Question 7.P.20: Two pipes of 0.35 m and 0.25 m dia and length 2000 m and 150......
Two pipes of 0.35 m and 0.25 m dia and length 2000 m and 1500 m with f values 0.021 and 0.018 connected is series carry water from a reservoir to a supply system, the head available being 8 m. Determine the flow quantity neglecting minor losses.
The head available should be equal to the sum of the frictional losses in the two pipes.
Neglecting loss at sudden contraction
\delta=[(0.021\times2000\times u_{1}^{\ 2})/(2\times9.81\times0.35)] \\ +\left[(0.018\times1500\times u_{2}^{\ 2})/(2\times9.81\times0.25)\right]
From continuity equation, we get
[(\pi\times0.35^{2})/4]\times u_{1}=[(\pi\times0.25^{2})/4]u_{2} \\ ∴ u_{2}=(0.35/0.25)^{2}\;u_{1}\quad\mathrm{or}\quad u_{2}^{\ 2}=(0.35/0.25)^{4}\;u_{1}^{\ 2}Substituting, and simplifying and solving,
u_{1} = 0.542 \mathrm{m/s}, u_{2} = 1.062 \mathrm{m/s}
flow rate = (0.542 × π × 0.35²)/4 = 0.0521 m³/s or 187.7 m³/hr
check the frictional drop:
h_{f} = [(0.021 × 2000 × 0.542²)/(2 × 9.81 × 0.35)] \\ + [(0.018 × 1500 × 1.062²)/(2 × 9.81 × 0.25)] \\ h_{f} = 1.8 + 6.2 = 8 m.