Bill Schertz wrote:
Everyone,
I discovered an error in my calculations that I presented at the Rotary
Roundup on Saturday. I have enclosed the revised graphs and sample
calculation as a PDF file that everyone should be able to open.
Thanks to those that questioned my numbers -- the methodology was correct,
but my data entry wasn't so good.
please use the attached figures in liu of the ones in the handout.
Bill Schertz
KIS Cruiser # 4045
Bill will be sending us a complete PDF of his entire excellent talk shortly I
suspect.
It is a unified cooling theory useful for predicting the specifications of a
complete
water side rotary engine cooling system.
Paul Lamar
Paul, how do you stuff 30,000 cubic feet of air in one minute through a 260
sq inches radiator? Seems a lot?
Bulent
You don't operate at that delta T. You operate at a much higher delta T.
Bill is still working on that part of the analysis. This last chart is a little
misleading. In the final analysis the air required will probably be much higher
than this but so too will be the delta T between the rad metal and the cooling
air coming in from the scoop.
An airplane flies at at least 200 FPS so if it is one square foot air intake
that is 200 times 60 or 12,000 cubic feet per minute at 200 HP. That is an air delta T of
only 30 degrees F.
Paul Lamar
I am afraid I added to the confusion here. Kill this sentence of mine in the first paragraph above
"In the final analysis the air required will probably be much higher than this but so too will be the delta T between the rad metal and the cooling air coming in from the scoop."
I'll give it another try.
What Bill's chart depicts is the increase in temperature of a range of air flows
necessary to carry away that much heat for X amount of HP. In other words if all you
can get through the rad is 6000 cubic feet per minute with a 200 HP engine the rad must
be able to raise its temperature by 50 degrees Fahrenheit. Lets say the outside air
coming in is 100 F therefore the air coming out the back side of the rad needs to be 150 F.
I assume that is engine HP Bill? Is that correct?
Paul, you are correct -- that is the ENGINE horsepower, assuming 28% of the energy in the fuel is converted to mechanical work.
Bill
The volume of air per minute you are able to get through the rad depends on its area and
the velocity going through the rad. You need to measure the velocity going
in and the velocity coming out knowing the area of the duct at its smallest point.
Also it needs to be measured all over the face of that area.
In general this velocity of air will not only depend on the forward
speed of the airplane but will also depend on the porosity of the
rad. So we are not out of the woods yet.
Then it must conform to this equation (P1 X V1)/T1 = (P2 X V2)/T2.
P1 is the pressure going in.
V1 is the volume going in.
T1 is the temperature going in expressed in Kelvin
P2 is the pressure coming out.
V2 is the volume coming out.
T2 is the temperature coming out expressed in Kelvin
For zero drag P2 should equal P1 or be greater.
In general it won't be. V2 will depend on the area of the exit duct.
T2 will be something like 20 to 50 degrees F hotter than T1.
This explanation is better -- sorry I couldn't respond sooner, but you have it right. (Lots of crap at work this week -- new boss)
Bill Schertz
KIS Cruiser # 4045
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