The 5252 constant is for units of ft*lb of torque and horsepower. Newton meters and kilowatts use a different constant and will have a different value when the two numbers happen to be equal.
Quote:
Originally Posted by dkjones96
Brings up another question though. In my particular case, with two peaks, which one is more efficient?

Maybe both, maybe neither. These charts don't show what is needed to make that determination. What is needed is a fuel consumption chart showing grams of fuel consumed, power produced, engine rpm, and at several different percentages of load. Then add gearing charts to indicate when shift points should be to minimize fuel use.
Interesting to know, but impractical to put in practice.
Here's a link to a thread discussing the BSFC for the 1Z engine in my white wagon. The BMEP is cylinder pressure, think of it as the amount of torque being produced. Also note that half the torque requires more than half the fuel. One more comment: Constant speed operation does not require maximum torque. 1900 rpm and 150 lb*ft at full load, makes nearly 55 hp. 1900 rpm in 5th is 55 mph. Steady 55 mph in 5th requires about 11 hp or 20% of load.
http://forums.tdiclub.com/showthread...=208125&page=3
Have you looked at the chart and digested it a bit?
At full load (14 bar pressure, making about 150 lb*ft of torque) at 1900 rpm the fuel needed is right around 200 grams per kilowatt each hour. 200 grams times 41 kW is a 8200 grams per hour consumption rate.
Requiring less power, say only 1/5 of full load, means the pressure is 2.8 bar and torque is 30 lb*ft. At 1900 rpm that's 8.2 kW. Notice the fuel use is less efficient per kW at this load with 280 grams required per kW. Nearly 40% more fuel per kW is needed, but since the number of kW is 80% less the result is a 2300 grams per hour consumption rate.
Full load is more efficient, but the lower load produces more distance per volume of fuel.