flyrotary@lancaironline.net Mailing List Archive

Right you are, Dave

Below is one semi-official definition of BTU in English units. 1 BTU is amount of heat to raise 1 lb of water 1 degree Fahrenheit.

So with Tracy's 30 gpm flow of water = 240 lbs/min. Since its temperature is raised 10 degree F we have

BTU = 240 * 10 * 1 = 2400 BTU/min

I know I'm ancient and I should move into the new metric world, but at least I didn't do it in Stones and Furlongs {:>)

The Columbia Encyclopedia, Sixth Edition. 2001.

British thermal unit

abbr. Btu, unit for measuring heat quantity in the customary system of English units of measurement, equal to the amount of heat required to raise the temperature of one pound of water at its maximum density [which occurs at a temperature of 39.1 degrees Fahrenheit (°F) ] by 1°F. The Btu may also be defined for the temperature difference between 59°F and 60°F. One Btu is approximately equivalent to the following: 251.9 calories; 778.26 foot-pounds; 1055 joules; 107.5 kilogram-meters; 0.0002928 kilowatt-hours. A pound (0.454 kilogram) of good coal when burned should yield 14,000 to 15,000 Btu; a pound of gasoline or other .

Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, NC

----- Original Message -----

From: DaveLeonard

To: Rotary motors in aircraft

Sent: Thursday, August 12, 2004 8:12 PM

Subject: [FlyRotary] Re: DeltaT Coolant was : [FlyRotary] Re: coolant temps

Ed, are those units right. I know that the specific heat of water is 1.0 cal/(deg Celsius*gram). Does that also work out to 1.0 BTU/(deg. Farhengight * Lb.) ?

Dave Leonard

Tracy my calculations shows your coolant temp drop is where it should be:

My calculations show that at 7 gph fuel burn you need to get rid of 2369 BTU/Min through your coolant/radiators. I rounded it off to 2400 BTU/min.

Q = W*DeltaT*Cp Basic Heat/Mass Flow equation With water as the mass with a weight of 8 lbs/ gallon and a specific heat of 1.0

Q = BTU/min of heat removed by coolant mass flow

Assuming 30 GPM coolant flow = 30*8 = 240 lb/min mass flow. specific heat of water Cp = 1.0

Solving for DeltaT = Q/(W*Cp) = 2400/(240*1) = 2400/240 = 10 or your delta T for the parameters specified should be around 10F

Assuming a 50/50 coolant mix with a Cp of 0.7 you would have approx 2400/(240 *0.7) = 2400/168 = 14.2F so I would say you do not fly with

a 50/50 coolant mix but something closer to pure water. But in any case, certainly in the ball park.

You reported 10-12F under those conditions, so I would say condition is 4. Normal operation

Ed

Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, NC