Cuantos caballos?

In my last post I tried to calculate the top speed of my Land Rover LR3 HSE. In doing so, I used the horsepower found in the vehicle's specifications, i.e., 300 horsepower. Earlier in my musings about the LR3, I was marginally successful in some calculations relating to the car's (truck's?) fuel consumption. Now I'd like to calculate the vehicle's horsepower using known facts. These facts are the engine displacement and red line r.p.m. (6200)and the r.p.m. at which it was rated as specified (5500).



I'll use the method that was employed before, i.e., I'll determine how much fluid volume (air is a fluid) is going through the engine at those r.p.m.'s and how much fuel would be in this fluid. From there, I'll use the energetic content of the fuel and some thermodynamic calculations of efficiency to see what's available at the flywheel. Here we go. One advantage I'll have is that I can estimate the efficiency by figuring out how much heat energy per second comes from putting air/fuel mixture through the engine at 5500 r.p.m. and utilizing the rated brake horsepower of 300 at that r.p.m. This should give me an approximation of how many joules per second go to the flywheel versus how many are discarded to the environment.



A straight ratio of (6200/5500)*300 tells me that I should be able to produce 338 horsepower at 6200 r.p.m. As mentioned in the earlier post linked above, Car and Driver stated that the LR3 is governor limited to a top speed of 121 m.p.h. Let's see what kind of overall efficiency is indicated if 5500 r.p.m. produces 300 horsepower. To do so, I need an estimate of manifold pressure - for starters I'll assume wide open throttle and maybe something like 0.2 p.s.i. losses for an absolute pressure of about 14.3 p.s.i.



So at 5500 r.p.m., the engine (as per my policy, I'll spare readers many of the actual calculations) will pump about 0.0147 kilograms of fuel air mixture through the engine each second. This takes into account manifold pressure of 14.3 p.s.i. as mentioned above. Oxidation of this mass of gasoline will release about 705,000 joules of energy each second. Since joules/second are watts, a measure of power, we can convert to horsepower. Doing so, if the heat of oxidation of that amount of gasoline could be converted to mechanical energy with 100% efficiency, we would develop 946 horsepower. The rated horsepower at 5500 r.p.m. is 300, implying an efficiency of about 32%. This seems very reasonable.



OK, so what's available at 6200 r.p.m.? Well, since all these calculations are based on mass flow of fuel air mixture through the engine, the simple calculation above gets close, at 338 horsepower. However, I assume that there's a slight improvement in volumetric efficiency, in other words, a slightly higher manifold pressure than at 5500 r.p.m. Making this assumption and calculating as above, it seems possible that, if full throttle at redline can be achieved, an absolute maximum of 358 horsepower could be measured at the flywheel.