Forces in a piston engine

Atomic_Sheep

Assuming you have a uniform expansion of gasses upon their ignition from the top most piston position to the bottom, would the force generated on the crankshaft vary at different points in the down cycle? My logic tells me that when you are in the top most part of the cycle, you would have a tiny component of force directed downwards and most of the force being exerted perpendicular to where you want it to be moving, then as you reach the middle of the cycle, the gases are finally pushing the rod directly downwards and hence peak force and then it tapers off again. Is this correct?

Also since a turbine engine is... well different in design, does this difference mean they are more efficient? Yes piston engines also have 4 strokes so they lose efficiency in that regard also but let's disregard that for a second.

Ranger Mike

initial assumption is incorrect. Maximum combustion pressure occurs just after top dead center and drops off rapidly after about 45 degrees past TDC. I do not understand your thinking on force NOT being applied to the piston and forcing it to bottom of the stroke.

Atomic_Sheep

By uniform expansion of gases, I was trying to state that I was trying to work in an environment where combustion pressure was uniform from the top of the stroke to the bottom (so working in a prefect world scenario). So I am trying to figure out whether the vector force component pointing directly down caused by the expanding gases which was pointing directly down (under the environment of uniformly expanding gases), changed at the various points of the down stroke. The way I see it, if we have a circular path of motion, I imagine that only at the top, the 90 degree point and the bottom parts of the stroke will the downward force equal and there be no lateral component. At other points of the stroke, lateral vector forces would be acting, thereby reducing the magnitude of the vertical force component of the expanding gases.

The second part of the question removes the assumption of uniformly expanding gases.

Q_Goest

The torque produced as the piston moves through it's stroke will vary for the reasons you've stated. At first there is a small change in cylinder volume (dV) per degree of rotation. After 90 degrees, there's a relatively large dV per degree of rotation. And at 180 degrees, there's a small dV per degree of rotation. But that doesn't mean there is energy wasted that might somehow be recovered if all the energy was exerted in that 90 degree range. Think of it this way, for a piston in a cylinder, and looking only at the gas, how do you find the work output? Forget the crankshaft for this thought excersize.

panapal

Please allow me to ask a PISTON / Engine question to which I do not know the answer. I am working on a project for college and need some brain power to assist me. Thank you.

What I need to calculate (I don't know the physics equations) is this. If I built a mid-sized car that seats 4 to 6 people and the cars empty weight is 1,200 - 1,500 lbs., how do I calculate how much force is needed to move the pistons up and down to rotate the crank shaft fast enough so the car can be driven at the same average acceleration and speed as a typical mid-sized car?

I can have as many pistons as needed (8 to 12 preferred). I can not use any discussion of fuel or combustion (assume the pistons simply "move"). I need an answer in "force necessary to accelerate the car" at "normal" speeds.

Maybe I'm not asking this correctly. Please help! Thanks.

Kozy

You have the right idea, the torque produced is the product of both the cylinder pressure (diminishing) and the mechanical leverage on the crank (rising and falling in a roughly sinusoidal manner). The peaks of both factors are not in the same locations so you end up with a slightly different torque curve to what you might expect over a full cycle.

Here is a simulated torque output for a single cylinder over 360 degrees of crank rotation:

The Chase

This question is very vague.

First of all, a kerb weight of 550-650 kgs for a 4-6 passenger vehicle is inappropriate, this is far too light. Secondly you have not stated what opposing forces are to be taken into account, for example frictional losses in the drivetrain or wheels to road, wind resistance or road incline.

Assuming you are totally neglecting these factors other than vehicle mass (which is far too light) then the equations should be relativly simple. The torque produced by the crankshaft will be the total amount of cylinders force*radius of the crank arm. A Toyota Previer for example produces 220 Newtons per meter out of its four cylinder engine with a 96mm stroke (which translates to a 48mm radius). Therefore;

220/4 = 55 N/m per cylinder,

1000mm/48mm = 20.833 (conversion factor),

20.833*55N = 1145.83 Newtons of force per cylinder

This is pretty typical of a large passenger vehicle. Hope this helps.

Ranger Mike

I suggest a review of the mechanics that occur is in order. The crankpin is not 90 degrees from TDC or BDC when the piston is at mid stroke. In fact when the crankpin is at 90 degrees the piston travels more than half stroke from TDC. Because the piston travels farther during the " first " 90 degrees, of crank rotation the piston velocity is higher from TDC to the mid point than it is during the last half of crank rotation.

SEE post gas pressure in internal combustion gasoline engine 22 Nov 09 on this forum

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