reply

Please log in using one of these methods to post your comment: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: