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I found out values of my 3 phase delta connected induction motor for full load torque, maximum torque and startup torque.

My maximum torque was higher than my full load torque, but the thing I'm wondering about is whether I'm right or not is that my startup torque is higher than my full load torque. Is that right, or is it supposed to be lower than full load? If it's supposed to be lower than it shows I need to look over my calculations again.

Full load torque= 25.68Nm

Max torque= 322.4Nm

Startup torque= 38.53Nm

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enter image description here

If 25.68Nm is your full-load torque, 38.53Nm (150%) as startup torque makes perfect sense for a Design A or Design B squirrel cage induction motor.

But a max or breakdown torque of 322.4Nm (1255%) is much too high, which should max out in the 200% to 250% range.

The torque the motor supplies depends upon the load. You'd have to compare your numbers to nameplate data to determine full-load torque.

$$T = \frac {9.55 P_m} {n}$$

Where T = Torque, \$P_m\$ = mechanical power out and n = speed in rpm.

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  • \$\begingroup\$ I don't know if it is a squirrel cage induction motor. All I know is that it's a 3 phase delta connected the inductiion motor. Would that have different values? \$\endgroup\$ – Sophie Jun 4 at 15:05
  • \$\begingroup\$ If it's an induction motor, it is a squirrel cage. I gather, no nameplate. How are you measuring torque? \$\endgroup\$ – StainlessSteelRat Jun 4 at 15:29
  • \$\begingroup\$ I use the following equation for torque= (m(N2/N2)/2 x pi x ns) x (s x E1^2 x R2/R2^2+(sX2)^2) \$\endgroup\$ – Sophie Jun 4 at 15:43
  • \$\begingroup\$ That equation for power neglects stator copper and iron losses and mechanical losses. It may not be a squirrel-cage motor. It could be a wound-rotor motor with little or no external rotor resistance. I suspect this is a teaching-laboratory motor rather than a standard industrial motor. \$\endgroup\$ – Charles Cowie Jun 4 at 16:34
  • \$\begingroup\$ @Sophie If there are wires going to rotor via slip-rings and brushes, you have a wound rotor. No wires, SCIM. \$\endgroup\$ – StainlessSteelRat Jun 4 at 16:49
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The startup torque that you have is a reasonable value for the full=load torque. However the maximum torque is much higher than would be expected for most motors. Something like 250% to 350% would be more usual.

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  • \$\begingroup\$ Looking at someone else calculations they didn't use the whole number, e.g 95.94433325 just 95.9, during the calculations. Would you recommend that or would you recommend using the whole number during each step of the calculation? \$\endgroup\$ – Sophie Jun 4 at 14:55
  • \$\begingroup\$ Because I get different results if I do. The above was using the whole number. Here are my results rounding: full load torque= 71.03Nm. Which would you recommend as I always thought the full load torque should be higher than the start-up torque or have I got that wrong? \$\endgroup\$ – Sophie Jun 4 at 14:58
  • \$\begingroup\$ The full-load torque is a continuous torque rating. Maximum torque and starting (locked rotor) torque can only be provided for a short time without overheating the motor. At locked rotor and maximum torque, the current is quite high, in the area of 5 or 6 times rated current. That is why both start up torque and maximum torque are higher than full-load torque. It is much less expensive to design a motor that can only operate continuously in a small part of the torque vs. speed curve. The rest of the curve is only for starting, accelerating to full speed and tolerating short overloads. \$\endgroup\$ – Charles Cowie Jun 4 at 16:12
  • \$\begingroup\$ I can not see how rounding a number to three significant figures would make a significant difference unless you are using the difference between two similar numbers for something. If you have a much higher full-load torque with recalculation, I suspect a calculation error. If you are working with a motor that is not designed to international standards, you could get results that would not be considered "typical," but it is hard to believe that any induction motor would have a maximum torque more than 350% of full-load torque. Starting torque could be either above or below full-load torque. \$\endgroup\$ – Charles Cowie Jun 4 at 16:24
  • \$\begingroup\$ Thanks for explaining why startup torque and maximum torque are higher than full load torque. I'll definitely check my results again to make sure I've got it right. I'll definitely recheck my maximum torque value. \$\endgroup\$ – Sophie Jun 4 at 17:08

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