Craig Spence wrote:

Can anyone advise me of the required (optimum) thermal rejection rate (BTUs)
for a Mazda 13B rotary oil cooler for an aircraft installation?
Regards
Craig Spence
Cobra Aviation - Australia

The oil is one third of the total cooling requirements. You can determine
that from this WOT heat balance chart for a 13B. For example; assume
the basic engine is 200 HP net representing 28% of the total heat in a gallon
of gas. Total theoretical heat in the fuel would then .28 x Total Heat = 200 HP
Total Heat in the fuel is therefore 200 /.28 or 714 HP.

>From the chart the oil cooling is about 8% of that so .08 times 714 HP
is 57 HP. One HP is equal to 2545 BTU's per hour. Oil cooling is therefore
145,065 BTU's per hour.

BTW turbo compounding is a hot topic lately so 45% of the 714 HP
or 321 HP goes out the exhaust in the form of waste heat. 

Paul Lamar
 
 
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