Problem#$5(8$ marks$)$

SAE $10$ W oil is to be delivered through the system shown. The $3$ cm cast iron pipe discharges to atmosphere. How much air pressure is required in the tank, $p_o,$ to deliver a $1(10{)}^{-3}\frac{m^3}{s}$ of oil through the pipe system?

Useful Formulae

Conservation of Mass:

$\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{t}}+$grad$*(V)=0$

Navier$-$Stokes Equation:

$\frac{\text{d}\text{e}\text{l}\text{V}}{\text{d}\text{e}\text{l}\text{t}}+V*gradV=-\frac{1}{}$gradp$+$

$ugra{d}^{2}V+g$

Conservation of Energy:

$p_1+\frac{1}{2}{V}_1^2+g{z}_1=p_2+\frac{1}{2}{V}_2^2+g{z}_2+h_l-W_{\text{p}\text{u}\text{m}\text{p}}^{}$

Hydrostatic pressure:

$\frac{dp}{dz}=-g$

Haaland Equation:

$\frac{1}{\sqrt[\phantom{2}]{f}}=-1\mathrm{.}8log[(\frac{lo\frac{n}{D}}{3\mathrm{.}7}{)}^{1\mathrm{.}11}+\frac{6\mathrm{.}9}{Re}]$

Cast iron pipe roughness:

$lon=0\mathrm{.}26mm$

Loss coefficient $($K$)$:

$90$ bend $=1\mathrm{.}5$

Physical properties and constants:

${}_{\text{w}\text{a}\text{t}\text{e}\text{r}}={10}^{3}kg*m^{-3},{}_{\text{w}\text{a}\text{t}\text{e}\text{r}}=1\mathrm{.}03{10}^{-3}Pa*s,g=-9\mathrm{.}81km*s^{-2}$

${}_{\text{a}\text{i}\text{r}}=1\mathrm{.}225kg*m^{-3},{}_{\text{a}\text{i}\text{r}}=1\mathrm{.}79894{10}^{-5}Pa*s$

${}_{(SAE10W)}=1{10}^{-1}Pa*s$

Unit conversion

$1hp=745\mathrm{.}7W$

$1$ bar $=1{10}^{5}Pa$

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