Transmission and Distribution Objective Questions With Easy Explanation Part-3

 21. By increasing the transmission voltage double of its original value, the same power can be dispatched keeping the line loss

(a) equal to its original value.

(b) half of original value. 

(d) one-fourth of the original value.

(e) double the original value.

 

Answer: (d) one-fourth of the original value.

 

Explanation:

The line losses are inversely proportional to the square of voltage and power factor.

PL  1/V2

Now voltage is doubled

V2 = 2V1

PL1/PL2 = (V2/V1)2

= (2V1/V1)2

PL2 = PL2/4

 

22. If a fixed amount of power is to be transmitted over certain length with fixed power loss, it can be said that volume of conductor is

(a) inversely proportional to magnitude of the voltage and that of power factor of the load.

(b) inversely proportional to square of the voltage and square of power factor of the load.

(c) proportional to square of voltage and that of power factor of the load.

(d) proportional to magnitude of the voltage only.

 

Answer: (b) inversely proportional to square of the voltage and square of power factor of the load.

 

Explanation:

  • The volume of copper required is inversely proportional to the square of the transmission voltage and the power factor.
  • Thus greater the transmission voltage level, lesser is the volume of copper required i.e. the weight of copper used for the conductors. The conductor material required is less, for higher transmission voltage.

 

23. For the same voltage drop, increasing the voltage of a distributor n-times

(a) reduces the x-section of the conductor by n times.

(b) increases the x-section of the conductor by n times,

(c) reduces the x-section of the conductor by n2 times.

(d) increases the x-section of the conductor by n2 times.

 

Answer: (a) reduces the x-section of the conductor by n times.

 

Explanation:


R=ρlA

R = resistance, l = length, A = area of cross-section and ρ = resistivity

SI unit of resistance is the ohm (Ω).

Voltage drop = I R

Active power = V I cos ϕ

 

Application:

Given V2 = n V1

 V2 / V1 = n

Since there is no change in the connected load, to maintain constant power transfer,

P2 = P1

V2 I2 cos ϕ2 = V1 I1 cos ϕ1

Since the type of connected load is also the same, the power factor is also constant.

 I2 = I1 / n

Given that the voltage drop is same,

I2 R2 = I1 R1

R2 = n R1

ρl / A2 = nρl / A1

A2 = A1 / n

Hence, reduces the x-section of the conductor by n-times

 

24. In a transmission system, the weight of copper used is proportional to

(a) E2. 

(b) E.  

(c) 1/E2.          

(d) 1/E.

(e) none of the above.

 

Answer: (c) 1/E2.

 

Explanation:

  • If power is transmitted, length of the line and loss in the transmission line is constant then the volume of required material will be inversely proportional to the square of supply voltage and power factor.
  • Hence, the volume of conductor material = 1/E2

 

25. The volume of copper required for an ac transmission line is inversely proportional to

(a) current.     

(b) voltage.

(c) pf. 

(d) both (b) and (c).

 

Answer: (d) both (b) and (c).

 

Explanation:

P = 3 (P / √3 V Cos θ)2 x [ρ (l /A)]

The volume of copper required for an AC transmission line is inversely proportional to voltage, power factor and proportion to the current.

 

26. Improving pf

(a) reduces current for a given output.

(b) increases losses in line.

(c) increases the cost of station equipment

 

Answer: (a) reduces current for a given output.

 

Explanation:

Improving the power factor results in less current being drawn, therefore less electricity costs, less heat and greater longevity of the electrical system. the installation, thus reduces the maximum demand tariff and thereby reducing your power costs.

 

27. For a given amount of power to be transmitted over a certain distance with fixed power loss, the volume of copper required is

(a) directly proportional to voltage.

(b) inversely proportional to voltage.

(c) inversely proportional to the square of voltage and pf of the load.

(d) directly proportional to the square of the voltage and pf of the load.

 

Answer: (c) inversely proportional to the square of voltage and pf of the load.

 

Explanation:

P = 3 (P / √3 V Cos θ)2 x [ρ (l /A)]

P = 3 (P2 / 3 V2 Cosθ2) x ρ (l /A)           

P = W = P2 ρ l / V2 Cosθ2 A

If power transmitted, length of the line, and loss in the transmission line is constant then the volume of required material will be inversely proportional to the square of supply voltage and power factor.

 

28. For the same conductor length, same amount of power, same losses and same maximum voltage to earth, which system requires minimum conductor area?

(a) single phase ac     

(b) 3 phase ac

(c) 2 wire ac   

(d) 3 wire dc

 

Answer: (d) 3 wire dc

 

Explanation:

  • For the same conductor length, same amount of power, same losses and same maximum voltage to earth, 3 wire DC system requires minimum conductor area.
  • For transmitting the same amount of power at the same voltage, a three-phase transmission line requires less conductor material than a single-phase line. The three-phase transmission system is so cheaper.
  • For a given amount of power transmitted through a system, the three-phase system requires conductors with a smaller cross-sectional area. This means a saving of copper and thus the original installation costs are less.

 

Important Point:

Below is given the table which shows the ratio of conductor-material in any system compared with that in the corresponding 2-wire DC system. Cos φ is the power factor in an AC system.

System

Same maximum voltage to earth

Same maximum voltage between conductors

DC system: Two wire

1

1

DC: Two wire mid-point earthed

0.25

1

DC: 3 wire

0.3125

1.25

Single phase: 2 wire

2/cos2ϕ

2/cos2ϕ

Single phase: 2 wire mid-point earthed

0.5/cos2ϕ

2/cos2ϕ

Single phase: 3 wire

0.625/cos2ϕ

2.5/cos2ϕ

2-phase: 4 wire

0.5/cos2ϕ

2/cos2ϕ

2-phase: 3 wire

1.457/cos2ϕ

2.914/cos2ϕ

3 phase, 3 wire

0.5/cos2ϕ

1.5/cos2ϕ

3 phase, 3 wire

0.583/cos2ϕ

1.75/cos2ϕ

 

29. Which of the following distribution systems is preferred for good efficiency and high economy?

(a) Single-phase, 2-wire system. 

(b) 2-phase, 3-wire system. 

(c) 3-phase, 3-wire system.   

(d) 3-phase, 4-wire system.

 

Answer: (d) 3-phase, 4-wire system.

 

Explanation:

  • There is a great saving in conductor material if DC system is adopted for transmission of electric power. However, due to technical difficulties, DC system is not used for transmission.
  • Considering the AC system, the 3-phase AC system is most suitable for transmission as well as distribution due to two reasons. Firstly, there is a considerable saving in conductor material. Secondly, this system is convenient and efficient.

 

30. The approximate cost ratio of a 220 kV underground cable transmission and 220 kV overhead transmission is

(a) 50  

(b) 25 

(c) 13  

(d) 5

 

Answer: (c) 13

 

Explanation:

  • The approximate cost ratio of a 220 kV underground cable transmission and 220 kV overhead transmission is 13.

For 11KV System

  • Underground network installation is more expensive than OH lines since the cost of cables includes cable charges along with road restoration charges which make the per-unit cost of the UG cabling system several times greater than the overhead system.
  • The estimated cost of the UG cabling system is about 3-4 times than the equivalent OH system ( like the Est cost of 11 kV OH S/C line with dog conductor is around Rs 5-6 Lakh/ km while the Est cost of 1 km of 3 x300 sq-mm 11 kV cabling system would be around Rs 20 Lakh/km).



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