Math & science behind oil coolers
The engine oil cooler specified by Land Rover for the P38a truck with the 4.6L motor is ESR 3205 (not to be confused with the smaller version for the 4.0L motor). It is a 5 plate “stacked plate” design with dimensions (of the actual plate assembly) 22.5”L x 3”H x 1” thick. Best estimate is that it is capable of dissipating about 24,000 BTU/hr of heat and is suitable for engines around 200 - 220HP. Pressure drop across the cooler estimated to be about 5 PSI.
The transmission cooler for the P38a is an ESR 2276. It is a 10-plate “stacked plate” design with dimensions 21.5”L x 6”H x 1.25” thick of the plate assembly. Best estimate of dissipation potential is 31,000 BTU/hr and have a pressure drop of 0.75 PSI.
Note that of the three designs of cooler - tube & fin, plate & fin and stacked plate - the coolers above (stacked plate) are the most efficient and effective for their size but also more expensive than the other two. Tube & fin designs, such as the standard RRC “bogbrush” transmission cooler (ESR1703), is the least expensive and the least efficient of the three designs.
All the oil lines (both engine and transmission) are a mix of 1/2” diameter hardline or flexible reinforced rubber hosing of 1/2” internal diameter. When converting to Earl’s Perform-O-Flex line, this translates to #8 line and -8AN fittings.
The math for the engine oil cooler:
We have the sizes and BTU/hr ratings of the coolers and therefore we should be able to calculate oil temperature out of the cooler by making a few assumptions. I am assuming the engine oil flow rate is 5 gpm, that we are using 20/50 oil, that we have a reference 100 deg F inlet temperature difference (ITD), specific heat of oil = 0.5205 BTU/lb-degF at 230degF and density of engine oil is 54.93lb/cu.ft (ATF fluid is slightly less at 52.66lb.cuft). Inlet air temp of 80 degF. Lets assume the engine oil inlet temp is 200degF and the outlet is 170degF to see if we’re correctly sized.
We have make some unit conversions since we have density in lb/cu.ft, and we need flow rate in gals/min, we use the fact there are 231 cu.ins in a US gallon and 1728 cu.ins in a cu.ft:
Oil mass flow rate (lb/min) = gpm x (231cu.in/gal) x (1cu.ft/1728cu.in) x oil density (lb/cu.ft)
= 5 x 231 x 1/1728 x 54.93
= 36.715 lb/min
On the next page, we solve for heat transfer.