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How will an electromotor 150HP, 2-pole, intended for 460V, 60Hz, 3510 RPM, 197 AMP be affected by working on an installation where 440V/60Hz is the voltage?

With regards to number of revolutions and torque, and power in the end?

Can you estimate what will be the reduce in power due to difference in voltage?

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  • \$\begingroup\$ This is one of those "if you have to ask, don't do it" questions -- at least, I wouldn't mess around with 150HP electric motors even if I did have a PE license, until I had lots of experience. \$\endgroup\$ – TimWescott Feb 7 at 20:28
  • \$\begingroup\$ @TimWescott - I don't have a PE license, I'm not even an electrical engineer. I'm a marine engineer (professional), trying to figure out some things with regards to a pump connected to an EM, which is described above. So I was hoping for a quick answer or two, which would really help with my side of the problem. And it's been a while since I've last read up on electrical engineering :) \$\endgroup\$ – Rook Feb 7 at 20:30
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    \$\begingroup\$ You didn't mention frequency. 440V is within the expected range of voltages for a nominal 460V line. Compared to exactly 460V, if you gave the motor exactly 440V then you can expect the current to be 5% or so higher for a given torque, with a bit more slip for the same torque (meaning it'd run slightly slower). So you'd need to derate the horsepower by about 5% to match. \$\endgroup\$ – TimWescott Feb 7 at 20:31
  • \$\begingroup\$ @TimWescott - Sorry, 60 Hz. \$\endgroup\$ – Rook Feb 7 at 20:32
  • \$\begingroup\$ If the motor's not fizzing and smoking at 460V, it probably won't fizz and smoke at 440V. I would bet that if you do get a PE on line (with relevant experience!) they would tell you to derate the thing by 5% and have a nice day. \$\endgroup\$ – TimWescott Feb 7 at 20:32
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Disclaimer: I'm not a PE. The biggest motor I've designed against was 300 watts or so, and the biggest motor that I've been an Electricians Monkey for was 10HP, and that a long time ago.

What I believe: It'll be fine. At worst, you'll need to derate it by 5%, to match the 5% drop in input voltage.

Why:

Any motor that's rated at a particular nominal voltage and frequency is going to be pretty forgiving of excursions around those rated conditions, or the motor company is going to have some pretty pissed off customers. Bigger motors will be less so, because if you can afford a 150HP motor and a place to put it, you can afford to have some pretty capable people tending it, but that doesn't mean you won't have it installed into some concrete mill off in the mountains someplace.

So, based on what I do know about induction machines and motors in general:

  • You're dropping the voltage by around 5%. That's not even a noticeable power drop for most electrical services. So that's a good sign, and my argument for "it'll be fine".
  • An induction machine is pretty laid back, and turns at a fairly constant rate with respect to the power supplied -- that's why they're popular. So if you drive it with voltage that's a bit low, then it'll slip a bit more (that 3510 RPM comes about because the motor wants to turn at 3600RPM = 60Hz exactly, but slips a bit).
  • If you have good reason to believe that a motor is operating efficiently, you can make a lot of predictions of that motor's behavior from conservation of energy arguments. It's an induction motor, so it'll turn at, essentially, 3600RPM. Power in equals power out, and power in equals current times voltage. So if you drop the input voltage by 5%, and the output speed doesn't change significantly (it won't), the input current has to go up by 5%.
  • By and large, the amount of power the motor can handle is a function of a do-not-exceed input current and a do-not-exceed input voltage. You're certainly not exceeding the input voltage, but you need to limit the input current -- you need to do this by derating your installation by 5% (or by counting on 440V being close enough). There will probably be increased current in the rotor, but it shouldn't be much -- and see my comment about a mere 5% change in input voltage.
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the standard NEMA design specifications for ac INDUCTION motors are that they will deliver rated HP at +-10% of rated voltage.

Peak torque capability of the motor will decrease at the square of the voltage difference. So 440/460 = 95.7% (well within the 10% tolerance), squared means the peak torque will drop to about 91% of normal. Motors only come in certain sizes, so at some point someone picked that 150HP motor for a specific amount of torque that it produced at rated speed, because the next size down, 125HP, was not going to provide enough torque. In the vast majority of cases, the amount you need is less than the amount the motor is capable of, so the chances that the original design engineer picked a motor that was less than 9% larger than necessary is very slim. I would not hesitate to use it.

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  • \$\begingroup\$ With the reduction in voltage, will the RPM of the motor/pump be affected? Could you write the basic equations of how from the reduction in voltage, the reduction in pump capacity is calculated? \$\endgroup\$ – Rook Feb 13 at 8:05

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