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
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
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.
- posted 14 years ago