# Thread: deriving fluid mechanics equations

1. ## deriving fluid mechanics equations

her0'es a question on fluid mechanics:
i need help in showing how to arrive at the expressions

3 a)
Air obeying Boyles law [imath] p=k \rho [/imath] is in motin in a uniform tube of small
cross-sectional area. show that if [imath] \rho [/imath] is the density and u is the velocity
at a distance x from a fixed point a; and t is time , this is true:
$\frac{ \partial^2 \rho}{\partial t^2} = \frac {\partial^2 \left[ (u^2 + k) \rho \right] }{\partial x^2}$

and here's another , any clues on how to start?
3 b. A steam is rushing from a boiler throught a conical pipe, the diameters of the ends a
being D and d . if v and u are the corresponding velocities of the steam an if the motion is
supposed to be that of divergence from the vortex of the cone prove that
$\frac{u}{v} = \frac {D^2} {d^2} e^{\frac{u^2-v^2}{2k} }$
where k is the pressure divide by the density and its a constant ie [imath] k= \frac{p}{\rho} [/imath]
note its getting in at one end with a velocity [imath] v [/imath] and density [imath] \rho_1 [/imath]
and out the other side with [imath] u [/imath] and [imath] \rho_2 [/imath]

i had posted the here mathbin.net/37252 but i managed the others this is whats troubling me

2. Originally Posted by phycdude
her0'es a question on fluid mechanics:
i need help in showing how to arrive at the expressions

3 a)
Air obeying Boyles law [imath] p=k \rho [/imath] is in motin in a uniform tube of small
cross-sectional area. show that if [imath] \rho [/imath] is the density and u is the velocity
at a distance x from a fixed point a; and t is time , this is true:
$\frac{ \partial^2 \rho}{\partial t^2} = \frac {\partial^2 \left[ (u^2 + k) \rho \right] }{\partial x^2}$

and here's another , any clues on how to start?
3 b. A steam is rushing from a boiler throught a conical pipe, the diameters of the ends a
being D and d . if v and u are the corresponding velocities of the steam an if the motion is
supposed to be that of divergence from the vortex of the cone prove that
$\frac{u}{v} = \frac {D^2} {d^2} e^{\frac{u^2-v^2}{2k} }$
where k is the pressure divide by the density and its a constant ie [imath] k= \frac{p}{\rho} [/imath]
note its getting in at one end with a velocity [imath] v [/imath] and density [imath] \rho_1 [/imath]
and out the other side with [imath] u [/imath] and [imath] \rho_2 [/imath]

i had posted the here mathbin.net/37252 but i managed the others this is whats troubling me
Here's the first. The coninuity and momentum equations are

$
\rho_t + (\rho u)_x = 0\;\;(1)
$

$
u_t + u u_x + \frac{p_x}{\rho} = 0
$

which from $P = k \rho$ the latter becomes

$
u_t + u u_x + k \frac{\rho_x}{\rho} = 0.\;\;(2)
$

By the following calculation $u \times (1) + \rho \times (2)$ gives

$
\left(\rho u \right)_t + (\rho u^2 + k \rho)_x = 0. \;\;(3)
$

Eliminating $\rho u$ from (1) and (3) gives the desired result.

3. Thanks, would u mind recomending a textbook on this kind of material?