I am also new here (but I have an engineering background) so I thought I would run through this problem to help my understanding of the technology. The physics/math has already been shown in the 2014 posting but in another form.
Thinking of this as the heating of a water tank with the engine as the heater. The rate of temperature change is the heat transfer rate divided by the mass times specific heat of water: change temp/min = (Btu/min) / (m x specific heat). The specific heat of water is 0.8784 BTU/(lb-F deg) and heat input is amount of HP rejected in BTU/min. One HP is equivalent to 42.41 BTU/min.
Working out the suggested duty cycle of 2 min at half throttle and 1.5 min at full throttle and using the case listed in the 2014 posting by Harold Bettes, a 20 gallon tank with 33% to 40% of HP generated by the engine is rejected to the coolant, a 500 HP maximum engine using the 40% rejects 200 Hp or 8482 BTU/min to the coolant. At one-half throttle, the heat input would be halved or 4241 BTU/min.
For a 20 gallon tank there is a mass of water of 166 lbs. The temp change for the first part of problem is therefor 2 min x 4241/(166 x 0.8784) = 58 F. The the second part, 1.5 min x 8482/(166 x 8784) = 87 F. Overall temperature change is 145 F deg. For an ambient temperature of 70 deg this results in an over temp of 215 degs.
At over 4000 ft elevation, the boiling point of water at Bonneville is 205 F which suggests a pressurized cooling system or a larger water tank. Because the coolant temperature varies so much over a run, is may be necessary to employ a thermostat to maximize power utilization. The effect of a 25 gallon tank on the temperature change can be easily calculated by proportion : 20 x 145 / 25 = 116 F increase, resulting in a final temp for 70 ambient of 186 F.
