Medical Office Tenant Improvement: Battery and NAC Voltage-Drop Check
A 3-story, 28,000 sq ft medical office building has a 24 V addressable FACP with existing 10 Ah batteries and a 1.5 A charger. The verified standby load is 0.42 A. The existing alarm load is 2.1 A. A tenant improvement adds 22 horn/strobes at 0.145 A each and 4 control relay modules at 0.025 A each. The contractor wants to reuse the existing batteries and charger and add the new devices to two existing NACs. One NAC is 450 ft of 14 AWG with 12 new horn/strobes; the other is 280 ft of 14 AWG with 10. The FACP is rated for 6 A total alarm current and each NAC is rated 2 A. The owner wants a 24-hour standby system with 5 minutes of alarm. The contractor's submittal says “10 Ah batteries are adequate because the existing system passed its annual test.”
1. Calculate the required battery capacity using 24-hour standby, 5-minute alarm, a 20% aging factor, and the 110% charger criterion. Is the existing 10 Ah battery adequate?
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New alarm load: existing 2.1 A + 22 × 0.145 A = 3.19 A for horn/strobes + 4 × 0.025 A = 0.10 A for relays = 5.39 A. Standby energy = 0.42 A × 24 h = 10.08 Ah. Alarm energy = 5.39 A × (5/60 h) = 0.449 Ah. Subtotal = 10.53 Ah. Apply 20% aging: 10.53 × 1.20 = 12.64 Ah. The 10 Ah battery is not adequate. The next standard size is typically 18 Ah (a 14 Ah may be marginal and must be checked against the exact manufacturer's derating). Also check the charger: the existing 1.5 A charger may not restore the larger battery within the required time; the designer must verify charger rating and calculate recharge. The fact that the old system passed an annual test is irrelevant because the load changed. Show the calculation in the submittal with the actual appliance currents from the manufacturer's data.
2. Check the two NACs. NAC 1 is 450 ft of 14 AWG with 12 new horn/strobes at 0.145 A each; NAC 2 is 280 ft of 14 AWG with 10. Each NAC is rated 2 A. What problems do you see?
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Current is within the 2 A NAC rating: NAC 1 = 12 × 0.145 = 1.74 A; NAC 2 = 10 × 0.145 = 1.45 A. But the panel total is 5.39 A against a 6 A rating, leaving only 0.61 A margin; check whether existing loads, strobe inrush, and relay currents are fully included. Voltage drop is the bigger problem. 14 AWG is about 3.07 Ω/1,000 ft. NAC 1 has 900 ft of conductor, about 2.76 Ω. At 1.74 A the drop is about 4.8 V. On a 24 V NAC that is about 20%, well beyond the usual 10% maximum, and the end devices may not operate. NAC 2: 560 ft × 3.07/1,000 = 1.72 Ω; at 1.45 A drop is about 2.5 V (10.4%), also marginal. Options are 12 AWG, splitting the circuits, or a remote power supply closer to the load. Recalculate using the actual end-of-line voltage and the appliance minimum operating voltage, not just nominal 24 V.
3. What should the contractor do about the FACP's 6 A alarm rating and the existing charger?
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The 5.39 A calculated alarm load is only 0.61 A below the 6 A panel rating, so any omitted device, strobe inrush, or future tenant load could overload it. The designer should produce a load schedule that accounts for all appliances, including existing devices and any synchronized strobe currents, and either replace the panel with a higher-rated one or use a remote power supply to remove NAC load from the panel. For the charger, a 1.5 A charger may be inadequate for an 18 Ah battery; verify against NFPA 72 and the manufacturer's recharge formula and the 110% criterion. Also confirm the battery enclosure can accept the larger battery and that the charger is listed for the battery type. Do not simply install a larger battery on an undersized charger; that creates a false sense of compliance and can fail the LAFD acceptance test.