anonymous
  • anonymous
Suppose the density ρ of a fluid varies from point to point as well as with time, that is, ρ = ρ(x,y,z,t). If we follow the fluid along a streamline, then x, y, z are functions of t such that the fluid velocity is V = i dx/dt + j dy/dt + k dz/dt. Show that then dρ/dt = ∂ρ/∂t + v (dot) ∇ρ. Combine this equation with ∇ (dot) v + ∂ρ/∂t = 0 to get ρ∇ (dot) v + dρ/dt = 0. (Physically, dρ/dt is the rate of change of density with time as we follow the fluid along a streamline; ∂ρ/∂t is the corresponding rate at a fixed point.) For a steady state (that is, time-independent), ∂ρ/∂t=0, but dρ/dt is
Physics
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chestercat
  • chestercat
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anonymous
  • anonymous
not necessarily zero. For an incompressible fluid, dp/dt=0; show that then ∇ (dot) v = 0. (Note that incompressible does not necessarily mean constant density since dp/dt = 0 does not imply either time or space independence of ρ; consider, for example, a flow of water mixed with blobs of oil.)
anonymous
  • anonymous
I got this far. dρ/dt = ∂ρ/∂x dx/dt + ∂ρ/∂y dy/dt + ∂ρ/∂z dz/dt + ∂ρ/∂t dt/dt \[\frac{ d \rho }{ dt } = \frac{∂\rho}{∂x}v _{x} +\frac{∂\rho}{∂y}v _{y} +\frac{∂\rho}{∂z}v _{z} +\frac{∂\rho}{∂t}\] \[\frac{d \rho}{dt} = (v _{x} + v _{y} + v _{z}) \cdot (\frac{∂\rho}{∂x} +\frac{∂\rho}{∂y} +\frac{∂\rho}{∂z}) +\frac{∂\rho}{∂t}\] \[\frac{d \rho}{dt} = \vec{v} \cdot \vec{∇} \rho +\frac{∂\rho}{∂t} \]
BAdhi
  • BAdhi
jst wandering isnt \[\nabla\rho = \left\langle\frac{\partial \rho}{\partial x},\frac{\partial\rho}{\partial y},\frac{\partial \rho}{\partial z},\frac{\partial \rho}{\partial t}\right\rangle \] ?

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anonymous
  • anonymous
No, time is the parameter of the function. So the x, y and z components are all functions of the one variable t.
IrishBoy123
  • IrishBoy123
firstly to clarify that \(\vec v\) is \(\vec v\) at all times in this question, and and not \(\vec V\) where \(\vec V = \rho \vec v\) ....as is used in a common version of the continuity equation: \[\nabla ·\vec V + \frac{\partial ρ}{\partial t} = 0.\]

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