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A 10,000 KVA, 50 Hz generator has reactance of 30 %, 10 %, 5%, to positive, negative and zero sequence currents respectively. It is connected to a line comprising three conductors of 1cm. diameter arranged in equilateral triangular spacing of 5 m side. The generator is excited to give 30 KV on open circuit. Find the current in the line when two lines are short circuited at a distance of 20 Km along the line. Neglect resistance and capacitance.

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A 10,000 KVA, 50 Hz generator has reactance of 30 %, 10 %, 5%, to posi...
Given data:
- Generator power rating: 10,000 KVA
- Generator frequency: 50 Hz
- Reactance: 30%, 10%, 5% for positive, negative, and zero sequence currents respectively
- Line conductors: Three conductors of 1cm diameter arranged in an equilateral triangular spacing of 5m side
- Generator open circuit voltage: 30 KV
- Two lines short-circuited at a distance of 20 Km

Calculation:
1. Reactance Calculation:
The reactance values are given as percentages. To calculate the actual reactance, we need to convert them to per unit values.
- Positive sequence reactance (X1) = 30% of generator reactance = 0.30 * 10,000 = 3,000 KVA
- Negative sequence reactance (X2) = 10% of generator reactance = 0.10 * 10,000 = 1,000 KVA
- Zero sequence reactance (X0) = 5% of generator reactance = 0.05 * 10,000 = 500 KVA

2. Short Circuit Current Calculation:
To calculate the short circuit current, we need to use the reactance values and the generator voltage.
- Positive sequence short circuit current (I1) = Generator voltage / X1 = 30,000 V / 3,000 KVA = 10 A
- Negative sequence short circuit current (I2) = Generator voltage / X2 = 30,000 V / 1,000 KVA = 30 A
- Zero sequence short circuit current (I0) = Generator voltage / X0 = 30,000 V / 500 KVA = 60 A

3. Line Configuration Calculation:
The line is composed of three conductors arranged in an equilateral triangular spacing. The distance between each conductor is 5m.
To calculate the line impedance, we need to calculate the inductance per unit length (L) of the conductor.
- Inductance per unit length (L) = (2 * 10^-7) * (ln(2s/d) + 0.25) = (2 * 10^-7) * (ln(2 * 5/0.01) + 0.25) = 2.4 * 10^-7 H/m

4. Short Circuit Distance Calculation:
The two lines are short-circuited at a distance of 20 Km (20,000 m). To calculate the line impedance at this distance, we need to calculate the total inductance (L_total).
- Total inductance (L_total) = L * (distance / conductor spacing) = 2.4 * 10^-7 * (20,000 / 5) = 9.6 * 10^-4 H

5. Short Circuit Current Calculation at 20 Km:
Using the total inductance, we can calculate the short circuit current at 20 Km using the reactance values.
- Positive sequence short circuit current at 20 Km (I1_20Km) = I1 * (X1 / (X1 + jL_total)) = 10 *
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it depends on the length of the conductor the capacitance of the line is proportional to the length of the transmission line their effect is negligible on the performance of short having a length less than 80 km and low voltage transmission accidents of the transmission line along with the conductances forms the shunted mittens the conductance and the transmission line is because of the leakage over the surface of the conductor considered a line consisting of two conductors and be each of radius are the distance between the conductors being Des shown in the diagram below minus the potential difference between the conductors and via's work QA charge on conductor QB charge on conductor vvab pencil difference between conductor and the Epsilon minus absolute primitivity QA plus QV = 0 so that QA equals QB - equals DBA equals data equals DB equals our substituting these values and voltage equation we get the capacitance between the conductors is cab is referred to as lying to line capacitance if the two conductors are in VR oppositely charge then the potential difference between them is zero then the potential of each conductor is given by one half bath the capacitance between each conductor and point of zero potential and is capacitive CN is called the capacitance to neut or capacitance to ground capacitance cab is the combination of two equal capacity and VN series thus capacitance to neutral is twice the capacitance between the conductors IE CN equals to Cave the absolute primitivity Epsilon is given by Epsilon equals epsilono Epsilon are where epsilano is the permittivity of the free space and Epsilon or is the relative primitivity of the medium prayer capacitance reactants between one conductor and neutral capacitance of the symmetrical three phase line let a balanced system of voltage be applied to a symmetrical three-phase line shown below the phasor diagram of the three phase line with equilateral spacing is shown below take the voltage of conductor to neutral as a reference phaser the potential difference between conductor and we can be written the similarly potential difference between conductors and sea is on adding equations one and two we get also combining equation three and four from equation 6 and 7 the line to neutral capacitance the capacitance of symmetrical three phase line is same as that of the two wire line Related: Capacitance of Transmission Lines?

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A 10,000 KVA, 50 Hz generator has reactance of 30 %, 10 %, 5%, to positive, negative and zero sequence currents respectively. It is connected to a line comprising three conductors of 1cm. diameter arranged in equilateral triangular spacing of 5 m side. The generator is excited to give 30 KV on open circuit. Find the current in the line when two lines are short circuited at a distance of 20 Km along the line. Neglect resistance and capacitance.?
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A 10,000 KVA, 50 Hz generator has reactance of 30 %, 10 %, 5%, to positive, negative and zero sequence currents respectively. It is connected to a line comprising three conductors of 1cm. diameter arranged in equilateral triangular spacing of 5 m side. The generator is excited to give 30 KV on open circuit. Find the current in the line when two lines are short circuited at a distance of 20 Km along the line. Neglect resistance and capacitance.? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A 10,000 KVA, 50 Hz generator has reactance of 30 %, 10 %, 5%, to positive, negative and zero sequence currents respectively. It is connected to a line comprising three conductors of 1cm. diameter arranged in equilateral triangular spacing of 5 m side. The generator is excited to give 30 KV on open circuit. Find the current in the line when two lines are short circuited at a distance of 20 Km along the line. Neglect resistance and capacitance.? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10,000 KVA, 50 Hz generator has reactance of 30 %, 10 %, 5%, to positive, negative and zero sequence currents respectively. It is connected to a line comprising three conductors of 1cm. diameter arranged in equilateral triangular spacing of 5 m side. The generator is excited to give 30 KV on open circuit. Find the current in the line when two lines are short circuited at a distance of 20 Km along the line. Neglect resistance and capacitance.?.
Solutions for A 10,000 KVA, 50 Hz generator has reactance of 30 %, 10 %, 5%, to positive, negative and zero sequence currents respectively. It is connected to a line comprising three conductors of 1cm. diameter arranged in equilateral triangular spacing of 5 m side. The generator is excited to give 30 KV on open circuit. Find the current in the line when two lines are short circuited at a distance of 20 Km along the line. Neglect resistance and capacitance.? in English & in Hindi are available as part of our courses for Electrical Engineering (EE). Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
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