Figure $4\mathrm{.}3\mathrm{.}6$ Pipe system for example $4\mathrm{.}3\mathrm{.}5.$

${\displaystyle \underset{?}{\overset{?}{}}}{h}_{L_m}=h_{L_{\text{e}\text{n}\text{t}\text{r}\text{a}\text{n}\text{c}\text{e}}}+h_{L_{\text{e}\text{l}\text{b}\text{o}\text{w}}}+h_{L_{\text{e}\text{l}\text{b}\text{o}\text{w}}}+h_{L_{\text{g}\text{l}\text{o}\text{b}\text{e}\text{v}\text{a}\text{l}\text{v}\text{e}}}+h_{L_{\text{e}\text{x}\text{i}\text{t}}}$

$=K_e\frac{V^2}{2g}+2K_{\text{e}\text{l}\text{b}\text{o}\text{w}}\frac{V^2}{2g}+K_{\text{v}\text{a}\text{l}\text{v}\text{e}}\frac{V^2}{2g}+\frac{V^2}{2g}$

$=(K_e+2K_{\text{e}\text{l}\text{b}\text{o}\text{w}}+K_{\text{v}\text{a}\text{l}\text{v}\text{e}}+1)\frac{V^2}{2g}$

where $K_e=0\mathrm{.}5,K_{\text{e}\text{l}\text{b}\text{o}\text{w}}=0\mathrm{.}25,$ and $K_{\text{v}\text{a}\text{l}\text{v}\text{e}}=1\mathrm{.}5.$ The energy equation is now expressed as

$50=(0\mathrm{.}5+20\mathrm{.}25+1\mathrm{.}5+1\mathrm{.}0)\frac{V^2}{2g}+9\mathrm{.}32f{V}^2$

$50=3\mathrm{.}5\frac{V^2}{2g}+9\mathrm{.}32f{V}^2$

$50=(0\mathrm{.}054+9\mathrm{.}32f)V^2$

$V=\sqrt[\phantom{2}]{\frac{50}{0\mathrm{.}054+9\mathrm{.}32f}}$

Assuming fully turbulent flow and using $\frac{k_s}{D}=\frac{0\mathrm{.}0005}{1}=0\mathrm{.}0005,$ we get $f=0\mathrm{.}0165$ from Figure $4\mathrm{.}3\mathrm{.}5,$ then

$V=\sqrt[\phantom{2}]{\frac{50}{0\mathrm{.}054+9\mathrm{.}320\mathrm{.}0165}}$

$V=15\mathrm{.}51f\frac{t}{s}$

Compute $R_e=\frac{VD}{v}=\frac{15\mathrm{.}511}{0\mathrm{.}609{10}^{-5}}=2\mathrm{.}55{10}^6.$ Referring to Figure $4\mathrm{.}3\mathrm{.}5($Moody diagram$),$ we see that the value of $f$ is $OK.$ Now use the continuity equation to determine $Q$ :

$Q=AV=[\frac{(\frac{12}{12}{)}^2}{4}](15\mathrm{.}51f\frac{t}{s})=12\mathrm{.}18f\frac{t^3}{s}$