Warehouse NAC Addition: Battery Capacity and Voltage Drop
A 24 VDC addressable fire alarm control unit protects a 60,000 ft² warehouse in Sparks, Nevada. The owner adds 14 wall-mounted horn/strobes to one existing NAC for a new office mezzanine. Each appliance nameplate lists 95 mA at 24 V in alarm and a listed operating range of 16–33 V. The NAC is rated 2.5 A and is wired with 300 ft of 18 AWG stranded copper, one-way. The panel’s standby load, including new monitor modules, is 0.35 A; alarm load is 0.35 A plus the new appliances. Existing batteries are 24 V, 7 Ah, dated six years ago; the charger is rated 1.0 A. The installer proposes to reuse the batteries because the panel shows AC and no trouble. The AHJ requires 24 h standby plus 5 min alarm, with batteries derated to 80%.
1. Are the existing 7 Ah batteries adequate for the added load? Show the calculation and state the minimum replacement size.
Read the worked answer
No. The standby load is 0.35 A × 24 h = 8.4 Ah. The alarm load is 0.35 A + (14 × 0.095 A) = 1.68 A; for 5 min this is 1.68 A × (5/60) = 0.14 Ah. Total required is 8.54 Ah before derating. At 80% usable capacity, the battery must be at least 8.54 / 0.80 = 10.7 Ah. A 7 Ah battery provides only 7 × 0.80 = 5.6 Ah, so it fails. A 12 Ah battery provides 9.6 Ah, which still fails. The next common size, 18 Ah, provides 14.4 Ah and is adequate. The six-year age is also a problem: sealed lead-acid batteries are typically replaced at about five years or when they fail a load test, whichever comes first. Reusing them without a load test is not acceptable. Verify the 1.0 A charger can recharge the selected battery within the manufacturer’s and NFPA 72 requirements.
2. Calculate the end-of-line voltage at normal and low-battery conditions and determine whether the circuit is acceptable. What correction is needed?
Read the worked answer
Alarm current is 14 × 0.095 A = 1.33 A.18 AWG copper is about 6.385 Ω per 1,000 ft per conductor. The loop is 2 × 300 ft = 600 ft, so R = 600 × 6.385 / 1,000 = 3.83 Ω. Voltage drop = 1.33 A × 3.83 Ω = 5.09 V. At a 24 V source, the end of the circuit is about 18.9 V. At the low-battery condition of 20.4 V, it is about 15.3 V, below the 16 V minimum listed for the appliances. The circuit is therefore not acceptable. Correct by using a larger conductor, splitting the 14 appliances across two NACs, or locating a listed remote power supply nearer the load, then recalculating. A 14 AWG circuit would reduce resistance substantially, but the calculation must be redone with the actual conductor and appliance data. NFPA 72 requires notification appliances to operate within their listed voltage range at the end of the circuit under worst-case supply conditions.
3. What documentation and acceptance testing should be completed before the AHJ signs off?
Read the worked answer
Provide a battery calculation showing 24 h standby plus 5 min alarm with the 80% derate, the new battery size, charger rating, and recharge time. Provide a voltage-drop calculation for each modified NAC using the actual conductor size, length, and appliance current. Update the record of completion and the as-built drawings to show the new devices, circuit routing, and power-supply calculations. During acceptance testing, verify each appliance operates in alarm, measure voltage at the end of the circuit with the panel on battery, and confirm the panel annunciates the added devices correctly. Perform a battery load test and record the results. Label the panel and batteries, and leave the owner the manufacturer’s instructions plus the testing and maintenance schedule.