The huge surface area and the high air flow rates through a

conventional radiator are necessary because the heat transfer

coefficient between the tubing and air is so low.

Air-particle fluidization increases the air side HX coefficient by an

order of magnitude thereby dramatically reducing materials costs, drag

and pumping losses.

Installing a fluidized bed radiator would only require removing the

fan, adding a blower and using the same water pump and hoses.

Assume the heat energy leaving the radiator is about equal to the

mechanical energy leaving the crank, an 80 hp engine would need to

sink 60 kW or 60 BTUs/sec. If ambient air is 100 F and the radiator

averages 200 F then the delta T driving the heat transfer is about 100

F.

The heat transfer coefficient between the fluidized bed and coolant

tubing is 500 BTUs/hr ft^2 F so a 100 F delta T => 50,000 BTUs/hr ft^2

or 14 BTUs/sec ft^2.

For 60 BTUs/sec the coolant tubing exposed to the bed only needs a

surface area a little over 4 ft^2 or 600 in^2.

Assume the HX coefficient is about the same on the liquid side of the

tube so double this area.

1/4" tubing has a circumference of about 1" so 100' of unfinned 1/4"

tubing is all that is required to move 60kW in a fluidized bed

radiator. 50' rolls of 1/4" cu tubing sell for $18 when used as a

draw at plumbing supply stores. Aluminum would be much cheaper on a

production run basis.

To keep the flow rate up and pumping losses down -- the coolant flow

might be 20 gallons/min on the freeway -- the tubing should be cut

into 1 or 2' lengths and parallel coiled inside of a 8" high 8" dia

cardboard cylinder or box filled 1/3 with light particles. A 12 volt

200 watt blower -- somewhat larger than the ac blower -- is ducted to

the bottom.

Hotter air would be rejected from the fluidized bed radiator because

less air and lower pumping losses in the form of a big fan and high

profile radiator drag are required to move 60 kW.

Bret Cahill

- posted 14 years ago