24 VAC Damper Actuators on a Long Control Run
A 1960s Dover school is replacing pneumatic damper actuators with 24 VAC electric actuators. The control transformer is in a mechanical room and feeds three actuators plus a 24 VAC controller. The longest run is 280 ft one-way of existing 18/2 stranded cable. Each actuator is rated 24 VAC, 0.32 A running, 1.1 A inrush, and a minimum operating voltage of 21.6 V. The transformer is 75 VA with a 120 V primary. At commissioning, the panel output is 24.1 V. At the farthest actuator, idle voltage is 22.9 V; when that actuator starts, voltage at its terminals falls to 17.8 V, the actuator chatters, and a second actuator sometimes fails to open fully. The contractor proposes replacing the 75 VA transformer with a 100 VA unit and leaving the 18/2 cable in place. The owner wants a reliable fix before ceiling close-in.
1. Calculate the steady-state and inrush voltage drop on the 280 ft 18/2 run and the transformer loading. Does the 100 VA transformer solve the problem? Show your assumptions.
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Assume 18 AWG copper is about 6.4 ohms per 1,000 ft. The loop is 560 ft, so R = 6.4 x 0.560 = 3.58 ohms. At 0.32 A running, drop = 1.15 V, leaving 24.1 - 1.15 = 22.95 V, which matches the idle reading closely. At 1.1 A inrush, drop = 3.94 V, leaving 20.16 V, already below the 21.6 V minimum; the measured 17.8 V indicates a shared common conductor or additional load on the same pair. Transformer load: three actuators at 0.32 A = 0.96 A, plus 0.1 A controller = 1.06 A, or about 25.4 VA at 24 V. The 75 VA transformer is adequate in steady state. A 100 VA transformer may improve momentary inrush capacity, but it cannot correct voltage drop in the 18 AWG conductors. The fix must address conductor size or circuit arrangement.
2. What wiring remedy would you specify for the farthest actuator, and what voltage would result at inrush? Consider 16 AWG and 14 AWG.
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For 16 AWG, use about 4.0 ohms per 1,000 ft. Loop R = 4.0 x 0.560 = 2.24 ohms. At 1.1 A, drop = 2.46 V, leaving 21.64 V from 24.1 V. That is only 0.04 V above the actuator minimum and is not a reliable margin. At 0.32 A, the drop is 0.72 V, leaving 23.38 V. For 14 AWG, use about 2.53 ohms per 1,000 ft. Loop R = 1.42 ohms; at 1.1 A, drop = 1.56 V, leaving 22.54 V; at 0.32 A, drop = 0.45 V, leaving 23.65 V. I would run a dedicated 14/2 circuit to the farthest actuator or provide a local 24 V transformer/relay at the actuator. Do not rely on 16 AWG with almost no inrush margin. Also verify that the common conductor is not shared with actuators that start simultaneously.
3. The contractor asks whether the 24 VAC circuit is Class 2 and whether the 18/2 cable can be run in the same raceway as a 120 V switching circuit. Answer both parts.
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A 24 VAC, 75 VA control source is a Class 2 circuit under Article 725 because it is within the 100 VA limit for that voltage range. Class 2 circuits must be separated from electric light and power circuits unless the wiring method or a barrier is specifically listed for the combination. The existing 18/2 cable cannot simply be pulled into the same raceway as a 120 V switching circuit. Use a separate raceway or a listed divider/barrier, and keep the Class 2 cable and power conductors physically separated. If a 100 VA transformer is installed, it remains at the Class 2 limit; do not assume the larger transformer permits mixing with power conductors.