Time to compress fluid in cylinder

[byxrube]

I have an application where I need to determine the time needed to compress a gas in a piston cylinder arrangement a given distance (ie. the time for the piston to travel x inches). Would anyone happen to know the methodology and equations used to find this?

 

[BigInch]

Its got more to do with the rpm of the driver and linkage geometery to the piston more than anything else. There is no time component in the equation of state.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[zdas04]

It looks like a simple geometry problem to me also.

On the other hand, the amount of hp required to meet your time requirements is a fluids question that some us might be able to help with if that is part of your question.

David

 

[byxrube]

I am looking for the time it takes to travel a distance given a constant force on the piston. The pressure on the other side of the piston will increase as it is compressed, therefore dampening the motion more. So given a constant force, I want to know what equation would give you the time to travel a given distance. I am assuming that you would need to take into account the change in volume and pressure, but I am not sure what equation relates it to time.

 

[israelkk]

This is a dynamic case. You have to write the differential equation/s that uses the force, inertia (mass), friction of the seals, pressure change in the cylinder as a function of the piston movement (decrease of the volume), etc. and solve it numerically.

 

[racookpe1978]

Unless you're compressing gas2 (the low pressure gas behind the piston) with some unknown gas1 (a high pressure gas doing the pushing) the piston WILL MOVE according to the mechanical linkage driving the piston.

Time (movement) of the piston is strictly geometrical - as noted above. The gas WILL COMPRESS as the piston moves, it has no choice.

if you have a gas-driving-a-piston, then the time to fill the upper piston will depend on its valve's characteristics, the length and diameter and umber of bends in connecting piping, and the relative sizes of the piston and initial gas flow.

Give us a drawing of your piston's driving mechanism - Your replies, unfortunately, are not yet clear to those of us reading them.

 

[byxrube]

I am looking at a gas spring rod. The piston is moved by an axial load on the rod. I am looking at the case when the force is constant. The force is not changing to get a desired rate, I want to know how fast the piston moves for a fixed value axial force.

 

[BigInch]

"So given a constant force,". Piston force won't be constant if the gas pressure changes. How are you going to compress the gas by increasing pressure on it, if the force stays constant and maintains the same pressure on the gas? The only way you could increase pressure with a constant force is to decrease the cross section of the piston.

israelkk you're saying the same thing as everyone else. Once you write the dV/dt function, you know the time it takes and all you have to do (with an ideal gas) is to use V1 P1/V2 to get the pressure P2.

If you have the power available to reach P2, its just a matter of driver rpm and geometry.

BTW BYX are you a student? This doesn't sound like a real-world question.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[zdas04]

One thing that is not real clear in the discussion above is that a compressible fluid acts just like a spring. If you have a mechanical spring with a higher spring constant than the compressible fluid then the pistion will travel to the stop (and you need a gorilla to retract it). If the spring constant is lower than the spring constant of the gas at some point in the travel then the piston will stop at that point.

Bottom line is that you are not going to do gas compression with a spring.

David

 

[BigInch]

That is true, but is the driver so underpowered to let that effect govern. My bottom line is do you want the driver controlling the rate of compression or the spring constant of the fluid to slow down the driver enough that it has control. I'd assure you have enough power to pressure the fluid in the time you have available, rather than the reverse.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[BigInch]

That's why I said the question doesn't sound real to me. Or its obviously not an oil/gas scenario.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[JStephen]

You could assume the gas stays at constant temperature, use ideal gas laws to figure pressure as a function of volume and thus geometry, and then integrate the resulting force to find the work required.

In actuality, the gas would heat up as it is compressed, which would throw the results off some. It's been a while since I took thermodynamics, though, and I'd have to hunt up how to figure that heating, or even if it is a specific amount.

 

[zdas04]

The adiabatic equation for temperature rise is:

T(out)=T(in)*(K(out)/P(in)^((k-1)/k)

all temperatures and pressures are in absolute units. "k" is the ratio of specific heats.

David

 

[BigInch]

You could use any thermodynamic process you want, but all those T still are not time, so please don't run off the track discussing all of those. The question won't get answered that way.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[zdas04]

Good point, sorry I'm easily dragged into the weeds.

David

 

[BigInch]

Its a jungle out there.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[mjpetrag]

I guess a little over simplified...let me know if I'm wrong but I'll take a stab at it

P2 = (P1V1)/(V2)

P2 = F2/A
F2/A = {P1V1)/(V2)

F2 = (m of piston)*(acceleration of piston) = m*a

(m*a)=(P1*V1*A)/(V2)
a = P1*V1*A/(V2*m)

(dU/dt) = P1*V1*A/(V2*m)
dU = int(P1*V1*A/(V2*m) dt)
U = (P1*V1*A*t)/(V2*m)

(dx/dt) = (P1*V1*A*t)/(V2*m)
x = int((P1*V1*A*t)/(V2*m)dt)
x = (P1V1A*t^2)/(2*V2*m)

t = sqrt[(x*2*V2*m)/(P1*V1*A)]

this does NOT take into account friction from the piston walls, compressibility, temperature, etc. etc.

-Mike

 

[BigInch]

If you know the acceleration of the piston, you must now the velocity at any given time, therefore you know the answer without all the PV stuff.

F2 = m * a is an instaneous force, yet P1, V1 and V2 are descrete values at two different times t0 and t1.

Even if you let F2 be the average force and a be the average accelaration, you must know the amount of time passed between having V1 and V2.

Bringing us back to, if you know the length of time between t0 and t1, why are you trying to solve for it? Which BTW I don't think you ever do.

U seems to be a bit of a red herring.

But getting back to the problem, here's velocity again in the form of dx/dt. This integral must be evaluated over dt, so oops, if we don't know t0 and t1, how can we integrate?



**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[mjpetrag]

right, there are no initial/final conditions. this was the only equation i could come up with to relate time with position (using indefinite integrals). fortunately the units work out right.

-Mike

 

[iainuts]

Right, you don't solve for time, you make that one of your given variables. Start with an initial pressure, initial spring force. Start as a given, the piston mass, the fluid state (pressure and temperature), initial fluid volume, piston diameter, fluid type if using a computer database for properties, time step (dt), a value for friction coefficient of the seals, seal height (to determine frictional seal load), and maybe the polytropic exponent if you're using that method to determine fluid states (alternatively you can use your fluids database and fluid properties such as entropy, internal energy, density, etc... to determine fluid state. You will also need heat flux if you want to go to those lengths. Heat flux is a function of dT across the cylinder wall and frictional heating for reciprocating pistons).

I would run the calculations across a row in a spreadsheet with dt in the first column. Next columns, put pressure, temperature, piston pressure load, piston spring load, calculate acceleration, calculate velocity, calculate position, calculate new fluid properties, etc...

This is essentially the Simpson rule, though you'll find you can often do some averaging, so you essentially turn the integration into a trapazoidal rule.

Make a bunch of graphs that show the location of the piston, velocity, etc...

Make dt a variable, then change it, making it smaller. You should find that for too large a value of dt, the piston location and velocity are completely eratic. As you make the dt variable smaller and smaller, the graphs all stop changing. You're done.

I'd be interested in seeing how others approach this problem.