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2.1 Determine the value of the phase voltage in a delta-connected system if the line voltage is 380 V - NSC Electrical Technology Power Systems - Question 2 - 2016 - Paper 1

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2.1 Determine the value of the phase voltage in a delta-connected system if the line voltage is 380 V. 2.2 Draw a neat, labelled voltage phasor diagram that represe... show full transcript

Worked Solution & Example Answer:2.1 Determine the value of the phase voltage in a delta-connected system if the line voltage is 380 V - NSC Electrical Technology Power Systems - Question 2 - 2016 - Paper 1

Step 1

Determine the value of the phase voltage in a delta-connected system if the line voltage is 380 V.

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Answer

In a delta-connected system, the relationship between the line voltage (VLV_L) and the phase voltage (VphV_{ph}) is given by:

Vph=VLV_{ph} = V_L

Therefore, the phase voltage is:

Vph=380extVV_{ph} = 380 ext{ V}

Step 2

Draw a neat, labelled voltage phasor diagram that represents a three-phase delta-connected system.

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Answer

A well-labelled phasor diagram should be drawn representing the three phases. The phases (VL1V_L1, VL2V_L2, and VL3V_L3) should be at 120-degree intervals, and each should be represented by a vector starting from a common point. The diagram should clearly indicate each phase's direction and angle.

Step 3

Current delivered by the alternator at full load.

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Answer

To calculate the current delivered at full load, we use the formula:

IL=S3VLI_{L} = \frac{S}{\sqrt{3} V_L}

Substituting the values: IL=20×1033×380I_{L} = \frac{20 \times 10^3}{\sqrt{3} \times 380}

This gives: IL30,39AI_{L} \approx 30,39 A

Step 4

Power rating of the alternator.

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Answer

The power rating can be calculated using:

P=3imesVLimesILimescos(θ)P = \sqrt{3} imes V_L imes I_L imes \cos(\theta)

Where (\cos(\theta)) is the power factor. Substituting the known values:

P=3×380×30,39×0,87P = \sqrt{3} \times 380 \times 30,39 \times 0,87

This results in: P17,4kWP \approx 17,4 kW

Step 5

State the function of a kilowatt-hour meter.

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Answer

The function of a kilowatt-hour (kWh) meter is to measure the amount of energy consumed by a load over a given period of time. It records energy consumption in kilowatt-hours.

Step 6

State TWO methods used to improve the power factor of a resistive inductive load.

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Answer

  1. Adding power-factor correcting capacitors in parallel with the load.
  2. Using synchronous motors to improve power factor.

Step 7

Calculate the power consumed by the load.

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Answer

The total power consumed by the load can be found by:

Pn=P1+P2P_n = P_1 + P_2

Substituting the given values:

Pn=120+50=170WP_n = 120 + 50 = 170 W

Step 8

TWO advantages of using the two-wattmeter method when measuring power of a balanced load can be determined.

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Answer

  1. Only two watt meters are required instead of three, simplifying the measurement apparatus.
  2. The method allows for the power of balanced and unbalanced loads to be accurately measured.

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