Choosing passage speed against a bunker reserve
MV Hooghly Star is on a 4,600 nm passage with 420 t of VLSFO on board, of which 12 t is unusable sludge and retention. At 15.0 kn she burns 30 t/day, of which 70% (21 t/day) is hull-related and varies as the cube of speed through the water, while 9 t/day is fixed machinery and auxiliary load. The charter party requires arrival at the discharge port with at least 45 t of fuel remaining, and there is no bunkering opportunity on the route. Routeing advice gives one 24-hour period of adverse current averaging 1.0 kn on the bow, with head seas that add 8% to consumption during that period if speed is maintained; the rest of the passage is treated as still water in fine weather. At 13.5, 14.0 or 15.0 kn the ship can meet the laycan, but slower speeds cost the charterer time. Company policy states that no speed instruction may be accepted if the reserve falls below the charter party figure.
1. Calculate whether the passage can be made at 15.0 kn within the reserve.
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Usable fuel = 420 − 12 = 408 t; the reserve must be at least 45 t, so the passage may burn no more than 408 − 45 = 363 t.
At 15.0 kn: during the 24 hours of adverse current the advance over the ground is (15.0 − 1.0) × 24 = 336 nm. Remaining distance = 4,600 − 336 = 4,264 nm, at 15.0 kn = 284.3 h = 11.85 days; total time = 12.85 days. Fuel = 30 × 12.85 = 385.4 t, plus the head-sea penalty for that one day (30 × 0.08 = 2.4 t) = 387.8 t. Arrival fuel = 408 − 387.8 = 20.2 t, which is about 25 t short of the required reserve.
So 15.0 kn is not acceptable, even though the ship would physically arrive with fuel in the tanks. The reserve, not the tank contents, is the governing figure.
2. Calculate the arrival fuel at 14.0 kn and at 13.5 kn, and identify the fastest compliant speed.
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At 14.0 kn: hull component = 21 × (14.0/15.0)³ = 21 × 0.813 = 17.07 t/day; total = 17.07 + 9 = 26.07 t/day. Current day advance = 13.0 × 24 = 312 nm; remaining 4,288 nm at 14.0 kn = 306.3 h = 12.76 days; total 13.76 days. Fuel = 26.07 × 13.76 = 358.7 t, plus 26.07 × 0.08 = 2.1 t of head-sea penalty = 360.8 t. Arrival = 408 − 360.8 = 47.2 t - compliant, but by only about 2 t.
At 13.5 kn: hull component = 21 × (13.5/15.0)³ = 21 × 0.729 = 15.31 t/day; total = 24.31 t/day. Current day advance = 12.5 × 24 = 300 nm; remaining 4,300 nm at 13.5 kn = 318.5 h = 13.27 days; total 14.27 days. Fuel = 24.31 × 14.27 = 346.9 t, plus 1.9 t = 348.8 t. Arrival = 59.2 t.
The fastest nominally compliant speed is 14.0 kn.
3. Make the recommendation to the master and set out the factors that should override a purely arithmetic answer.
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Recommendation. Order 13.5 kn, not 14.0 kn. The 14.0 kn case leaves 2.2 t of margin on a 408 t budget - under 0.6% - which is smaller than the uncertainty in the 8% head-sea allowance, the fuel density and the day-to-day weather. The cube law also cuts the wrong way: any unexpected foul weather at 14 kn costs disproportionately more fuel than at 13.5 kn.
Factors beyond the arithmetic. First, the charter party speed and consumption warranty: if the ship is warranted to make 14 kn on a stated consumption, a voluntary slow speed may trigger a performance claim or an off-hire argument, so the master should notify the charterer and the company in writing and cite the reserve clause. Second, ETA obligations at the discharge port and any notices the agent must give. Third, the fuel itself: the last parcel's quality and whether it can be burned cleanly at reduced load. Fourth, the human factor: a slow passage still needs the same watchkeeping and rest hours, so the saving must not be taken out of the crew. Record the calculation and the decision in the log and the noon reports.