Question

A ship has a length of 150m and wetted area of 3000m2.A model of this ship 5m in length when towed in fresh water at 2m/s produces a resistance of 39.24N.The ship surface has skin resistance of 49.05 N/m2 at a velocity of 3 m/s and the velocity index is 1.85.the corresponding values for the model surface towed in fresh water are respectively 16.38N/m2,3m/s and 1.90. calculate:

(1).The corresponding speed of ship
(2).The shaft power required to propel the ship at this speed through sea water (p=1030kg/m3).Take the propeller efficiency as 75%.

Answer 1

Ans 1) We know, Froude number will remain same, therefore

V_m / (g L_m)^1/2 = V_s / (g L_s)^1/2

=> 2 / ( 9.81 x 5)^1/2 = Vs / (9.81 x 150)^1/2

=> Vs = 0.285 x 38.36

=> Vs = 10.93 m/s

Ans 2) We know,

R_m / R_s = $\rho$ V_m² x S_m / 2g Vs²x S_s

where, R_m and R_s = resistance of model and ship respectively

Rs and R_m = resistance of ship and model respectively

S_s and S_m = wetted perimeter of ship and model respectively

Scale = L_s / L_m

= 150 / 5 = 30

Wetted area for model = 3000 /30 = 100 m²

Putting values,

39.24 / R_s = 1030 x 2² x 100 / 10.93² x 3000 x 2 x 9.81

=> R_s = 1845.70 kN

Power required = Rs x Vs

= 20173.5 kW

Efficiency =75% or 0.75

=> Actual shaft power required = 20173..5 / 0.75

= 26898 kW