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