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