Ed,
Please humor me (a non-engineer) while I ask a dumb question.  If it takes 1BTU to raise 1lb of water 1 degree, and you factor in 30 gpm flow to come up with a 2400 BTU requirement for a 10 degree rise for 1 lb of water, where does the number of pounds of water figure into the equation, or do we just ignore that issue?  Water is 8.34 lbs/gal, and say you have 2 gallons of coolant, that would be 16.68 lbs.  Seems that we would need to multiply the 2400 figure by 16.68 to arrive at a total system requirement of 40,032 BTU/min.  What am I missing here?

Mark S.


     At 09:58 PM 8/12/2004 -0400, you wrote:

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 {:>)
 
Ed
 
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