we know that the mass balance/ continuity equation states $$ {\partial \rho \over \partial t} + {\nabla \cdot \left(\rho \mathbf{u} \right)} = {\partial \rho \over \partial t} + {\nabla \rho \cdot \mathbf{u}} + {\rho \left(\nabla \cdot \mathbf{u} \right)} = 0$$ and for incompressible flows the material derivative is zero $${D \rho \over Dt} = {\partial \rho \over \partial t} + {\nabla \rho \cdot \mathbf{u}}=0 \implies \nabla \cdot \mathbf{u}=0 $$ But for non-isothermal incompressible flows is the material derivative still assumed to be zero?
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$\begingroup$ The continuity equation pertains to conservation of mass. Why would it matter if the flow is isothermal or non-isothermal? $\endgroup$– Bob DCommented Feb 2, 2020 at 18:23
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$\begingroup$ I was using the continuity equation to prove for incompressible flows $$ \nabla \cdot u =0$$ becz the material derivative is zero so the remaining term is zero. but for non-isothermal flows i would not assume the material derivative of density to be zero as the density can change with temperature yet we are still assuming that the divergence of the velocity field is zero $\endgroup$– ChemEngCommented Feb 2, 2020 at 22:35
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$\begingroup$ Good question. The definition of an incompressible material also includes no thermal expansion. $\endgroup$– Chet MillerCommented Feb 3, 2020 at 4:30
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I have written a lengthy post on different types of incompressibility assumptions some time ago, if you are interested in reading it. An incompressible flow is basically defined by a vanishing Lagrangian
$$\frac{D \rho}{D t} = \frac{\partial \rho}{\partial t} + \sum\limits_{j \in \mathcal{D}} u_j \frac{\partial \rho}{\partial x_j} = 0.$$
This basically means the density has to remain constant along any given stream line. The divergence free condition
$$\sum\limits_{j \in \mathcal{D}} \frac{\partial u_j}{\partial x_j} = 0,$$ is then a simple consequence of the continuity equation and thus holds for any incompressible flow.
The question here is: Can a non-isothermal flow be as well be an incompressible flow?
This depends on the equation of state: The ideal gas law for example guarantees for any point in space
$$p v = \frac{p}{\rho} = R_m T$$
and therefore the temperature would have to change linearly with pressure along the stream line (see e.g the values for density in function of temperature for air). Thus, in general a flow of an ideal gas can not be incompressible and non-isothermal at the same time.
The situation is different for other fluids such as liquids where the temperature plays only a minor role. The Tait equation for water
$$p - p_0 = C \left[ \left( \frac{\rho}{\rho_0} \right)^m - 1 \right] $$
for example allows the density to stay approximately constant while the temperature might change over several hundred degrees. In this case it is though likely that one can neglect changes in density across the entire fluid field resulting in an even simpler incompressible fluid assumption.
