The relationship between torque and horsepower is often misunderstood but is very important. To understand these two terms and how they are affected by other variables we have to start with definitions.

Torque is a twisting or turning ‘force such as the turning of a shaft in an electric motor. Horsepower is the same twisting or turning effect as related to time. Torque is usually measured in pound-feet (or foot-pounds) or some derivative such as ounce-inches or inch-pounds. In Europe the standard unit of measure for torque is Newton-Meters. Horsepower is pound feet per second (or minute). In terms of mechanics: one foot-pound of torque is the twisting effect created by applying a force of one pound against a lever, at a distance of one foot from the fulcrum or pivot point. Horsepower was established as the amount of effort that it takes to move a 3300 pound weight one foot in one minute. Note that torque has no time component, horsepower does.

The formula that relates these two to each other is as follows: T = HP X 5250 *I *rpm where T is torque, HP is horsepower, RPM is shaft speed expressed in minutes and 5250 is a mathematical constant. By looking at the formula we can see that:

1) For a given speed, if horsepower increases so does torque.

2) If rpm increases torque **DECREASES **for a given horsepower.

You may have noticed that for a given horsepower rating an 1800 rpm motor develops twice the torque that a 3600 rpm motor does at full load (3 ft-Lbs/hp vs. 1.5 ft-Lbs/hp). Likewise for a given rpm the way to get more torque is to increase horsepower.

So where does all of this good stuff come into play in the real world? Most commonly, when frequency changes are made in induction motors. The speed of an induction motor is determined by the number of electrical (not mechanical) poles and the frequency of the applied power. A common method of changing the speed of a three phase induction motor nowadays is by using a frequency inverter. The inverter typically starts with an A.C. input, rectifies it to a D.C. voltage and then, through a variety of methods, converts it back to A.C. through the use of triggering devices. The speed of the trigger determines the apparent frequency output of the inverter. So by turning an adjustment knob or reprogramming the output the frequency can be adjusted over a range, typically 0-90 Hz, although some inverters can go to 360 Hz. or beyond.

What happens to the voltage as the frequency changes?

Remember that the ratio of the voltage and frequency needs to remain constant. The volts per hertz ratio is adjustable in most inverters, but once it is set the ratio remains the same throughout the output range. This means that with an input voltage of 480 volts, 60 Hz. a properly adjusted inverter will have a volts per hertz ratio of 480 / 60 = 8. At 20 Hz. the voltage output will be 20 X 8 = 160 volts. At 57.2 Hz. the output voltage will be 57.2 X 8 = 457.6 volts. Because the strength of the magnetizing force remains the same, the torque exerted on the rotor (and therefore the output torque of the motor) remains constant throughout the speed range. That is until the inverter reaches 480 volts at 60 Hz. When the output is increased above the input rating the frequency still increases, the speed still increases, but the voltage can not increase above 480 volts. So at an output of 90 Hz. for example, the volts per hertz ratio is now 480 / 90, or 5.3 instead of 8. The strength of the magnetic attraction between the stator field and the rotor is now reduced and less torque is produced.

Jumping back to the torque/horsepower formula for a moment, we can see that for a 480VAC drive, from 0-60 Hz. the volts per hertz ratio remains constant, so the torque remains constant. At a setting of 50% the frequency is 30 Hz. the voltage is 240 volts, the speed is 50% of the nameplate value. The torque is unchanged, so the horsepower output is 50% of the nameplate rating. This means that as the speed increases from 0 to nameplate rated speed the motor is operating as a constant torque motor, and the horsepower goes from 0-100% of nameplate value.

After we pass the input voltage rating, the horsepower remains the same because the voltage remains the same, speed increases because the frequency increases, therefore the torque is reduced. At 90 Hz. the torque is only 66% of what it was at 60 hz (60/90). So the motor, above nameplate rpm, acts as a constant horsepower, variable torque motor.

Practical application time: The customer no longer needs to operate his machine at various speeds but instead would like to operate at a speed equal to 50% of nameplate. The motor was rated at 25 HP, 3600 rpm, 480 volts, 3 phase, 60 Hz. What motor does he need to accomplish 50% speed without over loading?

His constant frequency input voltage at 60 Hz is still 480 volts 3 phase. The speed at 50% would require a 4 pole motor (i.e. 1800 rpm), but his horsepower requirement is now only 50% or 12.5 HP. He would need a 15 HP motor, assuming a constant torque type of load. If he is pumping liquids with a centrifugal pump or moving air with a fan his load would be a variable torque load, requiring only a 7.5 HP motor.

Shop question: A customer calls and wants to know if you can rewind his 380 volt 50 Hz. 1500 rpm motor to operate on 480 volt 60 hz. What do you tell him?

You cautiously tell him that, depending on his load, he can probably use the motor as is. The volts per hertz ratio of 380/50 and 460/60 is still virtually the same. The one precaution would be to keep in mind that there will be an increase in speed from 1500 to 1800. The motor will need to produce 20% more horsepower for constant torque loads. If the motor was not fully loaded to start with, it will probably be okay. If the load is a variable torque load (i.e. pump or fan) it will probably not be sufficient to pull the load at the higher speed. Yes, the motor can be rewound to the new frequency and the customer will be very happy he doesn’t have to import a motor.

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