Electrical Power Transmission MCQs Part 22

Contents

- 1 The favorable economic factor involved in power transmission at high voltage is
- 2 EHV cables are filled with thin oil under pressure in order to
- 3 Range of accelerating factor is
- 4 The total diameter of a 3 layer ACSR conductor whose each strand has diameter d, then find the total diameter of ACSR conductor
- 5 The internal inductance of an ACSR conductor is 0.05 mH/Km for µr = 1. If the ACSR is replaced with another having µr = 2, then find the internal inductance

## The favorable economic factor involved in power transmission at high voltage is

The favorable economic factor involved in power transmission at high voltage is:

- Smaller size of generation plant
- Decreased insulation required for line conductors
- Reduction of conduction cross-section area
- Increased I²R losses

Correct answer: 3. Reduction of conduction cross-section area

## EHV cables are filled with thin oil under pressure in order to

EHV cables are filled with thin oil under pressure in order to:

- Prevent entry of moisture
- Prevent formation of voids
- To provide insulation
- To strengthen cable conductor

Correct answer: 2. Prevent formation of voids

## Range of accelerating factor is

Range of accelerating factor is

- 1 to 1.5
- 1.6 to 1.8
- 10.8 to 11.98
- 50 to 100

Correct answer: 2. 1.6 to 1.8

Explanation: Accelerating factor is used for reducing number of iterations using Gauss-Siedel method. The range of accelerating factor is between 1.6 to 1.8.

## The total diameter of a 3 layer ACSR conductor whose each strand has diameter d, then find the total diameter of ACSR conductor

The total diameter of a 3 layer ACSR conductor whose each strand has diameter d, then find the total diameter of ACSR conductor

- d
- 2d
- 3d
- 5d

Correct answer: 4. 5d

Explanation:

Total diameter of ACSR conductor D = (2X – 1) X d

Where,

X = layer number

d = diameter of each strand

Therefore,

Total diameter of ACSR conductor D = (2 X 3 – 1) X d = 5d

## The internal inductance of an ACSR conductor is 0.05 mH/Km for µ_{r} = 1. If the ACSR is replaced with another having µ_{r} = 2, then find the internal inductance

The internal inductance of an ACSR conductor is 0.05 mH/Km for µ_{r} = 1. If the ACSR is replaced with another having µ_{r} = 2, then find the internal inductance:

- 0.1 mH/Km
- 0.3 mH/Km
- 0.5 mH/Km
- 0.7 mH/Km

Correct answer: 1. 0.1 mH/Km

Explanation:

Formula for internal inductance L_{int} = µ_{r}/(2 x 10^{-7})

Therefore:

L_{int }∝ µ_{r}

µ_{r} = Relative permeability

L_{int2} = L_{int1} X (µ_{r2} / µ_{r1}) = 0.05 * 2/1 = 0.1 mH/Km