# 7.3.1: Energy Equation in Steady State

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The steady state situation provides several ways to reduce the complexity. The time derivative term can be eliminated since the time derivative is zero. The acceleration term must be eliminated for the obvious reason. Hence the energy equation is reduced to

$\label{ene:eq:govSTSF} \dot{Q} - \dot{W}_{shear} - \dot{W}_{shaft} = \int_S \left( h + \dfrac{U^2} {2\dfrac{}{}} + g\,z \right) U_{rn}\, \rho \,dA + \int_S P U_{bn} dA$

If the flow is uniform or can be estimated as uniform, equation (72) is reduced to

$\label{ene:eq:govSTSFU} \begin{array}{c} \dot{Q} - \dot{W}_{shear} - \dot{W}_{shaft} = \left( h + \dfrac{U^2} {2\dfrac{}{}} + g\,z \right) U_{rn}\, \rho A_{out} - \\ \left( h + \dfrac{U^2}{2\dfrac{}{}} + g\,z \right) U_{rn}\, \rho A_{in} + \displaystyle P\, U_{bn} A_{out} - \displaystyle P U_{bn} A_{in} \end{array}$

It can be noticed that last term in equation (73) for non-deformable control volume does not vanished. The reason is that while the velocity is constant, the pressure is different. For a stationary fix control volume the energy equation, under this simplification transformed to

$\label{ene:eq:govSTSFUfix} \dot{Q} - \dot{W}_{shear} - \dot{W}_{shaft} = \left( h + \dfrac{U^2} {2\dfrac{}{}} + g\,z \right) U_{rn}\, \rho A_{out} - \\ \left( h + \dfrac{U^2} {2\dfrac{}{}} + g\,z \right) U_{rn}\, \rho A_{in}$

Dividing equation the mass flow rate provides

Steady State Equation, Fix $$\dot{m}$$ & uniform

$\label{ene:eq:govSTSFUfixMass} \dot{q} - \dot{w}_{shear} - \dot{w}_{shaft} = \left.\left( h + \dfrac{U^2} {2\dfrac{}{}} + g\,z \right)\right|_{out} - \left.\left( h + \dfrac{U^2} {2\dfrac{}{}} + g\,z \right)\right|_{in}$