GMDSS Reserve Battery That Passes on Paper and Fails in a Cold Compartment
The GMDSS reserve source of energy on a 24 V DC vessel is a single lead-acid battery bank: 12 cells, nameplate 120 Ah at the 20-hour rate, installed in a compartment that reaches 0 °C in winter, now six years old. The ship's GMDSS documents require the reserve to carry all GMDSS loads for 6 hours, and the approved load figure in the manual is 14 A at 24 V. A capacity test at 20 °C, discharging at 14 A into a load bank, delivered 96 Ah before the bank reached its end voltage of 21.0 V. One cell's electrolyte reads specific gravity 1.150 against 1.225–1.230 in the others, and its voltage under load is 1.85 V against 1.98–2.00 V. During the same visit, the technician notes that the masthead VHF dropped out during a blackout drill: the transceiver terminals read 19.8 V while the battery terminals read 23.5 V, through 2 × 12 m of 6 mm² cable.
1. Does the bank meet the 6-hour reserve requirement? Show the arithmetic and the reasoning behind each allowance.
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Required energy: 14 A × 6 h = 84 Ah. That is what the reserve must deliver in service, not what the nameplate says.
What the bank actually delivers:
- At 20 °C the test gave 96 Ah, i.e. 96/84 = 114% of requirement. On that number alone it passes - but only just, and only at room temperature.
- Temperature: a lead-acid battery loses roughly 15–25% of its capacity at 0 °C compared with 20 °C. Take 20%: 96 × 0.8 ≈ 77 Ah, below the 84 Ah requirement. In the compartment this vessel actually has, the reserve fails.
- Age and the failed cell: six years old, with one cell at SG 1.150 and 1.85 V under load, is end of life regardless of arithmetic. The string is limited by its weakest cell.
- Rating basis: 120 Ah at the 20-hour rate means a 6 A discharge. The reserve duty is 14 A - a much heavier rate - and available capacity always falls as the discharge rate rises. Comparing a 20-hour rating with a 6-hour duty flatters the battery.
Conclusion: the bank does not meet the requirement under the conditions in which it would be used. Replace it, and size the replacement on its 6-hour-rate capacity, with margin for ageing and for 0 °C, rather than on a 20-hour nameplate figure.
2. Explain the VHF dropout at 19.8 V and say what you would check and change.
Read the worked answer
Calculate what the drop should be, then compare with what was measured.
Two 12 m runs = 24 m of conductor, 6 mm² copper, resistivity ≈ 0.0175 Ω·mm²/m:
R = 0.0175 × 24 / 6 = 0.07 Ω
At the VHF's transmit current of about 7 A, the expected drop is 0.07 × 7 ≈ 0.5 V. The measured drop is 23.5 − 19.8 = 3.7 V, roughly seven times the calculated value.
That is not a cable-gauge problem - 6 mm² is adequate for this load over this run. It is a resistance problem: a corroded or loose terminal, a green crimp, a fuse holder, a junction box, or a return path running through a bonding conductor with a bad joint. Inspect and feel for heat at each joint, then measure the voltage drop across each terminal while the load is on; the guilty connection will show most of the 3.7 V.
The symptom follows directly: the VHF has a low-voltage cutout, so it shut down. Below about 21.5 V the MF/HF transmitter will also fold back. With 22 A of transmit current, the same faulty path that drops 3.7 V at 7 A would drop far more, cutting range exactly when it matters. Fix the joints, re-measure the drop under full load, and only then accept the installation.
3. What do the specific gravity and cell-voltage readings mean, and what is the correct corrective action, including the charging system?
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One cell at SG 1.150 against 1.225–1.230 in its neighbours, and 1.85 V against 1.98–2.00 V under load, is a failed cell - most likely a partial short or heavy sulfation. It is not a watering problem and it will not be fixed by topping up.
An equalising (boost) charge can recover a stratified or chronically undercharged cell, so it is worth one attempt on a bank whose history is unknown. It will not recover a cell sitting 0.08 SG below the rest, and in a six-year-old string the remaining cells are near end of life anyway. Replacing one cell in an old series string is a short-lived patch: the string charges and discharges as a unit, the old cells limit the new one, and the new cell is usually destroyed within a season.
Correct action:
- Replace the bank as a complete unit, sized on the 6-hour rate at 0 °C with margin.
- Verify the charging system. For lead-acid the float voltage should sit near 2.25 V per cell - about 27.0 V for a 24 V bank at 25 °C - with temperature compensation, so somewhat higher in a 0 °C compartment. Check the charger's current limit and confirm the voltage sense point is at the battery terminals, not at the charger output; sensing at the charger hides exactly the kind of cable drop that caused the VHF dropout.
- Record the capacity test, the failing cell readings, the replacement and the charging check in the radio log, and inform the master and whoever is responsible for GMDSS maintenance that the reserve source was below requirement.