Energy Equation

We now apply the Reynolds Transport theorem (Eqn. 3.27) to derive an equation for energy conservation in a control volume. Now we have,

$$N~=~E,~~~ \texttt{and}~~~\eta ~=~{E \over m} =~e$$ (3.57)

On the LHS we have ${dE} \over {dt}$, which from the First Law of Thermodynamics is

$${{dE} \over {dt}}~=~ \dot{Q}~~-~~\dot{W}$$ (3.58)

Where $\dot{Q}$ is the rate at which heat is added to the system and $\dot{W}$ is the rate at which work is done on/by the system.

Substituting in Eqn.3.27 we have

$$~ \dot{Q}~~-~~\dot{W}~=~~{\partial \over \partial t} \int_{CV} e \rho d ~ \forall~~+~~ \int_{CS}~e \rho \overrightarrow{V}.d\overrightarrow{A}$$ (3.59)

In the above equation, e should include all forms of energy - internal, potential, kinetic and others. The others category will include nuclear, electromagnetic and other sources of energy. But for simple fluid flows these are not important. Fields such as Magneto Hydrodynamics and Relativistic Fluid Dynamics will involve these forms of energy too. We have then

$$e~=~u~+gz~+~{1 \over 2} V^2$$ (3.60)

Concerning work, we have different kinds - shaft work, W_s, work done by pressure, W_p and work due to shear forces on the control surface. Shaft work includes any work that is directly added to the system by means of a pump, piston etc. Work done by pressure is calculated as

$$d\dot{W_p}~=~-p~dA~V_n~=~p ~dA \cdot V$$ (3.61)

where dA ia an elemental area over the control surface, the velocity V_n is into the control volume (hence gets a negative sign). This equation is integrated over the control surface to obtain the total work due to pressure. Thus,

$$\dot{W_p}~=~\int_{CS}~p ~dA \cdot V$$ (3.62)

Work due to shear forces is small and is usually neglected. Heat added $\dot{Q}$ becomes important only occasionally in problems involving heat transfer. Upon substituting for various terms we have,

$$~ \dot{Q}-\dot{W_s}-\dot{W_p}~ =~{\partial \over \partial t} \int_{CV} (u~+gz~+~{1 \over 2} V^2... ...int_{CS}~(u~+gz~+~{1 \over 2} V^2) \rho \overrightarrow{V}.d\overrightarrow{A}$$

$$~ \dot{Q}~-~\dot{W_s}~-\int_{CS}~p ~dA \cdot V~ =~{\partial \over \partial t} \int_{CV} (u~+gz~+~{1 \over 2} V^2... ...int_{CS}~(u~+gz~+~{1 \over 2} V^2) \rho \overrightarrow{V}.d\overrightarrow{A}$$

$$~ \dot{Q}~-~\dot{W_s} =~{\partial \over \partial t} \int_{CV} (u~+gz~+~{1 \over 2} V^2... ... \over \rho}+gz~+~{1 \over 2} V^2) \rho \overrightarrow{V}.d\overrightarrow{A}$$

$$~ \dot{Q}~-~\dot{W_s} =~{\partial \over \partial t} \int_{CV} (u~+gz~+~{1 \over 2} V^2... ...int_{CS}~(h~+gz~+~{1 \over 2} V^2) \rho \overrightarrow{V}.d\overrightarrow{A}$$ (3.63)

where h is specific enthalpy given by $u~+~{p \over \rho}$. Equation 3.63 is the general form of the Energy Equation for a control volume.

(c) Aerospace, Mechanical & Mechatronic Engg. 2005
University of Sydney