Use Bernoulli equation: It can act on enormous scales such as naval vessels. Wat

Question

Use Bernoulli equation: It can act on enormous scales such as

naval vessels. Water is forced to flow faster around the ship and under it, especially when the ship is

operating in shallow water or around other vessels creating apparent suction toward a nearby surface.

The former is called the squat effect and must be accounted for when estimating the clearance under

the keel of a fast moving ship. On the 1st of November 2009, the largest ship cruise ship in the world at the time, the MS Oasis of the Seas, used the squat effect to increase clearance by 0.3 m while passing under a bridge in Denmark (total clearance

was only 0.6 m). To achieve this, she had to travel at a speed of 20 knots in calm weather. What is t

depth of the shipping channel at that point? The ship has a beam (width) of 47 m and draught,

from keel to waterline, of 9.3 m. Assume that the ship's cross section is a rectangular box 230 m long

at waterline. Assume that 10% of the displaced water flows under the ship (the other half forced to the

sides) and pressure on the bottom is uniform. Neglect the end effects and dimensions.

It can act on enormous scales such as naval vessels. Water is forced to flow faster around the ship and under it, especially when the ship is operating in shallow water or around other vessels creating apparent suction toward a nearby surface. The former is called the squat effect and must be accounted for when estimating the clearance under the keel of a fast moving ship. On the 1st of November 2009, the largest ship cruise ship in the world at the time, the MS Oasis of the Seas, used the squat effect to increase clearance by 0.3 m while passing under a bridge in Denmark (total clearance was only 0.6 m). To achieve this, she had to travel at a speed of 20 knots in calm weather. What is t depth of the shipping channel at that point? The ship has a beam (width) of 47 m and draught, from keel to waterline, of 9.3 m. Assume that the ship's cross section is a rectangular box 230 m long at waterline. Assume that 10% of the displaced water flows under the ship (the other half forced to the sides) and pressure on the bottom is uniform. Neglect the end effects and dimensions.

Explanation / Answer

The bottom of the ship can be assumed to be a rectangular cross-section of 230x47 m...

So the pressure of the water under the vessel is draught*density of water*gravity

Ignore gravity,

So, the bernoulli's equation stands as,

P1+0.5*d*v1^2= P2+ 0.5*d*v2^2 (d is the density of water)

P1= h1*d*g (h1 is the initial draught)

P2= h2*d*g (h2 is the final draught)

ignore v1 in comparison to v2. So the final equation stands as,

v2^2=2(h1-h2)g

The rate of volume of water displaced is, Whv (W is the width, D is the draught, v is the velocity of the ship)

Now 10% of this volume is force under the ship, so net volumetric flow under the ship=0.1*Whv

So, if I assume a rectangular cross-section under the ship, the area of the cross-section would be W*a (where a is the depth of the shipping channel)

the volumetric flow through this cross-section is Wa(v2)

And this is equal to the water displaced from above

So, Wa(v2)=0.1*Whv

v2=(0.1h/a)*v

Substitute this in the bernoulli's equation, to get the value of a,

h1-h2=0.3 m

h=draught=9.3 m

20 knots=10 m/s

The answer would come to 3.83 m......

If you have any doubts plz comment, i will reply asap.....But plz do rate....it took a lot of effort...

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