Physics problem....Help

[Robert]

This is probably the wrong forum, but...

Maybe there's a math-head here? Anyway, here's my problem:

I have a closed chamber whose volume = 40in^3. In it I have a vacuum of 10inHg (inches of mercury). Also I have a port hole of 1/10 of an inch to the outside world. If I open this port to the outside, how long will it take for 10inHg of vacuum to return to 1 ATM (14.7lbs/in^2)?

Is there a f(t) equation?

Robert

[Bravus]

First step is to get your pressures in consistent units. As I recall 1 atm = about 32 in Hg, but you should check that 'cos it's been a while...

If it was me I'd be going to metric units, but that's just my Aussie background. In imperial, work in either inHg or pounds per square inch.

You also need to know whether the porthole is circular and whether the 1/10 inch is the radius or diameter. Assuming it's the diameter of a circle I think you need to calculate the area of that circle.

You'll also need some sort of constant terms that describe the 'viscosity' (for want of a better word) and the density of air.

But yeah, there should definitely be a function that describes the solution. it will involve calculus because the rate of air flow will decrease as the difference between the internal and external pressures decreases.

[bevin]

You ask "how long to return to 1ATM", so I am assuming the low pressure side is a constant 1ATM. The answer to that question is easy - never. Basically as the pressure on the high pressure side reduces, it decreases the push through the hole...

So you are going to have to set a higher-than-one-ATM goal

You are releasing a gas through a tube. The radius and LENGTH of the tube are critical, as is the question of whether the flow is smooth or turbulent.

Smooth flow is proportional to the 4th power of the radius, turbulent to the 5th power.

XXX inches of mercury * 70.73 = YYY pounds/sq. ft.

The equations you want are

http://www.efunda.com/formulae/fluids/overview.cfm

and I see they have a free download of a pipe flow solver that might meet your needs

Basically constant = pressure x velocity x distance

which is going to give you a differential equation

which you are going to have to solve

/Bevin

[Bravus]

Hmm - I was imagining it as a simple container at a pressure significantly lower than atmospheric pressure (10 inHg as compared to the 29.9 inHg of atmospheric pressure (I looked it up!)), and it's just being opened out to the outside air, not into another container. So air rushes in from a virtually infinite reservoir, until the pressure inside and outside the container is the same.

In the scenario I think you're imagining, bevin, you have a limited total volume, i.e. a closed system.

And in the case I think Robert's asking about there's not a pipe, just the thickness of the wall of the container. For the purposes of a problem like this, unless it's specified, I think we'd assume that's zero.

[Amelia]

Danger, danger Will Robinson...headache on the horizon.

[Nan]

I think we might need another foreign language forum for those who speak Mathish.

[Robert]

And in the case I think Robert's asking about there's not a pipe, just the thickness of the wall of the container.

Right....and that hole has a dia. of 1/10 inch

[bevin]

How thick are the walls? Are they much less than 1/10 inch? They would have to be for the above formulae not to apply.

Can't assume to be 0, because then the D in the K=PVD is zero, resulting in an infinite V...

The reason the time is infinite to exactly equalize is because the P = f(T) function has an asymptotic limit of Patm. In practise, because air is made of discrete molecules, it isn't really infinite - but instead is probabilistic and they are never quite equal.

I was assuming the 1ATM was constant - it doesn't matter whether that is the low or the high P

[bevin]

http://www.pipeflowcalculations.com/naturalgas/index.htm

[Robert]

How thick are the walls? Are they much less than 1/10 inch?

The wall is about 2mm or .079 inch. Basically here's what I have:

Outside the cylinder 1 ATM or 14.7 PSI. Inside I have 10inHg or 4.91 PSI. That's a difference of 9.79 PSI.

It's reasonable to assume that when the hole/port is opened that initially the airflow into the vacuum will be at its max rate and then die down exponentially. Something like d^2y/dt^2 = (1-e^[-f(t)])...where d^2y/dt^2 is the change in air speed per change in time (decel) and "e" (2.718) to the negative function of time after equilibrium has been reached equals zero. You know like the charge on capacitor, i.e., v = V(1-e^(-t/cr))....But as to an actual formula, I have no idea???

[Bravus]

I wonder whether you need a flow equation at all. Why not just one in terms of pressure directly? dP/dt, the rate of change of pressure inside the container with respect to time. When that reaches zero, P(in)=P(out). The analogy with a discharging capacitor, or with Newton's laws of heating and cooling in particular, is relevant: a cold object in the atmostphere takes up heat from the environment and ends up at the ambient temperature in the same way. bevin is right that it theoretically takes an infinite time because it's asymptotic, but my cold drinks seem to get to room temperature in less than infinite time.

The law of heating is discussed here: http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/diffeqs/cool.html

You should be able to work through it by analogy pretty directly.

[Robert]

constant = pressure x velocity x distance which is going to give you a differential equation which you are going to have to solve

Not sure what C = P x ds/dt x s refers to?

To get a differential eqution I need something like the 2nd order format, P(t)y" + Q(t)y' + R(t)y = G(t) where y" (2nd derivate) = f(t,y,y'). Or a first order....But again, how do I get an equation based on f(t)? That's my problem.

Robert

[Bravus]

Not gonna do your homework *for* you... but if you go to the link above and work through it you'll at least definitely end up with a differential equation. I don't think you actually have to solve it, right? ('Cos I don't think you have enough info to fully evaluate the constant.)

[Robert]

Actually it's not homework....I've been out of college for 20 years.

I recently repaired my cruise control in my Camry....Part of the CC consists of an actuator, which is basically a regulated vacuum chamber that engages the throttle linkage via a flexible diaphragm.

The chamber has basically two ports: One for vacuum and one for outside air (hence 14.7 PSI). The Electro-valve is controlled by pulse width modulation (PWM).

Anyway, when you turn on the CC and press/twist "set" the CC computer places into its memory the frequency coming from the "speed sensor" at the end of transmission. It then compares this value with the speed sensor and varies the PWM to achieve the speed that was "set" into memory. The PWM is essentially the "error" like in PID circuit.

That got me thinking....I guess everyone has felt that momentary dip in speed after engaging the CC? Well, that time delay has to do with the time it takes the diaphragm to pull the throttle linkage back. In other words it takes time to get to a certain value of vacuum....The same if true when you turn it off - it takes some interval of time so that inside and outside pressure on the diaphragm reaches equilibrium and releases the throttle.

Now maybe you can see where my question originates?

Robert

[Bravus]

Ahhh - a practical situation! Cool... but that might make it tougher than if it was just a homework question! That means you really will have to try to figure out some kind of value for the constant term that takes into account the nature of air.

For pressure, I get:

P(t) = Pa + (Po - Pa)e^-kt where P(t) is the pressure at time t, Pa is the ambient pressure and Po is the original pressure. So if we sub in 1 atm for Pa and 0.334 atm for Po, we get:

P(t) = 1 + (0.334-1)e^-kt

= 1 - 0.666^-kt (whoa, the equation of the beast!)

Trouble is, we're still stuck with two unknowns. If we had a few values we could plug them in and solve simultaneously to find the value of k, but we don't. And the 1/10 of an inch aperture is one factor but there are others...

[Amelia]

Can't you just fix the car without all that? Or is it a kit from a box? hehehehe

[Bravus]

Curiosity driven, I think, Amelia: believe it or not, some sick individuals find this fun.

[Robert]

Can't you just fix the car without all that? Or is it a kit from a box? hehehehe

The cruise control is fixed. I replaced two things: 1 vacuum switch that tells the CC computer that the vacuum pump is putting out 10inHg of vacuum and 2, the actuator (the mechanism we are discussing in detail).

I'm just trying to understand how they designed the thing....

[bevin]

Your hole is 0.1" dia and 0.1" (approx) long, so it does need to be treated as a pipe.

Your topic is actually one that has had, and continue to have, a lot of physics, math, and computer power thrown at it

Physics and Math

http://en.wikipedia.org/wiki/Navier-Stokes_equations

http://www.navier-stokes.net/

If it was a liquid (ie: incompressible) what you would want is

http://en.wikipedia.org/wiki/Poiseuille's_law

Computers (CFD)

http://en.wikipedia.org/wiki/Computational_fluid_dynamics

http://www.fluent.com/

I don't think decay is going to truely exponential - that is what you would get if the flow was linearly proportional to the pressure difference, and I suspect that it is not, due to boundary conditions.

As Bravus suggests, you can play the math, or you could try one of the freeware packages at the end of the wiki CFD article

[Robert]

P(t) = Pa + (Po - Pa)e^-kt where P(t) is the pressure at time t, Pa is the ambient pressure and Po is the original pressure. So if we sub in 1 atm for Pa and 0.334 atm for Po, we get: P(t) = 1 + (0.334-1)e^-kt = 1 - 0.666^-kt (whoa, the equation of the beast!)

And then

t = -(1/k)ln[1-P(t)/0.666]

But, as you suggested, k is unknown.

Robert

[Amelia]

I am soooo glad I am simple.

[Dottie]

I agree wholeheartedly, Amelia. My husband has his Master's in physics, and is an engineer doing electro-optics. We only discuss mundane things!

[Bravus]

Dang, boys, our plumage display to impress the girls seems not to have worked as we'd hoped.

[bevin]

The power of CFD and other computer modelling has to be seen to be believed.

There is an interesting video of a computer simulating driving a SUV into a corner of a brick wall. You can see the grill, fender, and hood crumpling. More impressive, you can see the hood getting bent from the impact of the battery which has broken loose.

These simulations enable engineers to build lighter safer cheaper cars, airplanes that use a lot less fuel, refrigerators that use less energy...

/Bevin

[Robert]

My husband has his Master's in physics....

Good, now maybe you could ask him to get me an equation that takes into account the parameters I have given.

All I want is, t =

:duno