For the last 1200 m water is drawn by service pipe at uniform rate of $0.1m^3/sec$ per 300 m. Find the head lost in the last 1200 m length of pipe. Take friction factor as 0.04 and velocity is zero at dead end.

Data:-

$D=0.6m,H=150m$

$L=5000m$

Rate of water drawing$=0.1 \frac{m^3}{s}$ per 300 m

$q*=\frac{0.1}{300}=3.33\times 10^{-4}\frac{m^3/s}{m}$

$f=0.04,L_0=1200m, \text{vel. at dead end=0}$

To find:-

$h_f$ in last 1200 m of pipe=?

Solution:

Total distance $Q_0=\frac{1200}{300}\times 0.1=0.4m^3/sec$

-As the water is drawn at uniform rate, therefore deriving expression for head loss for pipe having uniform water drawing rate.

-Consider a section at distance 'x' from the start of uniform withdrawal at q* per meter length

$\therefore$ Discharge from x length

$Q_x=Q_0-(q*.x)$

in a small distance 'dx; head loss is $dh_f$

$\therefore dh_f=\frac{f.dx.v_x^2}{2gD}$

but $v_x=\frac{Q_x}{A}$

$\therefore v_x=\frac{\Delta Q_x}{\pi D^2}$

$\therefore v_x^2=[\frac{4(Q_0-q*.x)}{\pi D^2}]^2$

$\therefore dh_f=\frac{f}{2gD}\times \frac{16(Q_0-q*x)^2}{\pi ^2D^2}.dx$

$dh_f=\frac{8f}{\pi ^2gD^5}\times (Q_0-q*x)^2dx$

Integrating

$h_f=\int _0^{L_0}\frac{8f}{\pi ^2gD^5}(Q_0-q*x)^2dx$

$=\frac{-8f}{\pi ^2.g.D^5}\times \frac{1}{3q*}[(Q_0-q*L_0)^3-Q_0^3]$

$\therefore h_f=\frac{8f}{3\pi ^2.g.D^5}\times \frac{1}{q*}[(Q_0^3-(Q_0-q*L_0)^3]$.........(1)

Equation (1) is expression for pipe in which water is drawn at uniform rate

Now, for last 1200 m, $D=0.6 m,q*=3.33\times 10^{-4}\frac{m^3/s}{m},Q_0=0.4m^3/s,L_0=1200 m $ & $f=0.04$

$\therefore$ equation (1) becomes

$h_f=\frac{8\times 0.04}{3\pi ^2\times 9.81\times 0.6^5}\times \frac{1}{3.33\times 10^{-4}}[0.4^3-(3.33\times 10^{-4}\times 1200)^3]=2.7m$