Tom Bishop, PE
EASA Senior Technical Support Specialist 

Often when we are dealing with three-phase squirrel cage motors we ask ourselves “Is there a standard or a rule of thumb for this motor characteristic?” The 10 characteristics that follow apply to conditions associated with motor starting as well as steady state operation. Examples are provided for each of the characteristics. Where applicable, the source NEMA or IEC standard clauses are cited in Table 1.

Table 1: Motor Characteristics with NEMA and IEC References
Topic	NEMA MG 00001	IEC 60034-1
1. Voltage variation	12.44 & 20.14	7.3
2. Voltage unbalance	12.45, 12.36.2 & 20.24	none
3. Current variation	12.47	none
4. Current unbalance	14.36.5	none
5. Full load current	12.47	none
6. Locked rotor current	10.37.2	*
7. Instantaneous current at starting	12.36	none
8. Locked rotor stall time	12.49	none
9. Number of starts	12.54 & 20.12	none
10. Slip (synchronous rpm – rated rpm)	12.46	12.1
*IEC 60034-12 provides values for locked rotor apparent power.

 

1. Voltage variation
This is a variation of plus/minus 10% at rated frequency for induction motors. 

Example:
A 60Hz motor is rated 230 volts. The motor is used on a 208 volt system and the supply voltage at the motor is 203 volts. Is the supply voltage within the NEMA and IEC tolerances? 

Evaluate using formulas:
Acceptable variation from rated voltage = Rated voltage x (0.90 or 1.10) 

230 x 0.90 (100 - 10 = 90%) = 207 volts 
230 x 1.10 (100 + 10 = 110%) = 253 volts 

The 203 volt supply voltage value is below the acceptable range of 207 to 253 volts. Thus, it is not an acceptable value. The motor may need to be derated or replaced with a motor rated 200 volts for use on a 208 volt supply. It is not uncommon for voltages at the motor to be as low as 190 volt s o n a 2 0 8 v olt system. 

2. Voltage unbalance
There is a maximum 1% voltage unbalance at motor terminals without derating. 

Example:
A 50 hp (37 kW) motor is used on a 480 volt system and the line-to-line voltages at the motor are 452, 463 and 471 volts. Is this voltage unbalance within the NEMA tolerance? 

Evaluate using formulas: 
Average voltage = Sum of line-to-line voltages / 3 = (451 + 458 + 474) / 3 = 1383 / 3 = 461 

Determine acceptable voltage unbalance range = Average voltage x (0.99 or 1.01) 

Use absolute values for results 

Deviation from average = (461 – 451) = 10 
Deviation from average = (461 – 458) = 3 
Deviation from average = (461 – 474) = 13 
Maximum deviation from average = 13

Percent voltage unbalance = (Maximum deviation from average voltage / average voltage) x 100 = (13/ 461) x 100 = 0.028 x 100 = 2.8% 

The 2.8% voltage unbalance exceeds the NEMA 1.0% limit and requires derating (see Figure 1). The derating factor for 2.8% unbalanced voltage is about 0.9, meaning that the motor should be derated to 0.9 of its nameplate rated power. 

3. Current variation
There is a maximum 10% from rated current with rated voltage, frequency and at output power rating. 

Example:
Using the previous example 100 hp (75 kW) motor rated 115 amps at 460 volts. Determine the maximum current variation. 

Evaluate using formulas:
Acceptable variation from rated current = Rated current x (0.90 or 1.10) 

115 x 0.90 (100 - 10 = 90%) = 103.5 amps 
115 x 1.10 (100 + 10 = 110%) = 126.5 amps 

Note that the results are the same as for full load amps variation. (See # 5.) 

4. Current unbalance
At normal operating speed, the percent current unbalance will be approximately 6 to 10 times the percent voltage unbalance. 

Example:
Use the value 2.8% voltage unbalance from the voltage unbalance example. 

Evaluate using formulas:
Approximate current unbalance at six times the voltage unbalance = 6 x 2.8% = 16.8% 
Approximate current unbalance at 10 times the voltage unbalance = 10 x 2.8% = 28.0% 

5. Full load current
This can vary +/- 10% from rated. 

Example:
A 100 hp (75 kW), 460 volt motor is rated 115 amps. At full load it is drawing 121 amps. Is it overloaded, or is the current within the NEMA tolerance? 

Evaluate using formulas:
Acceptable current range at full load = Rated full load amps x (0.90 or 1.10) 

115 x 0.90 (100 - 10 = 90%) = 103.5 amps
115 x 1.10 (100 + 10 = 110%) = 126.5 amps

The 121 amp value is between the 103.5 and 126.5 amp values, thus it is acceptable. Note though that it could indicate an overload condition since it is above the nameplate rating. 

6. Locked rotor current
Use the ranges specified for the kVA code. 

Example:
The 100 hp (75 kW) motor in the example above has a kVA code letter G, indicating 5.6 – 6.3 kVA per horsepower. The measured locked rotor current is 777 amps. Is that value within the NEMA tolerance? 

Evaluate using formulas:
Locked rotor amps (LRA) = (starting kVA/hp x hp x 1000) / (volts x 1.732*)
*For a three-phase motor 

LRA = (5.6 x 100 x 1000) / (460 x 1.732) = 703 amps 
LRA = (6.3 x 100 x 1000) / (460 x 1.732) = 791 amps 

The measured 777 amp value is between 703 and 791 amps, thus it is acceptable. 

7. Instantaneous current at starting
This can be 1.8-2.8 times locked rotor amps. 

Example:
The 100 hp (75 kW) motor in the examples above has a kVA code letter G and is tripping the circuit protection at startup. The initial momentary starting current is 1,311 amps as measured with an ammeter capable of measuring instantaneous current. Is that value within the NEMA tolerance? 

Evaluate using formulas:
Using the values from the previous example, the LRA should be between 703 and 791 amps.

Minimum instantaneous starting current = 703 x 1.8 = 1,265 amps 
Maximum instantaneous starting current = 791 x 2.8 = 2,215 amps 

The instantaneous current of 1,311 amps is between 1,265 and 2,215 amps, thus it is acceptable. 

8. Locked rotor stall time
The minimum capability is not less than 12 seconds for motors not exceeding 500 hp (375 kW).

Exception: 2 pole Design BE motors time limit is eight seconds 

Example:
A 400 hp (300 kW) 4 pole Design A motor should be able to withstand locked rotor current for not less than 12 seconds. It is also required that the inertia of the load not exceed the values in MG 00001, Table 12-7. 

A 200 hp (150 kW) 2 pole Design BE motor should be able to withstand locked rotor current for not less than eight seconds. It is also required that the inertia of the load not exceed the values in MG 00001, Table 12-7. 

9. Number of starts
Two successive starts with motor initially at ambient; coast to rest between starts. 

Or, one start with motor at temperature not exceeding its rated load operating temperature. 

There is no power rating limit for this standard. 

Example:
Note that the limit for the number of starts could be applied to a 1 hp (0.75 kW) rating just the same as to a 1000 hp (750 kW) rating. However, the 1000 hp (750 kW) example motor is much more likely to have winding temperature detectors, which can be used to determine rated load operating temperature and winding temperature prior to restarting. 

10. Slip (slip = synchronous rpm – rated rpm) 

This can vary 20% from rated speed slip. 

Example:
A 50Hz, 4 pole motor has a synchronous speed of 1500 rpm and a nameplate full load speed rating of 1475 rpm. 

Slip rpm = synchronous speed – rated speed rpm = 1500 - 1475 = 25 rpm 

Acceptable slip rpm limit at full load = Slip rpm +/- 20% slip rpm 

Slip rpm maximum = 25 + (20% x 25) = 25 + 5 = 30 rpm
Slip rpm minimum = 25 - (20% x 25) = 25 - 5 = 20 rpm 

Acceptable speed range = synchronous speed minus minimum or maximum slip 

= 1500 -20 = 1480 rpm
= 1500 - 30 = 1470 rpm 

Note that any full load speed below the rated 1470 rpm could indicate an overload condition.

AVAILABLE IN SPANISH