Correct Answer - Option 1 : 29.7 W
Concept:
Power required in a flow is given by:
\(Power = \rho gQ{h_f}\)
Where ρ = Density,
Q = Discharge = A × V
hf = Head loss due to friction = \(\frac{{32\mu VL}}{{\rho g{d^2}}}\)
Calculation:
Given:
ρ = 870 kg/m3, µ = 0.104 kg/m-s , Q = 1.1 m3/hour = \(\frac{{1.1}}{{3600}}\) = 3.05 × 10-4 m3/sec, d = 2 cm = 0.02 m , L = 12 m
\(Power = \rho gQ \times \frac{{32\mu VL}}{{\rho g{d^2}}}\)
\(Power = Q \times \frac{{32\mu QL}}{{A{d^2}}}\;\)
\(P = \frac{{32\mu {Q^2}L}}{{\left( {\frac{{\pi \;}}{4}{d^2}} \right){d^2}}}\)
\(P = \frac{{32\; \times\; 0.104\; \times \;{{\left( {3.05\; \times {{10}^{ - 4}}} \right)}^2} \times \;12}}{{\left( {\frac{\pi }{4} \times {{0.02}^2}} \right)\;\times\;{{0.02}^2}}}\)
Power = 29.7 W