Question 6.2: As a second example, consider the situation shown in Figure ......

Defining the control volume as shown by the dashed lines, we may evaluate equation (6-10) term by term as follows:

{\frac{\delta Q}{d t}}-{\frac{\delta W_{\mathrm{s}}}{d t}}=\iint_{\mathrm{c.s.}}\left(e+{\frac{P}{\rho}}\right)\rho(\mathbf{v}\cdot\mathbf{n})\,d A+\,{\frac{\partial}{\partial t}}\int\!\!\int\!\int_{{\mathrm{c.v.}}}e\rho\,d V+{\frac{\delta W_{\mu}}{d t}} (6-10)

{\frac{\delta Q}{d t}}=0

-\frac{\delta W_{\mathrm{s}}}{d t}=(3\,h p)(2545\,{\mathrm{Btu}}/h p-h)(778\,{\mathrm{ft}}-{\mathrm{lb}}_{\mathrm{f}}/\mathrm{Btu})(h/3600\,\mathrm{s})

\quad=\:1650\,\mathrm{ft}\,\mathrm{lb}_{\mathrm{f}}/\mathrm{s}

\iint_{c.s.}(e+{\frac{P}{\rho}})\rho(\mathbf{v}\cdot\mathbf{n})\,d A=\iint_{A_{2}}{\Big(}e+{\frac{P}{\rho}}{\Big)}\rho(\mathbf{v}\cdot\mathbf{n})\,d A-\iint_{A_{1}}{\Big(}e+{\frac{P}{\rho}}{\Big)}\rho(\mathbf{v}\cdot\mathbf{n})\,d A

=\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})

=\left[\frac{v_{2}^{2}-v_{1}^{2}}{2}+g(y_{2}-y_{1})+\left(u_{2}-u_{1}\right)+\left(\frac{P_{2}}{\rho_{2}}-\frac{P_{1}}{\rho_{1}}\right)\right]\left(\rho v A\right)

Here it may be noted that the pressure measured at station (1) is the static pressure while the pressure measured at station (2) is measured using a pressure port that is oriented normal to the oncoming flow—that is, where the velocity has been reduced to zero. Such a pressure is designated the stagnation pressure, which is greater than the static pressure by an amount equivalent to the change in kinetic energy of the flow. The stagnation pressure is thus expressed as

P_{\mathrm{stagnation}}=P_{0}=P_{\mathrm{static}}+{\frac{1}{2}}\rho v^{2}

for incompressible flow; hence, the energy flux term may be rewritten as

\iint_{{ c}.{ s}.{}}\left(e+{\frac{P}{\rho}}\right)\!\rho({\bf v}\cdot{\bf n})d{ A}\;\;=\;\left({\frac{P_{0_{2}}-P_{1}}{\rho}}-{\frac{v_{1}^{2}}{2}}\right)(\rho v{ A})

=\left\{{\frac{6(1-1/13.6)\operatorname*{in.}\operatorname{Hg}(14.7\ lb/in^{2})(144\operatorname{in}^{2}/\mathrm{ft}^{2})}{(62.4\operatorname{lb}_{\mathrm{m}}/\operatorname{ft}^{3})(29.92\operatorname{in}.\operatorname{Hg})}}\right.

-\frac{v_{1}^{2}}{64.4(\mathrm{lb_{m}ft/} {\mathrm{s}}^{2}\mathrm{lb_{f}})}\Bigg\}\{(62.4\ \mathrm{lb_{m}/f t}^{3})\big(v_{1}\big)\big(\pi/4\mathrm{~ft}^{2}\big)\}

=\left(6.30-{\frac{v_{1}^{2}}{64.4}}\right)(49\,v_{1})\,\mathrm{ft}\,\mathrm{lb}_{f}/\mathrm{s}

\frac{\partial}{\partial t}\iiint_{\mathrm{c.v.}}e\rho\,d V=0

\frac{\delta W_{\mu}}{d t}\;=0

In the evaluation of the surface integral, the choice of the control volume coincided with the location of the pressure taps at sections (1) and (2). The pressure sensed at section (1) is the static pressure, as the manometer opening is parallel to the fluid-flow direction. At section (2), however, the manometer opening is normal to the flowing fluid stream. The pressure measured by such an arrangement includes both the static fluid pressure and the pressure resulting as a fluid flowing with velocity v_{2} is brought to rest. The sum of these two quantities is known as the impact or stagnation pressure.

The potential energy change is zero between sections (1) and (2) and as we consider the flow to be isothermal, the variation in internal energy is also zero. Hence, the surface integral reduces to the simple form indicated.

The flow rate of water necessary for the stated conditions to exist is achieved by solving the resulting cubic equation. The solution is

v_{1}\,=\,16.59\,\mathrm{ft}/\mathrm{s}(5.057\,\mathrm{m}/\mathrm{s})

\dot{m}\:=\:\rho A v=813\:\mathrm{lb}_{\mathrm{m}}/{\mathrm{s}}(370\,\mathrm{kg}/\mathrm{s})