Question

A ship of length 125 m has a design speed of 25 knots (1 knot = 0.5144 m/s). A 5.0 m long geometrically similar model with wetted surface area of 4 m\(^2\) has a coefficient of residuary resistance of \(1.346 \times 10^{-3}\) at the corresponding speed. The ship's residuary resistance in kN (in sea water of density 1025 kg/m\(^3\)), and the model speed in knots (round off to the nearest integer) respectively are

• A. 285 and 5
• B. 17 and 5
• C. 285 and 1
• D. 17 and 1

Hint:

For scaling problems based on Froude number similarity: - Speeds scale with the square root of the length scale: \( V_s/V_m = \sqrt{\lambda} \). - Areas scale with the square of the length scale: \( S_s/S_m = \lambda^2 \). - Forces (like resistance) scale with the cube of the length scale: \( R_s/R_m = \lambda^3 \) (assuming same fluid density).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem involves ship model testing and scaling laws. To predict the resistance of a full-scale ship from a model test, we use Froude's law of similarity. This involves two main steps: 1. Determine the "corresponding speed" for the model test, which is the speed that gives the same Froude number for both model and ship.
2. Scale the residuary resistance from the model to the ship. Frictional resistance is scaled separately, but here we only need to deal with the residuary component.
Step 2: Key Formula or Approach:
Let subscripts 's' denote the ship and 'm' denote the model.
1. Froude Number Similarity: The corresponding speed is found by equating the Froude numbers: \( Fr_s = Fr_m \).
\[ \frac{V_s}{\sqrt{gL_s}} = \frac{V_m}{\sqrt{gL_m}} \implies V_m = V_s \sqrt{\frac{L_m}{L_s}} \] 2. Residuary Resistance Scaling: The coefficient of residuary resistance (\(C_R\)) is assumed to be the same for both the model and the ship at corresponding speeds: \( (C_R)_s = (C_R)_m \).
The resistance is given by \( R = C . \frac{1}{2}\rho S V^2 \), where S is the wetted surface area. So, the ship's residuary resistance is: \[ (R_R)_s = (C_R)_s . \frac{1}{2}\rho_s S_s V_s^2 \] 3. Scaling Wetted Surface Area: For geometrically similar bodies, the ratio of surface areas is the square of the ratio of lengths: \( \frac{S_s}{S_m} = \left(\frac{L_s}{L_m}\right)^2 \).
Step 3: Detailed Calculation:
Part 1: Calculate the model speed
- Ship speed, \(V_s = 25\) knots.
- Ship length, \(L_s = 125\) m.
- Model length, \(L_m = 5.0\) m.
The length scale ratio is \( \lambda = L_s/L_m = 125/5 = 25 \).
\[ V_m = V_s \sqrt{\frac{L_m}{L_s}} = V_s \frac{1}{\sqrt{\lambda}} = 25 \text{ knots} \times \frac{1}{\sqrt{25}} = 25 \times \frac{1}{5} = 5 \text{ knots} \] Rounding to the nearest integer, the model speed is 5 knots.
Part 2: Calculate the ship's residuary resistance
- Model residuary resistance coefficient, \( (C_R)_m = 1.346 \times 10^{-3} \).
- So, \( (C_R)_s = 1.346 \times 10^{-3} \).
- Density of sea water, \( \rho_s = 1025 \) kg/m\(^3\).
- Model wetted surface area, \( S_m = 4 \) m\(^2\).
- Ship speed, \(V_s = 25 \text{ knots} = 25 \times 0.5144 = 12.86 \) m/s.
First, scale the wetted surface area:
\[ S_s = S_m \left(\frac{L_s}{L_m}\right)^2 = 4 \text{ m}^2 \times (25)^2 = 4 \times 625 = 2500 \text{ m}^2 \] Now, calculate the ship's residuary resistance: \[ (R_R)_s = (C_R)_s \times \frac{1}{2} \rho_s S_s V_s^2 \] \[ (R_R)_s = (1.346 \times 10^{-3}) \times \frac{1}{2} \times 1025 \times 2500 \times (12.86)^2 \] \[ (R_R)_s \approx (1.346 \times 10^{-3}) \times 0.5 \times 1025 \times 2500 \times 165.3796 \] \[ (R_R)_s \approx 284567 \text{ N} \] Convert to kN: \[ (R_R)_s = 284.567 \text{ kN} \] Rounding to the nearest integer, the ship's residuary resistance is 285 kN. Step 4: Final Answer:
The ship's residuary resistance is 285 kN and the model speed is 5 knots. This corresponds to option (A).