Question 11.10: Water flows through a pipe line of 300 mm in diameter and 20...
Water flows through a pipe line of 300 mm in diameter and 20 km long in a horizontal plane. At a point B, the pipe is branched off into two parallel pipes each of 150 mm diameter and 3.5 km long as shown in Fig. 11.21. In one of the these pipes, water is completely drained off from side tappings at a constant rate of 0.01 litre/s per metre length of the pipe. Determine the flow rate and loss of head in the main pipe. (Take friction factor for all the pipes as 0.012.)
Let Q be the flow rate through the pipe AB and be divided at B into Q_{1} \text { and } Q_{2} for the pipes BC and BD respectively. Then from continuity,
Q=Q_{1}+Q_{2}
Since the entire flow at inlet to the pipe BD is drained off through side tappings at a constant rate of 0.01 litre per metre length,
Q_{2}=0.01 \times 3500=35 \text { litre } / s =0.035 m^{3} / s
Hence, average velocity at inlet to pipe BD
=\frac{4 \times 0.035}{\pi(0.15)^{2}}=1.98 m/s
The loss of head in BD can be written with the help of Eq. (11.36b) as
\begin{aligned}h_{f}=& {\left[0.02 \times \frac{400}{0.4} \times \frac{(7.96)^{2}}{2 g}+0.02 \times \frac{200}{0.2} \times \frac{(31.83)^{2}}{2 g}\right.} \\&\left.+0.02 \times \frac{300}{0.3} \times \frac{(14.15)^{2}}{2 g}\right] Q^{2} \\=& 1301.46 Q^{2}\end{aligned} (11.36)
h_{f B D}=\frac{1}{3} \times 0.012 \times \frac{3500}{0.15} \times \frac{(1.98)^{2}}{2 \times 9.81}=18.65 m
Since B is a common point and C and D are at the same horizontal level and have the same pressure which is equal to that of the atmosphere, the loss of head in the parallel pipes BC and BD are equal.
Therefore, h_{f B C}=h_{f B D}=18.65 m (11.51)
Average flow velocity in pipe B C=\frac{4 Q_{1}}{\pi(0.15)^{2}}=56.59 Q_{1}
Equating h_{f B C} given by Eq. (11.51) with the different losses taking place in pipe BC, we can write
h_{f B C}=18.65=0.012 \times \frac{3500}{0.15} \times \frac{(56.59)^{2}}{2 g} Q_{1}^{2}+\frac{(56.59)^{2}}{2 g} Q_{1}^{2}
=45865.6 Q_{1}^{2}
which gives Q_{1}=0.02 m ^{3} / s
Hence, Q=0.035+0.02+0.055 m ^{3} / s
Velocity in the main pipe A B=\frac{4 \times 0.055}{\pi(0.3)^{2}}=0.78 m/s
The loss of head in the main pipe A B=0.012 \times \frac{20000}{0.30} \frac{(0.78)^{2}}{2 g}=24.81 m