5. Problem 10.5 (10 pts) During each period the power will be: 3 P = (0.602 kg/m³)m.4v = (0.602 kg/m³) × (0.37) × (3.14) × (10 m)²,³ = (69.9 kg/m) v³

For the different periods the power available will be: for 4 h at 2 m/s →→ (69.9 kg/m) x (2 m/s)³ = 560 W for 16h at 8 m/s →→ (69.9 kg/m) x (8 m/s) = 35790 W etc. etc. The energy generated during these periods is the power times the duration as for 4 h at 2 m/s → (0.560 kW) x (4 h) = 2.2 kWh for 16h at 8 m/s → (35.79 kW) × (16 h) = 572.6 kWh etc., etc. Adding these gives the total energy over the 24 h period is E= 1494 kWh The average power is, therefore P = 62.2 kW