(a) Correcting a Negative GM

A negative metacentric height (GM) means the metacentre (M) lies below the centre of gravity (G), making the ship unstable. If the ship is heeled by an external force, it will continue to heel further instead of returning to the upright position.

To correct a negative GM:

1. Lower the Centre of Gravity (G):
   • Transfer weights from higher positions to lower positions in the ship.
   • Load heavy ballast into double-bottom or low ballast tanks.
   • Remove or lower heavy weights carried on deck or in high spaces.
2. Avoid Raising G Further:
   • Minimise loading of heavy cargo on deck.
   • Avoid lifting heavy loads with ship's cranes unless necessary.
3. Reduce Free Surface Effect:
   • Press up slack tanks completely or empty them where possible.
   • This reduces the virtual rise of G caused by free surface.
4. Check Stability Calculations:
   • Recalculate the ship's GM after ballast transfer or cargo redistribution.
   • Ensure a positive GM is achieved before sailing.

Result:

When the centre of gravity is lowered below the metacentre, GM becomes positive, restoring the ship's stability.

(b) Shaft Power Required for a Similar 140 m Ship

Given Data

Step 1: Effective Power of 120 m Ship

$$EP = PC \times SP$$

$$EP = 0.6 \times 4100 = 2460\;kW$$

Step 2: Total Resistance

Since,

$$EP = R_T \times V$$

Convert 15 knots to m/s:

$$V = 15 \times 0.514 = 7.71\;m/s$$

$$R_T = \frac{2460}{7.71} = 319.066\;kN$$

Step 3: Frictional and Residual Resistance

Frictional resistance:

$$R_F = 0.55R_T$$

$$R_F = 0.55 \times 319.066 = 175.486\;kN$$

Residual resistance:

$$R_R = R_T - R_F$$

$$R_R = 319.066 - 175.486 = 143.57\;kN$$

Step 4: Scale Wetted Surface Area

For similar ships,

$$S \propto L^2$$

$$S_2 = S_1\left(\frac{L_2}{L_1}\right)^2$$

$$S_2 = 3000\left(\frac{140}{120}\right)^2$$

$$S_2 = 4083.33\;m^2$$

Step 5: Corresponding Speed

For similar ships,

$$V \propto \sqrt{L}$$

$$V_2 = 15\sqrt{\frac{140}{120}}$$

$$V_2 = 16.20\;knots$$

Step 6: Frictional Resistance of 140 m Ship

$$R_F = fSV^n$$

$$R_F = 0.42 \times 4083.33 \times 16.20^{1.825}$$

$$R_F = 227.984\;kN$$

Step 7: Residual Resistance of 140 m Ship

For similar ships,

$$R_R \propto L^3$$

$$R_{R2} = R_{R1}\left(\frac{L_2}{L_1}\right)^3$$

$$R_{R2} = 143.57\left(\frac{140}{120}\right)^3$$

$$R_{R2} = 227.984\;kN$$

Step 8: Total Resistance

$$R_T = R_F + R_R$$

$$R_T = 276.457 + 227.984 = 504.441\;kN$$

Step 9: Effective Power

Convert 16.20 knots to m/s:

$$V = 16.20 \times 0.514 = 8.33\;m/s$$

$$EP = R_T \times V$$

$$EP = 504.441 \times 8.33$$

$$EP = 4200.346\;kW$$

Step 10: Shaft Power

$$SP = \frac{EP}{PC}$$

$$SP = \frac{4200.346}{0.6}$$

$$\boxed{SP \approx 7000.57\;kW}$$

Answer

The shaft power required for the similar 140 m ship at the corresponding speed is approximately

7001 kW (≈ 7.0 MW).