In the sudden enlargement shown below in Figure 6.10, the pressure acting at section (1) is considered uniform with value P_{1}. Find the change in internal energy between stations (1) and (2) for steady, incompressible flow. Neglect shear stress at the walls and express u_{2} – u_{1} in terms of v_{1}, A_{1}, and A_{2}. The control volume selected is indicated by the dotted line.
Conservation of Mass
\iint_{{\bf c.s.}}\!\rho({\bf v}\cdot{\bf n})\,d A+\frac{\partial}{\partial t}\int\!\!\int\!\!\int_{{\bf c.v.}}\!\rho\,d V=0
If we select station (2), a considerable distance downstream from the sudden enlargement, the continuity expression, for steady, incompressible flow, becomes
\rho_{1}v_{1}A_{1}=\rho_{2}v_{2}A_{2}
or
v_{2}=v_{1}{\frac{A_{1}}{A_{2}}} (6-12)
Momentum
\Sigma F=\iiint_{\mathrm{c.s.}}\rho\mathbf{v}(\mathbf{v}\cdot\mathbf{n})\,d A+{\frac{\partial}{\partial t}}\iiint_{\mathrm{c.v.}}\rho\mathbf{v}\,d{{V}}
and thus
P_{1}A_{2}-P_{2}A_{2}=\rho v_{2}^{2}A_{2}-\rho v_{1}^{2}A_{1}
or
{\frac{P_{1}-P_{2}}{\rho}}=v_{2}^{2}-v_{1}^{2}\biggl({\frac{A_{1}}{A_{2}}}\biggr) (6-13)
Energy
{\frac{\delta Q}{\partial t}}-{\frac{\delta W_{\mathrm{{s}}}}{d t}}=\iint_{\mathrm{c.s.}}\!\rho\left(e+{\frac{P}{\rho}}\right)(\mathbf{v}\cdot\mathbf{n})\,d A+{\frac{\partial}{\partial t}}\iiint_{\mathrm{c.v.}}\!\rho\,e\,d V+{\frac{\delta W_{\mu}}{d t}}
Thus,
\left(e_{1}+{\frac{P_{1}}{\rho}}\right)(\rho v_{1}A_{1})=\left(e_{2}+{\frac{P_{2}}{\rho}}\right)(\rho v_{2}A_{2})
or, since \rho v_{1}A_{1}=\rho v_{2}A_{2},
e_{1}+{\frac{P_{1}}{\rho}}=e_{2}+{\frac{P_{2}}{\rho}}
The specific energy is
e={\frac{v^{2}}{2}}+g y+u
Thus, our energy expression becomes
{\frac{v_{1}^{2}}{2}}+g y_{1}+u_{1}+{\frac{P_{1}}{\rho}}={\frac{v_{2}^{2}}{2}}+g y_{2}+u_{2}+{\frac{P_{2}}{\rho}} (6-14)
The three control-volume expressions may now be combined to evaluate u_{2} – u_{1}. From equation (6-14), we have
u_{2}-u_{1}={\frac{P_{1}-P_{2}}{\rho}}+{\frac{v_{1}^{2}-v_{2}^{2}}{2}}+g(y_{1}-y_{2}) (6-14a)
Substituting equation (6-13) for (P_1 – P_2)/ρ and equation (6-12) for v_{2} and noting that y_{1} = y_{2}, we obtain
u_{2}-u_{1}=v_{1}^{2}\biggl(\frac{A_{1}}{A_{2}}\biggr)^{2}-v_{1}^{2}\frac{A_{1}}{A_{2}}+\frac{v_{1}^{2}}{2}-\frac{v_{1}^{2}}{2}\biggl(\frac{A_{1}}{A_{2}}\biggr)^{2}
={\frac{v_{1}^{2}}{2}}\left[1-2{\frac{A_{1}}{A_{2}}}+\left({\frac{A_{1}}{A_{2}}}\right)^{2}\right]={\frac{v_{1}^{2}}{2}}\left[1-{\frac{A_{1}}{A_{2}}}\right]^{2} (16-5)
Equation (6-15) shows that the internal energy increases in a sudden enlargement. The temperature change corresponding to this change in internal energy is insignificant, but from equation (6-14a) it can be seen that the change in total head,
\left(\frac{P_{1}}{\rho}+\frac{v_{1}^{2}}{2}+g y_{1}\right)-\left(\frac{P_{2}}{\rho}+\frac{v_{2}^{2}}{2}+g y_{2}\right)
is equal to the internal energy change. Accordingly, the internal energy change in an incompressible flow is designated as the head loss, h_{L}, and the energy equation for steady, adiabatic, incompressible flow in a stream tube is written as
{\frac{P_{1}}{\rho g}}+{\frac{v_{1}^{2}}{2g}}+y_{1}=h_{L}+{\frac{P_{2}}{\rho g}}+{\frac{v_{2}^{2}}{2g}}+y_{2} (6-16)
Note the similarity to equation (6–11).
g y_{1}+{\frac{v_{1}^{2}}{2}}+{\frac{P_{1}}{\rho}}=g y_{2}+{\frac{v_{2}^{2}}{2}}+{\frac{P_{2}}{\rho}} (6-11a)
y_{1}+{\frac{v_{1}^{2}}{2g}}+{\frac{P_{1}}{\rho g}}=y_{2}+{\frac{v_{2}^{2}}{2g}}+{\frac{P_{2}}{\rho g}} (6-11b)