JNTU
JNTU B.Tech III POWER SYSTEMS II Previous Question Paper (September 2021 - R13)
Answer any five questions. All questions carry equal marks.
Main
Write a short note on geometric mean distance and geometric mean radius.
Answer
Geometric Mean Radius (GMR), also known as self-GMD, represents the effective radius of a conductor that accounts for internal flux linkages. For a solid cylindrical conductor of radius r, the GMR is given by r' = r * e^(-1/4) = 0.7788r. It is a measure of the internal magnetic field effect; when calculating inductance, the actual radius is replaced by GMR to simplify the formula by assuming no internal flux.
Geometric Mean Distance (GMD) is the equivalent spacing between two groups of conductors. For a group of conductors m and n, the GMD is the m*n-th root of the product of all possible distances between each conductor in group m and each conductor in group n. It is used to determine the mutual inductance between circuits by representing the complex physical arrangement as an equivalent single-spacing configuration.
Two conductors of a single phase line, each of 1cm diameters, are arranged in a vertical plane with one conductor mounted 1m above the other. A second identical line is mounted at the same height as the first and spaced horizontally 0.25 m apart from it. The two upper and the two lower conductors are connected in parallel. Determine the inductance per km of the resulting double circuit line.
Answer
Let the conductors be labeled: 1 (top-left), 2 (top-right), 3 (bottom-left), 4 (bottom-right). The distance between the two lines is 0.25 m and vertical separation is 1 m.
Coordinates: 1(0, 1), 2(0.25, 1), 3(0, 0), 4(0.25, 0). Radius r = 0.005 m. GMR D_s = 0.7788 * 0.005 = 0.003894 m.
Distances: D_12 = D_34 = 0.25 m; D_13 = D_24 = 1 m; D_14 = D_23 = sqrt(1^2 + 0.25^2) = 1.0308 m.
Equivalent GMD (D_m): D_m = (D_13 * D_14 * D_23 * D_24)^(1/4) = (1 * 1.0308 * 1.0308 * 1)^(1/4) = 1.0153 m.
Equivalent GMR (D_s_eq): D_s_eq = (D_s * D_12 * D_s * D_34)^(1/4) = (0.003894 * 0.25 * 0.003894 * 0.25)^(1/4) = sqrt(0.003894 * 0.25) = 0.0312 m.
Inductance per phase per km: L = 2 * 10^-7 * ln(D_m / D_s_eq) H/m.
Converting to mH/km: L = 0.696 mH/km.
Derive an expression for the capacitance per km of a three phase line taking into account the effect of ground.
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A three-phase overhead transmission line has its conductors arranged at the corners of an equilateral triangle of 2m side. Calculate the capacitance of each line conductor per km if the diameter of the conductor is 1.25 cm.
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Derive the expression for A, B, C, D parameters of nominal-π medium length transmission line.
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A 3-ϕ transmission line has the following parameters per km: R=0.1Ω, L=1.117mH and C = 0.01 μF. Determine regulation and the efficiency of the line using nominal-T method if the line is 80km long and delivers 20 MW at 66 kV, 0.8 p.f lagging.
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Define the regulation and efficiency of transmission line.
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A single circuit 50 Hz, 3-ϕ transmission line has the following parameters per km: R=0.2 Ω, L=1.3 mH and C=0.01μF. Determine regulation and the efficiency of the line if the line is 120 km long and delivers 40MW at 132 kV and 0.8 p.f. lagging.
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Discuss about the reflection and refraction at T-junction in a travelling wave.
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A surge of 200 kV traveling in a line of natural impedance 400 ohms arrives at a junction with two lines of impedances 500 ohms and 300 ohms respectively. Find the surge voltages and currents transmitted into each branch line. Also find the reflected surge voltage and current.
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Define and derive the expression for the critical disruptive voltage.
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Explain Ferranti effect.
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Derive the expression for sag when the supports are at unequal heights.
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In a 33kV overhead line, there are three units in the string of insulators. If the capacitance between each insulator pin and earth is 11% of self capacitance of each insulator, find (i) The distribution of voltage over 3 insulators and (ii) String Efficiency.
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Explain in detail the construction of cables.
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The maximum and minimum stresses in the dielectric of a single core cable are 40 kV/cm and 10 kV/cm respectively. If the conductor diameter is 2 cm find thickness of insulation and operating voltage.
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