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03 February 2024

195.NA August 2022 Q.9(b)

February 03, 2024 Posted by AK No comments

 The length of a ship is 7.6 times the breadth, while the breadth is 2.85 times the draught. The block coefficient is 0.69, prismatic coefficient 0.735, waterplane area coefficient 0.81 and the wetted surface area 7000 m2. The wetted surface area S is given by Denny's formula S = 1.7 Ld + (∇/d) Calculate: 

(a) displacement in tonne 

(b) area of immersed midship section 

(c) waterplane area


Given

L = 7.6b

b = 2.85d ; d = b/2.85

Cb = 0.69

Cp = 0.735

Cw = 0.81

S = 7000 m2


To find

1.Displacement

2. Am

3.Aw


Solution

We know that Cb = ∇ / (L x b x d)

 ∇ = Cb x L x b x d

    = 0.69 x 7.6b x b x b/2.85

  = 1.84b3  


S = 1.7 Ld + (∇/d) 

7000 = 1.7 x 7.6b x b/2.85 + (1.84b / (b/2.85)) 

         = 4.53b2   + 5.24b2

          = 9.77b2

       b = 26.76 m

L = 7.6 x 26.76 = 203.38 m

d = 26.76/2.85 = 9.389 m


∇ = 1.84 x b3

   = 35259.5


1. Displacement =  ∇ x density

                           = 35259.5 x 1.025

                           = 36140 t


2. Midship coefficient Cm = Cb/Cp

                                         = 0.69/0.735

                                         = 0.938


We know that Cm = Am / b x d

                       Am = Cm x b x d

                             = 0.938 x 26.76 x 9.389

                             = 235.9 m2


3. Cw = Aw / L x b

    Aw = Cw x L x b

          = 0.81 x 203.38 x 26.76

          = 4408m2


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