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Answer: Although this question does not pertain to our course directly, still we would try to answer this question. The voltage rating of the fan capacitors can have some pre-specified values only and these values are given in appropriate Indian Standards. One such standard is IS : 1709 - 1984. Any capacitor manufacturer would alsways follow this standard and the capacitors are manufactured according to the specifications mentioned in this (and may be other related) standards.
Answer: The basic concept of this is quite clearly explianed in many basic electrical engineering textbooks.
Answer: No, not at all. The HVDC converter performance does not depend on the turns ratio of the converter transformer.
Answer: Click the module "stability Analysis" at the left pane of the page of the course. A list of different topics under this module would be shown to you. The last item in this list is "Appendix-A system data". Click that. A .pdf file containing the data would emerge.
Answer: This is clearly explained in the lecture notes of Newton - Raphson (Polar) load flow method.
Answer: If you take instantaneous current i = A*sin(wt), then the instantaneous voltage across an induactance L would be v = L*(di/dt) = AwL*cos(wt). Thus, v leads i by 90 degree. If the current phasor I is taken to be the reference phasor, then the voltage vector can be written as V = jI*wL = I*(jwL) = I*(jX). Thus, the effective impedance is Z = jX (i.e. it is oriented at 90 degree ahead from the reference vector) and that's why the reactance is always shown along the imaginary axis.
Answer: As I understand, there is not much of a technical reason for this practice. In fact, this is not followed in many other countries. For example, in USA, the voltage levels are not in multiple of 11. I guess, this practice was adopted by British and we have simply continued with that. Moreover, this is true only upto 220 kV. Beyond 220 kV (e.g. 400 kV, 765 kV) the oltages are not in the multiple of 11.
Answer: It is difficult to graphically represent the elements of Zbus matrix of a three phase system. However, with an analogy of the Zbus matrix of a single phase system, we can understand the nature of the elements of Zbus matrix of a three phase system. For a single phase system, Z_mn relates V_m to I_n, for i = 1, 2, ... N, where N is the total number of buses. Now, let us consider a two bus three phase system. The bus voltage vector is [V_1a V_1b ..... V_2c]^T and the bus injection current vector is [I_1a I_1b ..... I_2c]^T. The Zbus matrix is a (6x6) matrix with the elements Z_ijpq, where, i, j = 1, 2 and p, q = a, b, c. Thus, the element Z_ijpq relates the voltage V_ip to the current I_jq. Similar observation is true for any N bus three phase system.
Answer: As a principle, we do not encourage any of our students to use any program developed by somebody else. There are two reasons for it. Firstly, if you use any program developed by somebody else, then you really can not have the total command on the concept involved. After all, when you are developing your program, you have to make it sure that you understand the underlying concept completely (otherwise your program would not work). Therefore, once you successfully develop any program, it can be safely assumed that your understanding of the underlying concept is complete. Secondly, it is very difficult to fully comprehend a program written by somebody else. Therefore, if you need to make any advancement to the code, it would be next to impossible for you. On the other hand, you can always make amendments and improvements on your own program as you have total command over it. Because of the above two reasons, we strongly suggest that you please develop your own code.
Answer: We do so for taking a little pessimistic view about the rating of circuit breaker. Let me explain a bit more clearly. Because of the loads, in any system, the pre-fault voltage magnitudes are usually less than one per unit. If we use these voltage magnitudes for fault calculation, then the magnitudes of the fault currents would be little less as compared to the case when we do the calculations with voltages equal to (1+j0). We take the second view, i.e. voltages equal to (1+j0) and calculate the fault currents to be little higher so that the rating of the circuit breakers would be sufficient enough to tackle any scenario effectively.
Answer: For analysing the transient stability for unbalanced fault, we can no longer assume the system to be balanced. Therefore, in this case, the system has to be represented in its three phase domain. In the literature, one or two approaches are discussed for transient stability analysis with unbalanced fault. One such approach is to use the dynamic phasors. Obiously, the calculation procedure would be much more complicated as compared to the case of balanced fault. In this regard, we also should remember that, if the system is stable for a balanced fault, it would, by all probability, be also stable for any unbalanced fault. Therefore, it is usually sufficient for us to analyze the transient stability for balanced fault.
Answer: Yes. Thanks for pointing out the typing errors. 1. The correct data for the transmission line connected between buses 3 and 4 are as follows: R = 0.024 p.u., X = 0.136 p.u. and B/2 = 0.082 p.u. 2. True, the results are valid when (3*pi/Xcr) = (3*pi/Xci) = 3.0 ohms. This is also true for the results of 30 bus system (i.e. the results given in the lecture note for 30 bus system are valid when (3*pi/Xcr) = (3*pi/Xci) = 3.0 ohms. Further, in Table 2.38, the values of V4 and V5 are incorrectly typed. The correct values are: V4 = 0.77122 p.u. and V5 = 0.87058 p.u.
