Motor Calculations, based on 2017 NEC^® - Part 2

Figure 01

For EC&M Magazine
By Mike Holt, NEC^® Consultant

Click here to review Motor Calculations, based on the 2017 NEC Part 1.

Do you know how to size branch-circuit short-circuit and ground-fault protective devices for motor circuits?

Each motor branch circuit must be protected against short circuit and ground faults by an overcurrent device sized no greater than the percentages listed in Table 430.52. A branch-circuit short-circuit and ground-fault protective device protects the motor, the motor control apparatus, and the conductors against short circuits or ground faults, but not against overload [430.51].

The motor branch-circuit short-circuit and ground-fault protective device must comply with 430.52(B) and 430.52(C). A motor branch-circuit short-circuit and ground-fault protective device must be capable of carrying the motor's starting current. Motor branch-circuit conductors must comply with 430.22.

Question: What size conductor and inverse time circuit breaker are required for a 1 hp, 115V, single-phase continuous-duty motor?
(a) 12 AWG, 30A breaker     (b) 12 AWG, 35A breaker
(c) 12 AWG, 40A breaker     (d) 12 AWG, 45A breaker
Answer: (c) 12 AWG, 40A breaker

Step 1: Determine the branch-circuit conductor, 125% of FLC
[110.14(C)(1)(a)(1), Table 310.15(B)(16), 430.22, and Table 430.248]:
16A × 1.25 = 20A, 12 AWG rated 20A at 60°C
Step 2: Determine the branch-circuit protection, 250% of FLC
[240.6(A), 430.52(C)(1), and Table 430.248]:
16A × 2.50 = 40A

What if the motor branch-circuit short-circuit and ground-fault protective device values derived from Table 430.52 don't correspond with the standard overcurrent device ratings listed in 240.6(A)? In that case, you can use the next higher overcurrent device rating [430.52(C)(1) Ex 1].

Branch-circuit conductors to a single motor must have an ampacity of at least 125 percent of the motor FLC as listed in Tables 430.247 through 430.250 [430.6(A)].

The branch-circuit short-circuit and ground-fault protection devices are sized by considering the type of motor and the type of protection device and applying the percentage of the motor's FLC as listed in Table 430.52.

Question: If an inverse time circuit breaker is used for short-circuit and ground-fault protection, what size circuit breaker and conductor is required for a 7-1/2 hp, 230V, three-phase motor? ‚
(a) 10 AWG, 50A breaker     (b) 10 AWG, 60A breaker
(c) 10 AWG, 70A breaker     (d) 8 AWG, 70A breaker
Answer: (b) 10 AWG, 70A breaker

Step 1: Determine the branch-circuit conductor, 125% of FLC.
[110.14(C)(1)(a)(1), Table 310.15(B)(16), 430.22, and Table 430.250]:
22A × 1.25 = 28A, 10 AWG rated 30A at 60°eC
Step 2: Determine the branch-circuit protection, 250% of FLC.
[240.6(A), 430.52(C)(1), and Table 430.250]:
22A × 2.50 = 55A
Next size up, use 60A inverse-time breaker

Combined Overcurrent Protection
A motor can be protected against overload, short circuit, and ground faults by a single overcurrent device sized to the overload requirements contained in 430.32 [430.55].

Question: What size dual-element fuse is permitted to protect a 5 hp, 230V, single-phase motor with a service factor of 1.15 and a nameplate current rating of 23.50A from overloads as well as short circuits and ground faults?‚
(a) 20A fuse     (b) 25A fuse     (c) 30A fuse     (d) 35A fuse
Answer: (b) 25A fuse
Overload Protection [430.32(A)(1) and 430.55]
23.50A × 1.25 = 29.40A
The fuse can't be larger than 29.40A [430.32(A)(1)], use a 25A fuse [240.6(A)]

Feeder short-circuit and ground-fault protection
Feeder conductors must be protected against short circuits and ground faults by a protective device sized not more than the largest rating of the branch-circuit short-circuit and ground-fault protective device for any motor, plus the sum of the full-load currents of the other motors in the group [430.62(A)]

Question: What size inverse-time circuit breaker is required for feeder protection for the following motors? Both motor branch circuits have inverse-time circuit breakers.
Motor 1: 7-1/2 hp, 230V, three-phase: FLC = 22A [Table 430.250]
Motor 2: 5 hp, 230V, three-phase: FLC = 15.20A [Table 430.250]
(a) 70A breaker      (b) 80A breaker
(c) 90A breaker      (d) 90A breaker
Answer: (a) 70A breaker

Feeder protection [430.62(A)] is to be sized no greater than the largest branch-circuit ground-fault and short-circuit overcurrent device plus the other motors' FLC.
Step 1: Determine the largest branch-circuit ground-fault and short-circuit overcurrent device [430.52(C)(1) Ex].
7 1/2 hp Motor = 22A × 2.50 = 55A, next size up 60A
5 hp Motor = 15.20A × 2.50 = 38, next size up = 40A
Step 2: Determine the size feeder protection.
Not more than 60A + 15.20A = 75.20A, next size down = 70A [240.6(A)]
The next size up protection rule for branch circuits [430.52(C)(1) Ex 1] doesn't apply to the motor feeder overcurrent device rating.

Question: What size inverse-time circuit breaker is required for feeder protection required for the following motors? All motors branch circuits have inverse-time circuit breakers. Figure 01
Motor 1: 7½ hp, 208V, three-phase: FLC = 24.20A [Table 430.250]
Motor 2: 5 hp, 208V, single-phase: FLC = 30.80A [Table 430.248]
Motor 3: 1 hp, 115V, single-phase: FLC = 16A [Table 430.248]
(a) 70A breaker      (b) 80A breaker
(c) 90A breaker      (d) 100A breaker
Answer: (d) 100A breaker

Size your feeder protection [430.62(A)] no greater than the largest branch-circuit ground-fault and short-circuit overcurrent device plus the other motors' FLC in the same group.
Step 1: Determine the largest branch-circuit ground-fault and short-circuit overcurrent protection device [430.52(C)(1) Ex].
7½ hp Motor = 24.20A × 2.50 = 60.50, round up to 70A
5 hp Motor = 30.80A × 2.50 = 77A, round up to 80A
1 hp Motor = 16A × 2.50 = 40A
Step 2: Determine the size feeder protection based on the largest group of motors. To determine the group, you need to balance out the motors between L1, L2, and L3.
Not more than 80A + 24.20A = 104.20A
Next size down = 100A [240.6(A)]
The next size up protection rule for branch circuits [430.52(C)(1) Ex 1] doesn't apply to the motor feeder overcurrent protection device rating.

VA calculations
The measure of the mechanical output of a motor is horsepower. Horsepower can be converted to watts by multiplying the horsepower rating by 746W per horsepower. Since this is a measure of a motor's output, don't confuse it with the input power required by a motor.

Question: What are the output watts for a dual voltage, 1 hp motor rated 115/230V?
(a) 746W     (b) 1,000W     (c) 1,400W     (d) 1,840W
Answer: (a) 746W

Output = 746W × hp
Output = 1 hp × 746W
Output = 746W

The input VA of a motor is determined by multiplying the motor volts by the motor amperes. To determine the single-phase motor VA rating, you can use the following formula:
Single-Phase Motor VA = Motor Volt Rating × Motor Ampere Rating†

Question: What's the input VA for a dual voltage, 1 hp motor rated 115/230V?‚
(a) 920 VA at 230V     (b) 1,840 VA at 115V     (c) 1,840 VA at 230V     (d) b and c
Answer: (d) b and c

Table 430.248, 115V FLC = 16A, 230V FLC = 8A
VA at 230V = 230V × 8A
VA at 230V = 1,840 VA
VA at 115V = 115V × 16A
VA at 115V = 1,840 VA

Many people believe a 230V motor consumes less power than a 115V motor, but both motors consume the same amount of power.
To determine the three-phase motor VA rating, the following formula can be used:
Three-Phase VA = Motor Volt Rating × Motor Ampere Rating × 1.732

Question: What's the input VA for a 5 hp, 230V, three-phase motor?‚
(a) 3,730 VA     (b) 6,055 VA     (c) 6,440 VA     (d) 8,050 VA
Answer: (b) 6,055 VA [Table 430.250]

Table 430.250, FLC = 15.20A
Motor VA = 230V × 15.20A × 1.732
Motor VA = 6,055 VA

Adjustable speed drives
Circuit conductors supplying power conversion equipment included as part of an adjustable-speed drive system must have an ampacity at least 125 percent of the rated input current to the power conversion equipment [430.122(A)]. Overcurrent protection for adjustable-speed drive systems must be per manufacturer's instructions [430.130(A)(2)].

Question: What size branch-circuit conductors are required for an adjustable-speed drive system with a rated input of 25A when the terminals of the adjustable speed drive are rated 60DegrC?‚
(a) 14 AWG     (b) 12 AWG     (c) 10 AWG     (d) 8 AWG
Answer: (d) 8 AWG

Rated input from adjustable speed drive = 25A
Conductor = 25A × 1.25
Conductor = 31.25A, 8 AWG rated 40A at 60DegrC
[Table 310.15(B)(16)]

Avoiding problems
You can see that correctly sizing the branch-circuit short-circuit and ground-fault protective devices for motor circuits depends on first correctly characterizing the motor. Assuming you've done that and you've correctly sized the motor overcurrent protection, everything should work just fine. Right?

Yes, but only if many other assumptions are also correct. For example, it's a good practice to double check the motor application criteria to ensure you have the correct motor for the application. You save time if you do that before performing the calculations.

If later it turns out the branch-circuit protective device seems to nuisance trip, the first assumption should not be it's undersized. Something in the power supply (e.g., failed power factor capacitor) or the load (e.g., gearbox problem) is likely causing that. Motor applications are complicated for these and many other reasons. Work methodically when performing those calculations, so that your calculation work isn't the point of failure.