# Engine Heat Input Vs Torque Vs Power

1. Dec 5, 2012

### Peter.B

Hi all,

I am trying to compare the heat input into an engine with useful work out. I have taken a measurement for Torque, and have calculated heat input per cylinder based on combustion calculations i.e throttle % x cal value of fuel x fuel mass.

I am happy with the heat input calculation per cylinder, but am having difficulty establishing how overall work done and heat input per cylinder align.

* 3000 RPM
* Torque = 27.7
* Revs /Sec = 50

Work Done

Power = (Torque*((2*3.14)*Revs))

=(27.7*((2*3.14)*50))

= 8698 Watts

However when I compare this to the heat input, the heat input is much lower?

Heat input per cylinder per 2 cycle (i.e 2 revolutions of crank) = 237 Joules (I am 99% confident that this value is correct).

Heat input Power = (236 x 50 ) / 2 (Divide by 2 because heat input every 2 revs)

= 5926 Watts ??

The question I have is... Should I be taking into account the heat input to all cylinders? or is there some other fundamental thing which I am missing?

If anyone could point me in the right direction it would be much appreciated.

Thanks,

Peter

2. Dec 5, 2012

### SteamKing

Staff Emeritus
Yes, both cylinders produce the torque output of the engine.

3. Dec 5, 2012

### Peter.B

Thanks very much for the response Steam King.

So the value I currently have of 5925 joules heat input, should I multiply this by 2 or 4. (It is a 4 stroke engine).

Multiply by 2 = 73% efficiency

Multiply by 4 = 36% efficiency

If you could confirm that would be great.

Thanks again.

4. Dec 5, 2012

### rcgldr

What are basing heat input per cylinder? Is this an internal combustion engine or some other type of engine? If it's an internal combustion engine, then energy input is the chemical potential energy of the average amount of fuel consumed per revolution (and at 50 revs / second at the throttle setting you're using).

5. Dec 5, 2012

### Peter.B

Thank you rcgldr.

Yes it's an internal combustion engine.

I am basing the heat input per cylinder on the amount of fuel which is used in 1 full cycle (i.e 2 revoltions on the crank angle encoder).

I believe you have answered my question.

I believe it is (237 x 2 cylinders) x 50 rev/s

To give 23,700 Watts Heat Input Vs 8698 Watts of work out

= 37% efficient.

It's now making sense, thank you very much for your assistance! (Although if my logic is still incorrect, please let me know).

6. Dec 5, 2012

### rcgldr

Why "x 2 cylinders"? A 4 stroke engine only draws in fuel once every 2 revolutions. Are you basing this on fuel / air mixture and then the displacement per cylinder (cylinders don't get completely "filled"). Is there some way you can determine fuel flow in terms of volume (or mass) per unit of time at some specific rpm?

7. Dec 6, 2012

### Peter.B

Yes but as you have suggested, I am now calculating the average fuel input over all cylinders per second. I have checked with my lecturer, and the figures I have are correct.

I am calculating fuel input based on working cylinder volume, throttle %, air /fuel ratio, subsequent air / fuel mass and the calorific value of the fuel.

I am now happy with everything, and the figure of 37% efficiency is in line with what would be expected from a typical otto cycle.

Thank you

8. Dec 6, 2012

### rcgldr

I was just wondering about the "x 2 cylinders", when previously the math was per single cylinder.

I'm also wondering what type of load you had on the engine in order to measure the output power.