It sounds to me like you're wanting brake specific fuel consumption?
http://en.wikipedia.org/wiki/Brake_specific_fuel_consumption
Don't know EXACTLY how you'd do it, but heres some ideas that might be of help...
You'll almost certainly need the torque curve (torque verses rpm) for the engine (you can produce it from power verses rpm), as efficiency varies from a maximum at peak torque....
calculating BMEPs might be of some help
http://www.xplorer.co.za/articles/bmep.htm
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If your wanting the fuel consumption of an engine used for transportation, you need to specify the torque/rpm that will be asked of the engine.
You could perhaps provide an estimation of fuel consumption by working "backwards" from a "neat trick" I came across which uses a car's mass as it's own dynometer (power meter)
weigh the car, and then find several different (steady/long) gradients. ...If you know the mass of car, and measure what steady speed it attains while out of gear, going down a number of hills with KNOWN gradients,you can find the power that's required for the car to move on the flat at any speed (going down a hill "adds power" to the car, going up a hill takes it away. You can power a car by a downhill slope...). NB Acceleration functions much like going up a hill (local gravity "rotates", "g" will change too, because of the addition of vectors
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I forgotten the other method was called a rolldown test.
My idea is a little bit different, (and relates power requirement to speed more directly), but does use some of the same ideas...
I figure that as power can be calculated by power=mass*g*height_raised /time if you know how fast a car loses height while in neutral (ie drops altitude:- can be figured by knowledge of trig, speed and gradient) , once acceleration is 0 (ie speed has peaked ) you can calculate quite simply how how much power that vehicle needs to overcome it's drag forces, and maintain it's speed. NB the power "robbed" by climbing the same hill at that speed would be the negation of the power provided by "rolling" down the hill. Relating power requirements to speed allows drag coefficients to be figured out.
Consider the direction gravity acts on a vehicle. It is the same as the direction of a pendulum hung from the vehicle's roof. Fix the protractor to the roof of the car, so when the car is at rest, on the level it reads 90 deg (and goes towards 180 deg when the car is climbing a hill,towards 0 when going down a hill)
Go down a hill without accelerating, and the line of action of gravity is in the direction of travel (ie towards 0 deg). Force pulling the car along the road (and down the hill) due to gravity is f=m*g*cos angle. (If using a metric system power then found by p=f*v)
I haven't quite worked out the details of it out but here's the guist of my idea as to how to factor in power requirements for acceleration/consumed by braking.
Now consider the rate at which energy needs to be removed from a vehicle to brake it(ie the power the brakes must transfer into heat). Keep the same reference frame/protractor/pendulum to measure angle, and you find when braking the pendulum once again moves forward. A given angle of pendulum means v*m*g*cos angle units of power are transferred. pendulum moves towards 180 under acceleration/hill climbing
I think you then need to make the sum of the power transfers equal to 0(the vehicle doesn't really gain/lose power) to figure out the direction of the power transfers, but I'm struggling with that step, as something non-intuitive seems to be going on.