Calculating Energy
Calculating your car engines energy is very straight forward.
Buy or borrow a infrared thermometer that works from a distance. Google "ir thermometer" you will find lots of choices. They are available for under $20.
Measure your exhaust manifold temperature at idle, after 10 minutes of running the engine. Follow your exhaust pipe to the engine, see where it connects, then measure all over that for the hottest spot.
For a small for cylinder, 150°C was measured, when it was 30°C outside. So 120°C rise. To calculate backwards from an expansion ratio of 10, the heat ratio for an expansion of 10 is 10-.4, which is 0.4 (coincidence that it matches exponent). Divide 120°C/.4 to get 300°C. That is the LOWER bound of the initial heat inside the cylinder. The air has cooled a lot by the time it gets that far, much of the cooling inside the temperature The 0.4 factor is from a 1 to 10 expansion.
300 C, or 600 Kelvin, happens to be the double the absolute temperature outside. The ideal gas law says pressure will double when absolute temperature doubles.
Back-of-the-envelope
The calculation to the right is perfectly accurate for no specific engine, but typical of all water cooled engines.
A different back-of-the-envlope calculation is to estimate the power needed to run 35 and 70 mph, and back calculate the efficiency. That calculation yields 8% and 24% predicted efficiency based on equal city and highway mileage. So, similar in that both predict efficiency goes up with engine speed, but the predictions based on educated guesses of actual wind and rolling resistance, or rate of engine heat loss differ slightly. Both models show engine efficiency drops to zero and stays near zero at any low engine speed.
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How much power is that?
To idle
The work is the area under the curve, or (.1-.4-1)/.4 or 3.8 atmosphere*volume units. Set the volume scale as 1 liter, then the piston movement is .1 to 1 or .9 liters. To get atmosphere*volume units for 1 liter = 3.8/0.9=4.2 atmosphere*liters. If the engine is a 2.3 liter engine, 1.15 liters per revolution "fire" with a power stroke. So scale to 1.15 liter from .9, yields
1.15*4.2*10 Newton's/sq centimeter * 10 sq centimeter-meters per liter or
485 newton meters, aka joules aka watt-seconds per stroke.
600 rpm is 10 strokes per second, 1000 rpm about 17 strokes per second. So the total power is 485 * 17 or 8.25 kilowatts, or 11.1 horsepower. That's the converted portion of the heat. 40% of the total heat is unconverted. 11.1 horsepower is .6 or 60% of the total energy. So the total energy consumed is 11.1/.6 = 18.4 horsepower, just to idle, to do nothing.
To go 30 mph
To produce an extra 2 horsepower, the temperature goes up slightly, but mainly the RPM's go up 10 to 20%. If rpm's go to approximately 1200, that's about 3 extra horsepower burned about 21.4, and finally produces a NET 1.8 horsepower. So 8% fuel conversion at 30 mph.
At maximum power
The same engine advertises 135 horsepower. and 2.3 liters, at 6000 rpm. 100 revolutions per second. Working backwards, 135 hp is 100 kilowatts, is 1 kilowatt per stroke. Needs
1000*/1.15/10 newton's per cm (per atmosphere)/10 cm-meters per liter or 8.7 atmospheres.
1 "ambient" temperature, 300 C, adds 4.2 work units per liter. To find how much heat to add, 8.7/4.2= 2.1 units or 630 degrees C. (to correct this figure for friction and engine work, increase it about 10%. Maximum temp is determined by maximum average pressure, which is about 9.5, or 2.3 x ambient).
No power is 300 degrees, full power is 630 degrees. (Recall the RPMS go up too, so there is a multiplier of about 5 or 6 that is not obvious.)
At full power, the peak temperature difference between air and engine is about 1.6 higher. So it is "losing" about 35 horsepower of the heat in radiated heat. This also means it is "losing" about 480 degrees worth (1.6*300), but spreading that over 5 or 6 strokes. Why? The strokes are much faster, so the amount of loss per stroke is 480/5 or about 100 C . Instead of "wasting" 300 C, each stroke is now "wasting" 100 C, due to heat leaking, otherwise known as engine cooling.
Not figured into this analysis is added blow-by or added friction, both of which increase in proportion to peak pressures. The primary loss is direct thermal loss, which is shown in the next paragraph.
So at full power, the temperature efficiency is approximately (630-100)/630 * Work Ratio 0.6 = 50%. There is typically 10% difference between braking horsepower and calculated, due to friction and work in the water pump and valves. So this engine should see a braking horsepower efficiency of about 45%, ignoring pressure related losses. The manufacturer's claim for the engine is 40%.
This "back of the envelope" analysis shows the primary energy loss mechanism for fuel efficiency in water cooled engines, is the water cooled part, not friction or blow-by.
To go 70 mph
To go 70 mph requires only a few more horsepower than going 30 mph, about 3 times. Most modern cars with 4 speed transmissions can maintain 70 mph at under 2400 rpm or 20 strokes per second. The temperature loss load goes up from idle, but much less than at full power. To produce 6 horsepower or 4.4 kilowatts in 20 strokes requires a net 214 joules per stroke. This corresponds to less than half the 485 joules produced by 300 C, one can scale it to 300*214/485 or 132 degrees. If we estimate the heat loss is scaled up linearly about 60 degrees additional heat loss is incurred. This leaves 60%*132 or 80 degrees, for fuel units of 300+60+132 or 492 degrees. So to go 70 with 6 horsepower, efficiency goes up to 16%, (80/492).
So even though the manufacturer likes to think the engine is 40% efficient, it only reaches that at an engine speed that the car is never run at for more than a few seconds at a time, and then only when someone "floors" the accelerator. Yep, jackrabbit starts are more efficient than those responsible steady accelerations. Go figure.
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