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Math Help - deriving fluid mechanics equations

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by phycdude View Post
    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

     <br />
\rho_t + (\rho u)_x = 0\;\;(1)<br />

     <br />
u_t + u u_x + \frac{p_x}{\rho} = 0<br />

    which from P = k \rho the latter becomes

     <br />
u_t + u u_x + k \frac{\rho_x}{\rho} = 0.\;\;(2)<br />

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

     <br />
\left(\rho u \right)_t + (\rho u^2 + k \rho)_x = 0. \;\;(3)<br />

    Eliminating \rho u from (1) and (3) gives the desired result.
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  3. #3
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    Oct 2009
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    Thanks, would u mind recomending a textbook on this kind of material?
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