92. NA March 2020 Q.6(b)

 The length of a ship is 18 times the draught, while the breadth is 2.1 times the draught. At the load water plane, the water plane area co-efficient is 0.83 and the difference between the TPC in sea water and the TPC in fresh water is 0.7. Determine the length of the ship and TPC in fresh water.

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Given 

Take Length of ship = L
Draught of ship = L/18
Breadth of ship B = 2.1 x Draught
                          = 2.1 x L/18
Cw = 0.83
TPC Sw - TPC Fw = 0.7

To find

1.Length of ship L
2.TPC Fw

Solution

TPC = (Area of water plane x Density of water) / 100

TPC Sw = (Aw x 1.025) / 100
TPC Fw = (Aw x 1.000) / 100

It is given that
TPC Sw - TPC Fw = 0.7

(Aw x 1.025) / 100 - (Aw x 1.000) / 100 = 0.7
Aw (1.025 - 1.000) = 70
Aw = 2800 m2

1. As we know that
    Water plane area co-efficient = Area of water plane / (L x B)

    0.83 = 2800 /(L x 2.1 x L/18)
    0.0968 L x L = 2800
    L x L = 28915.66
    L = √ 28915.66

    Length of ship L = 170.0 m 

2. TPC Fw = (Aw x 1.000) / 100
                  = (2800 x 1.000) / 100

    TPC in Fresh water = 28