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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 phase voltage ( V_{ph}) is equal to the line voltage ( V_L). Therefore, the phase voltage is:

Vph=VL=380extVV_{ph} = V_L = 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

The voltage phasor diagram for a delta-connected system consists of three vectors ( V_L1, V_L2, V_L3) separated by 120°. It can be represented as follows:

  • Vector V_L1 is at 0°.
  • Vector V_L2 is at 120°.
  • Vector V_L3 is at 240°.

Each vector has a magnitude of 380 V.

Step 3

Current delivered by the alternator at full load

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Answer

To calculate the current delivered by the alternator, use the formula:

I_L = rac{S}{ ext{sqrt}(3) imes V_L}

Substituting the values:

S=20imes103extVA,VL=380extVS = 20 imes 10^3 ext{ VA}, V_L = 380 ext{ V}

Thus,

I_L = 30.39 ext{ A} $$

Step 4

Power rating of the alternator

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Answer

The power delivered can be calculated using:

P=extsqrt(3)imesVLimesILimesextCOS(θ)P = ext{sqrt}(3) imes V_L imes I_L imes ext{COS}(θ)

Given the power factor (p.f.) is 0.87, we can substitute:

P = 17.4 ext{ kW} $$

Step 5

State the function of a kilowatt-hour meter.

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Answer

The function of a kilowatt-hour meter is to measure the amount of energy consumed by a consumer over a period of time. Alternatively, it measures the energy usage of electrical systems.

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 the overall power factor.

Step 7

Calculate the power consumed by the load.

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Answer

To find the power consumed by the load, use the formula:

Pn=P1+P2P_n = P_1 + P_2

Substituting the given values:

Pn=120+50=170extWP_n = 120 + 50 = 170 ext{ W}

Step 8

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

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Answer

  1. Only two watt meters are required instead of three for balanced loads.
  2. The power of a balanced and unbalanced load can be measured accurately.

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