 # engine efficiency question

I was thinking how to rate an engine’s efficiency by a ratio of it’s output over its input. Since air is free and inexhaustable, I would use a flow meter on the fuel line to provide an instantaneous rate of fuel consumption as the input variable. Should I use torque or horsepower as my output variable?

It can’t be torque as torque is not a unit of power or energy. Although the units of torque seem to add up to energy, you need to add the revolution factor, which is unitless in this case. So if you use torque, you need to also measure rpm to get to your horsepower ratings.

Horsepower seems OK to use.

If I was going to rate the efficiency though, I would compare output power (horsepower) or energy (joules, hoursepower hours, what ever) to heat produced. This saves me the trouble of figuring out the exact amount of chemical energy in the fuel.

You need to use input power divided by output power.

You put the flow meter in the fuel line, and find that it’s using, say, 10 millilitres per second (0.01 litres per second).

Gasoline has a combustion energy of 32,000 kilojoules per litre. So in this example, 0.01 Litres per second times 32,000 kilojoules per litre equals 320 kilojoules per second entering the engine. 320 kilojoules per second equals 320 kilowatts.

Lets say you’re measuring 64 kilowatts at the wheels.

Then the overall engine and drivetrain efficiency is 64 kilowatts divided by 320 kilowatts equals 0.2, or 20%.

If you’re measuring in gallons per hour and horsepower, so you’ll need a few conversions along the way.