Correct Answer - Option 4 : 1.67 m
Concept:
Head loss due to sudden enlargement of pipe is given by,
\({\left( {{h_L}} \right)_{exp}} = \frac{{{{\left( {{v_1} - {v_2}} \right)}^2}}}{{2g}}\)
Where, v1 is the velocity in smaller section and v2 is the velocity in enlarged section
Calculation:
Given,
Q = 60 L/s = 0.06 m3/s
d1 = 100 mm = 0.1 m,
\({A_1} = \frac{\pi }{4}d_1^2 = 0.00785\;{m^2}\)
d2 = 200 mm = 0.2 mm
\({A_2} = \frac{\pi }{4}d_2^2 = 0.0314\;{m^2}\)
\({v_1} = \frac{Q}{{{A_1}}} = \frac{{0.06}}{{0.00785}} = 7.643\;m/s\)
\({v_2} = \frac{Q}{{{A_2}}} = \frac{{0.06}}{{0.0314}} = 1.91\;m/s\)
\({\left( {{h_L}} \right)_{exp}} = \frac{{{{\left( {{7.643} - {1.91}} \right)}^2}}}{{2\times 9.81}}=1.675\;m\)
Loss of energy per unit weight of water is given by 1.675 m