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Power engineering questions

  1. May 14, 2015 #1
    Hello, I've solved some questions for my exam tomorrow. Since I do not have any answers to the questions, could you please confirm if it is correct?
    I have questions on the last problem. Also I will post more questions, just I am working on them now.

    Question 1:
    a) [itex] 6 Pole ->3 pairs. N_{speed} = 1000 rpm;
    w_r = w_o (1-s)
    w_r = 950 rpm
    [itex]V_a = E + R_a x I_a -> E = 205V at 1000 rpm [/itex]
    [itex] at 1100 rpm E = \frac{1000}{1100} x 205 = 186V [/itex]
    [itex]So,~ from~the~formula~for~V_a~we~get,~ 220 = 186 + I_a x 0.5 -> I_a = 67.27 [/itex]

    Switches are put on the Live wire, then when it is open the load is connected to the neutral line (0v) , so anyone touching it would NOT get a shock.
    If the switch is on the neutral wire, then when it is open the load is connected to the live wire, so anyone touching it would get a shock.

    Question 2:
    Using the equation
    [tex] S = Pcos(\theta)~ and~ Q = Ptan(\theta)[/tex]
    i)Total Real Power = [tex]9kW+20kW+12kW = 41kW[/tex]
    ii)[tex]2.96kVAr+9kVAr = 11.96kVAr
    iii) [tex] S = \sqrt{41^2+11.96^2} = 42.7kVA [/tex]
    iv) [tex] S = VxI[/tex] So, [tex] I = 185.7A[/tex]
    v) Total power factor [tex] \frac{kW}{kVA} = \frac{41}{42.7} = 0.96[/tex]

    [tex]Q = \frac{V^2}{Xc}[/tex] [tex]Xc = 0.01923 Ohms [/tex][tex] C = 166μF[/tex]
    Here I have some ideas, but the value for P gets really big so I am confused if it correct. If we use the same equation from the beginning of the question [tex]Q = Ptan(\theta), P = 685kW[/tex] compared to 41kW .

    Capacitors are used to supply reactive power Q locally, rather than deliver it from the source over the transmission and distribution network. This reduces transmission losses and voltage drops.

    Question 3:

    c) [tex]\frac { V_2}{V_1} = \frac{N_2}{N_1} = 11 000 x N_1 = 230 x 3000 ~->~ N_1 = 63~ turns[/tex]
    d) Here I don't know how to find the [tex]I_L[/tex] and [tex]I_{phase}[/tex]
    the only thing that I can relate is the fact that the primary has delta connection, so [tex] I_L = \sqrt{3}I_{phase} [/tex]
    Which means that the line currents of secondary and primary should be the same and just the Phase current of the primary will be different -> 34.64 A

    Question 4:
    To synchronise, the generator voltage must have the same magnitude, frequency and phase as the grid.
    It has to be speed up to the 375rpm,
    And then increase the field current If until the generator output = 11 kV (line). [Note that the generator is open circuit at this stage, therefore E = Vgenerator = 11 kV.] Check for correct phase, then close circuit breaker S1 to connect to grid. The generator is now synchronised, but no power is being transferred yet.

    By the angle between phase voltage and phase EMF.

    I have similar example in the notes but it uses the 11kV as a Line voltage, which doesn't make any sense to me. Why should they ask to calculate it, when it is given. It must be something else.
    d) the power factor can be adjusted by [tex]I_f[/tex]

    Any feedback is much appreciated!
    Thank you !

    Attached Files:

    Last edited: May 14, 2015
  2. jcsd
  3. May 16, 2015 #2


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    Staff: Mentor

    A better approach is to work with another student in your class, each solving some of the problems and swapping answers with the other. Only post to PF about those one or two problems that neither of you can solve.
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