By Stephen J Chapma
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Excerpt from difficulties in electric EngineeringThis selection of difficulties has been ready for using scholars on the Massachusetts Institute of know-how, yet because the publication can be used in different technical faculties it kind of feels top to kingdom what flooring the issues are meant to hide. on the Institute the booklet can be utilized by the 3rd yr scholars in electric Engineering, and through the 3rd and fourth 12 months scholars within the classes of Civil, Mechanical, Mining and Chem ical Engineering.
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04 kvar Notice that the real and reactive powers supplied by the two transformers are radically different, put the apparent power supplied by each transformer is the same. Also, notice that the total power PA + PB supplied by the transformers is equal to the power consumed by the loads (within roundoff error), while the total reactive power Q A + QB supplied by the transformers is equal to the reactive power consumed by the loads. 46 2-14. 2-kV single-phase generator supplies power to a load through a transmission line.
85 PF lagging, draw the phasor diagram of one phase of the transformer. (b) What is the voltage regulation of the transformer bank under these conditions? (c) Sketch the equivalent circuit referred to the low-voltage side of one phase of this transformer. Calculate all of the transformer impedances referred to the low-voltage side. 85 PF lagging. 0 The base impedance of the transformer referred to the low-voltage side is: 3 Vφ ,base 2 3(115 kV ) = = 397 Ω Sbase 100 MVA 2 Z base = Each per-unit impedance is converted to actual ohms referred to the low-voltage side by multiplying it by this base impedance.
85 PF lagging. 85 PF lagging. 01% 7967 Note: It is much easier to solve problems of this sort in the per-unit system, as we shall see in the next problem. (c) This sort of repetitive operation is best performed with MATLAB. 85 leading. These calculations are done using an equivalent circuit referred to the primary side. 7; % Equivalent X (ohms) % % % % % Calculate the current values for the three power factors. The first row of I contains the lagging currents, the second row contains the unity currents, and the third row contains the leading currents.