Oil cooler math
So far, we’ve solved for oil mass flow rate. Based on some assumptions, we believe the oil mass flow rate is around 36.715 lb/min.
Solving for the worst case heat transfer is next. The formula for heat transfer is:
Q = Cp x W x delta T
where: Q = heat lost or gained (BTU/min)
Cp = specific heat of substance (in this case, oil) in BTU/lb-degF
W = mass flow rate of oil (lb/min)
delta T = T1-T2 (degF) where T1 = 200degF (oil inlet) and T2 = 170degF (oil outlet)
So: Q (BTU/min) = (BTU/lb-degF) x (lb/min) x (T1-T2)
540.28 BTU/min = 0.5205 x 36.715 x (200-170)
Thus we have a heat rejection of 573.3 BTU/min.
Next we calculate the inlet temperature differential (ITD) in BTU/min relative to the reference figure of 100deg F.
Q/ITD = (BTU/min) x 100/(oil temp in - air temp in).
Assuming our air temp is 80 deg F, we get:
Q/ITD = 573.3 x 100/(200-80) = 477.75 BTU/min @ 100degF ITD.
When multiplied by 60 to get BTU/hr, we get (337.675 x 60) = 28,665 BTU/hr.
Given that the engine oil cooler (ESR3205) only has a capacity of about 24,000BTU/hr, we can expect less than a 30 deg drop in temp between the inlet and the outlet. If we solve the equations for a 24,000BTU/hr cooler, we find that we can expect only a 25.11 deg drop from inlet to outlet.
(Note that some cooler manufacturers use a ITD ref of 130degF).