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 a 80 hp engine would need to sink
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.

1/4" tubing has a circumference of about 1" so 50' of unfinned 1/4"
tubing is all that is required to move 60kW. 50' rolls of 1/4" cu
tubing sell for $18 at some plumbing supply stores.

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.

The huge -- 10X larger -- surface area of a conventional radiator is
necessary, not because the air is warmed very much but because the
heat transfer coefficient is an order of magnitude less.

Hotter air would be rejected from the fluidized bed radiator because
less air and less pumping losses in the form of high profile radiator
drag are required to move 60 kW.


Bret Cahill