220/1/50 @ 60Hz- Effect and Consequences 3

[Mechomatic]

Hello EEs-

For reasons that are somewhere between "trivial" and "pipe-dream", I am trying to learn about the viability of utilizing a centrifugal pump with a nameplate rating different than the electrical supply. I've poked around this forum and gleaned some understanding, but I'm afraid the single class/lab in EE required for my BSME did not provide me with the tools I need to decipher the mathematics at play. Just to have an example, let's take the Taco 0010-F3 centrifugal circulator pump, which has a permanent split-capacitor, impedance protected motor. Nameplate data indicates: 115/1/60, 1.1A, 3250 RPM, 1/8 HP. It also has available motors in 220/1/50, 220/1/60, 230/1/60, and 100-110/1/50-60.

My understanding is that motors are designed for a specific V/Hz ratio (proportional to magnetic flux, right?). Operating at a different frequency with the same ratio (as with VFD) produces a linear relationship between frequency and motor torque. Operating at a different V/Hz ratio can result in problems, such as drawing too much current and burning up the motor. Am I right that operating a 60Hz motor at 50Hz with constant voltage (so at a higher V/Hz) is worse than operating a 50Hz motor at 60Hz? I say this under the impression that operating at a higher-than-rated V/Hz will result in too high current for the windings. Is there a quick way to predict what the actual current through the motor will be at 60Hz? I also guess that operating the motor at 60Hz will result in less torque, higher speed, and thus a longer start up time at increased start-up current. With the motor on a dedicated breaker, my guess would be that the breaker might have to be oversized (more than normal) so it doesn't produce nuisance trips during the prolonged start-up time?

Stemming from operating at non-design condition, what effect can I expect in terms of motor/pump output? The nameplate speed indiates ~10% slip. But with the higher speed and lower torque, will there be additional slip and how can I go about calculating that (analytically)? Based on

Torque = HP * 5250 / RPM

is it a safe assumption that the torque output of the motor will be 5/6 of the rated motor torque? Or will there also be an effect on the horsepower?


I know this is a lot of (probably) basic stuff I'm asking, but I seem to be going in circles on my own and hoped that one or some of you could clear the haze and set me on the right track. Thank you for your consideration.

 

[deltawhy]

GTME, check the FAQs area, there is a well written article on this exact topic.

 

[waross]

With the same V/Hz ratio and a frequency/voltage change:
Torque stays the same.
Speed changes in the ratio of the frequency change.
HP changes in the ratio of the frequency change.
Centrifugal pumps do not respond well to speed changes.
The capacitor may need to be changed on the PSC motor to develop maximum torque. The impedance of the capacitor and of the phase shifted winding change in opposite directions when the frequency is changed.
Read the FAQ

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[Mechomatic]

Thank you. I'll admit that I have not used the FAQs before and did not realize that the website search does not automatically include them when searching for a topic. I'll be sure to check there first in the future.

 

[Mechomatic]

So reading the FAQ, my take away is:

--50Hz motor to 60Hz supply, constant voltage- generally you're ok as far as the motor goes as long as the increased load HP requirement due to higher RPM doesn't cause the full load current to rise above the current shown on dataplate. Biggest issue here is your PF will drop unless you add capacitors to counteract the increased reactive power load (?).
--60Hz motor to 50Hz supply, constant voltage- probably shouldn't do it.

waross- can you elaborate on what you said about changing the capacitor?

This application would be for a heat transfer fluid circulator pump that has non-critical flow requirements. Slight decrease in flow rate is ok, increase in flow rate is just fine. I'll run this by the pump engineering forum, but- while I'm here- is there anything you guys readily see as a problem with this application? We're talking ~5 ft pressure head, between 20-30 gpm flow of 1.0 specific gravity fluid, so not a high load requirement.

 

[Mechomatic]

I'll pile this on here, if you gentlemen don't mind... I'm afraid my tiny company has no electrical engineers around for me to pester, so I'm left to bother you all in the hope you'll grace me with some much needed knowledge.

Trying to calculate PF

115V/1Ph/60Hz motor, 1.10A, 3250 RPM, 1/8 HP.

If I want to find the reactive load, then I take 1/8 HP * 745.9 W/HP = 93.2 W, which is the (max) working power used to develop torque in the motor.

93.2W / 115V = 0.809A, current used for torque.

sqrt( 1.1^2 - 0.809^2 ) = 0.745 A = current used to develop magnetic field in the core- is that correct? That's the current in the KVAR?

115V * 1.10A = 126.5 W, which is apparent power (KA) at full load? Or should it be 93.2 W (KW) + (0.745 A * 115 V) (KVAR) = 178.9 W (KA)


I know PF = KW / KA. Did I understand correctly (from a different FAQ) that the KVAR will be constant for a given motor at a given power supply- so current to develop the magnetic field is steady no matter the load applied to the motor?

 

[waross]

Power factor for that size motor is easy:
Forget it, ignore it, and don't think about it again.
I am sure that you have experienced a single phase motor trying to start with a problem in the start circuit. It hums and doesn't turn. It needs a rotating magnetic field to start turning.
A three phase motor is easy,- the three phases develop a rotating field. Other methods must be used to develop a rotating field to start a single phase motor.
A start winding is used and the phase angle of the current is altered to provide phase shift.
The phase angle of the current and the resulting magnetic field is determined by the inductive reactance of the winding and the resistance of the winding. As the ratio of resistance to reactance in the winding changes so also does the phase angle of the current change.
The main winding is highly reactive during starting and the current is shifted towards 90 degrees lagging.
The start winding is arranged to have less phase shift. This may be done in several ways.
1>- The start winding is designed with more resistance relative to the inductive reactance. This changes the ratio and the phase angle of the current shifts away from 90 degrees towards zero degrees. This develops enough of a rotating field to start the motor.
2>- A capacitor is placed in series with the start winding. The capacitive reactance cancels part or all of the inductive reactance of the start winding. This may develop a greater phase shift than the resistance method and often generates less heat in the starting winding.
3> The start winding capacitor is sized so that angular difference between the main winding and the start winding is close to the physical displacement angle between them. (Usually 90 degrees.) The start winding is left in the circuit when the motor is running and adds to the torque developed.
Reactance is a description of the action of a capacitor or inductor in an AC circuit.
Inductive reactance is proportional to the frequency. Capacitive reactance is inversely proportional to the frequency.
When the frequency is increased, the angle that the current is shifted changes and may no longer be optimum. The motor will probably still start and run. It may work well or it may overheat. You may consider changing the capacitor to one with (5/6[sup]2[/sup] = 70%) of the original rating. Due to the resistance of the winding this will not be exact but it should be very close. I would expect the % error to be less than the % tolerance of the capacitor.
The pump? If the motor is drawing close to rated amps now you may have to restrict the flow at 60 Hz. If the motor is drawing less than full load current it may work fine.
Solution:
Supply the correct voltage for 60 Hz operation using a transformer if needed. (Auto transformer connection recommended.)
Change the capacitor.
restrict the discharge until the current is at or below rated current.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[itsmoked]

As Bill states in his last sentence you're lucky you have a load you can easily alter without effecting its function. Once you have it running check the current. If it's a high just throttle the flow until it's within the motor's FLA. That will actually put it back, (pretty much), to your original flow.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[Mechomatic]

waross- Thank you very much. I greatly appreciate your taking the time to dumb it down for me. I'll look into getting an autotransformer to step up the voltage ~20% and swapping out the capacitor for one with ~70% rated capacity.

itsmoked- Yes, I am . I'll admit that I probably would not be so adventurous (or cavalier, depending on your viewpoint) with potentially installing a motor with a different supply frequency if the output were a critical system parameter. But, I believe I now have (at least some of) the tools to go about doing so in the future, if such a need arises. I certainly will put in the effort and time required, though, in a critical application. I'm fortunate this first time venturing into tinkering with nameplate electrical supply is a "relaxed" opportunity to learn and experiment.

Again, thank you all for your guidance and expertise.

 

[electricpete]

I know PF = KW / KA. Did I understand correctly (from a different FAQ) that the KVAR will be constant for a given motor at a given power supply- so current to develop the magnetic field is steady no matter the load applied to the motor?

Comparing steady state operation at different loads (let's say no-load and 100%) with constant voltage and supply frequency:
The kvar can vary substantially. My rough thumbrule a factor of 2 higher at full load. The reason for the change is var consumed in the leakage elements of the equivalent circuit.

Many folks assume (incorrectly) that vars are roughly constant as load changes. That approach probably originates from focusing only on vars consumed by the magnetizing element of equivalent circuit (which are almost constant with load) and neglecting vars consumed by leakage elements (which vary with load squared). Such approximation is pretty close for a transformer but not nearly as close for a motor.

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(2B)+(2B)' ?

 

[electricpete]

For single phase motor as Bill says there may be other considerations for example run cap in the circuit.

=====================================
(2B)+(2B)' ?