Fuel Endurance Decision Before Transit
You command a destroyer with 260 tons of F76 fuel remaining out of a 600-ton capacity. Standing orders require arrival in port with at least 10 percent of capacity plus 24 hours of emergency generator fuel at 1.0 ton per hour. The distance to port is 720 nautical miles. At 18 knots the main engines burn 3.2 tons per hour; at 21 knots, 4.6; at 24 knots, 6.0. Ship's service generators burn 1.0 ton per hour at all speeds. The navigator proposes 21 knots to meet the operational schedule. The chief engineer wants 18 knots. A weather route adds no distance. There is no opportunity to refuel before arrival. The executive officer asks you for a fuel-endurance recommendation that protects the reserve and states the margin.
1. Calculate whether the ship can transit at 21 knots and still meet the reserve.
Read the worked answer
Required reserve = 10% of 600 = 60 t, plus 24 h at 1.0 t/h = 24 t, total 84 t.
Usable fuel = 260 - 84 = 176 t.
At 21 kn: time = 720 / 21 = 34.29 h. Main burn = 4.6 x 34.29 = 157.7 t. Generator burn = 1.0 x 34.29 = 34.3 t. Total burn = 192.0 t. Arrival fuel = 260 - 192.0 = 68.0 t.
This is 16.0 t below the 84 t requirement, so 21 knots is not acceptable on fuel alone, before any weather or engineering margin.
2. Which speed among 18, 21, and 24 knots meets the reserve? Show the comparison.
Read the worked answer
At 18 kn: time = 720 / 18 = 40 h; main = 3.2 x 40 = 128 t; generator = 40 t; total = 168 t; arrival = 92 t; margin over 84 t = 8 t.
At 21 kn: total = 192 t; arrival = 68 t; short 16 t.
At 24 kn: time = 30 h; main = 6.0 x 30 = 180 t; generator = 30 t; total = 210 t; arrival = 50 t; short 34 t.
Only 18 knots meets the stated reserve. The margin is 8 t, about 1.9 hours at the 18-knot combined burn of 4.2 t/h, so it is thin. I would recommend 18 knots unless the operational commander accepts a reduced reserve in writing or confirms a refueling opportunity.
3. The commander directs 24 knots for four hours to avoid a squall, then 18 knots. Recalculate and state the risk.
Read the worked answer
First four hours at 24 kn: distance = 96 nm; burn = (6.0 + 1.0) x 4 = 28 t.
Remaining distance = 720 - 96 = 624 nm. At 18 kn: time = 624 / 18 = 34.67 h. Burn = (3.2 + 1.0) x 34.67 = 145.6 t. Total burn = 173.6 t. Arrival fuel = 260 - 173.6 = 86.4 t. Required reserve = 84 t. Margin = 2.4 t, about 34 minutes at 4.2 t/h.
The plan is legal on paper but has almost no allowance for current, generator load, or weather. I would report the 2.4 t margin, recommend cutting the 24-knot period to the minimum needed, and require a fuel-status report every hour. If the squall can be avoided at 21 knots or by course change, the margin improves.