Recent content by LagCompensator

  1. L

    Why are salient pole rotors preferred in hydro plants?

    Thanks, it makes sense. So basically a cylindrical rotor would be preferred in all cases due to uniform airgap, and hence reduce distortion in the airgap flux? However due to the slow operating speed of i.e. hydro plants we need to use salient rotors, because we would require a lot of poles to...
  2. L

    Why are salient pole rotors preferred in hydro plants?

    Hi all, I tried to post in this post https://www.physicsforums.com/threads/salient-and-non-salient-pole-synchronous-generator.878198/ but I now realise that creating a new thread might have been better. So... Salient rotors are not used in high rotational speed applications, and we must use...
  3. L

    LaTeX How can I get this chapter heading style (screenshot in post)

    Hi, Do anyone know how I can get this following chapter heading style? I've searched, but have not been able to find this exact one.
  4. L

    Parallel Resonance in Power System

    Hi, imagine we have the following circuit shown in the figure below. There are installed a passive filter to filter the 5th harmonic. Now, after installation this creates a parallel path with a generator connected, lets say that this parallel circuit has a resonance frequency at the 7th...
  5. L

    5th and 7th Harmonic mitigation (D-yd transformer)

    Bump, could any kind soul check if my last post (#8) is correct so we can settle for an answer.
  6. L

    5th and 7th Harmonic mitigation (D-yd transformer)

    Hi, I think that full cancellation is theoretical possible if the loads are equal, here's why: In the delta-delta the currents are the same at primary and secondary. a1 = x1 = sin(5 \omega t)\\b1 = y1 = sin(5 \omega t - 120^{\circ})\\c1 = z1 = sin(5 \omega t + 120^{\circ}) for the delta-wye we...
  7. L

    Circulating currents in a delta winding

    Thanks, for your time. No extra elaboration is needed @cabraham. By the way I found some Spice simulations for the pretty much the same problem, can't believe I have not seen that page before. http://www.allaboutcircuits.com/textbook/alternating-current/chpt-10/harmonics-polyphase-power-systems/
  8. L

    Circulating currents in a delta winding

    Hi, I have also been struggling with this topic, I have some questions. If we think of a 3 phase load producing tripling harmonics (current sources in parallel with load, injecting harmonic currents into the system). (I've read that harmonics generated due to non-linear loads can be thought of...
  9. L

    Odd Harmonics in Power System - reduction

    @jim hardy - Amazing, fantastic explaination.
  10. L

    Odd Harmonics in Power System - reduction

    Hi, sorry for interrupting your thread @OliskaP. @jim hardy, How exactly do a zig zag transformer manage to shift 60 degrees? I neither fully understand the zig zag transformer, and can not really seem to understand how this 60 degree phase shift is created. best regards.
  11. L

    Frequency Control in Power Plants -- Droop

    Hi, how do power operators change the droop setting in modern power plants? Back in the days it was with help of a flyball governor(?). In todays power plants do operators adjust some sort of gain electronically via a scada system or something?
  12. L

    Synchronous generator phasor diagram

    1. Problem Statement: A 4-pole, star-connected, 50 Hz, 11kV, 40 MVA turbogenerator, with a synchronous reactance of 0.8 p.i., is connected to a power network. This power network can be represented by 11-kV infinite bus with a series reactance of j 0.5 Ω. A voltage regulator adjusts the field...
  13. L

    Power supplied by capacitor bank

    When I calculated and got the wrong sign I did the following: (V_2-V_{Ground}) \times Y_c When I do it like this I am assuming the current leaves the bus, which should not be the case, therefore I should do it like this: (V_{Ground}-V_2) \times Y_c Doing it this way assumes that current flows...
  14. L

    Power supplied by capacitor bank

    j0.25 is the admittance not impedance, so if I solve the value for C, I get a positive value. Y = \frac{1}{Z} \rightarrow Z = \frac{1}{Y} = \frac{1}{j0.25} = -j4 \\ Z = \frac{1}{j\omega C} = -j4 \rightarrow C = \frac{1}{4 \times \omega} I did not write that it was the admittance, I just...
Top