Physics Question

[Al]

It will increase. IF We postulate that heat is added isothermally (s-l-o-w-l-y). Angular momentum is the integral of angular velocity times mass from the center of the cylinder to the outside surface. As we compress the gas the mass increases and the angular momentum. This the simple solution. But the problem is that may be impossible to compress it isothermally.

The reasons for this is that the gas movement is not frictionless and although the "heat of pressure" can be removed isothermally the friction of gas molecules dictate that heat will go away (isothermally) but in the process the velocity will slow down and the anguler momentum will decrease.

But also note even before pressure is added it will slow down due to this phenomenon.

In effect you would have to make the outlandish assumption that molecule movement is frictionless. The other reason there is no correct solution is that it violates the Second Law. lol..I'll check back tomorrow. I suspect that the OP already knows this.

 

[Mazomopar]

For 29.95 I can give you the answer.

 

[A_Harman]

It will increase. IF We postulate that heat is added isothermally (s-l-o-w-l-y). Angular momentum is the integral of angular velocity times mass from the center of the cylinder to the outside surface. As we compress the gas the mass increases and the angular momentum. This the simple solution. But the problem is that may be impossible to compress it isothermally.

The reasons for this is that the gas movement is not frictionless and although the "heat of pressure" can be removed isothermally the friction of gas molecules dictate that heat will go away (isothermally) but in the process the velocity will slow down and the anguler momentum will decrease.

But also note even before pressure is added it will slow down due to this phenomenon.

In effect you would have to make the outlandish assumption that molecule movement is frictionless. The other reason there is no correct solution is that it violates the Second Law. lol..I'll check back tomorrow. I suspect that the OP already knows this.

You can't assume that the heat is added isothermally. The compression stroke in internal combustion engines is adiabatic, meaning no heat transfer out of the gas.

 

[userfriendly]

Maybe off topic, but a team of engine designers could spend months perfecting piston and combustion chamber shapes.
With the advant of direct injection gasoline engines, fuel seperation from the intake mixture and valve overlap
are not so a much a concern as before, because the fuel is added during the last half of the compression stroke.

 

[JT20]

It sounds like you all are talking about Brownian motion. ?

 

[JT20]

Again, "closed cylinder" ( that puts a hard stop in there)

At that point the "gas" would be at a zero potential state thus no movement ( other that the nano second it would take for all previous movement to normalize)

Then starting "compression", it would "increase" as the piston moves forward ( changing the volume of the area) affecting the density of the gas.

Anything beyond that would have to have more info on the gas

Wouldn't the cylinder blow out if compression gets too great ?

Or, some weaker point, since in an automotive engine, a cylinder has a gasket in the middle or near top.. a potential weak point?

Like bolts shooting off a boiler with too much steam in it

 

[ZeeOSix]

The head gasket would most likely blow out/leak, or a rod bend before a cylinder would blow-out (assuming you mean the cylinder wall would fail).

But does the piston actually stop at TDC?

 

[A_Harman]

The head gasket would most likely blow out/leak, or a rod bend before a cylinder would blow-out (assuming you mean the cylinder wall would fail).

But does the piston actually stop at TDC?

It's impossible for the piston to go from ascending to descending without stopping.
If you want to split hairs by pointing out that the piston might rock in the bore as the rod goes over center, and is therefore not stationary, I would like to point out that.........

Wankels go hummmmmmmmmmmmmmmmm, pistons go boing-boing-boing!

 

[Al]

You can't assume that the heat is added isothermally. The compression stroke in internal combustion engines is adiabatic, meaning no heat transfer out of the gas.

This is not an IC compression. Like any engineering/physics problem (as you know) you usually need to make assumptions. I chose to compress for no particular. The solution will be different if adiabatic.

 

[Al]

BTW..I have no confidence that my answer is correct. I am guessing it is wrong. Wish Mola would give us the correct answer. I have argued him in the past. But when it comes to this stuff he definitely knows his sh..

 

[ZeeOSix]

It's impossible for the piston to go from ascending to descending without stopping.
If you want to split hairs by pointing out that the piston might rock in the bore as the rod goes over center, and is therefore not stationary, I would like to point out that.........

Wankels go hummmmmmmmmmmmmmmmm, pistons go boing-boing-boing!

Guess you missed that whole thread ... it was tongue in cheek (hence the emojis) for those who did see it.

 

[JT20]

It is a good question.. Is this cylinder one of the holes in an IC engine or ICE? Or is it some other kind of cylinder, like for an Acetylene torch..

Gas (or, more accurately, atomized gas, not sure if that is the vapor of) is compressed to the point of combustion, at that point the cylinder I think is closed or the exhaust valve is opening. So I would imagine there would be increase up to the point of that controlled explosion... Wait, that's diesel where the compression ignites. So... Hmmm. Spark. Yeah, is this an ICE cylinder?

 

[javacontour]

What is the orientation of the rotation of the gas? Is it parallel or perpendicular to the piston?

If the radius of the rotation is closed off as the piston decreases the volume of the cylinder, which impacts the velocity more, the increased velocity as it seeks to conserve the existing momentum or the friction from more frequent contact with the sides of the smaller volume cylinder.

I would imagine, off the top of my head, it depends on what the plane of the rotation is compared to the top of the piston. I.E. is the gas rotating along the X or Y axis of the cylinder?

Say you have a closed off cylinder with a gas inside. The gas is spinning in the cylinder at a given angular velocity. You start to compress the gas in the cylinder with a piston. Would the compression increase, decrease, or have no effect on the angular momentum and angular velocity of the gas?

 

[ABN_CBT_ENGR]

Wouldn't the cylinder blow out if compression gets too great ?

Or, some weaker point, since in an automotive engine, a cylinder has a gasket in the middle or near top.. a potential weak point?

Like bolts shooting off a boiler with too much steam in it

something will give for sure if the force exceeds the strength of the materials

 

[ABN_CBT_ENGR]

What is the orientation of the rotation of the gas? Is it parallel or perpendicular to the piston?

The gas would not be rotating once closed and compression started

 

[CR94]

What is the orientation of the rotation of the gas? Is it parallel or perpendicular to the piston?
...
I would imagine, off the top of my head, it depends on what the plane of the rotation is compared to the top of the piston. I.E. is the gas rotating along the X or Y axis of the cylinder?

Can't we assume it's rotating about the center-line of the cylinder?

I think people are overlooking the fact that the mass and net angular momentum theoretically remain the same regardless of changes in density, volume, temperature, etc. (ignoring viscous drag that will slow the rotation).

 

[javacontour]

Why, if it came from a valve, and began a tumble, it's rotation would be perpendicular to the manner you described, meaning as the cylinder was compressed, the gas would begin to hit the piston or be pushed up possibly hitting the top of the cylinder.

At least from the original problem statement, I saw no indication of the orientation of the rotation.

Can't we assume it's rotating about the center-line of the cylinder?

I think people are overlooking the fact that the mass and net angular momentum theoretically remain the same regardless of changes in density, volume, temperature, etc. (ignoring viscous drag that will slow the rotation).

 

[javacontour]

I was thinking the same. The ice skater example. As she tucks her body in, the spins faster. Conservation of angular momentum.

But the difference here is the effect of the friction of the air around her is miniscule compared to the friction of the gas making contact with cylinder walls or the piston.

If you have a cylinder, you have a body that is not trivial to the problem and contact with it must be considered. How much of the angular momentum is lost to contact with the cylinder?

Can't we assume it's rotating about the center-line of the cylinder?

I think people are overlooking the fact that the mass and net angular momentum theoretically remain the same regardless of changes in density, volume, temperature, etc. (ignoring viscous drag that will slow the rotation).

 

[CR94]

Why, if it came from a valve, and began a tumble, it's rotation would be perpendicular to the manner you described, meaning as the cylinder was compressed, the gas would begin to hit the piston or be pushed up possibly hitting the top of the cylinder.

At least from the original problem statement, I saw no indication of the orientation of the rotation.

The original hypothetical question only specified "The gas is spinning in the cylinder at a given angular velocity," nothing about valves, tumbling, or other real-world complications. We have to make simplifying assumptions in order to have any hope for a neat, simple answer.

 

[javacontour]

The original hypothetical question only specified "The gas is spinning in the cylinder at a given angular velocity," nothing about valves, tumbling, or other real-world complications. We have to make simplifying assumptions in order to have any hope for a neat, simple answer.

Can make some assumptions, but the orientation of the spin is a pretty important clarifying question. Rotation which axis makes a difference here.