The Admiralty coefficient is defined as C = Delta^(2/3) V^3 / P, where Delta is displacement (tonnes), V ship speed, and P shaft power.

Chosen the parent ship: Delta=12000 t. Compute C at each trial speed.

Speed V=10: numerator = 12000^(2/3) x 10^3. 12000^(2/3)=524, 10^3=1000, so 524,000/880 = 595.

V=11: 524x1331=697,444; /1155 = 604.

V=12: 524x1728=905,472; /1520 = 596.

V=13: 524x2197=1,151,228; /2010 = 573.

V=14: 524x2744=1,437,856; /2670 = 538.

V=15: 524x3375=1,768,500; /3600 = 491.

Plotting C against V gives a curve that is roughly flat (595-604) at low speed and falls progressively at higher speed (to about 490 at 15 kn), because wave-making resistance and other power terms rise faster than V^3 at high Froude numbers. The curve demonstrates that the constant-Admiralty-coefficient assumption holds only near moderate speeds.

For geometrically similar ships the displacement scales as the cube of length:

Delta140 = 12000 x (140/120)^3 = 12000 x 1.5876 = 19051 t.

So Delta140^(2/3) = 19051^(2/3) = 714.

Corresponding speed: for similarity, V2/V1 = sqrt(L2/L1) = sqrt(1.1667) = 1.0801. For the 140 m ship running at 14 knots, the corresponding speed of the parent (120 m) ship is 14/1.0801 = 12.96 knots, which lies between the 13-knot trial point (C=573) - read C is about 570 from the curve.

Using the Admiralty coefficient at that point, C=570:

P140 = Delta140^(2/3) x V^3 / C = 714 x 14^3 / 570 = 714 x 2744 / 570 = 1,959,216 / 570 = 3437 kW.

Answer: the shaft power required is about 3440 kW (approximately 3.4 MW).