Given: propeller diameter D = 6 m, pitch ratio = 0.95, so pitch P = 0.95 x 6 = 5.7 m.

Propeller speed = 1.8 rev/s.

Real slip = 34% = 0.34, Apparent slip = 7% = 0.07.

Shaft power = 10000 kW, SFC = 0.22 kg/kW-hr.

Pitch speed (theoretical speed of advance) = P x rev/s = 5.7 x 1.8 = 10.26 m/s.

Apparent slip = (Pitch speed - V)/Pitch speed, where V is the ship speed.

So V = Pitch speed x (1 - apparent slip) = 10.26 x (1 - 0.07) = 10.26 x 0.93 = 9.542 m/s.

Ship speed in knots = 9.542 x 3600/1852 = 34351/1852 = 18.55 knots.

Answer: Ship speed = 18.55 knots.

Real slip = (Pitch speed - Va)/Pitch speed, where Va is the speed of advance.

Va = Pitch speed x (1 - real slip) = 10.26 x (1 - 0.34) = 10.26 x 0.66 = 6.772 m/s.

Taylor wake fraction w = (V - Va)/V = (9.542 - 6.772)/9.542 = 2.770/9.542 = 0.2903.

Answer: Taylor wake fraction = 0.290 (29%).

Consumption varies as the cube of speed. To halve the consumption, the speed must be reduced by the cube root of 0.5.

Reduced speed = V x (0.5)^(1/3) = 18.55 x 0.7937 = 14.72 knots.

Answer: Reduced speed = 14.72 knots.

At normal speed 18.55 knots, the daily consumption = shaft power x 24 x SFC = 10000 x 24 x 0.22 = 52800 kg/day = 52.8 tonne/day.

At reduced speed, consumption per day = 52.8 x (0.5) = 26.4 tonne/day (since speed cubed halved).

Let the normal voyage time be T days. Distance = 18.55 x 24 x T (nautical miles).

At reduced speed 14.72 knots, the time taken = Distance/(14.72 x 24) = (18.55 x 24 x T)/(14.72 x 24) = 18.55/14.72 x T = 1.2602 T days.

The voyage takes 3 days longer, so 1.2602 T = T + 3, giving 0.2602 T = 3, T = 11.53 days.

Voyage consumption at reduced speed = 26.4 x 1.2602 x 11.53 = 26.4 x 14.53 = 383.6 tonne.

Answer: Voyage consumption at reduced speed = 383.6 tonne.

This is the same as (d), the total fuel for the voyage at reduced speed.

Answer: 383.6 tonne of fuel required for the voyage at reduced speed.