Cruise Ship Displacement and Evacuation Time Estimation

1. Displacement Calculation

To keep the cruise ship afloat, the weight of the water displaced by the ship must be equal to the weight of the ship itself. This follows from Archimedes' principle:

The weight of the ship is given as:

$$ \text{Weight of the ship} = 200,000 \, \text{tons} = 200,000 \times 1,000 \, \text{kg} = 200,000,000 \, \text{kg} $$

To stay afloat, the ship must displace an equivalent mass of seawater. Assuming the density of seawater is approximately:

$$ \text{Density of seawater} = 1,025 \, \text{kg/m}^3 $$

The volume of water displaced $ V $ can then be calculated as:

$$ V = \frac{\text{Mass of ship}}{\text{Density of seawater}} = \frac{200,000,000 \, \text{kg}}{1,025 \, \text{kg/m}^3} \approx 195,122 \, \text{m}^3 $$

Therefore, the ship must displace approximately 195,122 cubic meters of seawater to remain afloat.

2. Hypothetical Evacuation Time if Hull is Breached

Now, let us hypothesize the time available for evacuation if the hull is breached at the stern, near the ballast.

For simplicity, let's assume the following:

The breach is close to the waterline.
The breach has an area of $ A = 1 \, \text{m}^2 $.
The rate of water entering is determined by Torricelli’s theorem: $ v = \sqrt{2gh} $, where:
- $ g $ is the acceleration due to gravity (approx. $ 9.8 \, \text{m/s}^2 $)
- $ h $ is the height of the water column above the breach (let’s assume $ h = 5 \, \text{m} $)

The initial inflow velocity $ v $ is:

$$ v = \sqrt{2 \cdot 9.8 \cdot 5} \approx 9.9 \, \text{m/s} $$

Thus, the inflow rate $ Q $ (volume per second) through the breach is:

$$ Q = A \cdot v = 1 \, \text{m}^2 \times 9.9 \, \text{m/s} = 9.9 \, \text{m}^3/\text{s} $$

Given the total volume that must remain undisturbed to keep the ship afloat is $ 195,122 \, \text{m}^3 $, let’s hypothesize the time $ t $ to reach critical displacement if the rate of flooding remains constant:

$$ t = \frac{195,122 \, \text{m}^3}{9.9 \, \text{m}^3/\text{s}} \approx 19,719 \, \text{s} $$

Converting to hours:

$$ t \approx \frac{19,719}{3600} \approx 5.5 \, \text{hours} $$

Conclusion

This gives approximately 5.5 hours for evacuation under idealized conditions. In a real scenario, structural factors and safety measures would affect this estimate, but it provides a general idea of the time available in such an emergency.