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Homework Help: Electrical Machine - Shunt DC Motor Problem

  1. Nov 28, 2017 #1
    Hi guys, can someone please have a look at the following problem that I am struggling with;

    An 8.25 kW shunt DC motor is supplied with a terminal DC voltage of 300 V. The armature resistance of the motor Ra is 0.25 ohms, and the field resistance Rf is 40 ohms. The field winding consists of 1500 turns. An adjustable resistance Radj is connected in series to the field winding. Radj may be varied over the range from 0 to 80 ohms and is currently set at 20 ohms. Armature reaction may be ignored in this machine. The magnetization curve for this motor, taken at a speed of 1200 rpm, is shown in Figure Q1.

    (i) What is the speed of the motor when operating at rated load with an armature current of 30 A?

    (ii) If the motor is now unloaded with no changes in terminal voltage or Radj, what is the no-load speed of the motor?

    (iii) Iron core losses are 250 W. What are the copper losses and mechanical losses in the motor at rated load? (ignore brush and stray losses)

    (iv) What is the efficiency of the motor at rated load?

    (v) If the terminal voltage is reduced to 260 V and the armature current stays at 30 A, calculate the required Radj in order to restore the speed of the motor to the value in (i).






    I have attempted to calculate I(f), which I am getting a value of 5A.
    Also the mmf seems to be 7500 A.turns - (1500*5)

    But I do not know where to go from here, any help would be appreciated.
     

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    Last edited: Nov 28, 2017
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  3. Nov 28, 2017 #2

    cnh1995

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    You know the back emf generated with 1500A-T at 1200 rpm from the graph.
    Compare it with the actual emf induced in the motor. With constant flux, both the emfs are proportional to the speeds.
     
  4. Nov 29, 2017 #3
    So would it be correct if I used the formula of Ea1/Ea2 = nm1/nm2

    And Ea1 in this case would be 75V (reading from the graph at 1500 A-T) and Ea2 being 275V and nm1 is 1200 rpm.
    So nm2 would come out to be 4400 rpm?

    Thank you for your help so far.
     
  5. Nov 29, 2017 #4

    cnh1995

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    Yes.

    No. You have E1=275V (from graph) and N1=1200 rpm.
    You have to calculate E2 from the data given in the problem and use the above formula to get N2.
     
  6. Nov 29, 2017 #5
    okay so using the formula E2 = Vdc - Ia*Ra

    I would get a value of 292.5 V for E2.

    So therefore my nm2 would be 1276.36 rpm.

    Does that look right to you?
     
  7. Nov 29, 2017 #6

    cnh1995

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    Yes.
     
  8. Nov 29, 2017 #7
    Thank you so much with your help so far!

    My next confusion arises with the no load speed of the motor, I'm not sure what no load actually means, and so what assumptions I can make about the system?

    Any guidance would be a great help.
     
  9. Nov 29, 2017 #8

    cnh1995

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    For a motor, no load means zero load torque. Practically, the motor generates just enough torque so as to overcome friction (and windage.)

    To be accurate, you should first determine the frictional torque from the given data and calculate the no-load armature current. But that would come out to be very small. So just take no load armature current=0.
     
  10. Nov 29, 2017 #9
    Okay, so after doing that I am coming to an equation of n2 = (I(f)1*E2/I(f)2*E1) * n1

    my problem is how do I get a value of E1 and I(f)1 whilst knowing that n1 is 1200 rpm?

    Because I have calculated previosuly that the field current is 5A and I know E2 would be 300 V from the information you have given me.
     
  11. Nov 29, 2017 #10

    cnh1995

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    Yes, so you know E1=275V, N1=1200 rpm.
    Now you have E2=300V, what is N2?
    The field current (and mmf) is unchanged.
     
  12. Nov 29, 2017 #11
    Therefore n2 = (300/275) * 1200
    Which comes out to be 1309.09 rpm
    Correct?
     
  13. Nov 29, 2017 #12

    cnh1995

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    Yes.
     
  14. Nov 29, 2017 #13
    Thank you so much for your help, I feel indebted to you now!

    I will attempt (iii) now, will you be able to verify if my answer is correct please?
     
  15. Nov 29, 2017 #14
    Here is my attempt at the next question:

    The rated load armature copper loss is given as: Pacl = Ia^2 * Ra

    Which comes out to a value of 225 W

    The field copper loss is: Pfcl = I(f)^2 * (Rf + Radj)

    Which comes out to be 1500 W

    Now I'm not sure what to do with the information of Iron Core losses are 250 W?

    My attempt to get the mechanical loss is as follows:

    Power absorbed by the machine is: Pind = Ea * Ia ----> (267.5) * (30)

    And that comes out to be 8025 W

    Now I would subtract that by 8.25 kW but it would get me a negative answer, so I'm not sure what's going on.
     
  16. Nov 29, 2017 #15

    cnh1995

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    Leave the field loss out of the calculations, for it matters only when calculating the efficiency.

    Armature voltage is 300V, armature current is 30A. What is the back emf?(you've already done that part). What is the gross mechanical power? How much is the copper loss?
     
  17. Nov 29, 2017 #16
    The back emf value is 292.5 V?

    If so the power absorbed by the machine would be Ea * Ia = (292.5)(30) = 8775 W

    Copper loss would just be as above, no? Ia^2 * Ra = (30)^2 * (0.25) = 225 W

    Mechanical loss would be 8775 - 8250 = 525 W

    How do I calculate gross mechanical power?
     
  18. Nov 29, 2017 #17

    cnh1995

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    This is the gross mechical power.
    Yes.
    No, 525W is the fixed (or constant) loss. It is the sum of iron loss and friction loss.
    You can subtract the iron loss from the fixed loss to get the mechanical loss.
     
  19. Nov 29, 2017 #18
    Thank you!

    So for efficiency, as you stated we need to use field copper loss, which would be what I calculated above as 1500 W.

    But what I'm confused about is that ive seen an equation saying Power in can be calculated as follows: Vdc * (Ia + If)

    With this equation it seems like i dont need the other values?

    Can you please clarify this.

    Thanks again for helping me so much!
     
  20. Nov 29, 2017 #19

    cnh1995

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    That is the total input power to the motor, and 8.25kW is the output power. You can calculate efficiency from this.
     
  21. Nov 29, 2017 #20
    Thanks again!

    For the last part of the question i have to find a suitable Radj value so as the speed of the motor would be 1276.36 rpm when terminal voltage is 260 V instead of 300V.

    Would you point me in the right direction for the final time please.
     
    Last edited: Nov 29, 2017
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