As long as we are on the subject of motors.

Our shop was doing some experimenting to determine the horsepower needed to drill a hole. If the motor is drawing X amps before drilling and Y amps while drilling, I figured that a decent approximation for the actual drilling is: V*(Y-X)*efficiency. Is this valid?

I have no good idea myself, but I do have a question: Why would you subtract the before drilling amps? Wouldn't that load always be present and wouldn't you really just need to look at the loaded amperage in your calculation?

Steve

The test is for the drilling, not the motor. We have a gear train and other power transmission components to design. This is what is mainly for.

Ok, fair enough. But, still...if you are sizing your drive train and using motor amperage measure the work load, don't you calculate the capability of the drive train according to the required motor horsepower and then use a safety factor multiplier? If so, then it seems to me that the upper work load value is the only one you would be interested in.

Please keep in mind that I have never done this before so I may well be talking out my ***.

Steve

If you want to split hairs, to calculate the true ouput power of the motor you would need to simultaneously measure input voltage, current, and power factor. You also need to estimate the internal losses of the motor.
P = 1.73 x V x I x pf x eff x 0.001341
P - output power in horsepower
1.73 - square root of 3 (for three phase systems)
V - input voltage, line to line
I - input current
pf - power factor
eff - efficiency
0.001341 - a conversion from electrical power to mechanical horsepower

The motor current will vary as the input voltage varies, so for small load change applications (like dull drill bits) input power rather than current is a more true estimate of output power.
Power factor comes into play when a motor is oversized compared to the load. On a lightly loaded motor a small change in load may not be cause an increase in current at all.

Thanks, Jim. That is what I am looking for. We only need a close approximation.

Steve- I am only on the edge of this project. I asked some of the same questions, but did not get them answered, yet.

Measuring motor torque

Golly, it seems like I am repeating myself alot on this subject but, again, the best way to measure motor load is to forget amps and measure motor slip.

You do this by noting the nameplate speed. Assuming 60Hz, if the nameplate says, for example, 1765 rpm, you can tell immediately that this is a four pole motor with no load speed of 1800 rpm and full load speed of 1765 rpm. Output torque is a simple straight line proportion between these two points. Use a strobe light on the motor shaft or handheld speed meter to find actual motor speed while drilling.

Now, to get to actual ft-lbs torque at full rated load (and therefore, nameplate rpm), two pole motors develop 1.5 ft-lb torque per HP, four pole motors develop 3 ft-lbs torque per HP, six poles are 4.5 ft-lbs per, and eight poles are 6 ft-lbs per, and so on.

For example, we have a 10hp motor nameplated 1770 rpm. Since this is clearly a four pole motor with sync speed of 1800 rpm, we have 30 rpm of slip at full load. Since this is a 10hp four pole motor, full load torque will be 10 times 3 or 30 ft-lbs.

Now, if we place a load on the motor and observe the motor speed to be 1790 rpm, that is clearly 10 rpm slip. Since this is one-third of full load slip (1800-1770=30), we know that output torque is one third of rated torque. Since rated torque is 30 ft-lbs, our actual output torque must be one-third of that or 10 ft-lbs.

If we now further load the motor so its speed drops to 1780 rpm, this is 20 rpm slip which is 2/3 of rated slip which should give us 2/3 of rated torque or 20 ft-lbs.

Isn't that neat! Several days of this and you get so you can do it in your head!! Well, maybe not! But still, isn't this simpler!

Dick's way is probably most accurate - in fact several years ago a company, Rexnord I think, actually marketed a motor power monitor based on slip. It never seemed to catch on, however.

As several guys have pointed out monitoring amps alone is a lousy way to monitor power. No load amps on a motor can be as high as amps at 25% load for example. Voltage variations can also have an impact.

If you don't want to use Dick's method for some reason get a motor data sheet that has efficiency at 100%, 75%, and 25% load. Get a kW transmitter from Load Controls Inc, Cutler Hammer, or others. Then you can monitor the kW into the motor, and by applying the efficiency and unit conversion factor get pretty close to the hp out.

I have suggested Dick's method. Easy to do as we have strobe lights. This is just for an experiment.