SOme questions about rpm and Torque(I think?)

Since in most AC and DC are similar in current needs, is the torque equal.
I may have the wrong term, but will a DC motor of the same amperage pull the same apparatus as an AC motor will? If not, how can I determine an equal sized DC motor?

If I have an DC motor that runs 3000 RPM at 7 amps; I turn the power down to the point where the motor is running at 1750 RPM ; Will it take less amps to run or simply use the same at a lower RPM.

If you could please include any formula's so I may work my way thru this. I understand Ohm's Law to a point, but not sure if it applies here and if it does, then where does everything fit to the formula?


Many thanks for any info here.

When events look their worst, there is always a light at the end!

 

HERE's a nice little paper on torque calculations for DC and AC motors.

It seems that the generated torque is a function of maximum horsepower rating and RPM.

Lower down on the page is a section on selecting AC and DC motors that might directly answer your question.

There are tons of formulas on the page.

 

If the motors have similar efficiency then they will use similar power with the same load torque and RPM. Starting torque is a different story as that's highly dependent on the type of motor.

 

OK, by the formulas:

I have a 1/3 Hp 115VAC, 1750RPM (base Rate) , rated at 7A and should produce a torgue of 1 Lb-ft.

If I use a 1/2 HP 24VDC, 3000 RPM, rated at 7A and its torque is .87 lb-ft.,
Then I only need to slow it down to 1500 RPM to reach the 1.0 lb-ft torque.

Would this DC motor use less current (A) by reducing its speed to the 1500 RPM?

 

In a DC motor, voltage is speed (rpm) and current is torque.

 

ok,
in other words, the 24VDC is only 24V at 3000 and its 7 A current is only at .87lbs-ft.

thus, if you slow it down, to 1500 rpm, the volts will be less but the current (amperage) is larger since the lbs-ft is higher(1.0)?

At roughly half the 24V is 12V, making the current greater than 14A to only reach 0.13 lbs-ft. extra. not a good trade.

May have to go to a geared motor where more torque should be higher than the current?