Question 6.1: As a first example, let us choose a control volume as shown ......
As a first example, let us choose a control volume as shown in Figure 6.5 under the conditions of steady fluid flow and no frictional losses. For the specified conditions the overall energy expression, equation (6-10), becomes
{\frac{\delta Q}{d t}}-{\frac{\delta W_{\mathrm{S}}}{d t}}=\iint_{\mathrm{c.s.}}\rho\left(e+{\frac{P}{\rho}}\right)\left(\mathbf{v}\cdot\mathbf{n}\right)d A+{\frac{\partial}{\partial t}}\iiint_{\mathrm{c.v.}} 
Considering now the surface integral, we recognize the product \rho(\mathbf{v}\cdot\mathbf{n})~d A to be the mass flow rate with the sign of this product indicating whether mass flow is into or out of the control volume, dependent upon the sense of (\mathbf{v}\cdot\mathbf{n}). The factor by which the mass flow rate is multiplied, e+P/\rho,, represents the types of energy that may enter or leave the control volume per mass of fluid. The specific total energy, e, may be expanded to include the kinetic, potential, and internal energy contributions, so that
e+{\frac{P}{\rho}}=g y+{\frac{v^{2}}{2}}+u+{\frac{P}{\rho}}
As mass enters the control volume only at section (1) and leaves at section (2), the surface integral becomes
\iint_{\mathrm{c.s.}}\rho{\Big(}\mathrm{e}+{\frac{P}{\rho}}{\Big)}(\mathbf{v}\cdot\mathbf{n})\,d A=\left[{\frac{v_{2}^{2}}{2}}+g\mathbf{y}_{2}+u_{2}+{\frac{P_{2}}{\rho_{2}}}\right](\rho_{2}v_{2}A_{2})-\left[{\frac{v_{1}^{2}}{2}}+g\mathbf{y}_{1}+u_{1}+{\frac{P_{1}}{\rho_{1}}}\right](\rho_{1}v_{1}A_{1})\,
The energy expression for this example now becomes
{\frac{\delta Q}{d t}}-{\frac{\delta W_{s}}{d t}}=\left[{\frac{v_{2}^{2}}{2}}+g y_{2}+u_{2}+{\frac{P_{2}}{\rho_{2}}}\right](\rho_{2}v_{2}A_{2})-\left[{\frac{v_{1}^{2}}{2}}+g y_{1}+u_{1}+{\frac{P_{1}}{\rho_{1}}}\right](\rho_{1}v_{1}A_{1})
In Chapter 4, the mass balance for this same situation was found to be
\dot{m}=\rho_{1}v_{1}A_{1}=\rho_{1}v_{2}A_{2}
If each term in the above expression is now divided by the mass flow rate, we have
\frac{q-\dot{W}_{s}}{\dot{m}}=\left[\frac{v_{2}^{2}}{2}+g y_{2}+u_{2}+\frac{P_{2}}{\rho_{2}}\right]-\left[\frac{v_{1}^{2}}{2}+g y_{1}+u_{1}+\frac{P_{1}}{\rho_{1}}\right]
or, in more familiar form
\frac{v_{1}^{2}}{2}+g y_{1}+h_{1}+\frac{q}{\dot{m}}=\frac{v_{2}^{2}}{2}+g y_{2}+h_{2}+\frac{\dot{W}_{\mathrm{s}}}{\dot{m}}
where the sum of the internal energy and flow energy, u+P/\rho , has been replaced by the enthalpy, h, which is equal to the sum of these quantities by definition h\equiv u+P/\rho .