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3.1 Define the following terms: 3.1.1 Reactive power 3.1.2 Efficiency 3.2 Calculate the value of the phase voltage in a star-connected three-phase system if the line voltage is 380 V - NSC Electrical Technology Power Systems - Question 3 - 2020 - Paper 1

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3.1-Define-the-following-terms:--3.1.1-Reactive-power-3.1.2-Efficiency--3.2-Calculate-the-value-of-the-phase-voltage-in-a-star-connected-three-phase-system-if-the-line-voltage-is-380-V-NSC Electrical Technology Power Systems-Question 3-2020-Paper 1.png

3.1 Define the following terms: 3.1.1 Reactive power 3.1.2 Efficiency 3.2 Calculate the value of the phase voltage in a star-connected three-phase system if the li... show full transcript

Worked Solution & Example Answer:3.1 Define the following terms: 3.1.1 Reactive power 3.1.2 Efficiency 3.2 Calculate the value of the phase voltage in a star-connected three-phase system if the line voltage is 380 V - NSC Electrical Technology Power Systems - Question 3 - 2020 - Paper 1

Step 1

3.1.1 Reactive power

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Answer

Reactive power is the power in an AC circuit when the current is out of phase with the voltage due to either capacitive or inductive loads. It is measured in VAR (Volt-Ampere Reactive) and represents the power that does not contribute to the net work done.

Step 2

3.1.2 Efficiency

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Efficiency is the percentage ratio of the output power to the input power in a system. It can be calculated using the formula:

ext{Efficiency} = rac{ ext{Output Power}}{ ext{Input Power}} imes 100 ext{ %}

Step 3

3.2 Calculate the value of the phase voltage in a star-connected three-phase system if the line voltage is 380 V.

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Answer

For a star-connected system, the relationship between line voltage (VLV_L) and phase voltage (VPHV_{PH}) is given by:

V_{PH} = rac{V_L}{ ext{√3}} = rac{380 V}{ ext{√3}} \\ = 219.39 V

Step 4

3.3.1 Explain the effects a low power factor will have on a system.

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A low power factor can lead to several negative effects on an electrical system, including:

  1. Increased losses: Low power factor results in higher current, which increases resistive losses in conductors, leading to overheating.
  2. Reduced capacity: Low power factor reduces the overall capacity of the electrical system, requiring larger equipment and conductors.
  3. Voltage drop: Higher currents cause more significant voltage drops along transmission lines, leading to poor voltage quality at the load.
  4. Increased demand charges from utility providers.

Step 5

3.3.2 List THREE methods to improve the power factor of a resistive inductive load in an AC system.

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Methods to improve power factor include:

  1. Using power factor correction capacitors: By adding capacitors in parallel with the load, the reactive power is counteracted, thereby improving the power factor.
  2. Utilizing synchronous motors: These can be operated at leading power factors to compensate for lagging loads.
  3. Employing variable speed drives: These devices can help manage load conditions more effectively and improve the overall power factor.

Step 6

3.4 Name the instrument that is used to measure electrical power over a certain period of time.

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Answer

The instrument used to measure electrical power over a specific period of time is the Kilowatt-hour meter (KWh meter) or Energy meter.

Step 7

3.5.1 Line current delivered by the generator

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Answer

The line current (ILI_{L}) delivered by a delta-connected generator is calculated as follows:

IL=ext3imesIPH=ext3imes30A=51.96AI_{L} = ext{√3} imes I_{PH} = ext{√3} imes 30 A = 51.96 A

Step 8

3.5.2 Impedance of each phase

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The phase voltage (VPHV_{PH}) for a star-connected load is equal to the line voltage:

Z_{PH} = rac{V_{PH}}{I_{PH}} = rac{380 V}{30 A} = 12.67 Ω

Step 9

3.5.3 Active power generated

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Active power (PP) generated by the generator can be calculated using the formula:

P=ext3imesVLimesILimesextCosθP = ext{√3} imes V_{L} imes I_{L} imes ext{Cos} θ

In this case:

P=ext3imes380Vimes(3imes30A)imes0.9 =30.48kWP = ext{√3} imes 380 V imes (√3 imes 30 A) imes 0.9 \ = 30.48 kW

Step 10

3.6 Calculate the total input power.

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The total input power measured by the wattmeters is the sum of the individual readings:

Ptotal=P1+P2=10kW+3kW=13kWP_{total} = P_1 + P_2 = 10 kW + 3 kW = 13 kW

Step 11

3.7 Name ONE application of a wattmeter in three-phase AC systems.

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Answer

A wattmeter can be used to measure the input power/output power in an AC system, providing essential information for monitoring and managing energy consumption.

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