Pressure and Force

[MolaKule]

QOTD:
Just before the piston in an IC engine is on its downward stroke (TDC), the combustible mixture ignites and the cylinder pressure rises to 250 psi. The piston crown (face) is 3.5 inches in diameter.

1) At this instant, what is the force on the piston, in pounds,

2) what happens to the force on the top of the piston as the piston moves downward?

Prior respondents to today's QOTD are not eligible to answer.

 

[Belker]

1. 2,405.28 pounds;
2. I would expect the force to decrease as the piston moves downward.

 

[fsdork]

1. Same answer as @Belker : 2405.28 lbs, assuming the piston crown is flat. If not, surface area increases and so too does the force on the piston.
2. It depends upon how long and how rapidly the air/fuel mixture continues to combust. The gases will expand as long as the burn continues, so pressure could conceivably decrease, stay the same, or increase. The first couple Google results say pressure peaks at TDC and decreases during the power stroke due to increased volume in the cylinder, so - that.

 

[FowVay]

Let's see, the area of the piston face is πr². So πx1.75²= 9.621
9.621 sq. inches X 250 lbs per square inch = 2405.25 pounds

and of course, as the piston drops down the cylinder volume increases and the pressure is converted from potential energy to kinetic energy in the form of a rotating crankshaft.

 

[4WD]

2405
easier to use .7854 … nice circular key strokes

 

[MolaKule]

All correct answers and thanks for your replies.

The instantaneous force at time of combustion would be 2405 lbs, since as, FowVay said,"
the area of the piston face is πr². So πx1.75²= 9.621
9.621 sq. inches X 250 lbs per square inch = 2405.25 pounds."

Area A also equals π x (D/2)² so Area could have been determined this way.

9.621 in² X 250 Lbs/in² = 2405.25 Lbs. since the in² cancel.

According to the Ideal Gas Law, P(pressure) X Volume = n X R X T, where T is temperature, n is number of moles of the substance, R is the Gas Constant, so Pressure = nRT/V, but Volume is increasing so the Pressure is decreasing since the piston is traversing downward opening up more Volume, i.e., as V gets larger, Pressure has to get smaller.

So the force on the piston decreases as the piston is traversing downward.

What about Temperature? T at a 0 degree crank angle position (TDC at max compression) is about 2500 K but rises to about 3500K at the 20 degree crank angle, and then falls off to 2000K at 60 degrees crank angle, and continues to fall as the crank angle increases.

So even though the pressure in this example was 250 psi at TDC, Pressure increases to maximum at the 20 degree crank angle (after TDC) and falls rapidly as the crank angle increases. But the Volume is increasing at a greater rate than the other variables mentioned.