Alternative Forms of the Energy Equation

Internal Energy Formulation¶

Total internal energy is defined as

e_o=e+\frac{1}{2}\mathbf{v}\cdot\mathbf{v}

Inserted in Eqn. (3.57), this gives

\rho\frac{De}{Dt} + \rho\mathbf{v}\cdot\frac{D \mathbf{v}}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho

Now, let’s replace the substantial derivative $D\mathbf{v}/Dt$ using the momentum equation on non-conservation form (Eqn. (3.51)).

\rho\frac{De}{Dt} -\mathbf{v}\cdot\nabla p + \cancel{\rho\mathbf{f}\cdot\mathbf{v}} + \nabla\cdot(p\mathbf{v}) = \cancel{\rho\mathbf{f}\cdot\mathbf{v}} + \dot{q}\rho

Now, expand the term $\nabla\cdot(p\mathbf{v})$ gives

\rho\frac{De}{Dt} \cancel{-\mathbf{v}\cdot\nabla p} + \cancel{\mathbf{v}\cdot\nabla p} + p(\nabla\cdot\mathbf{v}) = \dot{q}\rho\Rightarrow \rho\frac{De}{Dt} + p(\nabla\cdot\mathbf{v}) = \dot{q}\rho

Divide by $\rho$

\frac{De}{Dt} + \frac{p}{\rho}(\nabla\cdot\mathbf{v}) = \dot{q}

Conservation of mass gives

\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0\Rightarrow \nabla\cdot\mathbf{v} = -\frac{1}{\rho}\frac{D\rho}{Dt}

Insert in Eqn. (3.62)

\frac{De}{Dt} - \frac{p}{\rho^2}\frac{D\rho}{Dt} = \dot{q}\Rightarrow \frac{De}{Dt} + p\frac{D}{Dt} \left(\frac{1}{\rho}\right)= \dot{q}

\frac{De}{Dt} + p\frac{D\nu}{Dt} = \dot{q}

Compare with the first law of thermodynamics: $de=\delta q-\delta w$

h=e+\frac{p}{\rho}\Rightarrow \frac{Dh}{Dt}=\frac{De}{Dt}+\frac{1}{\rho}\frac{Dp}{Dt}+p\frac{D}{Dt}\left(\frac{1}{\rho}\right)

with $De/Dt$ from Eqn. (3.62)

\frac{Dh}{Dt}=\dot{q} - \cancel{p\frac{D}{Dt}\left(\frac{1}{\rho}\right)} +\frac{1}{\rho}\frac{Dp}{Dt}+\cancel{p\frac{D}{Dt}\left(\frac{1}{\rho}\right)}

\frac{Dh}{Dt}=\dot{q} + \frac{1}{\rho}\frac{Dp}{Dt}

h_o=h+\frac{1}{2}\mathbf{v}\mathbf{v}\Rightarrow\frac{Dh_o}{Dt}=\frac{Dh}{Dt}+\mathbf{v}\cdot\frac{D\mathbf{v}}{Dt}

From the momentum equation (Eqn. (3.51))

\frac{D\mathbf{v}}{Dt}=\mathbf{f}-\frac{1}{\rho}\nabla p

which gives

\frac{Dh_o}{Dt}=\frac{Dh}{Dt}+\mathbf{v}\cdot\mathbf{f} -\frac{1}{\rho}\mathbf{v}\cdot\nabla p

Inserting $Dh/Dt$ from Eqn. (3.68) gives

\frac{Dh_o}{Dt}=\dot{q} + \frac{1}{\rho}\frac{Dp}{Dt}+\mathbf{v}\cdot\mathbf{f} -\frac{1}{\rho}\mathbf{v}\cdot\nabla p = \frac{1}{\rho}\left[\frac{Dp}{Dt}-\mathbf{v}\cdot\nabla p\right] + \dot{q} + \mathbf{v}\cdot\mathbf{f}

The substantial derivative operator applied to pressure

\frac{Dp}{Dt}=\frac{\partial p}{\partial t}+\mathbf{v}\cdot\nabla p

and thus

\frac{Dp}{Dt}-\mathbf{v}\cdot\nabla p=\frac{\partial p}{\partial t}

which gives

\frac{Dh_o}{Dt}=\frac{1}{\rho}\frac{\partial p}{\partial t} + \dot{q} + \mathbf{v}\cdot\mathbf{f}

If we assume adiabatic flow without body forces

\frac{Dh_o}{Dt}=\frac{1}{\rho}\frac{\partial p}{\partial t}

If we further assume the flow to be steady state we get

\frac{Dh_o}{Dt}=0

This means that in a steady-state adiabatic flow without body forces, total enthalpy is constant along a streamline.