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Posted for "Ed Anderson" <eanderson@carolina.rr.com>:
Hi Bill,
Might be able to get that delta T in the near future.
Very interesting plot. Just for your information, I assumed the air delta T
as 20F. In otherwords, the heat transfered to the air from the radiators
raised the temperature of the air by 20F. If the rejected heat raises the
temp more than that then the heat rejection would improve. I don't have any
real data (yet) on the delta T of the air. Also in my model the coolant
model assumes a 30F total drop in coolant temps across the two cores (from
210 to 180F) which in the case of the engine producing 160 HP but with an
airspeed only 100 mph cause for a coolant for of 25 GPM. However, this
provides sufficient coolant flow such that the BTU delivered by the pump is
always greater than the capacity of the radiators to shed. I know this is
not necessarily the case, but just a simplfication.
This is not actually figured in the cooling equation as I assume that the
coolant flow always provides sufficient heat transfer. In this model the
heat rejection is limited by the airflow through the radiators..
Now that I think about it, it would appear that I probably should have the
coolant temp drop across the radiators equal the air temp increase through
the radiators, in otherwords both = either 30F or 20F. Any suggestions?
I had just installed some thermocouples in an attempt to measure the coolant
Delta T through the radiators. Won't be able to see what they give until I
get gear box back on and fly again.
Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, NC
eanderson@carolina.rr.com
Ed's spread sheet is the basis for the following note and attached pdf file.
At 5600 rpm, he shows 3901 Btu/min being rejected from the Radiators.
SNIP
I would be very interested in the actual delta-T that Ed gets on his
engine/radiators.<<<
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