Acceleration torque at 50% voltage = 25% of full torque. If pump torque > motor accelerating torque, motor will not accelerate. This violates MG1-33 because the motor will stay at locked rotor current for >20 seconds, tripping overload.
(Motors and Generators) is the primary North American standard for the construction, performance, and testing of alternating current (AC) and direct current (DC) motors and generators. First published in the 1920s and updated regularly, MG 1 is harmonized with other international standards but retains unique North American practices, particularly regarding voltage, frequency, and enclosure types.
Introduction In the world of industrial electric motors, standards are not just recommendations—they are the backbone of safety, interoperability, and performance. Among the most frequently referenced yet often misunderstood sections of the NEMA (National Electrical Manufacturers Association) Standards Publication MG 1 are MG1-32 and MG1-33 , commonly searched together as "NEMA MG1-32 AMP-33" . nema mg1-32 amp- 33
In simpler terms, this section defines the standard methods for calculating the apparent power (kVA) that a motor draws from the line —specifically when using reduced-voltage starting methods such as autotransformers, part-winding, or wye-delta starters. Why is MG1-32 Critical? When an induction motor starts, it draws a high inrush current (typically 600% of full-load current) for a few cycles, followed by a starting current (typically 500–600% of full-load amps) until it reaches full speed. This current, multiplied by the voltage, gives the starting kVA .
The document is divided into "Parts" and then numbered "Sections." Sections 32 and 33 fall under (General Concerning Definite-Purpose Motors) and Part 31 (General Concerning Integral-horsepower Motors), respectively. Part 2: Deep Dive into NEMA MG1-32 What is NEMA MG1-32? NEMA MG1-32 is titled: "Determination of Motor Input kVA at Starting and During Acceleration for Reduced Voltage Starting." Acceleration torque at 50% voltage = 25% of full torque
| Starting Method | % of Full Voltage | % of Starting Current | % of Starting Torque | % of Starting kVA | |----------------|------------------|----------------------|----------------------|--------------------| | Full Voltage | 100% | 100% | 100% | 100% | | Autotransformer (80% tap) | 80% | 80% | 64% | 64% | | Autotransformer (65% tap) | 65% | 65% | 42% | 42% | | Wye-Delta (Star-Delta) | 58% | 33% | 33% | 33% | | Part-Winding (50-100% winding) | 100% | 50-70% | 20-45% | 50-70% | Problem: A 100 HP, 460V, three-phase motor has a locked rotor current of 600A (Code G motor). Calculate the starting kVA using a wye-delta starter.
Transformer 300 kVA cannot supply 1120 kVA. Voltage drop would exceed 30%. (Motors and Generators) is the primary North American
Starting kVA = 1120 × 0.25 = 280 kVA (acceptable for 300 kVA transformer)