Question

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 equation

This 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 1

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

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