EBK POWER SYSTEM ANALYSIS AND DESIGN
6th Edition
ISBN: 9781305886957
Author: Glover
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 5, Problem 5.42P
A three-phase power of 460 MW is transmitted to a substation located 500 km from the source of power. With
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Chapter 5 Solutions
EBK POWER SYSTEM ANALYSIS AND DESIGN
Ch. 5 - Representing a transmission line by the two-port...Ch. 5 - The maximum power flow for a lossy line is...Ch. 5 - Prob. 5.21MCQCh. 5 - A 30-km, 34.5-kV, 60-Hz, three-phase line has a...Ch. 5 - A 200-km, 230-kV, 60-Hz, three-phase line has a...Ch. 5 - The 100-km, 230-kV, 60-Hz, three-phase line in...Ch. 5 - The 500-kV, 60-Hz, three-phase line in Problems...Ch. 5 - A 40-km, 220-kV, 60-Hz, three-phase overhead...Ch. 5 - A 500-km, 500-kV, 60-Hz, uncompensated three-phase...Ch. 5 - The 500-kV, 60-Hz, three-phase line in Problems...
Ch. 5 - A 350-km, 500-kV, 60-Hz, three-phase uncompensated...Ch. 5 - Rated line voltage is applied to the sending end...Ch. 5 - A 500-kV, 300-km, 6()-Hz, three-phase overhead...Ch. 5 - The following parameters are based on a...Ch. 5 - Consider a long radial line terminated in its...Ch. 5 - For a lossless open-circuited line, express the...Ch. 5 - A three-phase power of 460 MW is transmitted to a...Ch. 5 - Prob. 5.55PCh. 5 - Consider the transmission line of Problem 5.18....Ch. 5 - Given the uncompensated line of Problem 5.18, let...
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- comparing Lenz's law and the left hand generator rule, which of these is the more important fundamental principle?arrow_forwardExample: Electric Field and Potential Inside a Charged Sphere Problem: A sphere of radius R = 0.2 m is uniformly charged with a total charge Q = 5 μC. The sphere is made of a dielectric material with relative permittivity € = 4. Calculate: 1. The electric field intensity E(r) inside and outside the sphere. 2. The electric potential (r) at any point inside the sphere. Solution: Step 1: Given Data Radius of the sphere: R = 0.2m, Total charge: Q-5 μC=5× 10° C. Step 2: Electric Field Inside the Sphere (< Using Gauss's Law:arrow_forwardplease remember to draw the circuitsarrow_forwardA balanced three-phase, A - connected induction motor consumes 3246 W when the l voltage is 208 V, and the line current is 10.6 A. Calculate: i. The motor's winding resistance. ii. The motor's winding reactance. 12 marrow_forwarda) An iron ring, having a mean circumference of 250 mm and a cross-sectional area of 400 mm², is wound with a coil of 70 turns. Using the following data, calculate the current required to set up a flux of 510µWb in the ring. H (A/m) 350 600 1250 B (T) 1.0 1.2 1.4 b) Calculate also: i. The inductance of the coil at the current obtained in Question 2 (a) above. ii. The self-induced e.m.f. if this current is switched off in 0.005 s. Assume that there is no residual flux.arrow_forwardA balanced three-phase, 1351-V, 60-Hz, A-connected source feeds a balanced Y- connected load with a per-phase impedance of 360 + j150 Q as shown in Figure 1. Calculate: i. The readings on each of the wattmeters ii. The power factor of the load using the wattmeter readings. NOTE: i. ii. Let VAN be the reference phasor, and the phase sequence be ABC anticlockwise. Assume the voltage-drop on the conductors between the source and the load to be zero volts. V b V₁ W 000 000 ; A 360 + j150 360 + j150 4 b 0000 000 B 360 + j150 C W₂ Figure 1arrow_forwarda) Three 30 2 resistors are arranged as shown in Figure 1 below. They are connected to a 480 V three-phase supply. The phase sequence is RYB anticlockwise. Calculate: i. The total power drawn by the circuit using the phase parameters. ii. The power read by each wattmeter. b) If Za, one of the 30 2 resistors, is now removed from the circuit, calculate: R- i. The line currents: IR, Iv, and la ii. The power read by each wattmeter. iii. The total power drawn by the two resistors. W₁ Be- W2 www 'R 22 12 B Figure 1arrow_forwardA certain magnetic circuit may be regarded as consisting of three parts, A, B and C in series, each one of which has a uniform cross-sectional area. Part A has a length of 300 mm and a cross- sectional area of 450 mm². Part B has a length of 120 mm and a cross-sectional area of 300 mm². Part C is an airgap 1.0 mm in length and of cross-sectional area 350 mm². The flux in the airgap is 0.35 mWb. Neglect magnetic leakage and fringing. The magnetic characteristic for parts A and B is given by: H (A/m) 400 560 800 1280 1800 B (T) 0.7 0.85 1.0 1.15 1.25 Calculate: i. The reluctance of each part, that is, of Part A, Part B, and Part C ii. The total reluctance of the magnetic circuit. iii. The total m.m.f.arrow_forwardHW_02.pdf EE 213-01 Assignments HW_#2 Toms are as muIcate uah.instructure.com b Answered: HW_#1 HW_01.pdf EE 213-01 Assignments P Pearson MyLab and Mastering uah.instructure.com P Course Home Watc... ✓ Download → Info × Close 1) (5 pts)For the circuit shown, find the value of Ia, Ib, Ic and Va: Ib 10 Ohms + Ic 40 Ohms 20 Ohms 70 Volts a Page 1 > of 2 - ZOOM + pui via Canvas Hint: use KCL to find Ic in terms of la and lb, use KVL around the right loop to find a relationship between la and lb, use KVL around the outer loop to solve for a current value (use ohms law for the voltage drops across the resistors) 2) (5 pts) – For problem 1, show that the power supplied (delivered) by the source equals the power absorbed by the resistors. 3) (5 pts) Determine the equivalent resistance looking into terminals a-b for the two circuits shown. Use series and parallel resistor combinations. Hint: work from the opposite end of terminals a-b towards terminals a-b a) 24 Ohms 10 Ohms 8 Ohms G a REQ b)…arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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Series compensation of long transmission lines; Author: Georg Schett;https://www.youtube.com/watch?v=smOqSxFBvVU;License: Standard Youtube License