Stay with me on this....I haven't done an engineering calculation like this since college
I went to bed thinking about my original question, "What is the temperature rise of the air in the engine compartment?). I started thinking about the worst case scenario. First assumption is that all energy in the gasoline is converted to heat. We know that combustion is not 100% efficient. The second assumption is that all this heat is 100% transfered through the front radiator, raising the temperature of the air passing through it. Last assumption is that only air through the radiator can be sucked in by the engine.
Again this is worst case. We know combustion heat is removed by the exhaust gases, removed by the engine oil system, radiated out by engine block heat, etc.
Then I looked at the performance of our cars. Cruising down the freeway at 60 mph getting 20 miles per gallon (3 gallons/hr). Looking on-line, gasoline has the specific energy of 120 MJ per USGal. We are consuming .05 gal/min, producing 6000KJ/min of heat.
I am not home to measure but I estimated our radiators to be about 2 feet by 1.5 feet. Approximately .335 square meters.
At 60mph this equates to 1,609.32 meters per min. So there is about 539 cubic meters of air going through the radiator every minute.
Here is the engineering lesson
dT = Q/Cp x m
Really important to make sure all the units match or the calculation is wrong.
dT is the change in Temperature (C)
Q is the heat produced (KJ/min)
Cp is the specific heat (air in this example; KJ/Kg x C). This is the amount of heat required to raise the temperature of a material by one degree.
m is the mass (again air through the radiator)
Q was estimated to be 6000KJ/min
Cp for air at 68F is 1.005 KJ/KG x C
m takes some calculation. air density is 1.205 Kg per cubic meter. We have 539 cubic meters per minute. So, the mass of air through the radiator = 649.64 Kg per min
dT = 6000 / (1.005 x 649.64) = 9.2 degrees C
One last conversion gives us 16.5 degrees Fahrenheit. This worst case example estimates the temperature of the ambient 68F air to rise to about 85 degrees inside the engine compartment.
Researching further, from Wikipedia:
Gasoline (petrol) engines
Modern gasoline engines have a maximum thermal efficiency of about 25% to 30% when used to power a car. In other words, even when the engine is operating at its point of maximum thermal efficiency, of the total heat energy released by the gasoline consumed, about 70-75% is rejected as heat without being turned into useful work, i.e. turning the crankshaft.[1] Approximately half of this rejected heat is carried away by the exhaust gases, and half passes through the cylinder walls or cylinder head into the engine cooling system, and is passed to the atmosphere via the cooling system radiator.[2] Some of the work generated is also lost as friction, noise, air turbulence, and work used to turn engine equipment and appliances such as water and oil pumps and the electrical generator, leaving only about 25-30% of the energy released by the fuel consumed available to move the vehicle.
If this Wiki article holds true, then about 38% of heat produced transfers to the radiator
This would result in (.38 x 16.5F) a temperature rise of only 6.5F, raising our engine intake air to just 75F!
Again, using the web. This theoritical rise in temperature from 68f to 75F results in a power reduction of only .8%!
