Density of Flow Along a Tube Under the Action of Advancing Piston

[erobz]

Homework Statement

Determine the density distribution $\rho (u)$ along the tube as the piston advances

Relevant Equations

Conservation of Mass

I'm trying to figure out how describe the mass of air between the piston face and the und of the tube ( position $o$) in the acompanying diagram.

At $t = 0$, the mass of air in the tube is $M_o$, and the system is static with tube length $l$. The $x$ coordinate describes how far along the tube the piston has advanced at time $t$. The $u$ coordinate ranges from the piston face to the end of the tube $0 \leq u \leq l-x$

Let the density, and velocity (both assumed to be uniformly distributed over the cross section) of the outflow be given by $\rho_o (t), v_o(t)$respectively.

Applying conservation of mass I get the following relationship:

$$ A \int_0^{ l-x(t) } \rho (u) ~du = M_o - A \int_0^t \rho_o (t) v_o (t) ~dt$$

Where $A$ is the cross sectional area of the tube, and the $x$ coordinate is given by:

$$x(t) = \int_0^t \dot x ~dt $$

My goal is to get $\rho (u)$ as a function of all the other variables. My instinct is to try transform $du \to k ~dt$ and then differentiate both sides w.r.t. $t$.

? Is it ok to write:

$$ \begin{aligned} u &= l-x \\ & = l - \int_0^t \dot x ~dt \\ \implies & \frac{du}{dt} = -\dot x \end{aligned} $$

Substituting for $du$ :

$$ -A \int_0^{ l-x(t) } \rho (u) ~\dot x ~dt = M_o - A \int_0^t \rho_o (t) v_o (t) ~dt$$

Now differentiate both sides w.r.t time $t$:

$$ -A \frac{d}{dt} \left( \int_0^{ l-x(t) } \rho (u) ~\dot x ~dt \right) = \cancel {\frac{d}{dt} M_o}^0 - A \frac{d}{dt} \left( \int_0^t \rho_o (t) v_o (t) ~dt \right) $$

$$ \boxed{ \implies \rho(u) \dot x = \rho_o(t) v_o(t) } $$

Possibly I've performed some unsavory mathematics?

 

[Chestermiller]

Suppose the piston were moving very slowly. Do you think the density would be changing with x or t?

 

[erobz]

Suppose the piston were moving very slowly. Do you think the density would be changing with x or t?

If the piston were moving with constant velocity I expect that the velocity $v_o(t) = \dot x$ and the density to be uniform over $u$. So $\rho(u) = \rho_o(t) = \text{const.}$ ( in the absence of viscous friction ), But this is just about conservation of mass. Forces are irrelevant at this point as far as I can tell.

 

[Chestermiller]

If the piston were moving with constant velocity I expect that the density $v_o(t) = \dot x$ and the density to be uniform over $u$. So $\rho(u) = \rho_o(t) = \text{const.}$

If the air is considered inviscid and there is no consideration of startup accelerations, that is correct. But, if the piston suddenly starts moving with constant velocity or the viscosity of the included, this would not be the case.

 

[erobz]

If the air is considered inviscid and there is no consideration of startup accelerations, that is correct. But, if the piston suddenly starts moving with constant velocity or the viscosity of the included, this would not be the case.

I agree with that, the various forces dictate the outcomes for $v_o(t), \rho(u),...$. But mass conservation applies independent of that. That is the only equation I have considered up to this point.

 

[Chestermiller]

I agree with that, the various forces dictate the outcomes for $v_o(t), \rho(u),...$. But mass conservation applies independent of that. That is the only equation I have considered up to this point.

Conservation of mass requires that $$\frac{\partial \rho}{\partial t}+\nabla \centerdot(\rho \bf{v})=0$$. If the fluid is viscous, pressure will be a function of x, as will be density (via the ideal gas law). Velocity will also be a function of x and r. Startup (transient) adds time t, so P = P(x,t), $\rho=\rho(x,t)$, and $v=v(x,r,t)$.

 

[erobz]

Conservation of mass requires that $$\frac{\partial \rho}{\partial t}+\nabla \centerdot(\rho \bf{v})=0$$. If the fluid is viscous, pressure will be a function of x, as will be density (via the ideal gas law). Velocity will also be a function of x and r. Startup (transient) adds time t, so P = P(x,t), $\rho=\rho(x,t)$, and $v=v(x,r,t)$.

I'm making the simplifying assumption that the properties are uniformly distributed over the cross-section. So, I'm neglecting the dependency on $r$.

to that point, is what I did not cohesive with what you suggest? If not, where do I go off, because I can't see what else it could be.

 

[Chestermiller]

I'm making the simplifying assumption that the properties are uniformly distributed over the cross-section. So, I'm neglecting the dependency on $r$.

If we include the viscous behavior of the air then the axial velocity must be a function of r because of the no-slip (zero wall velocity) boundary condition.

to that point, is what I did not cohesive with what you suggest? If not, where do I go off, because I can't see what else it could be.

If air compressibility is included (as it must be), and the piston is suddenly set into motion at constant velocity, a compression wave will travel down the tube at the speed of sound. Upstream of the leading edge of the compression wave, the velocity of the air will be on the order of the piston velocity and the density will be elevated above the initial density. Downstream of the compression wave, both the density and the air velocity will be unchanged. So, until the compression wave reaches the tube exit, the exit velocity will be zero.

 

[erobz]

If we include the viscous behavior of the air then the axial velocity must be a function of r because of the no-slip (zero wall velocity) boundary condition.

I didn't want to start that yet, but surely it can be fudged with a power law approximation for the dissipative forces?

If air compressibility is included (as it must be), and the piston is suddenly set into motion at constant velocity, a compression wave will travel down the tube at the speed of sound. Upstream of the leading edge of the compression wave, the velocity of the air will be on the order of the piston velocity and the density will be elevated above the initial density. Downstream of the compression wave, both the density and the air velocity will be unchanged. So, until the compression wave reaches the tube exit, the exit velocity will be zero.

$l$ in the application I have in mind is on the order of a meter. The period of time where the shockwave propagates should be ignorable.

 

[Chestermiller]

I didn't want to start that yet, but surely it can be fudged with a power law approximation for the dissipative forces?

$l$ in the application I have in mind is on the order of a meter. The period of time where the shockwave propagates should be ignorable.

I provided my best judgment based on >50 years of fluid dynamics (my thesis area) experience. I stand by what I said.

 

[erobz]

I provided my best judgment based on >50 years of fluid dynamics (my thesis area) experience. I stand by what I said.

I didn’t mean to imply you weren’t speaking from experience, (I’m not attacking your credentials, I have none)it’s just that I’m looking for a zeroth order model.

However, at 1 meter the transient shockwave is propagated in less than 0.003 s, so I don’t see what the argument is on that point.

My main concern is whether or not I could do what I did with the mathematics. I don’t feel that has been straightly answered yet. Perhaps I should repost in the math forum?

 

[erobz]

Homework Statement: Determine the density distribution $\rho (u)$ along the tube as the piston advances
Relevant Equations: Conservation of Mass

I'm trying to figure out how describe the mass of air between the piston face and the und of the tube ( position $o$) in the acompanying diagram.

View attachment 328837

At $t = 0$, the mass of air in the tube is $M_o$, and the system is static with tube length $l$. The $x$ coordinate describes how far along the tube the piston has advanced at time $t$. The $u$ coordinate ranges from the piston face to the end of the tube $0 \leq u \leq l-x$

Let the density, and velocity (both assumed to be uniformly distributed over the cross section) of the outflow be given by $\rho_o (t), v_o(t)$respectively.

Applying conservation of mass I get the following relationship:

$$ A \int_0^{ l-x(t) } \rho (u) ~du = M_o - A \int_0^t \rho_o (t) v_o (t) ~dt$$

Where $A$ is the cross sectional area of the tube, and the $x$ coordinate is given by:

$$x(t) = \int_0^t \dot x ~dt $$

I think this is ok up to here. The interpretation I'm imagining that some time $t$ passes, we freeze time. Integrate the RHS. The integral on the LHS must be equivalent when we integrate over the spatial coordinate at that time. This admittedly sounds like partial derivatives need to be involved.

My goal is to get $\rho (u)$ as a function of all the other variables. My instinct is to try transform $du \to k ~dt$ and then differentiate both sides w.r.t. $t$.

? Is it ok to write:

$$ \begin{aligned} u &= l-x \\ & = l - \int_0^t \dot x ~dt \\ \implies & \frac{du}{dt} = -\dot x \end{aligned} $$

Substituting for $du$ :

$$ -A \int_0^{ l-x(t) } \rho (u) ~\dot x ~dt = M_o - A \int_0^t \rho_o (t) v_o (t) ~dt$$

Now differentiate both sides w.r.t time $t$:

$$ -A \frac{d}{dt} \left( \int_0^{ l-x(t) } \rho (u) ~\dot x ~dt \right) = \cancel {\frac{d}{dt} M_o}^0 - A \frac{d}{dt} \left( \int_0^t \rho_o (t) v_o (t) ~dt \right) $$

$$ \boxed{ \implies \rho(u) \dot x = \rho_o(t) v_o(t) } $$

Possibly I've performed some unsavory mathematics?

I'm looking for help in figuring out why this part is "funny math". I not interested in the fluids model perse. Why I don't believe myself is that $u$ is a spatial coordinate, and I can't make sense of the end result because everything else is a function of time $t$, and with time "frozen" $u$ is also frozen. So it's got to be junk.