Question 13.1: Water at 59 F flows through a straight section of a 6-in-ID ......

The control volume in this case is the pipe and the water it encloses. Applying the energy equation to this control volume, we obtain

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

An evaluation of each term yields

{\frac{\delta Q}{d t}}=0~~~~~~~~~{\frac{\delta W_{s}}{d t}}={\dot{W}}

\iint_{\mathrm{c.s.}}\rho\!\left(e+\frac{P}{\rho}\right)({\bf v}\!\cdot\!{\bf n})\ d{A}=\rho A v_{\mathrm{avg}}\biggl(\frac{v_{2}^{2}}{2}+g y_{2}+\frac{P_{2}}{\rho}+u_{2}-\frac{v_{1}^{2}}{2}-g y_{1}-\frac{P_{1}}{\rho}-u_{1}\biggr)

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

and

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

The applicable form of the energy equation written on a unit mass basis is now

\dot{W}/\dot{m}=\frac{v_{1}^{2}-v_{2}^{2}}{2}+g(y_{1}-y_{2})+\frac{P_{1}-P_{2}}{\rho}+u_{1}-u_{2}

and with the internal energy change written as gh_{L}, the frictional head loss, the expression for w becomes

\dot{W}/\dot{m}=\frac{v_{1}^{2}-v_{2}^{2}}{2}+g(y_{1}-y_{2})+\frac{P_{1}-P_{2}}{\rho}-g h_{L}

Assuming the fluid at both ends of the control volume to be at atmospheric pressure, (P_{1}-P_{2})/\rho=0, and for a pipe of constant cross section (v_{1}^{2}-v_{2}^{2})/2=0, giving for {\dot{W}}/{\dot{m}}

{\dot{W}}/{\dot{m}}=g(y_{1}-y_{2})-g h_{L}

Evaluating h_{L}, we have

\mathrm{Re}={\frac{(\textstyle{\frac{1}{2}})(4)}{1.22\times10^{-5}}}=164,000

\frac{e}{D}=0.0017\quad\mathrm{(from\:Figure~13.2)}

{f_{f}}=0.0059~~~\left(\mathrm{from\,equation}\left(13-15\mathrm{a}\right)\right)

yielding

h_{L}={\frac{2(0.0059)(120\,\mathrm{ft})\left(16\,\mathrm{ft}^{2}/s^{2}\right)}{(0.5\,\mathrm{ft})(32.2\,\mathrm{ft}\!/s^{2})}}=1.401\,\mathrm{ft}

The power required to produce the specified flow conditions thus becomes

=0.300\,\mathrm{hp}