Question

What is the maximum power available at the downstream end of a pipeline 3 km long, 20 cm in diameter, if water enters at the upstream end at a pressure of 720 m of water, with taking pipe friction coefficient as 0.03 and g as 10 m/s²?
1. 770 mhp
2. 740 mhp
3. 700 mhp
4. 660 mhp

Answer 1

Correct Answer - Option 4 : 660 mhp

Concept:

Net available head at the outlet of the pipe is

\(h = H - {h_f}\)

where H is total head at the source and the \({h_f}\;\) be the head loss in the transmission.

\(h = H - \frac{{4fl{V^2}}}{{2gd}}\)

Power available at the pipe outlet is

\(P = wQh = w.\frac{\pi }{4}{d^2}V\left[ {H - \frac{{4fl{V^2}}}{{2gd}}} \right]\)

For maximum power

\(\frac{d}{{dV}}P = 0\; \to \;H = 3{h_f}\)

Calculation:

Given,

 f = 0.03, L = 3 km; D = 0.2 m ; Total Available head, H = 720 m of water; g = 10 m/s²

In case of maximum power transmission through pipe,

Head loss in friction = 1/3^rd of total available head

i.e. h_f = 720/3 = 240 m of water

We know that:

\({h_f} = \frac{{fL{V^2}}}{{2gD}}\)

\(240 = \frac{{0.03 \times 3000 \times {V^2}}}{{2 \times 10 \times 0.2}}\)

⇒ V = 3.27 m/s

Discharge through pipe, Q = AV

\({\rm{Q}} = \frac{{\rm{\pi }}}{4} \times {\rm{\;}}{0.2^2} \times 3.27 = 0.102{\rm{\;}}{{\rm{m}}^3}/{\rm{s}}\)

Net available head at downstream of pipe, H_net = Total head Available – Head loss through pipe

H_net = 720-240 = 480 m

Maximum power available at downstream end, P = ρQgH_net­

\({\rm{P}} = 1000 \times 0.102 \times 10 \times 480 = 489.6{\rm{\;KW}}\)

Further, we know that 1 hp or 1 mhp = 746 watt

⇒ P = 489.6/0.746 = 656.3 mhp

P ≈ 660 mhp