5.6 The generalized circuit constants of a transmission line areA = 0.93 + j0.016B = 20 + 140The load at the receiving-end is 60 MVA, 50 Hz, 0.8 power factorlagging. The voltage at the supply end is 220 kV. Calculate the loadvoltage. ig. 6.2.Y13The elements of this matrix, of course, agree with those previously calculatedin Eq. (6.3)Example 6.2Figure 6.5 shows the one-line diagram of a simple four-bus system. Table 6.1gives the line impedances identified by the buses on which these terminate. Theshunt admittance at all the buses is assumed negligible.(a) Find YBUS assuming that the line shown dotted is not connected.(b) What modifications need to be carried out in YBUs if the line shown dottedis connected.123Fig. 6.5 Sample system for Example 6.2Table 6.1Line,X, pubus to bus1-20.151-30.302-30.452-40.303-40.15R, pu0.050.100.150.100.05

Given data : Parameters values of given transmission line is shown below,…

5.6 The generalized circuit constants of a transmission line are A = 0.93 + j0.016 B = 20 + 140 The load at the receiving-end is 60 MVA, 50 Hz, 0.8 power factor lagging. The voltage at the supply end is 220 kV. Calculate the load voltage.

5.6 The generalized circuit constants of a transmission line are A = 0.93 + j0.016 B = 20 + 140 The load at the receiving-end is 60 MVA, 50 Hz, 0.8 power factor lagging. The voltage at the supply end is 220 kV. Calculate the load voltage.

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

See similar textbooks

Similar questions

For a single phase ac voltage controller/chopper, which is designed by using thyristors, with the series R (=8[Ohm]) and L(=30 [mH]) load, if the rms value of the current flowing through each thyristor in the chopper circuit is equal to 2.83[A], then calculate average power absorbed by the load. O 64.8529 W 128.1424 W O 38.5784W O 91.3782 Vw O 120.5604 w O 288.3204 w O 187.5384W O 52.3564 Vw
consider the following diagram of impedances in per unit of a common base a) get the YBUS admittance array and the ZBuz impedance array 213-j2 pu zt2-j0.15 zg2-j1.85 791-j0.9 211-0.1 1 212-j0.5pu 223-j0.5pu Va 1 p.u. Jv1-1p.u. 22-10 pu
Which termination would result to a phase change of 180 degrees at the load if Zo = 100 ohm? a. 50 ohm b. short circuit c. 75 ohm d. all of these What is the purpose of impedance matching? a. maximum return loss b. maximum SWR c. maximum load power d. maximum reflection A short piece of transmission line that may be open or shorted and used for impedance matching purposes. a. converter b. decoder c. stub d. transformer
The steady-state voltage drop between the load and the sending end of the line seen in (Figure 1) is excessive Suppose that V-4950/0° V (rms). A capacitor is placed in parallel with the 192 kVA load and is adjusted until the steady-state voltage at the sending end of the line has the same magnitude as the voltage at the load end, that is, 4950 V (rms). The 192 kVA load is operating at a power factor of 0.8 lag. Part A Calculate the size of the capacitor in microfarads if the circuit is operating at 60 Hz. In selecting the capacitance, use the value that results in the lowest possible power loss in the line. Express your answer in microfarads to three significant figures. ▸ View Available Hint(s) Figure 1 of 1 202 1100 192 kVA 0.8 lag ΜΕ ΑΣΦ. 11 vec C= 22.8 Submit Previous Answers Incorrect; Try Again; 5 attempts remaining Provide Feedback ? F
Derive the expression for the average and rms load voltage for a three phase HWCR operating with an RL type load. ? What will be the average power in the load for the circuit shown, when alpha = π/4 Assume SCR to be ideal. Supply voltage is 330 sin 314t. Also calculate the RMS power .?
For the system as shown in the following figure, convert the components to same base voltage and apparent power, by choosing 100 kVA as the base kVA and 2000 V as the base voltage for the generators. Redraw the single line diagram with the new per unit values. 15 KVA 2000 V Z=j0.3 pu 25 KVA 2000 V Z=j0.15 pu G1 G2 35 KVA 2000/8000 V Z=j0.2 pu Z=30+j100 Z=60+j120 pu O pu leee 35 KVA 8000/2000 V Z=j0.3 pu 35 kVA 2000/8000 V Z=j0.25 pu G3 10 KVA 2000V Z=j0.35 pu M 30 KVA 2000 V
In the figure, bus 1 is the reference bus, V1 = 1.06 p.u and the angle is 0.. A load with active power= 100 MW and reactive power = 50 Mvar is connected to bus 2. The line impedance is z12 = 0.12 + j0.16 p.u on a base of 100 MVA. Start with an initial estimate of |V2|0) = 1.0 p.u and 820) = 0•. Find |V2| () (one iteration). 100 MW 212 = 0.12 + j0.16 ++ 50 Mvar Vị = 1.0/0° О а. |V2| (1) = 0.4 p.u. Ob. |V2| (1) = 0.5 p.u. Oc. |V2| (1) = 0.7 p.u. d. (1) = 0.8 p.u. е. |V2| (1) = 0.6 p.u.
An RLE load is operating from a type A chopper. If the supply voltage of 600 V, L 0.08 H, E = 120 V and R = 0 . For a duty cycle of 0.3, the chopping frequency to limit the amplitude of load current excursion to 10A is Hz.
I need the answer quickly
3. A circuit consists of two impedances. Z1 =20230° and Z2 =25260° 2, in parallel, supplied by a source voltage V=1002600 volts. Determine the power triangle for each of the impedances and for the source.
Q. 13. Transmission systems that produce switching surges higher than lighting surges are the ones that operate at: A. 200 kV B. 300 kV C. 400 kV D. B and C Q. 14. When constructing volt time curve, if flashover occurs on the tail side of the wave, the point on the curve will have a voltage magnitude equal to: Whos A. The applied wave peak B. The applied wave flashover point C. The applied wave peak multiplied by a factor of 0.9 D. The applied wave flashover point multiplied by a factor of 0.9