You have a group of motors (a motor bank) that are all connected to a single feeder conductor — and you need to size that conductor correctly so it doesn’t overheat or trip the system when the motors run.
This follows CEC Rule 28-108 which gives us a method to size the feeder conductor based on the types of motor duty (continuous, short-time, or intermittent).
Example
Step 1: Handle Continuous Duty Motors
According to Rule 28-108(1)(a):
Take the largest continuous duty motor, multiply its FLA by 125%, then add the FLA of any other continuous duty motors.
- Largest continuous: 40A
- Multiply: 40 × 1.25 = 50A
- Add other continuous: +36A
→ Total for continuous motors = 86A
Step 2: Handle Short-Time and Intermittent Motors
Use multipliers from Table 27:
- Short-Time (STD) 5 min = 110% → 50 × 1.1 = 55A
- Intermittent (ID) 15 min = 85% → 60 × 0.85 = 51A
Step 3: Add Them All Together
Now add all the calculated values:
- Continuous motors: 86A
- Short-Time motor: 55A
- Intermittent motor: 51A
Total= 86 + 55 + 51 = 192A
Step 4: Choose the Right Conductor Size
Use CEC Table 2, 75°C column:
- You’re looking for a conductor with at least 192A ampacity
- From Table 2 → 3/0 AWG copper is rated 200A at 75°C
Answer: 3/0 AWG copper is your minimum feeder conductor size
Super Fun Fact!
Since this feeder conductor doesn’t directly touch any motors, Rule 4-006 (conductor correction/derating) may apply depending on your installation (e.g., ambient temp, bundling, etc.).