Question 6.4: In the sudden enlargement shown below in Figure 6.10, the pr......

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)