Fire Alarm NAC Voltage Drop After Adding Strobes
A school gym fire alarm upgrade: the existing 24 VDC NAC circuit has 16 horn/strobes at 70 mA each, wired with 14 AWG stranded, one-way length 520 ft. The contractor adds four synchronized strobes at 110 mA each at the far end. The panel NAC output is 24 VDC nominal, rated 2.5 A, and its low-battery voltage is 20.4 VDC. Appliances are listed for 16–33 VDC. The contractor plans to reuse the 14 AWG and set the panel output to 24 V. The battery calculation from the previous inspection used 24 h standby plus 5 min alarm; existing standby load is 0.35 A, alarm load excluding the NAC is 0.60 A, and the battery is 26 Ah. The AHJ requires voltage-drop and battery calculations at final inspection, and the school cannot lose the gym for more than one day.
1. Calculate the NAC voltage drop at low-battery voltage and determine whether the existing 14 AWG can support the added strobes.
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Total NAC current is 16 x 0.070 A plus 4 x 0.110 A, or 1.12 A plus 0.44 A, equal to 1.56 A. The circuit loop length is twice the one-way length: 2 x 520 ft = 1,040 ft. For 14 AWG stranded copper, use about 3.07 ohms per 1,000 ft. Loop resistance is 1,040/1,000 x 3.07 = 3.19 ohms. Voltage drop is 1.56 A x 3.19 ohms = 4.98 V. At the panel's low-battery voltage of 20.4 V, the end-of-line voltage is 20.4 - 4.98 = 15.42 V. That is below the appliance minimum of 16 V.
The calculation is conservative because it lumps all current at the far end, but it is the right check for a worst-case low-battery condition. The 14 AWG cannot be approved on the basis of a 24 V nominal panel setting. The panel must be evaluated at its minimum operating voltage, not just normal output.
2. What are the practical correction options, and which one fits the one-day shutdown constraint?
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Option one is to replace the 14 AWG with 12 AWG. For 12 AWG stranded copper at about 1.93 ohms per 1,000 ft, loop resistance is 2.01 ohms and drop is 1.56 x 2.01 = 3.13 V. At 20.4 V, the end voltage is 17.27 V, which is above 16 V and provides about 1.27 V of margin. Option two is to split the NAC into two circuits so each has less current and less drop. Option three is to reduce the load or relocate devices, but the added strobes are required.
Given the one-day shutdown, splitting the circuit may be faster if spare NAC outputs and wire paths exist. Replacing 520 ft of 14 AWG with 12 AWG is predictable but labor-intensive. The contractor should also verify synchronization compatibility and the panel's NAC rating of 2.5 A; 1.56 A is within rating. Before the shutdown, pre-pull the replacement cable or pre-wire the split circuit, pre-test the panel programming, and schedule the AHJ inspection for the same day.
3. Check the battery calculation and state whether the 26 Ah battery is adequate. What would change if the AHJ requires a 10% voltage margin?
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Standby energy is 0.35 A x 24 h = 8.4 Ah. Alarm energy is 0.60 A plus 1.56 A NAC current, or 2.16 A, for 5 minutes: 2.16 x 5/60 = 0.18 Ah. Total is 8.58 Ah. Applying a 1.25 aging/derating factor gives about 10.7 Ah. A 26 Ah battery has ample capacity for this load, assuming it is healthy, properly charged, and listed for the panel. The battery calculation should still be updated with the new NAC current and the actual appliance currents, not the old inspection values.
If the AHJ requires a 10% voltage margin, compare the end voltage with 20.4 V minus 2.04 V, or 18.36 V. The 12 AWG result of 17.27 V would not meet that stricter criterion. A 10 AWG replacement gives about 1.21 ohms per 1,000 ft, loop resistance 1.26 ohms, drop 1.96 V, and end voltage 18.44 V, only barely above 18.36 V. The safer design is to split the NAC or reduce the far-end load. Do not rely on a marginal calculation at final inspection.