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Peak vs Average Cylinder Pressure -> Torque

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In "Pressure Volume Cycle", you state that it's the 'peak' cylinder pressure that's responsible for the torque. Doesn't this imply all the work is done at the top of the stroke? Wouldn't a consistent pressure over the power-stroke perform more work, and hence generate more torque? (e.g. a slower flame-front, rather than a short-and-sharp expansion)

Don't recall the video, or if I watched it, but there are several things that affect instantaneous and peak and mean torque.

It's more common to refer to Brake Mean Effective Pressure - basically a fancy way of saying the average of the pressure. I expect it's a communication issue, but as an aside there can be infinite pressure at TDC and there would be ZERO torque, and for best BMEP, the peak pressure is usually reached around, IIRC(!), 10-12 degrees ATDC, when the connecting rod angularity and crank throw position means the force is directed to the now increasingly offset crank pin.

Hi Craig,

I can kind of see what you might be thinking, when the maximum pressure is reached the combustion is still occurring and gases expanding; the reason we want this to happen after tdc is that we want the force to push the piston down and this needs to be done after the piston has reached tdc for an efficient application of power. This is normally targeted in the OEM world at 8 degrees ATDC and is called Mass Fraction Burn. These is a link for a paper on combustion if it interests you.

(https://ilot.lukasiewicz.gov.pl/KONES/2002/02/str193.pdf)

Think of it as your pedalling a bicycle, applying maximum force at tdc wouldn't move the cranks around at all, only when you tip over the top of the cycle can you apply maximum force and then won't need as much as the rotation starts to continue the cranks rotating.

Getting the spark to start before the piston has reached tdc means that there is time for the air/fuel mixture to compress and as the flame propagates to reach maximum pressure just after tdc and maximum cylinder pressure can force piston back down generating torque. As you said about changing the flame speed is done with octane and how fast the flame can spread; also multiple points of ignition (see Jet ignition by Mahle)

Hope this helps and hasn't confused you any further.

Thanks for the replies - and the very interesting link!

I was sort of zeroing in on the 'peak pressure' statement - implying a large force, but over a small time (inventing numbers - say 1kn for 0.01s), versus trying to get pressure over a longer time (e.g. 700n for 0.03s).

I hadn't really factored in what is effectively lever-length - thanks for that! Definitely complicates matters. Now that I'm thinking about it, I can also see how that's going to have a big impact on forces throughout the engine - early in the stroke = shorter lever => more stress (for the same pressure), but also more work.

You want the fuel to burn over a controlled period, otherwise you get what's technically called an 'explosion'.

Not quite sure about the very last part of your response - work done is force through (times) a distance*, if you apply force without movement you do no work.

*This is one of my bug-bears.

Work done is F x d - the distance the force is applied through - in old terms is expressed as ft.lbs.

Torque is F x r - the perpendicular distance the force is being applied at from the centre of rotation - in old terms lbs.ft.

ft.lbs and lbs.ft look the same and many people use them interchangeably, but they actually mean different things.

An example. You put a 12" long wrench on the crank pulley bolt and it takes 10lbs force at 1 foot distance to turn the engine - that's ten lb.ft of torque, you rotate the engine through one revolution - that is 10 lbs through 1 foot x 2 Pi travel distance, you do ~6.28 ft.lbs of work. From those two statements, and the knowledge one (Watt's) horsepower is 33,000 ft.lbs of work done per minute, the switched on people will figure out how the ~5252 we use as a constant is derived.