Correct Answer - Option 2 : 17.4
Concept:
The pumping power fow water flowing required to pump is , P = ρgQH
and Head is, \(H=\frac{fLQ^2}{12D^5}\)
where, ρ = density of liquid, Q = discharge through pipe, H = required head, f = Darcy friction factor
Calculation:
Given:
L = 1 km = 1000 m, D = 200 mm = 0.2 m, Q = 0.07 m3/s, f = 0.02, ρ = 1000 kg/m3
\(∴ H=\frac{0.02~\times~1000~\times~0.07^2}{12~\times~0.2^5}\)
∴ H = 25.52 m
Pumping power = ρgQH
∴ 1000 × 9.81 × 0.07 × 25.52
∴ 17524.58 W = 17.52 kW