Question

A pipeline (diameter 0.3 m, length 3 km) carries water from point P to point R (see figure). The piezometric heads at P and R are to be maintained at 100 m and 80 m, respectively. To increase the discharge, a second pipe is added in parallel to the existing pipe from Q to R. The length of the additional pipe is also 2 km. Assume the friction factor, f = 0.04 for all pipes and ignore minor losses.


What is the increase in discharge if the additional pipe has same diameter (0.3 m)?

• A. 0%
• B. 33%
• C. 41%
• D. 67%

Answer 1

The correct option is C 41%

The difference between piezometric head between P and R is $100-80=20m$. Now, if the minor losses are neglected, then the only loss is the head loss
Let the initial discharge be $Q_1$

$\therefore h_L=\frac{fL{Q}_1^2}{12d^5}$

$\Rightarrow 20=\frac{0.04\times 3000\times Q_1^2}{12\times (0.3{)}^5}$

$\Rightarrow Q_1=0.0697m^3/s$


$h_L=h_{PQ}+h_{QR}$
$\therefore 20=\frac{f{L}_{PQ}{Q}_2^2}{12\times (0.3{)}^5}+\frac{f{L}_{QR}{\left(\frac{Q_2}{2}\right)}^2}{12\times (0.3{)}^5}$

$=\Rightarrow 20=\frac{0.04\times 1000\times Q_2^2}{12\times (0.3{)}^5}+\frac{0.04\times 2000\times Q_2^2}{4\times 12\times (0.3{)}^5}$

$\Rightarrow Q_2=0.0986m^3/s$
$\therefore$ % increase in discharge

$=\left(\frac{Q_2-Q_1}{Q_1}\right)\times 100$

$=\left(\frac{0.0986-0.0697}{0.0697}\right)\times 100$

$=41.4\%$