A balanced 3-phase load of 30 MW is supplied at 131.1 kV and 0.85 pf lagging by means of transmission line. The sending end voltage is 141.1 kV. The value of constant A is 0.8. Find the regulation of the transmission line. O-34.53 O 25.00 O 32.09 O 26.91
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- A short transmission line is represented by an impedance of 0.01 + jo.04. What is the voltage regulation for a unity source and full-load current of 1.934-8° per-unit? O a. 0.8% O b. 1.2% c. 2.8% d. 0.3% O OConsider a 10 kVA, 200/400 V, 50 HZ: single-phase tranformer has the folloving test results:0.C. test: 200 V, 0.6A, 63 W (L.V.side)S.C. test : 20 V, 25A, 85 W (H.V. side)Deternine:i- The eficiengy at 75 % offiull-load at 0.9 leading power factor.ii -The secondary terminal voltage on fill-load at tunity power factonThere is an installation with two equal three-phase loads of 150 kW each and a power factor of 0.9 in lagging. By connecting a pure reactive element in parallel, the reactive power at the input of the installation is 160 kVARS, with a lagging power factor. It can be said that:A)Capacitors were connected and the fp improved. B) Capacitors were connected and the fp improved. C) Reactors were connected and the fp improved. D)Reactors were connected and fp improved. E) None of the above
- The instantaneous power absorbed by the load in a single-phase ac circuit, for a general R LC load under sinusoidal-steady-state excitation. is (a) Nonzero constant (b) Zero (c) Containing double-frequency componentsFor the given circuit below a. Calculate E, IR, and IL, in phasor form.b. Calculate the total power factor (leading or lagging) and the average power delivered to the circuit.c. Draw the admittance diagram and draw the phasor diagram of the currents Is, IR, and IL, and the voltage E.a-Calculate and draw the load current for the first two-period interval b-Calculate and draw the source current and find its average value
- Two delta connected loads are connected in parallel and powered by a balanced Y-connectedsource. The smaller of two loads draws 10 kVA at a lagging power factor of 0.75 and the largerdraws 25 kVA at a leading power factor of 0.8. The line voltage is 400 V. Calculate a) The powerfactor at which the source is operating. b) The total power drawn by the two loads. c) The phasecurrent of each load.In matlab, To obtain voltage and current waveforms from transmission line which tool can be used O a. voltage measurement O b. Three-phase VI measurment c. Series RLC branch O d. Parallel RLC load1. Assume that a 2.5-kV 1-phase circuit feeds a load of 450 kW at a lagging load power factor and the load current of 215 A.If it is desired to improve the power factor determine the following. a. The new corrected p.f after installing a shunt capacitor unit with a rating of 350kvar. none 0.992 lag 0.9 lag 0.952 lag
- A single phase transmission line is delivering 500 kVA load at 2 kV. Its resistance is 0-2 0 and inductive reactance is 0-4 0. Determine the voltage regulation if the load power factor is 0-707 lagging. O 8.5 % O 7.2 % O 5.3 % O 6.4 %The circuit shows an unbalanced electrical installation powered by a positive sequence symmetrical three-phase network of 380 V compound voltage. The loads are single phase. Load 1 absorbs an active power of 950 W with f.d.p. 0.5 inductive. Load 2 absorbs an active power of 1,140 W with f.d.p. Unit. Load 3 absorbs from the network an active power of 760 W with f.d.p. 0.5 capacitive. Taking the network voltage URN as the phase reference, calculate: a) complex expressions of the line currents IR. Is. and ITThe circuit shows an unbalanced electrical installation powered by a positive sequence symmetrical three- phase network of 380 V compound voltage. The loads are single phase. Load 1 absorbs an active power of 950 W with f.d.p. 0.5 inductive. Load 2 absorbs an active power of 1,140 W with f.d.p. Unit. Load 3 absorbs from the network an active power of 760 W with f.d.p. 0.5 capacitive. Taking the network voltage URN as the phase reference, calculate: a) complex expressions of the line currents IR. Is. and IT RO- SO TO IR Is IT U₁-380 V CARGA 1 P₁ =950 W cos p1=0,5 ind. CARGA 2 P₂=1.140 W cos p₂=1 CARGA 3 P3=760 W cos p3=0,5 cap.