1.

The rate of change of energy in a moving model is \(\rho\frac{De}{Dt}\). In the final equation, this term is reduced to \(\frac{\partial(\rho e)}{\partial t}+\nabla.(\rho e\vec{V})\). Which of these equations is used for this reduction?(a) Equations of state(b) Stress-strain equation(c) Momentum equation(d) Continuity equationThis question was addressed to me during a job interview.My doubt stems from Energy Equation topic in section Governing Equations of Fluid Dynamics of Computational Fluid Dynamics

Answer»

Right option is (d) CONTINUITY equation

For explanation: Continuity equation is used as given below.

\(\rho\FRAC{De}{Dt}=\rho\frac{\PARTIAL e}{\partial t}+\rho\VEC{V}.\nabla e \)

But,

\(\rho\frac{\partial e}{\partial t}=\frac{\partial(\rho e)}{\partial t}-e\frac{\partial \rho}{\partial t}\)

And

\(\rho\vec{V}.\nabla e=\nabla.(\rho e\vec{V})-e\nabla.(\rho \vec{V})\)

Therefore,

\(\rho\frac{De}{Dt}=\frac{\partial(\rho e)}{\partial t}-e\frac{\partial \rho}{\partial t}+\nabla.(\rho e\vec{V})-e\nabla.(\rho \vec{V})\)

\(\rho\frac{De}{Dt}=\frac{\partial(\rho e)}{\partial t}-e(\frac{\partial \rho}{\partial t}+\nabla.(\rho \vec{V}))+\nabla.(\rho e\vec{V})\)

APPLYING the continuity equation, \(\frac{\partial \rho}{\partial t}+\nabla.(\rho \vec{V})=0\), and hence

\(\rho\frac{De}{Dt}=\frac{\partial(\rho e)}{\partial t}+\nabla.(\rho e\vec{V})\).


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