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A uniform pipeline, $5km$ in length, $100mm$ in diameter and with a roughness size of $0.03mm$, conveys water between two reservoirs.

The difference in water level between the reservoirs is $50m$.

In addition to the entry loss of $0.5\frac{v^2}{2g}$, a valve produces a head loss of $10\frac{v^2}{2g}$.

Determine the steady discharge between the reservoirs.

I know that the first stage should be to find a value for velocity $v$ using the Colebrook-White Formula:$$v=-2\sqrt{2gDs_{f}}log(\frac{k_{s}}{3.7D}+\frac{2.51u}{D\sqrt{2gDs_{f}}})$$

where $v=$ velocity, $g= 9.81$ m/s, $D=0.1$, $s_{f}=0.01$, $k_{s}=0.03$ x $10^{-3}$ and $u=1$ x $10^{-6}$.

Subbing these values in, I get velocity $v=2.312$ m/s (3 d.p.).

Now I assume I need to sum the entry and head losses $h_{l}$ $$h_{l}=0.5\frac{2.312^2}{2\times 9.81}+10\frac{2.312^2}{2\times 9.81}=2.861m$$

Now adjust $h$ to get $h_{f}=50-2.861=47.139$

Then find a new value for $s_{f}=47.139/5000=9.4278\times 10^{-3}$

And finally recalculate for $v$ using the Colebrook-White Formula:$$v=-2\sqrt{2gDs_{f}}log(\frac{k_{s}}{3.7D}+\frac{2.51u}{D\sqrt{2gDs_{f}}})$$

where $v=$ velocity, $g= 9.81$ m/s, $D=0.1$, $s_{f}=9.4278\times 10^{-3}$, $k_{s}=0.03$ x $10^{-3}$ and $u=1$ x $10^{-6}$.

Subbing these values in, I get velocity $v=2.240$ m/s (3 d.p.).

Now recalculating $$h_{l}=0.5\frac{2.240^2}{2\times 9.81}+10\frac{2.240^2}{2\times 9.81}=2.685m$$

$h_{l}+h_{f}=2.685+47.139=49.82\approx 50m$ which is sufficiently accurate to be acceptable.

Since $Q=vA \implies Q=2.240\times 0.05^{2}\pi=0.01759$ $m^{3}/s$

Have I answered this correctly? Thanks!

Jake Wrightson
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  • I'm a civil engineering student and this is a question from a past paper. I'm just struggling finding an answer to this particular question. – Jake Wrightson Sep 27 '18 at 18:56
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    @JeffreyJWeimer even if it is homework, the fact that he posted a detailed attempt at solution is generally deemed sufficient to warrant our studying and replying. We don't have to give the correct answer (if different), but could provide hints as to where errors might have been made. – Carl Witthoft Sep 27 '18 at 19:14
  • @CarlWitthoft Fair enough. My comment is withdrawn. – Jeffrey J Weimer Sep 27 '18 at 19:25
  • Propagation of uncertainties throughout based on g = 9.81 as the only limiting term might revel the final answer can only be precise to at best three significant digits. – Jeffrey J Weimer Sep 27 '18 at 19:28

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