A ship of 355190 tonne displacement is 325 m long, 56m wide and floats in sea water of density 1025 kg/m3 at a draught of 22.4 m. The propeller has a diameter of 7.4 m, a pitch ratio of 0.85, and when rotating at 1.5 rev/s the real slip is 48.88% and the fuel consumption is 165 tonne per day. The taylor wake fraction Wt=0.5Cb-0.05.
Calculate
a) The speed in knots
b) The reduced speed at which the ship should travel if the fuel consumption in a voyage is to be halved.
C) the length of the voyage if the extra time on passage is six days when travelling at the reduced speed.
d) The amount of fuel required onboard before commencing on the voyage at the reduced speed.
Given
Δ = 355190 t
L = 325 m
B = 56 m
ρ = 1.025
d = 22.4 m
D = 7.4 m
p = 0.85
n = 1.5 rps
Sr = 0.4888
FC = 165 t/day
Solution
We know that ∇ = Δ / ρ
= 355190 / 1.025
= 346526.8 m3
block coefficient Cb = ∇ / (L x B x d)
= 346526.8 / (325 x 56 x 22.4)
= 0.85
Wt = 0.5Cb - 0.05
= (0.5 x 0.85) - 0.05
= 0.375
pitch ratio p = P / D
Pitch P = p x D
= 0.85 x 7.4
= 6.29 m
Theoretical speed Vt = P x n
= 6.29 x 1.5
= 9.435 m/s
Real slip Sr = (Vt - Va) / Vt
0.4888 = (9.435 - Va) / 9.435
Va = 4.823 m/s
a) Wake fraction Wt = (V - Va) / V
0.375 = (V - 4.823) / V
V = 7.717 m/s
Ship speed V = 7.717 x 3600 / 1852
= 15 knots
b) Voyage consumption is proportional to speed2
VC1 / VC2 =( V1 / V2)2
It is given that VC2 = 0.5 VC1
So VC1 / 0.5 VC1 = (15 / V2)2
So V2 = 10.61 knots
c) let dist travelled = D
At normal speed time taken = D / (15 x 24)
At reduced speed time taken = [D / (10.61 x 24)] + 6 (6 days extra given in qns
therefore D / (15 x 24) = [D / (10.61 x 24)] + 6
D = 5215 nm
d) At service speed fuel consumed = FC per day x distance / speed x 24
= 165 x 5215 / 15 x 24
= 2390 t
At reduced speed fuel consumed = 0.5 x 2390
= 1195 t
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