Jawaharlal Nehru Technological University Hyderabad

Jawaharlal Nehru Technological University Hyderabad B.Tech III POWER SYSTEMS - II Previous Question Paper (August - 2022 - R16)

B.Tech | III | I | R16 | August - 2022

Answer PaperPaper Code: 135BJMaximum Marks: 75Time: 3 Hours

Answer any five questions

All questions carry equal marks

Main

115 marks
a

What is Aluminium Conductor Steel Reinforced conductor (ACSR)? Explain the advantages of ACSR conductors when used for overhead lines.

6 marks

Answer

An Aluminium Conductor Steel Reinforced (ACSR) conductor consists of a central core of galvanized steel wires surrounded by one or more layers of stranded aluminium wires. The steel core provides the necessary mechanical strength to support the conductor against tension and wind loads, while the outer aluminium layers provide the high electrical conductivity required for power transmission.

The advantages of using ACSR conductors include:

  • High Strength-to-Weight Ratio: The steel core allows for longer spans between transmission towers, reducing the total number of towers required.
  • Reduced Sag: Due to the high tensile strength of the steel core, the conductor experiences less sag, which allows for greater ground clearance.
  • Economical: By utilizing aluminium for current carrying, the cost is significantly lower than using copper, and the lighter weight reduces the structural requirements of the support towers.
  • Improved Corona Performance: The larger overall diameter of ACSR compared to an equivalent copper conductor reduces the surface voltage gradient, thereby minimizing corona loss and radio interference.
b

Calculate the capacitance per phase of a three phase, three wire system by considering earth effect, when the conductors are arranged in a horizontal plane with spacing D12=D23=3.5m, and D31=7m. The conductors are transposed and each has a diameter of 2.0 cm. Assume the transmission line is 4m above the ground level.

9 marks

Answer

Given: D12 = 3.5 m, D23 = 3.5 m, D31 = 7 m. Diameter d = 2 cm, so radius r = 0.01 m. Height h = 4 m. The capacitance of a three-phase line with earth effect is calculated using the method of images. The equivalent capacitance per phase is influenced by the proximity of the ground, which introduces image charges.

First, calculate the self and mutual distances including images:

H11 = H22 = H33 = 2h = 8 m
H12 = H23 = sqrt(D12^2 + (2h)^2) = sqrt(3.5^2 + 8^2) = 8.73 m
H31 = sqrt(D31^2 + (2h)^2) = sqrt(7^2 + 8^2) = 10.63 m

The potential coefficients are Pii = (1/(2πε)) * ln(2h/r) and Pij = (1/(2πε)) * ln(Hij/Dij). For a transposed line, the average potential coefficient P is:

P = (1/3)(P11 + P22 + P33) = (1/(2πε)) * ln(2h/r)
P = (18 * 10^9) * ln(8/0.01) = 18 * 10^9 * 6.6846 = 1.203 * 10^11

The average mutual potential coefficient P' is:

P' = (1/3)(ln(H12/D12) + ln(H23/D23) + ln(H31/D31)) * (1/(2πε))
P' = (18 * 10^9) * (ln(8.73/3.5) + ln(8.73/3.5) + ln(10.63/7)) / 3 = 18 * 10^9 * 0.772 = 0.139 * 10^11

The capacitance per phase C is given by:

C = 1 / (P - P') = 1 / (1.203 - 0.139) * 10^11 = 9.39 * 10^-12 F/m = 9.39 pF/m
215 marks
a

Clearly explain what do you understand by Geometrical Mean Distance (GMD) and Geometrical Mean Radius (GMR) of a transmission line?

6 marks

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b

Discuss about the bundled conductors.

4 marks

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c

What do you understand by transposition of lines?

5 marks

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315 marks
a

Derive the A, B, C and D constants for Nominal-T model.

6 marks

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b

Draw the vector diagrams of nominal-p and nominal-T models of medium transmission line. Derive the expression for voltage regulation of both the models.

9 marks

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415 marks
a

An overhead single phase delivers 1.1 MW at 33 kV at 0.9 power factor lagging. The total resistance of the line is 10 Ω and the total inductive reactance is 15 Ω. Determine (i) % voltage regulation (ii) sending end power factor (iii) transmission efficiency.

9 marks

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b

Explain the interpretation of the long line equations.

6 marks

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515 marks
a

How the corona forms in power systems and write its disadvantages?

6 marks

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b

A 500 KV surge travels on an overhead line of surge impedance 400 Ω towards its junction with a cable that has a surge impedance of 40 Ω. Find: i) transmitted voltage, ii) reflected voltage, iii) reflected current.

9 marks

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615 marks
a

Discuss about the attenuation in transmission lines and causes of it.

4 marks

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b

Explain Ferranti effect in power systems.

4 marks

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c

Derive expression for reflection and refraction coefficients for open circuit and short circuit lines.

7 marks

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715 marks
a

Explain how sag is determined for an overhead line conductor taking into account the effects of wind and ice loading.

5 marks

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b

Explain about the various methods to improve the string efficiency.

5 marks

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c

With neat sketch, explain about suspension type and strain type insulators.

5 marks

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815 marks
a

Explain the construction of 3-core belted type cable.

6 marks

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b

Explain about different methods of grading of cables.

9 marks

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This page contains the solved question paper with answers of POWER SYSTEMS - II from Jawaharlal Nehru Technological University Hyderabad for B.Tech EEE III I, August - 2022, R16.

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