Electronic – Torque measurement device calibration

One way to use the calibration is to make no physical adjustments to the device. Simply do a mathematical adjustment to all the measurements you take with the device.

For example, an anemometer measures windspeed. Say you have a 5 mph reference flow and a 10 mph reference flow. Say in the calibration measurement, the 5 mph flow measures as 6.5 mph and the 10 mph flow measures as 12 mph. You then determine that the measured speed (I'll call it v_m) is related to the real speed (v) by

\$v_m = 1.1v + 1\$

For your "live" measurements you'll reverse this formula to get the real speed in terms of the raw measurement:

\$v = (v_m - 1) / 1.1\$

So if you get a raw measurement of 8 mph, you use your calibration to estimate that the real speed was about 6.4 mph.

Of course the correction I described is based on a very simple, linear model of the instrument errors. In the real world, you might need a more complex model to get a correction formula that gives really accurate measurements.

One of the several important advantages of the three-phase, a.c. induction motor is the grace with which it delivers a variable torque. You did not mention the motor's number of poles or whether your power was of the 50-Hz, Old World kind or of the 60-Hz kind used in the New World. In the Old World, the maximum (not minimum) speed of a typically configured induction motor is 3000 rpm, and this only if it is a two-pole motor. Four-pole motors are more common: their maximum speed in the Old World is 1500 rpm.

For the sake of answering, let us suppose that you are in the Old World and that your motor has two poles: maximum speed, 3000 rpm. (If in the New World, you can scale this answer's speeds by [60 Hz]/[50 Hz] = 1.20: maximum speed, 3600 rpm.)

For moderate values of torque—usually up to about 115 percent of the motor's full rated load—the motor's speed will decrease slightly and approximately linearly as the mechanical load to which the motor delivers power demands torque. For example, if the motor turns at 2800 rpm at full rated load, then it will turn at about 2850 rpm at 75 percent of full rated load, 2900 rpm and 50 percent, 2950 rpm at 25 percent, and (almost) 3000 rpm when unloaded. At loads greater than 115 percent, the motor's speed will decrease more than the linear model predicts, until the motor's breakdown torque is reached (the exact value of breakdown torque depends on the motor, the ambient temperature and other factors, but 200 to 230 percent of full rated load might be a typical figure). At breakdown torque, the motor will spin down to a stop, thenceforth acting as an inductive winding and delivering the motor's rated locked rotor or starting torque to the load.

I give all this detail for three reasons. First, because it may contain the answer you seek. Second, to explain why there is no single torque, but are many torques involved. Third, to convey the crucial concept that, within the motor's operational domain, it is the mechanical load, not the motor, that determines the torque.

If you really need to measure the actual torque accurately under specific conditions, then use a dynamometer. However, if an approximate indication of torque is all you need, and if you are operating in the motor's normal, spun-up state, at no more than 115 percent of full load, then you can approximate the torque pretty accurately from the motor's measured speed and its nameplate data, using the linear relationship described. The approximation is so much easier than the dynamometer that I would recommend the approximation in most cases.

Note: A few, unusual three-phase a.c. induction motors do not have any breakdown torque. In other words, for a few induction motors, the locked-rotor torque is the maximum torque the motor can deliver. In the United States, such motors usually read "Design: D" on their nameplates. This is probably not the case for you, but I thought that I should at least mention the matter.