Question 11.7: Two reservoirs are connected through a 300 mm diameter pipe ...
Two reservoirs are connected through a 300 mm diameter pipe line, 1000 m long as shown in Fig. 11.19. At a point B, 300 m from the reservoir A, a valve is inserted on a short branch line which discharges to atmosphere. The valve may be regarded as a rounded orifice 75 mm diameter, C_{d} = 0.65. If friction factor f for all the pipes is 0.013, calculate the rate of discharge to the reservoir C when the valve at B is fully opened. Estimate the leakage through the short pipe line at B.
Let the flow rate through the first 300 m of the pipe be Q and the flow rates through the next 700 m of the pipe and the short branch line containing the valve be Q_{1} and Q_{2} respectively.
From continuity, Q=Q_{1}+Q_{2}
Velocity in the pipe B C=\frac{4 Q_{1}}{\pi(0.3)^{2}}=14.15 Q_{1}
Applying Bernoulli’s equation between points B and C,
\frac{p_{B}}{\rho g}+\frac{(14.15)^{2}}{2 g} Q_{1}^{2}=\frac{p_{ atm }}{\rho g}+0.013 \frac{700}{0.3} \frac{(14.15)^{2}}{2 g} Q_{1}^{2}+\frac{(14.15)^{2}}{2 g} Q_{1}^{2}
or \frac{p_{B}-p_{ atm }}{\rho g}=309.55 Q_{1}^{2} (11.45)
The discharge through the valve acting as an orifice can be written as
Q_{2}=0.65 \times \pi\left(\frac{0.075}{4}\right)^{2} \sqrt{\frac{2\left(p_{B}-p_{ atm }\right)}{\rho}} (11.46)
Using Eqs (11.45) and (11.46), we have
Q_{2}=0.65 \times \pi\left(\frac{0.075}{4}\right)^{2} \sqrt{2 \times 309.55 \times 9.81} Q_{1}
=0.224 Q_{1}
Hence, Q=1.224 Q_{1}
Applying Bernoulli’s equation between A and C through the path ABC, we have,
28=\left(0.5+0.013 \frac{300}{0.3}\right) \frac{(14.15 \times 1.224)^{2}}{2 g} Q_{1}^{2}
+\left(1+0.013 \frac{700}{0.3}\right) \frac{(14.15)^{2}}{2 g} Q_{1}^{2}
=526.16 Q_{1}^{2}
which gives Q_{1}=0.231 m ^{3} / s
Q_{2}=0.224 \times 0.231=0.052 m ^{3} / s