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Re: [FlyRotary] Re: Questions..
Ok, Bulent
I was not certain whether you
were talking area or volume. As I mentioned, the two GM cores give approx
2*(95*3.5) = 665 cubic inches. Not saying this is the minimum you will
need, however, that amount of area and core has been shown to be adequate for
cooling of the 13B. 458 cubic inches is about 31 % smaller in cubic volume
than the two cores, so you might be a leaning a bit on the low side
there. The smaller your core, the more crucial the ducting is to get
maximum cooling effectiveness, so 458 will likely put a premium on your duct and
diffuser. On the other hand, if 458 is adequate then we will have another
data point.
There is no doubt in my mind that your 458
cubic inches will easily cool a 13B at cruise airspeeds, its the climbout that
would concern me. My 0.02.
Ed
Ed Anderson
RV-6A N494BW Rotary
Powered
Matthews, NC
----- Original Message -----
Sent: Sunday, July 04, 2004 11:05
PM
Subject: [FlyRotary] Re:
Questions..
> 3 For the water radiator I'm
planning of using a core 3 times the size of
> the oil cooler. Does
anybody think that it will be too small?
If you are
talking core surface area the oil cooler is approx 19x4 = 76 sq inch.
so 3 times that would be 228 sq inch. Two
of the GM
evaporator cores will give you 95x2 = 190 sq inches so the area sounds OK.
How about the core volumn? Yes, thickness does play a part in cooling.
The two gm cores have a core volumn of 2x( 9x10x3.5) = 630 cubic
inches. Just for a point of
reference.
Ed,
The oil cooler has
2” thick core so, 76 sq inches x 2 =152 cubic inches. Rad. will be 152 x
3=458cubic inches.
Bulent
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