Question

A propeller rotating at a speed of 108 RPM behind the ship produces a thrust of 720 kN with a torque of 700 kNm, when it travels at a speed of 15 knots. In open water, this propeller rotating at the same speed, produces the same thrust at an advance speed of 12 knots, and develops the same torque at an advance speed of 12.3 knots. Then, the average of the wake fractions is ____________ (correct to two decimal places).

Hint:

Wake fraction depends on hull–propeller interaction: lower inflow speed behind hull means higher effective wake.

Answer 1

Correct Answer: 0.19

Wake fraction when thrust is matched: \[ w_T = 1 - \frac{V_a}{V_s} = 1 - \frac{12}{15} = 1 - 0.80 = 0.20 \] Wake fraction when torque is matched: \[ w_Q = 1 - \frac{12.3}{15} = 1 - 0.82 = 0.18 \] Average wake fraction: \[ w = \frac{w_T + w_Q}{2} = \frac{0.20 + 0.18}{2} = 0.19 \] Thus: \[ \boxed{0.19\ \text{to}\ 0.19} \]
Final Answer: 0.19