- #1
me24
- 3
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The Reynolds Transport theorem (RTT) is usually applied to derive the conservation of mass, momentum and energy in a fluid. But when I try to apply the RTT to other physical quantities, I get weird results. Can anyone see where I'm going wrong?
As a simple example, take the physical quantity to be position, [itex]x[/itex]. Assume that the fluid density [itex]\rho[/itex] and velocity [itex]u[/itex] are independent of time, so that the time derivative of the volume integral [itex]\int \rho\,x\,dV[/itex] is zero. Applying the RTT to [itex]\rho\,x[/itex] gives,
[tex]
\frac{\partial\,\rho\,x}{\partial\,t} + \nabla\cdot(\rho\,x\,u) = 0
[/tex]
which can be rearranged as,
[tex]
\left[\frac{\partial\,\rho}{\partial\,t} + \nabla\cdot(\rho\,u)\right]\,x
+ \rho\left\{\frac{\partial\,x}{\partial\,t} + u\cdot\nabla\,x\right\}
[/tex]
This can be simplified by applying the mass equation,
[tex]
\frac{\partial\,\rho}{\partial\,t} + \nabla\cdot(\rho\,u) = 0
[/tex]
to yield the result,
[tex]
\rho\,D_t(x) = 0
[/tex]
where [itex]D_t[/itex] is the material derivative,
[tex]
D_t = \frac{\partial}{\partial\,t} + u\cdot\nabla
[/tex]
I think that [itex]\partial\,x/\partial\,t = 0[/itex] and [itex]\nabla\,x = I[/itex], where [itex]I[/itex] is the identity matrix, in which case the result simplifies to,
[tex]
\rho\,u = 0
[/tex]
This is incorrect [itex]u[/itex] does not need to be zero in order for the time derivative of [itex]\int \rho\,x\,dV[/itex] to be zero.
As a simple example, take the physical quantity to be position, [itex]x[/itex]. Assume that the fluid density [itex]\rho[/itex] and velocity [itex]u[/itex] are independent of time, so that the time derivative of the volume integral [itex]\int \rho\,x\,dV[/itex] is zero. Applying the RTT to [itex]\rho\,x[/itex] gives,
[tex]
\frac{\partial\,\rho\,x}{\partial\,t} + \nabla\cdot(\rho\,x\,u) = 0
[/tex]
which can be rearranged as,
[tex]
\left[\frac{\partial\,\rho}{\partial\,t} + \nabla\cdot(\rho\,u)\right]\,x
+ \rho\left\{\frac{\partial\,x}{\partial\,t} + u\cdot\nabla\,x\right\}
[/tex]
This can be simplified by applying the mass equation,
[tex]
\frac{\partial\,\rho}{\partial\,t} + \nabla\cdot(\rho\,u) = 0
[/tex]
to yield the result,
[tex]
\rho\,D_t(x) = 0
[/tex]
where [itex]D_t[/itex] is the material derivative,
[tex]
D_t = \frac{\partial}{\partial\,t} + u\cdot\nabla
[/tex]
I think that [itex]\partial\,x/\partial\,t = 0[/itex] and [itex]\nabla\,x = I[/itex], where [itex]I[/itex] is the identity matrix, in which case the result simplifies to,
[tex]
\rho\,u = 0
[/tex]
This is incorrect [itex]u[/itex] does not need to be zero in order for the time derivative of [itex]\int \rho\,x\,dV[/itex] to be zero.