# the relationship of amps to rpms in an induction motor

when searching for a 2000 watt induction motor online the rpms seems to range from 3450-3600 Why? if you wanted to double rpms to say 7000 ,is that possible and if you are using 3 phase 220 V line, how would amps be effected. where can I study about this? I am a novice

• The rotation speed depends on the kind of motor. we have lots of motors. and electromagnetic equations are applied for each differently.
– AHB
Dec 6, 2015 at 15:41
• As SunnyBoy says, if you need to increase speed, a VFD is the answer. Current (amps) doesn't depend much on the speed, but depends A LOT on the torque load.
– user16324
Dec 6, 2015 at 16:21

Speed:

The speed is measured in RPM. 60 Hz excitation voltage corresponds to 3600 RPM. The mechanical speed can be further reduced by using a motor with higher magnetic pole count. Two pole motor has the same mechanical frequency as the excitation voltage. Four pole motor will have half the frequency, six-pole motor will have a third of the frequency.

Induction machines slip behind the excitation frequency by definition. Typical slips are 0-5%. The higher the mechanical loading, the higher the slip.

7000 RPM is not possible without either a variable frequency drive (VFD) or a gearbox.

Power

For simplicity, one can assume that the mechanical power out is about 90% of the electrical power in. 2 kW motor can therefore draw ~2.2 kW of real power. With 3-phase 220V line voltage the current would be:

$$V_{phase} = \frac{V_{line}}{\sqrt3} = 127 Vrms$$

$$I_{phase} = \frac{P}{3V} = \frac{2200 W}{127 V*3} = 5.8 Arms$$

Induction motor speed is a matter of poles and frequency - on 50 Hz you'll get 3000, 1500, 1000 as you go from 2-pole to 4 pole to 6 pole. On 60 Hz its 3600, 1800 and 1200 - on 400 Hz its 24,000, 12,000 and 8000.

A variable frequency drive can provide a wide range of speeds, but it can also provide more speed than a motor may be designed for - if a motor purely intended to run on 60 HZ is attached to a VFD and wound up to 120 Hz, you might find the mechanical limits of strength in the rotor and destroy it (loud noises, sudden stops, great expense.) There are motors specifically made for VFD operation which are rated for higher operation speeds - while expensive as compared to a normal motor, they are cheap as compared to what can happen when a normal motor comes apart in use.

The other common option for higher-speed motors is not an induction motor - rather a brushed type DC or universal motor. On average these are fairly loud as compared to an induction motor, and there will be sparks from the brushes (and the brushes wear and need to be replaced) but they are a common approach to getting high speed operation, as seen in most portable saws, drills and routers, or small hand-held rotary tools.

For an application requiring a fixed 7000 or 7200 RPM speed, a simple belt & pulley arrangement to double the motor shaft speed is likely to be the most cost-effective and power efficient approach (gearboxes are relatively expensive and can also be surprisingly inefficient.)

Please comment on the following including if possible potential commercial applications. We have just completed a working prototype that is comprised of two 1000W (1.34 hp each) induction type motors, which based on an efficiency of 0.885, equals to 1.18hp net for each motor. The technology is modular. The prototype contains two modules for proof of concept. The RPM from the first motor (3,450) is transferred to the second motor, and then aggregates with the RPM from the second motor. The combined RPM of about 7,000 is evidenced on the rotor of the second motor. This increase in RPM is not due to either modulation of frequency or by using different sized pulleys/gears. In addition to the accumulated RPM, the hp from both motors has also been aggregated. This has been made possible due to way in which the RPM has been increased and accumulated, however combining the motors, the efficiency goes down somewhat to 0.78 (0.885 x 0.885), or a combined hp of 2.1hp (0.78 x 2.68hp). The torque for the first motor is easily calculated using the formula ( hp x 5250)/RPM …. (1.18 x 5250) /3450 or 0.7754 ft-lb. The torque for the combined motors (2.1 x 5250) / 7000 equals 1.575 ft-lb, however the RPM has increased from 3,450 to about 7000 RPM.

In other words, the prototype demonstrates an increase in the second motor of both RPM, hp and torque.