Tangential force per unit length on the pipe due to viscous stress
Consider steady viscous flow under constant pressure gradient $frac{dp}{dz} = -P$ through a pipe that runs along the $z$-axis and has triangular cross-section. The sides are given by the planes $$x = 1 quad x + 2 -ysqrt 3 = 4 quad x +2 +ysqrt 3 = 0$$
Show that the axial flow speed $omega$ is a constant times $(x - 1)(x + 2 -ysqrt 3)(x +2 +ysqrt 3)$ and find the constant. Find the tangential force per unit length on the pipe due to viscous stress.
Attempt: I believe I found our constant which I will call $c$.
Using Navier Stokes:
$$begin{align} rhofrac{Dvec{u}}{Dt}=-nabla p-rhovec{g} +munabla^2vec{u} quad text{where the second term is $0$}end{align} $$
We have no flow left or right only up or down. So we only have deal with the $z$ component. Steady flow $implies$ derivative $frac{dv}{dz}=0$. We rewrite the N-S in axial flow:
$$ frac{Du}{Dt} =frac{partial u}{partial t}+(ucdot nabla)u=frac{partial u_z}{partial t} +underbrace{(u_zcdotnabla)u_z}_{text{must be } 0}$$
Using N-S
$$ begin{align} underbrace{0}_{text{steady flow}}rightarrowrhofrac{partial u_z}{partial t}= & frac{partial p}{partial z} + mu(partial_{xx}u_z+partial_{yy}u_z)=0 \ -p = &mu(partial_{xx}u_z+partial_{yy}u_z) \
frac{-p}{mu} =&partial_{xx}u_z+partial_{yy}u_z \ frac{-p}{mu} = & 6c(1+x)+(-6c(x-1)) \ frac{-p}{mu}=&6c+6cx-6cx+6c \ frac{-p}{mu}=& 12c \ c=frac{1}{12}frac{-p}{mu}end{align}$$
Now I'm not sure how to find the axial speed or the tangential force, any help is much appreciated.
fluid-dynamics
asked Nov 28 at 16:19
elcharlosmaster
1510
add a comment |
Consider steady viscous flow under constant pressure gradient $frac{dp}{dz} = -P$ through a pipe that runs along the $z$-axis and has triangular cross-section. The sides are given by the planes $$x = 1 quad x + 2 -ysqrt 3 = 4 quad x +2 +ysqrt 3 = 0$$
Show that the axial flow speed $omega$ is a constant times $(x - 1)(x + 2 -ysqrt 3)(x +2 +ysqrt 3)$ and find the constant. Find the tangential force per unit length on the pipe due to viscous stress.
Attempt: I believe I found our constant which I will call $c$.
Using Navier Stokes:
$$begin{align} rhofrac{Dvec{u}}{Dt}=-nabla p-rhovec{g} +munabla^2vec{u} quad text{where the second term is $0$}end{align} $$
We have no flow left or right only up or down. So we only have deal with the $z$ component. Steady flow $implies$ derivative $frac{dv}{dz}=0$. We rewrite the N-S in axial flow:
$$ frac{Du}{Dt} =frac{partial u}{partial t}+(ucdot nabla)u=frac{partial u_z}{partial t} +underbrace{(u_zcdotnabla)u_z}_{text{must be } 0}$$
Using N-S
$$ begin{align} underbrace{0}_{text{steady flow}}rightarrowrhofrac{partial u_z}{partial t}= & frac{partial p}{partial z} + mu(partial_{xx}u_z+partial_{yy}u_z)=0 \ -p = &mu(partial_{xx}u_z+partial_{yy}u_z) \
frac{-p}{mu} =&partial_{xx}u_z+partial_{yy}u_z \ frac{-p}{mu} = & 6c(1+x)+(-6c(x-1)) \ frac{-p}{mu}=&6c+6cx-6cx+6c \ frac{-p}{mu}=& 12c \ c=frac{1}{12}frac{-p}{mu}end{align}$$
Now I'm not sure how to find the axial speed or the tangential force, any help is much appreciated.
fluid-dynamics
asked Nov 28 at 16:19
elcharlosmaster
1510
add a comment |
Consider steady viscous flow under constant pressure gradient $frac{dp}{dz} = -P$ through a pipe that runs along the $z$-axis and has triangular cross-section. The sides are given by the planes $$x = 1 quad x + 2 -ysqrt 3 = 4 quad x +2 +ysqrt 3 = 0$$
Show that the axial flow speed $omega$ is a constant times $(x - 1)(x + 2 -ysqrt 3)(x +2 +ysqrt 3)$ and find the constant. Find the tangential force per unit length on the pipe due to viscous stress.
Attempt: I believe I found our constant which I will call $c$.
Using Navier Stokes:
$$begin{align} rhofrac{Dvec{u}}{Dt}=-nabla p-rhovec{g} +munabla^2vec{u} quad text{where the second term is $0$}end{align} $$
We have no flow left or right only up or down. So we only have deal with the $z$ component. Steady flow $implies$ derivative $frac{dv}{dz}=0$. We rewrite the N-S in axial flow:
$$ frac{Du}{Dt} =frac{partial u}{partial t}+(ucdot nabla)u=frac{partial u_z}{partial t} +underbrace{(u_zcdotnabla)u_z}_{text{must be } 0}$$
Using N-S
$$ begin{align} underbrace{0}_{text{steady flow}}rightarrowrhofrac{partial u_z}{partial t}= & frac{partial p}{partial z} + mu(partial_{xx}u_z+partial_{yy}u_z)=0 \ -p = &mu(partial_{xx}u_z+partial_{yy}u_z) \
frac{-p}{mu} =&partial_{xx}u_z+partial_{yy}u_z \ frac{-p}{mu} = & 6c(1+x)+(-6c(x-1)) \ frac{-p}{mu}=&6c+6cx-6cx+6c \ frac{-p}{mu}=& 12c \ c=frac{1}{12}frac{-p}{mu}end{align}$$
Now I'm not sure how to find the axial speed or the tangential force, any help is much appreciated.
fluid-dynamics
asked Nov 28 at 16:19
elcharlosmaster
1510
Consider steady viscous flow under constant pressure gradient $frac{dp}{dz} = -P$ through a pipe that runs along the $z$-axis and has triangular cross-section. The sides are given by the planes $$x = 1 quad x + 2 -ysqrt 3 = 4 quad x +2 +ysqrt 3 = 0$$
Show that the axial flow speed $omega$ is a constant times $(x - 1)(x + 2 -ysqrt 3)(x +2 +ysqrt 3)$ and find the constant. Find the tangential force per unit length on the pipe due to viscous stress.
Attempt: I believe I found our constant which I will call $c$.
Using Navier Stokes:
$$begin{align} rhofrac{Dvec{u}}{Dt}=-nabla p-rhovec{g} +munabla^2vec{u} quad text{where the second term is $0$}end{align} $$
We have no flow left or right only up or down. So we only have deal with the $z$ component. Steady flow $implies$ derivative $frac{dv}{dz}=0$. We rewrite the N-S in axial flow:
$$ frac{Du}{Dt} =frac{partial u}{partial t}+(ucdot nabla)u=frac{partial u_z}{partial t} +underbrace{(u_zcdotnabla)u_z}_{text{must be } 0}$$
Using N-S
$$ begin{align} underbrace{0}_{text{steady flow}}rightarrowrhofrac{partial u_z}{partial t}= & frac{partial p}{partial z} + mu(partial_{xx}u_z+partial_{yy}u_z)=0 \ -p = &mu(partial_{xx}u_z+partial_{yy}u_z) \
frac{-p}{mu} =&partial_{xx}u_z+partial_{yy}u_z \ frac{-p}{mu} = & 6c(1+x)+(-6c(x-1)) \ frac{-p}{mu}=&6c+6cx-6cx+6c \ frac{-p}{mu}=& 12c \ c=frac{1}{12}frac{-p}{mu}end{align}$$
Now I'm not sure how to find the axial speed or the tangential force, any help is much appreciated.
fluid-dynamics
fluid-dynamics
asked Nov 28 at 16:19
elcharlosmaster
1510
asked Nov 28 at 16:19
elcharlosmaster
1510
asked Nov 28 at 16:19
elcharlosmaster
1510
asked Nov 28 at 16:19
elcharlosmaster
1510
asked Nov 28 at 16:19
elcharlosmaster
1510
1510
add a comment |
add a comment |
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