Battery Set, Charger, and NAC Voltage Drop on a Tenant-Improvement Submittal
A tenant improvement adds 26 addressable photoelectric smoke detectors, 11 monitor modules and 4 relay modules to an existing 24 VDC addressable panel. The contractor's battery calculation shows a measured standby load of 0.410 A (panel quiescent plus every module and detector) and an alarm load of 2.90 A (panel plus every appliance and relay energized). The panel is listed for 24 hours of standby followed by 5 minutes of alarm, and the manufacturer's worksheet requires a 1.20 derating factor and a 1.25 aging factor. The contractor proposes reusing two 12 V, 12 Ah sealed lead-acid batteries already on site and states the existing 1.0 A charger is adequate. Separately, one NAC serves 16 appliances rated 16-33 V at 0.075 A each, with 300 ft of 18 AWG to the last device.
1. Is the proposed 12 Ah battery set adequate? Show the arithmetic and state what you would specify.
Read the worked answer
No - the 12 Ah set is short.
Standby energy: 0.410 A x 24 h = 9.84 Ah Alarm energy: 2.90 A x (5/60 h) = 0.242 Ah Subtotal: 10.08 Ah Apply derating: 10.08 x 1.20 = 12.10 Ah Apply aging: 12.10 x 1.25 = 15.1 Ah required
15.1 Ah > 12 Ah, so the proposal fails by roughly 25 percent. Note also that the two 12 V units are in series, so the set's capacity is 12 Ah, not 24 Ah - series adds voltage, not amp-hours. Do not propose paralleling small batteries to reach the number; use the panel manufacturer's listed battery (or listed battery cabinet) in the next standard size above 15.1 Ah, which is typically 18 Ah. Confirm the enclosure/cabinet is listed for the larger set, that the terminal voltage matches the panel's charging profile, and record the calculation basis on the submittal so the AHJ can check the factors used.
2. Is the 1.0 A charger adequate to restore the battery within the code-required recharge period?
Read the worked answer
It depends on the basis, and it is close enough that it must be documented.
Available charging current after carrying the standby load: 1.0 A - 0.410 A = 0.59 A
- Recharge basis using actual depleted energy (10.08 Ah): 10.08 / 0.59 = 17.1 h - acceptable.
- Recharge basis using the aged/derated capacity (15.1 Ah): 15.1 / 0.59 = 25.6 h - exceeds the 24-hour recharge requirement.
The defensible approach is the conservative one: size the charger so it restores the full derated capacity within 24 hours while supplying standby. That requires roughly 0.410 + (15.1 / 24) = 1.04 A, so a 1.0 A charger sits right at the line. Recommend a panel or charger rated at least 1.25 A, or obtain the manufacturer's published recharge table showing the 1.0 A unit meets the 24-hour requirement with the selected battery. Document which assumption was used; a charger that cannot restore the battery leaves the building on reduced standby after any outage.
3. Check the NAC voltage drop. Is the circuit acceptable?
Read the worked answer
Probably acceptable at nominal voltage, marginal at worst-case panel output - and the worst case is what governs.
Total load: 16 x 0.075 A = 1.20 A 18 AWG resistance ≈ 6.51 ohms per 1000 ft; round-trip length = 600 ft, so R = 3.91 ohms. Drop = 1.20 x 3.91 = 4.69 V (conservative - lumps all load at the far end).
- At 24.0 V nominal: 24.0 - 4.69 = 19.3 V - fine.
- At a battery-only panel output of about 20.4 V: 20.4 - 4.69 = 15.7 V - below the 16 V appliance minimum.
Voltage drop must be calculated at the panel's minimum rated output (battery-only), not at 24 V nominal, because that is the condition the circuit sees during the fire event that matters. Options: move to 16 AWG, split the circuit into two NACs, or redistribute appliances so the far-end current is lower. Recalculate using distributed loads if you want the credit, and put the calculation in the submittal.