45.NA April 2019 Q.9

Length between perpendiculars - 140m, breadth - 18.5m, draught 8.10m in sea water of 1025 Kg/m3. Displacement - 17025 tonnes, face-pitch ratio of propeller - 0.673, diameter of propeller - 4. 8m. The results of the speed trial show that the true slip may be regarded as constant over the range 9 to 13 knots and equal to 30%. The wake fraction may be calculated from the equation   w = 0.5 Cb – 0.05 ; where W = wake fraction, Cb = block coefficient. If the fuel used is 20 tonnes/day at 13 knots and the consumption/day varies as the cube of the  speed of the vessel, determine the fuel used/day when the propeller rotates at 110 RPM.

If you are noticing some error in problems kindly comment below.Thanks

Given

L = 140 m
B = 18.5 m
d = 8.10 m
Δ = 17025 t
p = 0.673
D = 4.8 m
w = 0.5 Cb – 0.05
C2 = 20 t/day
V2 = 13 knots
Real slip = 0.3
N = 110 rpm

To find

1. Fuel used/day when the propeller rotates at 110 RPM, C1

Solution

Block coefficient Cb = Under water volume  / (L x B x d)

Under water volume ∇ = Δ / Density of water 
                                     = 17025 / 1.025
                                     = 16609.75 m3

Block coefficient Cb = 16609.75 / (140 x 18.5 x 8.1)
                                  = 0.792

Wake fraction w = 0.5 Cb – 0.05
                           = (0.5 x 0.792) - 0.05
                           = 0.346

Face pitch ratio p = Pitch (P) / Diameter (D)

                    Pitch , P = p x D

                                  = 0.673 x 4.8

                                  = 3.23 m

      Theoretical speed Vt = P x N x 60 / 1852

                                        = 3.23 x 110 x 60 / 1852

                                        = 11.51 knots

       Real slip = (Vt - Va) / Vt

               0.3 = (11.51 - Va) / 11.51

           3.453 = 11.51 - Va

       Speed of advance Va = 8.06 knots

       Wake fraction w = (V - Va) / V

                        0.346 = (V - 8.06) / V 

                    0.346 V = V - 8.06

       Speed of ship V = 12.32 knots (Take this as V1)

       It is given that consumption/day varies as the cube of the speed of the vessel

       So C ∝ V3

       C1 / C2 = (V1 / V2)3

       

       C1 / 20 = (12.32 /13)3

       

       C1 = 17.02 t/day