In reviewing some of the foregoing calculations that followed from the information provided by Harold Bettes, there appears to be confusion arising from a slight misstatement of one of the formulas and the units of measurement required. In hopes of clarifying this, they are restated below with a little more explanation of what they represent and the units required.
The first formula basically calculates the flow rate required to achieve a given temperature rise given a stated energy input. That is, what flow rate is needed to keep the water temperature within desired bounds while going through an engine that is rejecting heat to the coolant at a particular rate.
GPM = 5.089*(Hp)/((c)*(To - Ti)) where:
GPM = gallons per minute
5.089 = (42.44 Btu/min)/(8.34 lb/gal (for water)) conversion factor for units
Hp = horsepower put into the coolant, about 1/3 of the engine’s flywheel horsepower
c = specific heat of the coolant (Btu/lb-F) = 1.0 for water
To = temperature of outflow (degrees F)
Ti = temperature of the inflow (degrees F)
Using this to recalculate the example situation from before, (HP = 200, To=180, Ti=100) it is found that the flowrate is 12.72 GPM. This is much more reasonable than the 106 result from before, as the 106 was not gpm, but lb/minute of water.
The second formula determines what volume of fluid is needed to keep its temperature rise within the desired limits given exposure to a heat input for a period of time. How big does the reservoir need to be?
V = 5.089*(Hp)*t/((c)* (To - Ti)) where variables are as identified above and t is the time of exposure in minutes and V is volume in gallons.
-or, if the variable quantities remain the same as the first formula, (and rather intuitively):
V = GPM*t
Again, using the example with t = 3 minutes, the volume is found to be 38.16 gallons, which agrees with the previously determined result. Interestingly, this method effectively assumes that the reservoir volume is only circulated through the engine once, that no mixing occurs in the reservoir, and that no heat is lost out of the system piping or tank.