Message-ID: <000901c3af7f$61a33860$1702a8c0@WorkGroup>
From: "Ed Anderson" <eanderson@carolina.rr.com>
To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
Subject: Ball Park waste Heat calculations
Date: Thu, 20 Nov 2003 11:00:12 -0500
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One lb of gasoline has approx 19,000-20,000 BTU of energy.

An intercombustion engine is generally given the ability to convert approx
25% to useful energy, 50% energy out the exhaust in form of heat, and 25%
waste heat that you need to get rid of using radiators and oil coolers.

A commonly accepted relationship between HP and fuel burn for a rotary is
given by:

HP = (Lbs/Hour of fuel)/0.55 or (Gallons/Hour *6)/0.55  (0.55 = BSFC for
rotary)

or looking at it the other way

GPH (fuel burn) = 0.55*HP/6

Example take 200HP

GPH = 0.55*200/6  = 18.33 gallons/hour required to support 200HP

So if you are burning 18.33 gallons/hour *6 (lbs/gallon) = 110 lbs of
fuel/hour

or 110/60 min = 1.833 lbs/minute.  Energy then is 19,000 BTU/Lbs*1.833
lbs/min= 34833 BTU/Min

If 25% of that is waste heat, then 0.25*34833 = 8708 BTU/Min of waste heat
you must get rid of.

Given that 2/3 is gotten rid of by radiators and 1/3 by oil cooler

We have 2/3*8708 = 5805 BTU/min by radiators and 2902 Btu/min by oil cooler
for a ball park figure of the waste heat you need to consider.

You of course, can calculate such waste heat figures for any HP or fuel flow
if different than 200HP.

Ed

Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, NC
eanderson@carolina.rr.com