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Calculate Engine Rotational Speed: Formula for Frequency, Voltage, and 4-Tooth Inductive Converter

smerfy 48554 6
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Treść została przetłumaczona polish » english Zobacz oryginalną wersję tematu
  • #1 2493412
    smerfy
    Level 13  
    Hello everyone, I have a big problem. Well, I need a formula for calculating the rotational speed of the motor, if we have a given frequency and voltage, and the inductive converter disc has 4 "teeth", maybe someone had such a lab and knows what this formula looks like, I mean the formula itself, I can handle the rest. Obviously, the measurement was made using the indirect method by measuring the frequency of the pulses generated by the transducer.
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  • #2 2493455
    Aleksander_01
    Level 43  
    Hello
    Formula for an ideal motor (no slip) n = frequency / number of pole pairs. Since this "cog" has four teeth, the impulse frequency will automatically be four times the value of the stator supply frequency.
    Regards

    If you want to get RPM, the formula looks like this: n = 60f / p, and in your particular case n = (4 * 60 * power frequency) / number of pole pairs.
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  • #3 2496901
    smerfy
    Level 13  
    Thanks a lot Aleksander_01 but it seems to me that this pattern will not be useful to me. I do not have the motor pole pairs given, the only data is the rotation frequency in kHz and the supply voltage of the system. Here is the link to the exercise manual: http: //www.ime.uz.zgora.pl/ssienkowski/apps/te/cw6a.pdf.
    I don't know how to bite it, maybe someone knows what pattern to use ??. I would be grateful
  • #4 2497028
    mksm
    Level 18  
    Hello
    If you have the pulse frequency from the sensor, then to get the rotor (rotation) frequency you only need to divide by 4.
    There is something in the manual about the plot ? = f (U), so it is 2?f where f is the rotor rotation frequency. So I don't know what the problem is.
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  • #5 2497108
    smerfy
    Level 13  
    mksm, but something is wrong because it is not possible for the motor to spin at a speed of 217228 with a supply voltage of 220 [V], because if I have a frequency f = 138.4kHz at Uz = 220 V and on the basis, it receives 138400/4 and then the spin speed will be = 6.28 × 34600 = 217228. It is almost impossible. And it seems that the pattern you gave is the pulsation and is equal to 1 / 2?f.
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  • Helpful post
    #6 2497448
    mksm
    Level 18  
    Hello
    Yes, it is impossible for it to have a speed of over 2,000,000 rpm, but it's not the pattern, but the frequency that a colleague gives. Is my friend sure that they are for sure kHz and not Hz? or is the comma in the right place? because it is enough that it is not 138.4kHz but 138.4Hz and the speed is 2076 rpm or if you prefer 215.3rd / s which is the most likely speed for the motor.
    The formula for ? is absolutely right.
  • #7 2497590
    smerfy
    Level 13  
    mksm, I have reviewed the files and I have to admit you are right the frequency is given in kHz but it is not 138.4 kHz but 0.138kHz, so the formula is good, thanks a lot, now it should fit. Thank you very much once again and best regards. :D

Topic summary

The discussion revolves around calculating the rotational speed of a motor using a formula that incorporates frequency and voltage, specifically for a system with a 4-tooth inductive converter. The initial formula provided is n = 60f / p, where 'n' is the RPM, 'f' is the frequency, and 'p' is the number of pole pairs. However, the user faced challenges due to the lack of information on pole pairs and initially misinterpreted the frequency value. After clarifications, it was established that the pulse frequency from the sensor should be divided by 4 to obtain the rotor frequency. The conversation highlights the importance of ensuring the correct unit of frequency (kHz vs. Hz) to avoid unrealistic RPM calculations.
Summary generated by the language model.
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