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Course work help

Daniel S Robinson
Hi

I was wondering if some one could help me with my assignment.

Normally I wouldn't ask but I'm struggling to find a formula

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I'm stuck on the 1st question, I have found two formulas   I= Tω/V   and  Ea = 2πNT/Ia   re aranged   Ia = 2πNT/Ea

Both give different answers and don't take into account the back EMF.


Can any one give me some pointers please. I have looked through all my books but can't find any answers.

Any help would be appreciated.

Thank You

Daniel Robinson


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  • Former Community Member
    0 Former Community Member
    Principles_and_Propertiees_of_high_dynamic_DC_Miniature_Motors.pdf

    Daniel,

    It's a bit of a trick question - the logical order is to calculate power first then current then resistance. The main thing you need to understand is the principle of operation (elec & mech). Assume a basic DC motor config: -

    Total power in is volts times current.

    Mech power out is torque times speed (convert rpm to radians/s)..

    Some power is lost as heat due to the winding resistance (I squuared times R).

    After voltage is applied the motor accelerates and current drops until the back emf (proportional to speed) plus the voltage across the winding resistance is equal to the input voltage.

     Thus, very approximately (mental arithmetic): -

    Mech power out 5000rpm (about 500 rad/s) times 40Nm is 20kW 

    (that is a lot for a DC motor)

    5V is lost across the winding resistance so the 20kW comes from 115V times current so the current is approx 170A. (power is volts times current)

     From Ohm's law the resistance is, therefore, 0.03Ohm.

    (this is not a realistic example -typical academic exercise).

    If you increase the load to 60Nm the speed must drop so that the back emf drops, allowing the current to increase. Torque is proportional to current so that the new current is 170A times 60/40  = 250A.

    The current through the resistance results in 0.03 times 210 = 7.5V.

    The remaining voltage (120 - 7.5 =112.5V) is the back emf (new equilibrium).

    The back emf is proportional to speed and so I estmate 5000rpm times112.5/120.

    I recommend the attached for further study.

    Chris



Reply
  • Former Community Member
    0 Former Community Member
    Principles_and_Propertiees_of_high_dynamic_DC_Miniature_Motors.pdf

    Daniel,

    It's a bit of a trick question - the logical order is to calculate power first then current then resistance. The main thing you need to understand is the principle of operation (elec & mech). Assume a basic DC motor config: -

    Total power in is volts times current.

    Mech power out is torque times speed (convert rpm to radians/s)..

    Some power is lost as heat due to the winding resistance (I squuared times R).

    After voltage is applied the motor accelerates and current drops until the back emf (proportional to speed) plus the voltage across the winding resistance is equal to the input voltage.

     Thus, very approximately (mental arithmetic): -

    Mech power out 5000rpm (about 500 rad/s) times 40Nm is 20kW 

    (that is a lot for a DC motor)

    5V is lost across the winding resistance so the 20kW comes from 115V times current so the current is approx 170A. (power is volts times current)

     From Ohm's law the resistance is, therefore, 0.03Ohm.

    (this is not a realistic example -typical academic exercise).

    If you increase the load to 60Nm the speed must drop so that the back emf drops, allowing the current to increase. Torque is proportional to current so that the new current is 170A times 60/40  = 250A.

    The current through the resistance results in 0.03 times 210 = 7.5V.

    The remaining voltage (120 - 7.5 =112.5V) is the back emf (new equilibrium).

    The back emf is proportional to speed and so I estmate 5000rpm times112.5/120.

    I recommend the attached for further study.

    Chris



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