Mixing
Inviscid Taylor Couette Flow
$begingroup$
I'm trying to solve a simple problem where I have an INVISCID fluid between two cylinders and they are rotating with some angular velocity $Omega_1$ and $Omega_2.$ The cylinders have radii $R_1$ and $R_2$ where the latter is larger than the former. I'm using polar coordinates and the Euler equations.
I'm trying to show that the Euler equations give a steady solution
$$textbf{U} = V(r) textbf{e}_theta$$
where $V(r)$ is just an arbitrary function of $r$. I am having trouble satisfying boundary conditions for this problem since the fluid is inviscid.
Would the conditions
$$u_theta = R_1 Omega_1, r=R_1 \
u_theta = R_2 Omega_2, r=R_2$$
apply here?
Using the Euler equation
$$frac{D textbf{u}}{Dt} = -nabla p$$
where $p=p(r)$
$$frac{V_theta^2}{r^2} = frac{1}{rho}frac{partial p}{partial r}$$
How can I solve for the velocity? The viscous term is zero so it removes the possibility to use the $theta$ momentum equation.
Schematic of Problem
fluid-dynamics
asked Jan 16 at 20:22
whiteiversonwhiteiverson
11
$endgroup$
|
show 3 more comments
$begingroup$
I'm trying to solve a simple problem where I have an INVISCID fluid between two cylinders and they are rotating with some angular velocity $Omega_1$ and $Omega_2.$ The cylinders have radii $R_1$ and $R_2$ where the latter is larger than the former. I'm using polar coordinates and the Euler equations.
I'm trying to show that the Euler equations give a steady solution
$$textbf{U} = V(r) textbf{e}_theta$$
where $V(r)$ is just an arbitrary function of $r$. I am having trouble satisfying boundary conditions for this problem since the fluid is inviscid.
Would the conditions
$$u_theta = R_1 Omega_1, r=R_1 \
u_theta = R_2 Omega_2, r=R_2$$
apply here?
Using the Euler equation
$$frac{D textbf{u}}{Dt} = -nabla p$$
where $p=p(r)$
$$frac{V_theta^2}{r^2} = frac{1}{rho}frac{partial p}{partial r}$$
How can I solve for the velocity? The viscous term is zero so it removes the possibility to use the $theta$ momentum equation.
Schematic of Problem
fluid-dynamics
asked Jan 16 at 20:22
whiteiversonwhiteiverson
11
$endgroup$
1
$begingroup$
Those boundary conditions hold in the viscous case. With the Euler equations you can impose $u_r = 0$ if the cylinder walls are impermeable. Specifying the no-slip condition generally over-determines the problem.
$endgroup$
– RRL
Jan 16 at 20:27
$begingroup$
To make some progress here you should show your attempt to solve the PDE up to applying boundary conditions or ask about this on Physics.SE.
$endgroup$
– RRL
Jan 16 at 20:32
$begingroup$
I don't understand how to impose $u_r = 0$ in this problem?
$endgroup$
– whiteiverson
Jan 16 at 21:50
$begingroup$
Well you already satisfied it by assuming unidirectional flow with $u_r = 0$ everywhere. Why are you trying to solve the Euler equation for this configuration? With full Navier-Stokes including the viscous terms you get a second-order linear differential equation with solution $u_{theta}(r) = Ar + B/r$. $A$ and $B$ are determined by the no-slip boundary conditions. I don't see this as well posed for the Euler equations.
$endgroup$
– RRL
Jan 16 at 22:10
$begingroup$
You have only a trivial solution of rigid rotation $u_theta = Omega r$ which can satisfy $frac{u_theta^2}{r^2} = frac{1}{rho} frac{partial p}{partial r}$ with a constant pressure gradient, but that won't satisfy the boundary conditions.
$endgroup$
– RRL
Jan 16 at 22:13
|
show 3 more comments
$begingroup$
I'm trying to solve a simple problem where I have an INVISCID fluid between two cylinders and they are rotating with some angular velocity $Omega_1$ and $Omega_2.$ The cylinders have radii $R_1$ and $R_2$ where the latter is larger than the former. I'm using polar coordinates and the Euler equations.
I'm trying to show that the Euler equations give a steady solution
$$textbf{U} = V(r) textbf{e}_theta$$
where $V(r)$ is just an arbitrary function of $r$. I am having trouble satisfying boundary conditions for this problem since the fluid is inviscid.
Would the conditions
$$u_theta = R_1 Omega_1, r=R_1 \
u_theta = R_2 Omega_2, r=R_2$$
apply here?
Using the Euler equation
$$frac{D textbf{u}}{Dt} = -nabla p$$
where $p=p(r)$
$$frac{V_theta^2}{r^2} = frac{1}{rho}frac{partial p}{partial r}$$
How can I solve for the velocity? The viscous term is zero so it removes the possibility to use the $theta$ momentum equation.
Schematic of Problem
fluid-dynamics
asked Jan 16 at 20:22
whiteiversonwhiteiverson
11
$endgroup$
I'm trying to solve a simple problem where I have an INVISCID fluid between two cylinders and they are rotating with some angular velocity $Omega_1$ and $Omega_2.$ The cylinders have radii $R_1$ and $R_2$ where the latter is larger than the former. I'm using polar coordinates and the Euler equations.
I'm trying to show that the Euler equations give a steady solution
$$textbf{U} = V(r) textbf{e}_theta$$
where $V(r)$ is just an arbitrary function of $r$. I am having trouble satisfying boundary conditions for this problem since the fluid is inviscid.
Would the conditions
$$u_theta = R_1 Omega_1, r=R_1 \
u_theta = R_2 Omega_2, r=R_2$$
apply here?
Using the Euler equation
$$frac{D textbf{u}}{Dt} = -nabla p$$
where $p=p(r)$
$$frac{V_theta^2}{r^2} = frac{1}{rho}frac{partial p}{partial r}$$
How can I solve for the velocity? The viscous term is zero so it removes the possibility to use the $theta$ momentum equation.
Schematic of Problem
fluid-dynamics
fluid-dynamics
asked Jan 16 at 20:22
whiteiversonwhiteiverson
11
asked Jan 16 at 20:22
whiteiversonwhiteiverson
11
edited Jan 16 at 21:49
whiteiverson
asked Jan 16 at 20:22
whiteiversonwhiteiverson
11
asked Jan 16 at 20:22
whiteiversonwhiteiverson
11
asked Jan 16 at 20:22
whiteiversonwhiteiverson
11
11
1
$begingroup$
Those boundary conditions hold in the viscous case. With the Euler equations you can impose $u_r = 0$ if the cylinder walls are impermeable. Specifying the no-slip condition generally over-determines the problem.
$endgroup$
– RRL
Jan 16 at 20:27
$begingroup$
To make some progress here you should show your attempt to solve the PDE up to applying boundary conditions or ask about this on Physics.SE.
$endgroup$
– RRL
Jan 16 at 20:32
$begingroup$
I don't understand how to impose $u_r = 0$ in this problem?
$endgroup$
– whiteiverson
Jan 16 at 21:50
$begingroup$
Well you already satisfied it by assuming unidirectional flow with $u_r = 0$ everywhere. Why are you trying to solve the Euler equation for this configuration? With full Navier-Stokes including the viscous terms you get a second-order linear differential equation with solution $u_{theta}(r) = Ar + B/r$. $A$ and $B$ are determined by the no-slip boundary conditions. I don't see this as well posed for the Euler equations.
$endgroup$
– RRL
Jan 16 at 22:10
$begingroup$
You have only a trivial solution of rigid rotation $u_theta = Omega r$ which can satisfy $frac{u_theta^2}{r^2} = frac{1}{rho} frac{partial p}{partial r}$ with a constant pressure gradient, but that won't satisfy the boundary conditions.
$endgroup$
– RRL
Jan 16 at 22:13
|
show 3 more comments
1
$begingroup$
Those boundary conditions hold in the viscous case. With the Euler equations you can impose $u_r = 0$ if the cylinder walls are impermeable. Specifying the no-slip condition generally over-determines the problem.
$endgroup$
– RRL
Jan 16 at 20:27
$begingroup$
To make some progress here you should show your attempt to solve the PDE up to applying boundary conditions or ask about this on Physics.SE.
$endgroup$
– RRL
Jan 16 at 20:32
$begingroup$
I don't understand how to impose $u_r = 0$ in this problem?
$endgroup$
– whiteiverson
Jan 16 at 21:50
$begingroup$
Well you already satisfied it by assuming unidirectional flow with $u_r = 0$ everywhere. Why are you trying to solve the Euler equation for this configuration? With full Navier-Stokes including the viscous terms you get a second-order linear differential equation with solution $u_{theta}(r) = Ar + B/r$. $A$ and $B$ are determined by the no-slip boundary conditions. I don't see this as well posed for the Euler equations.
$endgroup$
– RRL
Jan 16 at 22:10
$begingroup$
You have only a trivial solution of rigid rotation $u_theta = Omega r$ which can satisfy $frac{u_theta^2}{r^2} = frac{1}{rho} frac{partial p}{partial r}$ with a constant pressure gradient, but that won't satisfy the boundary conditions.
$endgroup$
– RRL
Jan 16 at 22:13
1
1
$begingroup$
Those boundary conditions hold in the viscous case. With the Euler equations you can impose $u_r = 0$ if the cylinder walls are impermeable. Specifying the no-slip condition generally over-determines the problem.
$endgroup$
– RRL
Jan 16 at 20:27
$begingroup$
Those boundary conditions hold in the viscous case. With the Euler equations you can impose $u_r = 0$ if the cylinder walls are impermeable. Specifying the no-slip condition generally over-determines the problem.
$endgroup$
– RRL
Jan 16 at 20:27
$begingroup$
To make some progress here you should show your attempt to solve the PDE up to applying boundary conditions or ask about this on Physics.SE.
$endgroup$
– RRL
Jan 16 at 20:32
$begingroup$
To make some progress here you should show your attempt to solve the PDE up to applying boundary conditions or ask about this on Physics.SE.
$endgroup$
– RRL
Jan 16 at 20:32
$begingroup$
I don't understand how to impose $u_r = 0$ in this problem?
$endgroup$
– whiteiverson
Jan 16 at 21:50
$begingroup$
I don't understand how to impose $u_r = 0$ in this problem?
$endgroup$
– whiteiverson
Jan 16 at 21:50
$begingroup$
Well you already satisfied it by assuming unidirectional flow with $u_r = 0$ everywhere. Why are you trying to solve the Euler equation for this configuration? With full Navier-Stokes including the viscous terms you get a second-order linear differential equation with solution $u_{theta}(r) = Ar + B/r$. $A$ and $B$ are determined by the no-slip boundary conditions. I don't see this as well posed for the Euler equations.
$endgroup$
– RRL
Jan 16 at 22:10
$begingroup$
Well you already satisfied it by assuming unidirectional flow with $u_r = 0$ everywhere. Why are you trying to solve the Euler equation for this configuration? With full Navier-Stokes including the viscous terms you get a second-order linear differential equation with solution $u_{theta}(r) = Ar + B/r$. $A$ and $B$ are determined by the no-slip boundary conditions. I don't see this as well posed for the Euler equations.
$endgroup$
– RRL
Jan 16 at 22:10
$begingroup$
You have only a trivial solution of rigid rotation $u_theta = Omega r$ which can satisfy $frac{u_theta^2}{r^2} = frac{1}{rho} frac{partial p}{partial r}$ with a constant pressure gradient, but that won't satisfy the boundary conditions.
$endgroup$
– RRL
Jan 16 at 22:13
$begingroup$
You have only a trivial solution of rigid rotation $u_theta = Omega r$ which can satisfy $frac{u_theta^2}{r^2} = frac{1}{rho} frac{partial p}{partial r}$ with a constant pressure gradient, but that won't satisfy the boundary conditions.
$endgroup$
– RRL
Jan 16 at 22:13
|
show 3 more comments
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$begingroup$
Those boundary conditions hold in the viscous case. With the Euler equations you can impose $u_r = 0$ if the cylinder walls are impermeable. Specifying the no-slip condition generally over-determines the problem.
$endgroup$
– RRL
Jan 16 at 20:27
$begingroup$
To make some progress here you should show your attempt to solve the PDE up to applying boundary conditions or ask about this on Physics.SE.
$endgroup$
– RRL
Jan 16 at 20:32
$begingroup$
I don't understand how to impose $u_r = 0$ in this problem?
$endgroup$
– whiteiverson
Jan 16 at 21:50
$begingroup$
Well you already satisfied it by assuming unidirectional flow with $u_r = 0$ everywhere. Why are you trying to solve the Euler equation for this configuration? With full Navier-Stokes including the viscous terms you get a second-order linear differential equation with solution $u_{theta}(r) = Ar + B/r$. $A$ and $B$ are determined by the no-slip boundary conditions. I don't see this as well posed for the Euler equations.
$endgroup$
– RRL
Jan 16 at 22:10
$begingroup$
You have only a trivial solution of rigid rotation $u_theta = Omega r$ which can satisfy $frac{u_theta^2}{r^2} = frac{1}{rho} frac{partial p}{partial r}$ with a constant pressure gradient, but that won't satisfy the boundary conditions.
$endgroup$
– RRL
Jan 16 at 22:13