APM346-2016F > APM346--Misc

Bonus Topic 2 (air pipe)

(1/1)

**Victor Ivrii**:

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

\begin{equation}

u_{tt}-ku_{xx}=0.

\end{equation}

**Luyu CEN**:

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 = ÏV = ÏSdx(6)The acceleration of the mass is

a = utBy 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))= ÏSdx utDivide both sides by Sdx

(p(x, t)- p(x + dx, t)/dx=Ïutwhich is

Ït+(Ïu)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 dxDivide both sides by dt dx S:

Ï (u(x, t) â€“ u(x + dx, t)) / dx = (Ï(x, t + dt)- Ï(x, t))/ dtThen

Ït+(Ïu)x=0 (2)

Unreadable!

**XinYu Zheng**:

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**:

OK. Just remarks

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

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