139.NA August 2021 Q.8(b)

A ship of 6000 tonne displacement has a wetted surface area of 2500 m2 and a speed of 15 knots.

i. Calculate the corresponding speed and wetted surface area of as similar ship of 2000 tonne displacement.

ii. If the skin resistance is of the form R = 0.45 S V^1.83 N; find the resistance of the 6000 tonne Ship

IF YOU FIND ANY ERROR KINDLY COMMENT BELOW

Given

Δ1 = 6000 t

S1 = 2500 m2

V1 = 15 knots

Δ2 = 2000 t

R = 0.45 S V^1.83 N

To find 

1. V2 and S2

2.R

Solution

1. We know that S ∝ L² ----------------------(1)

   Δ  ∝ L³ --------------------------(2) 

From (1) -------------- L ∝ S^1/2      -------------------------(3)

From (2) -------------- L ∝ Δ^1/3      -------------------------(4) 

Combining (3) and (4)

Δ^1/3 ∝ S^1/2

(Δ2 / Δ1)^1/3 = (S2 / S1)^1/2

(2000 / 6000)^1/3 = (S2 / 2500)^1/2

0.693 = (S2 / 2500)^1/2

0.693 = S2^1/2  / 50  

S2^1/2 = 50 x 0.693

        

S2^1/2 = 34.65

S2 = 1200.62 m2

We know that S ∝ L² ----------------------(1)

V ∝ L^1/2 -------------------------(2)   

From (1) -------------- L ∝ S^1/2      -------------------------(3)

From (2) -------------- L ∝ V²      -------------------------(4) 

Combining (3) and (4)

V² ∝ S^1/2       

V ∝ S^1/4

V2 / V1 = (S2 / S1)^1/4

V2 / 15 = (1200.62 / 2500)^1/4

V2 / 15 = 0.8324

V2 = 12.48 knots

2. It is given that , R = 0.45 S V^1.83 N

Substitute S1 and V1 in above equation 

R = 0.45 x 2500 x 15^1.83 N

R = 159734.79 N

    = 159.735 kN