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I am trying to understand the derivation of a general balance equation for an arbitrary quantity, as given by this textbook (pg. 13). Now, I have a decent knowledge of how to derive the continuity/Navier-Stokes equations in a conventional method, but this book's approach involves deriving a general balance equation which can be quickly transformed into conservation of mass/momentum/energy etc.

First we define $\psi$ as an arbitrary quantity per unit mass, $\rho$ as the fluid density, $\underline{\underline{J}}$ as the surface efflux of $\psi$ (tensor), and $\phi$ as a body source of $\psi$.

Firstly,

$$ \frac{d}{dt} \int_V \rho\psi dV=-\oint_A(\underline{n}\cdot \underline{\underline{J}})dA+\int_V \rho\phi dV $$

Using Reynolds Transport theoerm and Divergence theorem, we can transform this equation into the following form:

$$ \int_V \frac{\partial\rho\psi}{\partial t}dV+\int_V \nabla \cdot (\rho\psi \underline{u})dV= -\int_V(\nabla \cdot \underline{\underline{J}})dV+\int_V \rho\phi dV $$

Since this applies to any fixed volume, we can say that

$$ \frac{\partial\rho\psi}{\partial t}+ \nabla \cdot (\rho\psi \underline{u})= -(\nabla \cdot \underline{\underline{J}})+ \rho\phi $$

Which is deemed to be the general balance equation for single phase flow.

Now, to convert this to the mass conservation equation, we would set $\psi = 1$, $\phi=0$ (assuming no volume source), and $\underline{\underline{J}}=0$. To convert to the momentum conservation equation, we would set $\psi = \underline{u}$, $\phi=\underline{g}$, and $\underline{\underline{J}}=-\underline{\underline{T}}$ (shear stress tensor).

What I really don't understand about this is the physical meaning of $\underline{\underline{J}}$. The book states that it is a surface flux. Why then, for the mass equation, would we set it to zero? In the general balance equation, what is the physical difference between the convective term (2nd term on LHS) and the surface flux term (1st term on RHS)?

Also, why is it that we use $\underline{n}\cdot \underline{\underline{J}}$ instead of $\underline{\underline{J}} \cdot \underline{n}$? Since J is a tensor quantiy, aren't these two terms different unless it is a symmetric tensor? I have typically seen the latter convention used which in turn allows us to use Divergence theorem.

If anyone can help clarify this for me or point me towards some more explanatory literature, that would be greatly appreciated.

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J is a flux of mass, heat, or momentum over and above the mean convective flux of these quantities on the other side of the equation. In the case of the mass balance equation, if there is no diffusion occurring, all the transport of mass is carried by the convective flux. However, if there is diffusion of a species relative to its mean convective flow, then this mass transfer is captured by J. If there is heat transfer, J describes heat conduction over and above the mean convective flow of heat. In the case of momentum, by analogy, the stress tensor is sometimes regarded as a flux of momentum relative to the mean momentum convection.

With regard to the dot product of J with a normal, for diffusion and heat conduction, J is a vector, so there is no problem. In the case of momentum, J is -T, which is a symmetric 2nd order tensor, so, here again, no problem.

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  • $\begingroup$ Thank you for your reply. When using this general balance to find the energy conservation equation, we set J = q - T . u, where q is the heat flux (vector), T is the stress tensor, and u is the velocity vector. What exactly is the physical meaning of this term in regards to surface flux of energy? The q terms makes obvious sense but why do we need to subtract stress x velocity? This latter term appears to be power per unit area, does that mean that q would be in units of J/s / m2? Thanks again $\endgroup$ – PluckestChoice Mar 19 '17 at 19:06
  • $\begingroup$ That term is the rate at which work is being done at the boundary of the control volume. $\endgroup$ – Chet Miller Mar 19 '17 at 21:14

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