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Calculating Electric motor consumption when a VSD is fitted

  • 27-08-2018 7:26am
    #1
    Calahonda52
    Registered Users, Registered Users 2 Posts: 12,582 ✭✭✭✭


    Want to calculated the power consumed by an electric motor with a VSD, where the motor RPM is not available:

    Assumptions:
    1: Motor efficiency: 85%
    motor name plate output power, o/p, divided by this to give i/p: input power

    2: Load Factor: 70%
    i/p multiplied by this

    3: VSD efficiency: 95%
    Value from 2 above divide by this

    Then:

    If
    RPM = N = (Hz*(2/No of poles)*60
    And
    P1/P2 = (N1/N2)*3 cubed where P is the power consumed.
    Then
    P1/P2 = ((Hz1*(2/No of poles)*60)/ ((Hz2*(2/No of poles)*60))*3 cubed
    P1/P2 =(Hz1/Hz2)*3 cubed

    So P2= P1*(Hz2/Hz1)*3 cubed.
    Eg
    P at 20 Hz = (P at 50 Hz) * (20/50)*3 cubed
    = 3.47*(0.4*0.4*0.4)= 0.22 kW

    This make sense?

    “I can’t pay my staff or mortgage with instagram likes”.



Comments

  • #2
    ted1
    Registered Users, Registered Users 2 Posts: 23,658 ✭✭✭✭ted1


    Calahonda52 wrote: »
    Want to calculated the power consumed by an electric motor with a VSD, where the motor RPM is not available:

    Assumptions:
    1: Motor efficiency: 85%
    motor name plate output power, o/p, divided by this to give i/p: input power

    2: Load Factor: 70%
    i/p multiplied by this

    3: VSD efficiency: 95%
    Value from 2 above divide by this

    Then:

    If
    RPM = N = (Hz*(2/No of poles)*60
    And
    P1/P2 = (N1/N2)*3 cubed where P is the power consumed.
    Then
    P1/P2 = ((Hz1*(2/No of poles)*60)/ ((Hz2*(2/No of poles)*60))*3 cubed
    P1/P2 =(Hz1/Hz2)*3 cubed

    So P2= P1*(Hz2/Hz1)*3 cubed.
    Eg
    P at 20 Hz = (P at 50 Hz) * (20/50)*3 cubed
    = 3.47*(0.4*0.4*0.4)= 0.22 kW

    This make sense?
    Unless your doing an academic excercise it makes much more sense to sub Meter the load.


  • #3
    Calahonda52
    Registered Users, Registered Users 2 Posts: 12,582 ✭✭✭✭Calahonda52


    ted1 wrote: »
    Unless your doing an academic excercise it makes much more sense to sub Meter the load.

    This is much more than an academic exercise: I have 146 extract fans and 54 AHUs, all on VSDs

    “I can’t pay my staff or mortgage with instagram likes”.



  • #4
    Nukem
    Registered Users, Registered Users 2 Posts: 4,082 ✭✭✭Nukem


    If there is a VSD you should be able to see the amps (L1,L2,L3) & voltage onthe HMI (toggle down), so potentially do the V*I calc? Check the VSD spec for the likes of PF (e.g. >.98) and if the motor was matched correctly and programmed right, there should have the motor efficiency entered and accounted for

    Just check for 3ph as you may need to multiply by 3*SQRT (1.732)


  • #5
    Calahonda52
    Registered Users, Registered Users 2 Posts: 12,582 ✭✭✭✭Calahonda52


    Nukem wrote: »
    If there is a VSD you should be able to see the amps (L1,L2,L3) & voltage onthe HMI (toggle down), so potentially do the V*I calc? Check the VSD spec for the likes of PF (e.g. >.98) and if the motor was matched correctly and programmed right, there should have the motor efficiency entered and accounted for

    Just check for 3ph as you may need to multiply by 3*SQRT (1.732)
    Thanks for this.
    I have done that V*I math but I need to be able to calculate the power requirement as the VSDs on the AHUs ramp up when the extract fans come on so as to maintain + pressure and meet the ACH requirements

    “I can’t pay my staff or mortgage with instagram likes”.



  • #6
    2011
    Moderators, Home & Garden Moderators, Technology & Internet Moderators, Regional East Moderators Posts: 12,603 Mod ✭✭✭✭2011


    Calahonda52 wrote: »
    Thanks for this.
    I have done that V*I math but I need to be able to calculate the power requirement as the VSDs on the AHUs ramp up when the extract fans come on so as to maintain + pressure and meet the ACH requirements

    Well that is easy then, simply run the motor at different speeds and using a grip on ammeter measure the current drawn on the supply side of the VSD. You already know the line voltage and the power factor is going to be a constant so a power calculation is straight forward.
    It would then be possible to put a table together of RPM V power consumed. If you wanted to get really fancy you could graph it.


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  • #7
    Calahonda52
    Registered Users, Registered Users 2 Posts: 12,582 ✭✭✭✭Calahonda52


    Thanks for all the insights.
    As there are c 200 fan motors in this complex, ranging from (0.75 to 25) kW, I need to know is my math correct.

    “I can’t pay my staff or mortgage with instagram likes”.



  • #8
    2011
    Moderators, Home & Garden Moderators, Technology & Internet Moderators, Regional East Moderators Posts: 12,603 Mod ✭✭✭✭2011


    Calahonda52 wrote: »
    Thanks for all the insights.
    As there are c 200 fan motors in this complex, ranging from (0.75 to 25) kW, I need to know is my math correct.

    That’s easy.
    Power consumed = Root 3 x Line voltage x line current x power factor

    Now all you need to do is measure the current.


  • #9
    Calahonda52
    Registered Users, Registered Users 2 Posts: 12,582 ✭✭✭✭Calahonda52


    Calahonda52 wrote: »
    Want to calculated the power consumed by an electric motor with a VSD, where the motor RPM is not available:

    Assumptions:
    1: Motor efficiency: 85%
    motor name plate output power, o/p, divided by this to give i/p: input power

    2: Load Factor: 70%
    i/p multiplied by this

    3: VSD efficiency: 95%
    Value from 2 above divide by this

    Then:

    If
    RPM = N = (Hz*(2/No of poles)*60
    And
    P1/P2 = (N1/N2)*3 cubed where P is the power consumed.
    Then
    P1/P2 = ((Hz1*(2/No of poles)*60)/ ((Hz2*(2/No of poles)*60))*3 cubed
    P1/P2 =(Hz1/Hz2)*3 cubed

    So P2= P1*(Hz2/Hz1)*3 cubed.

    P1=P2*(Hz1/Hz2)*3 cubed
    Eg
    P at 20 Hz = (P at 50 Hz) * (20/50)*3 cubed
    = 3.47*(0.4*0.4*0.4)= 0.22 kW
    Error corrected

    “I can’t pay my staff or mortgage with instagram likes”.



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