Exersice 1. Single phase load&compensation I 100 10+j20 Two loads z1 = 100 and z2 = 10+j20 are connected across a 200 V rms, 50 Hz source as shown in figure. a. Find real, reactive powers and power factor at the source, find total current through the circuit. b. Find the capacitance of the capacitor across the loads to improve the overall power factor to 0.8 lagging. Exersice 2. Power and current calculations: balanced and unbalanced systems In the system given below, find Ia, Ib, Ic and apparent power supplied by the source, if; a. za = zb = zc = j 1.0 b. za = zb = j1.0 and zc = j 0.9 (znn’ = j0.1 ) Ia a 10 j 0.1 a’ za j 0.1 n’ n 1120 c 1-120 Ib b Ic Exersice 3. zb zc j 0.1 c’ b’ j 0.1 3-phase loads A three-phase line has an impedance of 2+j4 . The balanced line feeds two balanced load. The first load is Y connected and has an impedance of 30+j40 . The second load is connected and has an impedance of 60-j45 . The line is energized by a three-phase supply having a voltage of 207.85 V. Taking voltage of phase “a” as the reference, determine; a. The current through, reactive and active powers from the supply. b. c. d. e. The line voltage at which the loads are connected. The current per phase through each load. The total active and reactive powers in each load and the line. Power factor of combined loads. 2+j4 60-j45 VL= 207.85 V 2+j4 60-j45 60-j45 2+j4 30+j40 30+j40 30+j40 Exersice 4. Ideal, single phase two winding transformer A single-phase two winding transformer is rated at 20 kVA, 480/120 V, 50 Hz. A source connected to the 480 V winding supplies an impedance load connected to the 120 V winding. The load absorbs 15 kVA at 0.8 power factor lagging when the load voltage is 118 V. Assume that the transformer is ideal and calculate; a. b. c. d. The voltage across the 480 V winding The load impedance The load impedance transformed to the 480 V winding The real and reactive power supplied to the 480 V winding. (Exersices 5 to 9 are taken from “Power System Analysis & Design, SI Version”, J. Duncan Glover, Mulukutla S. Sarma and Thomas Overbye, Publisher: Cengage Learning,. 2012, Cengage Learning, Edition: 5th Revised edition. ISBN-10: 1111425795. ISBN-13: 9781111425791) Exersice 5. Transformer short circuit and open circuit tests A single-phase two winding transformer is rated at 20 kVA, 480/120 V, 50 Hz. During a short circuit test, where rated frequency is applied to the 480 V winding (winding 1), with the 120 V winding (winding 2) shorted, the following readings are obtained: V1 = 35 V, P1 = 300 W. During an open circuit test, where rated voltage is applied to the winding 2, with winding 1 is open, the following readings are obtained: I2 = 12 A, P2 = 200 W. a. From the short circuit test, determine the equivalent series impedance Zeq1 referred to the winding 1. Neglect the shunt admittance. b. From the open circuit test, determine the values of shunt elements. Referred to winding 1. Neglect the series impedance. Exersice 6. Per unit impedance: single phase transformer A single-phase two winding transformer is rated at 20 kVA, 480/120 V, 50 Hz. The equivalent leakage impedance of the transformer referred to the 120 V winding (winding 2)is Zeq2 = 0.052578.13 . Using the transformer ratings as base values, determine the per unit leakage impedance referred to winding 2 and winding 1. Exersice 7. Per unit circuit: three zone single phase network Three zones of a single-phase circuit are identified in Figure. The zones are connected by transformers T1 and T2, whose ratings are also shown. Using base values of 30 kVA and 240 volts in zone 1; draw the per-unit circuit and determine the per-unit impedances and the per-unit source voltage. Then calculate the load current both in per-unit and in amperes. Transformer winding resistances and shunt admittance branches are neglected. Exersice 8. Per unit and actual currents in balanced three phase network A balanced Y-connected voltage source with V = 480 0 volts is applied to a balanced load with Zload = 3040 . The line impedance between the source and load is Zline = 185 for each phase. Calculate the per-unit and actual current in phase a of the line using Sbase3 = 10 kVA and VbaseLL = 480 volts. Exersice 9. Voltage calculations : balanced Y-Y and -Y connection Three single-phase two-winding transformers, each rated 400 MVA, 13.8/199.2 kV, with leakage reactance Xeq = 0.10 per unit, are connected to form a three-phase bank. Winding resistances and exciting current are neglected. The high-voltage windings are connected in Y. A three-phase load operating under balanced positive-sequence conditions on the high-voltage side absorbs 1000 MVA at 0.90 p.f. lagging, with VAN = 199.2 kV. Determine the voltage Van at the low-voltage bus if the lowvoltage windings are connected a. in Y, b. in . V "1,(\ -=- - 4% 0 :::- Vz-/\ 0.. 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