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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

a99b51746862e37c3db292871f6eeeea-huge-image.png

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


  • 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



  • 0
    Evening All

    I thought I was doing ok until I read Christopher's post !

    72ddc36c7c5370f215d0da4643807429-original-image.png

    I'll have to have another look.

    Thank you for your help.
  • 0
    Hi Christopher

    Just looking through your post and noticed you have used 5000RPM instead of 500RPM. That would account for your higher readings.

    I might not be that wrong after all.
  • Former Community Member
    0 Former Community Member
    Daniel,

    Well spotted. I could claim it was deliberate but, unfortunately, I need some new specs. 5000RPM is just about feasible with good bearings: Tesla super car?

    Your part I calculation is not quite right - 2piNT is the correct mech power but the relevant voltage (Ea) is 115V: the other 5V is dropped across the resistance, resulting in heat loss.

    That is why I recommended to calculate mech power first. That part of the elec power which converts to mechanical is Ea times current.

    Thus IxEa=2piNT

    Chris
  • 0
    It's OK, my numbers were not quite the same as his either, your track is correct.


    500RPM is  8.3 revs second.

    Each rev is equivalent to a 40N force sweeping a full circle of 1m radius, so 6.2 metres *40N =  call it 250 joules per rev.

    But there are 8.33 rev/second  - 2010 joules per second = watts 

    call it 2 kilowatts of mechanical work.


    Now  5 of the volts are thermal the rest is the magnetic forces of the self generated opposition due to the rotation, so 

    2000 watts, 115V  (right - this is not that realistic ) 17.3 amps.

    So the armature resistance is 5V/17A =  ~ 0.33 ohms





  • 0
    Thank you for your feedback. So I need to rework my answers. Instead of using 120v I need to use 115volts on Each and try again.

  • 0
    Daniel S Robinson:

    . . .

    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.


    . . .

     


    Hello Daniel. You have had some useful replies and I don't propose to enlarge on any of these. However I would give you a general tip for when you tackle questions like this.


    Always be sure of the units you are dealing with.


    In the first formula, I = Tω/V, the ω  refers to speed in radians per second. It is important that you convert from revs per minute. Likewise in the second formula, Ea = 2πNT/Ia, N represents revolutions per second, and the 2π performs the conversion to radians per second. If you do not use the correct units you can expect different answers.


    Some of the other replies demonstrate this.


    We seem to be lacking some information in the  question. Is this a shunt wound motor (whose speed varies little with torque) or a series wound motor (whose speed varies a lot with torque)? I would guess from the data that it is shunt wound but we should not use guess work with  calculations like this.


    Zoomup's calculator includes a hash constant to do the conversion of non-SI units. This can be a useful cross-check, but in the work you submit you will be expected to explain your calculations.


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