Electronics

Electronics

Electro magnetic Power Vs Shaft power of motor

    • junaed.buet
      Subscriber

      Hello. I designed a 2-D FE BLDC motor and the windings are excited with an external circuit (fig 1).I tried to calculate the motor actual input power with three volt meter and three ammeter and I used the  formula to calculate the motor input power Pin_motor = Average (Ia*Va+Ib*Vb+Ic*Vc).The result is shown in fig 2. I assigned core loss in rotor and stator (result in fig3) and eddy effects in magnets and with estimated phase resistance (rph) I calculated copper loss with the formula Pcopper = Average ((Ia^2+Ib^2+Ic^2)*rph). The motor power value I receive by subtracting all losses (Coreloss, solidloss, Copperloss) from Pin_motor is close to the value of electromagnetic power which is Moving1.Torque*Moving1.speed. Should it not be equal to Shaft power? I learnt from a previous post in forum is that the shaft power is calculated by subtracting Coreloss from Moving1.Torque*Moving1.speed. But here the power I am getting by subtracting all losses from motor input power is a value close to electromagnetic power not Shaft power. Shouldn't be the output is shaft power? The moving1.torque curve is shown in figure 4 and I am summerizing all corresponding values with used formula in table 1. 

       

      Fig 1: External Circuit

      Fig 2: Motor input power                                                            Fig 3: Coreloss (Assigned in rotor and stator)

      Fig 4: Moving Torque

      Table: 1 

      You can download and check the calculation from the google drive link: Calculation_table

      Sorry for the long post. In one sentence,  the output power = input power - loss matches only if I consider the moving1.torque is shaft torque(which is motor output power). Kindly let me know where I may have mistaken.

    • HDLI
      Ansys Employee

      Hello junaed.buet,

            FEA uses different methods to calculate and get voltage*current and torque*speed, so we would not get the same values on the input and output power in the simulation. Additionally, we could not consider any mechanical losses such as windage and friction loss, core loss or stray loss in the input power calculation too.

            My suggestion is

            Output power = torque*speed*stacking factor - core loss - other mechanical/stary loss

            Input power = output power + all losses

      HDLI

          

    • junaed.buet
      Subscriber

      Thank you for your response. I specified the stacking factor for rotor and stator in the material properties as shown in the figure below. Do I have to multiply the stacking factor with the torque (moving. torque) again or in my calculation the stacking factor is used already?

    • HDLI
      Ansys Employee

      Hello junaed.buet,

            If specifying the stacking factor, you do not need to multiply it again. Thanks.

      HDLI

    • junaed.buet
      Subscriber

      Thanks. From the simulation I got the inverter input current from an ammeter at the DC voltage side .(Figure Attached). The inverter input current value is 24.56 A and the DC voltage is 48 Volt. Isn't inverter input power here is  inverter input current(24.56 A)*DC voltage(48 Volt)? Also is it wrong or not valid, the motor input calculation equation in my 1st post Pin_motor = Average (Ia*Va+Ib*Vb+Ic*Vc)? Sorry for bothering :( 

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