Author Topic: Bonus Topic 2 (air pipe)  (Read 837 times)

Victor Ivrii

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Bonus Topic 2 (air pipe)
« on: September 22, 2016, 08:03:52 AM »
Consider an air pipe and denote by $\rho =\rho(x,t)$ the density and by $u=u(x,t)$ a velocity of the air.
a.  Explain
\begin{align}
&\rho u_t = - p_x ,\\
&\rho_t  + (\rho u)_x=0
\end{align}
where $p=p(\rho)$ is a pressure.
b. Assuming that $\rho-\mu$ ($\mu$ is a constant) and $u$ are small linearize these equations to
\begin{align}
&\mu u_t = - k\rho_x ,\\
&\rho_t  + (\mu u)_x=0
\end{align}
where $k=p'(\mu)$.
c. Prove that then $u$ (and $\rho$) satisfy equation

u_{tt}-ku_{xx}=0.
« Last Edit: September 22, 2016, 08:16:11 AM by Victor Ivrii »

Luyu CEN

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Re: Bonus Topic 2 (air pipe)
« Reply #1 on: September 23, 2016, 03:24:33 PM »
For part(a) only
Let S be the surface area and dx a small distance along the pipe, for a fixed point (x,t)
The density of air around this point can be regarded as constant and by the density formula
m = ÏV = ÏSdx
(6)
The acceleration of the mass is
a = ut
By the pressure formula
F = S(p(x, t)- p(x + dx, t))
(7)

By Newtonâ€™s second law,  F = ma, plug in (6) and (7), we have
S(p(x, t)- p(x + dx, t))= ÏSdx ut
Divide both sides by Sdx
(p(x, t)- p(x + dx, t)/dx=Ïut
which is
Ït+(Ïu)x=0
(1)
Consider a change in mass for a fixed volume in the pipe, it equals to the difference of the mass that flowed in and the mass that flowed out
dm = dt Ï dx S (u(x, t) â€“ u(x + dx, t))
Also by the density formula
dm =  (Ï(x, t + dt)- Ï(x, t)) S dx
From the above two equations
dt Ï dx S (u(x, t) â€“ u(x + dx, t)) = (Ï(x, t + dt)- Ï(x, t)) S dx
Divide both sides by dt dx S:
Ï (u(x, t) â€“ u(x + dx, t)) / dx = (Ï(x, t + dt)- Ï(x, t))/ dt
Then
Ït+(Ïu)x=0
(2)

« Last Edit: September 27, 2016, 11:12:30 AM by luyu »

XinYu Zheng

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Re: Bonus Topic 2 (air pipe)
« Reply #2 on: October 14, 2016, 07:41:13 PM »
No one has taken up (b) and (c) for a while, so I will finish them.

(b). If $\rho-\mu$ is small, then $\rho= \mu$ to first order. Taking a Taylor expansion of $p(\rho)$ at $\mu$, we have $p(\rho)=p(\mu)+p'(\mu)(\rho-\mu)+O(\rho^2)$. Then $p_x=k\rho_x$ where $k=p'(\mu)$ to first order. Putting these back in the original equations yields what we want to show.

(c). Take a $t$ derivative in (3) and an $x$ derivative in (4) to obtain $u_{tt}=-(k/\mu)\rho_{xt}$ and $\rho_{tx}+\mu u_{xx}=0$. Combining these equations using $\rho_{xt}=\rho_{tx}$ gives $u_{tt}-ku_{xx}=0$.
Now take an $x$ derivative in (3) and a $t$ derivative in (4) to obtain $\mu u_{tx}=-k\rho_{xx}$ and $\rho_{tt}+\mu u_{xt}=0$. Combining these using $u_{xt}=u_{tx}$ gives $\rho_{tt}-k\rho_{xx}=0$.

Victor Ivrii

$\rho_t + (\rho u)_x=0$ is a continuity equation which is a mass conservation law in the differential form. $\rho u_t=-p_x$ is a dynamic equation.