Undergrad Filling a rigid tire from a pressure regulated compressed air line

[erobz]

It appears you interpreted v1=z˙ ect. so you have two fat pipes facing downward but ignore the connecting pipe. I don't know if that is valid? I suggest you ignore viscous effects for the ideal case first. Also, focus on the connecting pipe.

I ignore the viscous effects in the tanks, because they tanks are assumed to be large, and the kinetic head is going to be very small compared the kinetic head in the pipe. Thus very small viscous losses in the tanks. The head loss in the connecting pipe is what is considered significant with $v_p$.

If you want the "Bernoulli Solution", take $k$ to be zero in post 15. The only factor that contains $k$ is $\beta$.

$\beta = \sqrt{ \frac{2g}{ 1 + k \left( \frac{A_2}{A_p} \right)^2 - \left( \frac{A_2}{A_1} \right)^2 } }$

The entire effect of the pipe disappears ( because there are none), The solution that follows remains fundamentally unchanged by ignoring viscous effects as $\beta$ is just a constant.

 

[erobz]

What I would need to know, (in the least) on whether or not (13) has issues, is whether or not the Second Law dictates that given that the net rate of heat transfer $\dot Q = 0$ the change in enthalpy is always positive in the direction of flow. That is an assumption "I thought was ok to make" to get to (13) in post #20.

 

[sonhousen]

That would be the case if the tank pressure is just high enough higher than the regulator equilibrium would take a long time if not forever but that is not what he is asking. I guess he made the assumption the tank pressure would be much higher than the regulator output pressure.

 

[erobz]

The "Energy Equation" (1) and "Bernoulli Equation" (post 28) fail because both are derived under the assumption of steady flow. The flow in the tanks problem or the filling a tire problem will not be steady, so they do not apply. Unfortunately, analyzing (10) ( the First Law of Thermodynamics) is at least a way forward (of that much I am now certain).

 

[bob012345]

See if this relates;

https://www.physics.princeton.edu/~mcdonald/examples/tirepump.pdf

 

[erobz]

Ok, I think I found another route. It's the same type of derivation as (10) ( from Reynolds Transport Theorem ) but its known as the "Momentum Equation" in fluid mechanics ( It is Newtons Second Law for fluid systems ). Which is basically how I had started this whole thing before I led everyone on the wild goose chase!

$$ \sum F = \frac{d}{dt} \int_{cv} \boldsymbol v \rho d V\llap{-} +\int_{cs} \boldsymbol v \rho \boldsymbol V \cdot d \boldsymbol A $$

I'm not sure why there are different symbols for the velocity $\boldsymbol v, \boldsymbol V$ , but that is how it is in the textbook.

Anyhow the control volume is the pipe between the tanks

$$ \sum F = P_1 A - P_2 A -\tau \pi D l $$

Momentum Accumulation in the Control Volume:

Incompressible flow with constant cross sectional area $A$ implies that the velocity is not varying along the length of the pipe and can come outside of the integral along with the density:

$$ \frac{d}{dt} \int_{cv} \boldsymbol v \rho d V\llap{-} = \frac{d}{dt} \rho v A \int_{cv} dl = \rho A l \frac{dv}{dt}$$

The second integral is the Net Efflux of Momentum across the control surface, and because of my assumptions of uniform velocity distribution, incompressibility, and constant cross sectional area it is identically 0:

$$ \int_{cs} \boldsymbol v \rho \boldsymbol V \cdot d \boldsymbol A = 0 $$

That implies we have:

$$ \sum F = P_1 A - P_2 A -\tau \pi D l = \rho A l \frac{dv}{dt} $$

We've moved $P_1, P_2$ from the tank surface to the inlet and outlet of the pipe it follows that:

$$ P_1 = \rho g z = \rho g \frac{1}{A_1} \left( V\llap{-} - {V\llap{-}}_p - A_2 h \right) $$

$$ P_2 = \rho g h $$

$$ v_p = \frac{A_2}{A_p} \dot h $$


That means that finally, and I hope to the satisfaction of all involved in the hunt for oscillation, exponential decay we arrive at:

$$ \rho g \left(\frac{1}{A_1} \left( V\llap{-} - {V\llap{-}}_p - A_2 h \right) \right) - \rho g h - \tau \pi D l = \rho A l \ddot h $$

I'm not going to bother to simplify it more at this time.

It's a Second Order,Linear ODE of the Non-Homogeneous variety.

 

[bob012345]

$$ \rho g \left(\frac{1}{A_1} \left( V\llap{-} - {V\llap{-}}_p - A_2 h \right) \right) - \rho g h - \tau \pi D l = \rho A l \ddot h $$

I'm not going to bother to simplify it more at this time.

It's a Second Order,Linear ODE of the Non-Homogeneous variety.

But if I understand it correctly, the solution is just sinusoidal!

 

[erobz]

See if this relates;

https://www.physics.princeton.edu/~mcdonald/examples/tirepump.pdf

I think I've found what you are looking for in the latest approach.

 

[bob012345]

I think I've found what you are looking for in the latest approach.

$$\rho g \left(\frac{1}{A_1} \left( V\llap{-} - {V\llap{-}}_p - A_2 h \right) \right) - \rho g h - \tau \pi D l = \rho A l \ddot h$$
Unless you explain what the variables all are and how they depend on each other I can't make anything out of this. My comment assumes this was a differential equation in the variable $h$ and the other stuff was constants. If that is not the case then show the dependence on $h$ please.

 

[erobz]

But if I understand it correctly, the solution is just sinusoidal!

actually, I don't think so.$$ \rho g \left(\frac{1}{A_1} \left( V\llap{-} - {V\llap{-}}_p - A_2 h \right) \right) - \rho g h > 0 $$

Then we bring it over to the RHS:

$$ \ddot h - k h = -b $$

The solution to that ( I believe ) is:

$$ h = c_1 e^{kt} + c_2 e^{-kt} + C $$

You're going to try and make me solve it!

 

[erobz]

$$\rho g \left(\frac{1}{A_1} \left( V\llap{-} - {V\llap{-}}_p - A_2 h \right) \right) - \rho g h - \tau \pi D l = \rho A l \ddot h$$
Unless you explain what the variables all are and how they depend on each other I can't make anything out of this. My comment assumes this was a differential equation in the variable $h$ and the other stuff was constants. If that is not the case then show the dependence on $h$ please.

Thats is all explained in post #15

https://www.physicsforums.com/threa...ated-compressed-air-line.1015542/post-6636284

 

[erobz]

$$\rho g \left(\frac{1}{A_1} \left( V\llap{-} - {V\llap{-}}_p - A_2 h \right) \right) - \rho g h - \tau \pi D l = \rho A l \ddot h$$
Unless you explain what the variables all are and how they depend on each other I can't make anything out of this. My comment assumes this was a differential equation in the variable $h$ and the other stuff was constants. If that is not the case then show the dependence on $h$ please.

Yeah, its a Second order ODE with constant coefficients. It looks like it would be of the form from how I have it written:

$$ \ddot h + k h = -b $$

giving you the oscillating solution, but its actually $z > h$

So its actually of the form:

$$ \ddot h - k h = -b $$

Giving the exponential solution.

 

[bob012345]

Yeah, its a Second order ODE with constant coefficients. It looks like it would be of the form from how I have it written:

$$ \ddot h + k h = -b $$

giving you the oscillating solution, but its actually $z > h$

So its actually of the form:

$$ \ddot h - k h = -b $$

Giving the exponential solution.

Then the equation needs correcting to account for that because just solving what is written the solution oscillates. But the important point is you now agree it is an exponential solution as @Baluncore pointed out.

 

[erobz]

Then the equation needs correcting to account for that because just solving what is written the solution oscillates. But the important point is you now agree it is an exponential solution as @Baluncore pointed out.

maybe I did something wrong, I'll clean it up.

 

[bob012345]

maybe I did something wrong, I'll clean it up.

I still recommend for the water tank case to take a step back and simplify. Assume everything happens quasi-statically. Maybe do a simple case of water leaking from one tank first. Then add the pressure of the second tank opposing that flow.

 

[erobz]

Then the equation needs correcting to account for that because just solving what is written the solution oscillates. But the important point is you now agree it is an exponential solution as @Baluncore pointed out.

lol, but now that I've done something wrong I'm not sure! Don't hate me...

 

[bob012345]

lol, but now that I've done something wrong I'm not sure! Don't hate me...

Don't despair! If you have no viscous effects I think you should get an oscillation. Now if you add viscosity you should get a damped oscillation as @Baluncore discussed above. If it is critically damped you get the exponential decay.

 

[erobz]

Don't despair! If you have no viscous effects I think you should get an oscillation. Now if you add viscosity you should get a damped oscillation as @Baluncore discussed above. If it is critically damped you get the exponential decay.

Ohhh. I'm dispairing! Viscous effects are in the model!

These are the tanks and relevant parameters

This is the control volume inside the pipe joining the tanks:

The tanks are removed and replaced with their effective external forces on the control volume$F_{shear} = \tau \pi D l$ this is the viscous force that resists flow

$F_1 = \rho g z A$ hydrostatic pressure from tank 1 acting on $A$

$F_2 = \rho g h A$ hydrostatic pressure from tank 2 acting on $A$

$\rho A l$ is the mass of the control volume

I have that:

$$ \rho A l \frac{dv}{dt} = \rho g z A - \rho g h A - \tau \pi D l \tag{1} $$

The total Volume of the system $V\llap{-}$ is given by:

$$ A_1 z + A_2 h + {V\llap{-}}_{pipe} = V\llap{-} $$

$$ \Downarrow $$

$$ z = \frac{ V\llap{-} - {V\llap{-}}_{pipe} }{ A_1} - \frac{A_2}{A_1} h \tag{2}$$

$$ \Downarrow $$

$$ \dot z = -\frac{A_2}{A_1} \dot h \tag{3} $$

and applying continuity between the pipe and tank 2:

$$ v A = A_2 \dot h \tag{4}$$

So all together we have that:

$$ \rho l A \frac{ A_2}{A} \ddot h = \rho g A \left( z - h \right) - \tau \pi D l \tag{5}$$

The criterion here is that

$$ z \geq h $$

hence $z - h \geq 0$ must be satisfied for the solution to hold So whatever this term is, it is positive on the RHS and must be negative on the LHS of (5).

This implies that there are bounds on $h$ determined by (2)?

Other than that, I have no idea how to resolve this if its incorrect? Is there something that happens when the particular solution is added to the homogenous solution? I'm not seeing it if that is the case. Certainly stepping into the Twilight Zone a little bit here...

 

[bob012345]

It looks to me that the viscous effect you included is just a constant. In a LRC damped oscillator the resistance R is multiplied by $\dot{I}$. Maybe the shear force needs an $\dot{h}$? Did you include Poiseuille's Law for fluids? Also, it might be easier to make the variable $z-h$.

https://courses.physics.ucsd.edu/2015/Summer/physics2cl/documents/Phys2CL_Exp2_2010.pdf

 

[erobz]

It looks to me that the viscous effect you included is just a constant. In a LRC damped oscillator the resistance R is multiplied by $\dot{I}$. Maybe the shear force needs an $\dot{h}$?

I think you must be correct.

$$ \tau_o = -\mu \left. \frac{dV}{dr} \right|_{r_o} $$

Where

$\mu$ is the kinematic viscosity of the flow

$r$ is the distance from the pipe centerline normal to the flow velocity distribution

$\left. \frac{dV}{dr} \right|_{r_o}$ this bit is the evaluation of that flow distribution at the pipe walls.

for laminar flow this is going to have a linear proportionality "$\varphi V$". For other flow distributions it varies.

I was allowing myself to be fooled once again by a derivation assuming steady flow ... where the authors weren't integrating anything, and they casully dropped $\tau_o$ in there as if it were a constant... my bad

 

[bob012345]

What book is that? It's most likely correct for the problem it is doing which of course may be a bit different than yours.

 

[erobz]

Engineering Fluid Mechanics - Ninth Edition. Yeah, it's fine for what it is about to do in the rest of the derivation. I just didn't catch that steady flow assumption..., so for what they are doing (Deriving the Darcy Weisbach Equation for head loss in a pipe) the fact that $\tau$ is actually a function of $V$ is not relevant at that point. They explain $\tau$ a few pages later

Derivation grabbing...guilty as charged!

 

[erobz]

$$ \rho l A \frac{ A_2}{A} \ddot h = \rho g A \left( \frac{ V\llap{-} - {V\llap{-}}_{pipe} }{ A_1} - \left( \frac{A_2}{A_1} + 1 \right) h \right) - \mu \pi D l \frac{A_2}{A} \dot h \tag{6}$$

Thank you ( All ) for your continued skepticism!

 

[bob012345]

$$ \rho l A \frac{ A_2}{A} \ddot h = \rho g A \left( z - h \right) - \mu \pi D l \frac{A_2}{A} \dot h \tag{6}$$

Thank you ( All ) for your continued skepticism!

Great but one thing is what about $z$? I suspect you can frame this in terms of $z-h$ as the variable in the equation. Also, is it $\frac{ A_2}{A}$ or $\frac{ A_2}{A_1}$?

 

[erobz]

Great but one thing is what about $z$? I suspect you can frame this in terms of $z-h$ as the variable in the equation. Also, is it $\frac{ A_2}{A}$ or $\frac{ A_2}{A_1}$?

I edited it and put it all in terms of $h$

Its $A$ ( from the Area of the pipe) and there is some cancelation I can do...

 

[erobz]

Great but one thing is what about $z$? I suspect you can frame this in terms of $z-h$ as the variable in the equation.

$$ \rho l A_2 \ddot h = \rho g A \left( z - h \right) - \mu \pi D l \varphi \dot h \tag{5}$$

If we do the following substitution as you suggest from (5):

$$ \lambda = z-h \implies \dot \lambda = \dot z - \dot h $$

Then substituting

$$ \dot z = -\frac{A_2}{A_1} \dot h \implies \dot h = - \frac{A_1}{A_2 + A_1} \dot \lambda \implies \ddot h = - \frac{A_1}{A_2 + A_1} \ddot \lambda $$

Putting that all back into (5):

$$ -\rho l \frac{A_1 A_2}{A_2 + A_1} \ddot \lambda = \rho g A \lambda + \mu \pi D l \frac{A_1}{A_2 + A_1} \varphi \dot \lambda $$

Or:

$$ \rho l \frac{A_1 A_2}{A_2 + A_1} \ddot \lambda + \mu \pi D l \frac{A_1}{A_2 + A_1} \varphi \dot \lambda + \rho g A \lambda = 0 $$I'm pretty sure this now has the solution modes @Baluncore identified quite some time ago. Thanks for going the distance on this! We'll... from the original question I'm now suspecting it's more like 1/8 the distance! However, I think I can live with that...

 

[Chestermiller]

Going back to the original problem, if the volume of the tube is much less than the volume of the tire, the tube can be assumed to be flowing at quasi-steady state. At very long times, the tire pressure will be very close to the regulator pressure, and the flow through the tube will be laminar. From the Hagen-Poiseuille equation for laminar flow, it follows that $$\dot{m}=\frac{\rho_R\pi^2D^6}{512\mu L}(p_R-p_T)$$where $\dot{m}$ is the mass flow rate through the tube, $\rho_R$ is the gas density at the regulator, $p_R$ is the regulator pressure, and $p_T$ is the tire pressure. The rate of change of pressure in the tire is given by the ideal gas law: $$\frac{\dot{m}}{M}=\frac{V}{RT}\frac{dp_T}{dt}$$If we combine these two equations, we obtain an equation of the form: $$\frac{dp_T}{dt}=k(p_R-p_T)$$The solution to this equation is such that the difference between the regulator pressure and the tire pressure decreases as $$e^{-kt}$$So the tire pressure never quite reaches the regulator pressure, although, after a time on the order of say 4/k, it closely approaches the regulator pressure.

 

[erobz]

Going back to the original problem, if the volume of the tube is much less than the volume of the tire, the tube can be assumed to be flowing at quasi-steady state. At very long times, the tire pressure will be very close to the regulator pressure, and the flow through the tube will be laminar. From the Hagen-Poiseuille equation for laminar flow, it follows that $$\dot{m}=\frac{\rho_R\pi^2D^6}{512\mu L}(p_R-p_T)$$where $\dot{m}$ is the mass flow rate through the tube, $\rho_R$ is the gas density at the regulator, $p_R$ is the regulator pressure, and $p_T$ is the tire pressure. The rate of change of pressure in the tire is given by the ideal gas law: $$\frac{\dot{m}}{M}=\frac{V}{RT}\frac{dp_T}{dt}$$If we combine these two equations, we obtain an equation of the form: $$\frac{dp_T}{dt}=k(p_R-p_T)$$The solution to this equation is such that the difference between the regulator pressure and the tire pressure decreases as $$e^{-kt}$$So the tire pressure never quite reaches the regulator pressure, although, after a time on the order of say 4/k, it closely approaches the regulator pressure.

Well, now that we have an answer, I'm only half tempted to try and work up to it (or something like it) from first principles! Thanks Chester!

 

[bob012345]

Well, now that we have an answer, I'm only half tempted to try and work up to it (or something like it) from first principles! Thanks Chester!

It would make a nice completion to this thread! :)

 

[erobz]

It would make a nice completion to this thread! :)

I got to be honest bob, as much as I would like to be able to do it... upon first inspection it seems to be an entirely different beast. I can spitball some ideas on how I think the "Momentum Equation" plays out, and maybe I can be work through it with much help...I'll give it a try. Put on your skeptical hat!