Question 11.6: Two points A and B at the ground level were supplied equal q...
Two points A and B at the ground level were supplied equal quantity of water through branched pipes each 200 mm in diameter and 10 m long. Water supply is made from an overhead tank whose water level above the ground is 12 m, and the length and diameter of the pipe up to the junction point O are 14 m and 500 mm. The point O is also on the ground level as shown in Fig. 11.18. The connection of a new pipe of 200 mm diameter and 20 m length is to be made from O to C. The friction factor f for all the pipes is 0.016. Valves in the pipelines A and B are provided for controlling the flow rates.
Calculate, (i) the flow rates at A and B when the valves are fully open, before C was connected, (ii) the flow rates at A, B and C with valves fully open, (iii) the valve resistance coefficients on pipelines A and B so as to obtain equal flow rates at A, B and C, and the value of such flow rates (Neglect entry and bend losses).
(i) Let the flow rate through the main pipe from the overhead tank to the junction O be Q and those through the pipes OA and OB are Q_{1} \text { and } Q_{2}. From continuity,
Q_{1}+Q_{2}=Q
Since length, diameter and friction factor for the pipes OA and OB are equal,
Q_{1}=Q_{2}=Q / 2
Velocity in the main pipe from the tank to the pointO=\frac{4 Q}{\pi(0.5)^{2}}
= 5.09 Q
Velocity in the pipe OA = \frac{2 Q}{\pi(0.2)^{2}}=15.92 Q
Applying Bernoulli’s equation between a point at the water level in the overhead tank and the point A through the path connecting the main pipe and the pipe OA, we can write
12=0.016 \frac{14}{0.5} \frac{1}{2 g}(5.09 Q)^{2}+0.016 \frac{10}{0.2} \frac{1}{2 g}(15.92 Q)^{2}
or 12=10.92 Q^{2}
which gives Q=1.05 m ^{3} / s
Hence flow rates at A and B are
Q_{1}=Q_{2}=\frac{1.05}{2}=0.525 m ^{3} / s
(ii) Let Q be the flow rate in the main pipe and Q_{1}, Q_{2}, Q_{3} be the flow rates through the pipes OA, OB and OC respectively.
From continuity, Q=Q_{1}+Q_{2}+Q_{3} (11.42)
If the discharge pressures at A, B and C are equal, then the sum of the frictional loss and the velocity head (or the exit loss) through each pipe OA, OB and OC must be equal. Hence we can write
\left(1+0.016 \frac{10}{0.2}\right) \frac{16}{\pi^{2}(0.2)^{4}} Q_{1}^{2}=\left(1+0.016 \frac{10}{0.2}\right) \frac{16}{\pi^{2}(0.2)^{4}} Q_{2}^{2}
=\left(1+0.016 \frac{20}{0.2}\right) \frac{16}{\pi^{2}(0.2)^{4}} Q_{3}^{2}
whch gives Q_{2}=Q_{1} (11.43)
and Q_{3}=0.832 Q_{1} (11.44)
Therefore, from Eqs (11.42), (11.43) and (11.44) we get
Q=2.832 Q_{1}
Applying Bernoulli’s equation between a point at the water level in the overhead tank and the point A through the hydraulic path connecting the main pipe and the pipe OA, we can write,
12=0.016 \frac{14}{0.5} \frac{1}{2 g}\left(\frac{4 \times 2.832 Q_{1}}{\pi(0.5)^{2}}\right)^{2}+0.016 \frac{10}{0.2} \frac{1}{2 g}\left[\frac{4 Q_{1}}{\pi(0.2)^{2}}\right]^{2}
=46.06 Q_{1}^{2}
which gives Q_{1} = 0.51 m^{3}/s
and from (11.43) Q_{2} = 0.51 m^{3}/s
from (11.44) Q_{3} = 0.42 m^{3}/s
from (11.42) Q = 1.44 m^{3}/s
(iii) Let Q_{1} be the flow rate through OA, OB and OC. Then the flow rate through the main pipe Q = 3 Q_{1}.
Since the diameter of the pipes OA, OB and OC are same, the average velocity of flow through these pipes will also be the same. Let this velocity be V_{1}.
Then, V_{1}=\frac{4 Q_{1}}{\pi(0.2)^{2}}=31.83 Q_{1}
Velocity through the main pipe V=\frac{4 \times 3 Q_{1}}{\pi(0.5)^{2}}=15.28 Q_{1}
Let K be the valve resistance coefficient in pipe OA or OB.
Equating the total losses through two parallel hydraulic paths OC and any one of OA and OB, we have
\left(0.016 \times \frac{10}{0.2}+K+1\right) \frac{(31.83)^{2}}{28} Q_{1}^{2}=\left(0.016 \times \frac{20}{0.2}+1\right) \frac{(31.83)^{2}}{2 g} Q_{1}^{2}
or 1.8 + K = 2.6
Hence K = 0.8
Applying Bernoulli’s equation between a point at the water level in the overhead tank and the point A through the path connecting the main pipe and OA, we have
12=0.016 \frac{14}{0.5} \frac{(15.28)^{2}}{2 g} Q_{1}^{2}+\left[0.016 \frac{10}{0.2}+0.8+1\right] \frac{(31.83)^{2}}{2 g} Q_{1}^{2}
=139.76 Q_{1}^{2}
which gives Q_{1} = 0.293 m^{3}/s
and hence, Q = 3 × 0.293 = 0.879 m^{3}/s