I mentioned in my last post that I'm trying to understand the factors that make it impossible for me to achieve the fuel savings in my Land Rover LR3 that I did in my Jeep Grand Cherokee Limited. I'm beginning to think my understanding of the internal combustion engine is sadly lacking.
For example, in the post linked above, I tabulated some of the relevant numbers for each vehicle. The LR3 uses a smaller engine to produce a higher rated horsepower than the Jeep. In the highest gear (at freeway speeds) the engine rpm is lower as well so one would think that the smaller engine turning more slowly would burn less fuel and hence derive heat to perform work at a slower rate. I know that the LR3 has a higher compression ratio. Could this be the explanation?
It's well known that the maximum theoretical efficiency, E, of an engine using an idealized Otto cycle is E=1-r^(1-y) where r is the compression ration and y (should be be the Greek letter gamma) is the ratio of the constant pressure to constant volume heat capacities. For the LR3 with its 10.5:1 compression ratio, this works out to 0.61. For the Grand Cherokee Limited at 9.3:1 it is 0.59. So what does this mean?
It means that for a given amount of heat from burning fossil fuels, the LR3 engine would be able to do ((0.61-0.59)/0.59)*100%=3.4% more work per joule of heat from burning gasoline than the Jeep if they were both working at the maximum theoretical efficiency. In order to see what how this affects our consumption, let's see how much heat per second from burning fuel is available to each engine.
In one second, at 1750 rpm in the Jeep, the engine moves 68.5 liters ((1750/2)/60) * 4.7 liters of fuel/air mixture through the engine. (The division of 1750 rpm by two is necessary because the rpm readout is crankshaft rpm, the crankshaft in a four stroke engine revolves twice for each engine cycle). The LR3 at 1660 rpm will move 60.9 liters per second. In these mixtures will be fuel, and I will assume that the richness of the mixture is the same for each engine, since the ECS (engine control system) will try to maintain the so-called "stoichiometric" ratio (14.7:1 by air mass to fuel mass). This mystifies me because the LR3 should burn less fuel since it's moving a smaller volume of fuel/air mixture through the engine at a presumed identical mixture. Yet the Grand Cherokee indicates instant mileage of approximately 31 m.p.g, the LR3 shows about 21.5.
The density of air at typical temperatures, pressures, and relative humidities is about 1.16 kilograms/meter^3 or 0.00116 kilograms/liter. This density is reduced in the intake manifold due to throttling effects, in fact, that's how the throttle works. I have equipped my LR3 with a Scan Gauge II so that I can measure absolute manifold pressure. At a steady 55 m.p.h. on level ground, manifold pressure is 68% of ambient. Since density is proportional to pressure, the ambient density of 0.00116 kilograms/liter is reduced in the cylinders to 7.84*10^(-4) kilograms/liter. We'll assume that the mixture is air as an approximation, since it's about 94% air in reality. Therefore, in one second, the LR3 moves 0.0477 kilograms of mixture through the engine and in that fluid, there should be (1/15.7)*0.0477=0.00304 kg. of gasoline. Burning this gasoline will release about 143,000 joules of heat energy. Hats off to me, this is great information. There's only one problem.
Since a gallon of gasoline weighs about 2.65 kilograms, this implies that I'm burning (0.00304*3600)/2.65=4.13 gallons/hour. At 55 m.p.h., this is about 13.3 m.p.g. The LR3 is no economy car but it isn't as bad as that. As I stated earlier, I expect about 21.5 m.p.g. at 55 m.p.h. on the freeway. Clearly, something is wrong in my assumptions. I'm not sure what it is, the mass flow calculation seems pretty straightforward.