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A supply system feeds the following loads
(i) A lighting load of 500 kW
(ii) A load of 400 kW at a p.f. 0.707 lagging
(iii) A load of 800 kW at a p.f. 0.8 leading
(iv) A load of 500 kW at a p.f. 0.6 lagging
(v) A synchronous motor driving a 540 kW d.c. generator and having an overall efficiency of 90%
Calculate the power factor of the synchronous motor so that station power factor may become unity.
  • a)
    0.789 lagging
  • b)
    0.789 leading 
  • c)
    0.645 lagging 
  • d)
    0.645 leading
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A supply system feeds the following loads(i) A lighting load of 500 kW...
(i) The light load works at unity power factor.
P1 = 500 kW, Q1 = 0 KVAR
(ii) P2 = 400 kW, cosϕ2 = 0.707 lagging
Q2 = P tanϕ2 = 400 × tan cos-1 (0.707) = 400 KVAR (lagging)
(iii) P3 = 800 kW, cosϕ3 = 0.8 leadingQ3 = P3 tanϕ3 = 800 × tan cos-1(0.8) = 600 KVAR (leading)
(iv) P4 = 500 kW, cosϕ4 = 0.6 lagging
Q4 = P4 tanϕ4 = 500 × tan cos-1(0.6) = 666.67 KVAR (lagging)
Leading KVAR to be taken by synchronous motor
Q5 = Q4 + Q2 – Q3
Q5 = 400 + 666.67 – 600 = 466.66 KVAR

Cosϕ5 = 0.789 leading
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Most Upvoted Answer
A supply system feeds the following loads(i) A lighting load of 500 kW...
Given Data:

Lighting Load = 500 kW
Load 1 = 400 kW at a p.f. of 0.707 lagging
Load 2 = 800 kW at a p.f. of 0.8 leading
Load 3 = 500 kW at a p.f. of 0.6 lagging
Synchronous Motor Load = 540 kW
Overall Efficiency of Synchronous Motor = 90%

To find: Power factor of the synchronous motor so that the station power factor becomes unity.

Solution:

Step 1: Calculation of Real Power and Reactive Power for each Load

Real Power (P) = Apparent Power (S) x Power Factor (PF)
Reactive Power (Q) = Apparent Power (S) x sin(cos^-1(PF))

Load 1:
P1 = 400 kW
PF1 = 0.707 lagging
Q1 = S1 x sin(cos^-1(PF1)) = 400 x sin(cos^-1(0.707)) = 282.84 kVAR

Load 2:
P2 = 800 kW
PF2 = 0.8 leading
Q2 = S2 x sin(cos^-1(PF2)) = 800 x sin(cos^-1(0.8)) = -346.41 kVAR

Load 3:
P3 = 500 kW
PF3 = 0.6 lagging
Q3 = S3 x sin(cos^-1(PF3)) = 500 x sin(cos^-1(0.6)) = 400 kVAR

Step 2: Calculation of Total Real Power, Total Reactive Power, and Apparent Power

Total Real Power (P) = P1 + P2 + P3 + PL
where PL is the real power of the lighting load = 500 kW

P = 400 + 800 + 500 + 500 = 2200 kW

Total Reactive Power (Q) = Q1 + Q2 + Q3
Q = 282.84 - 346.41 + 400 = 336.43 kVAR

Total Apparent Power (S) = √(P^2 + Q^2)
S = √(2200^2 + 336.43^2) = 2221.3 kVA

Step 3: Calculation of Synchronous Motor Load

The synchronous motor load consists of a DC generator, so the total load on the synchronous motor will be the sum of the synchronous motor load and the DC generator load.

Synchronous Motor Load = 540 kW / 0.9 (Overall Efficiency)
Synchronous Motor Load = 600 kW

DC Generator Load = 540 kW - 600 kW = -60 kW
(Note that the negative sign indicates that the DC generator is supplying power back to the system)

Step 4: Calculation of Required Reactive Power to Improve Power Factor

To improve the power factor to unity, we need to supply reactive power to the system. The required reactive power can be calculated as:

Qreq = P x tan(cos^-1(PF))
where PF = 1 (Unity Power Factor)

Qreq = 2200 x tan(cos^-1(1)) = 0 kVAR

This means that the system should have zero reactive power to achieve unity power factor.

Step 5: Calculation of Required Power Factor for the Synchronous Motor

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A supply system feeds the following loads(i) A lighting load of 500 kW(ii) A load of 400 kW at a p.f. 0.707 lagging(iii) A load of 800 kW at a p.f. 0.8 leading(iv) A load of 500 kW at a p.f. 0.6 lagging(v) A synchronous motor driving a 540 kW d.c. generator and having an overall efficiency of 90%Calculate the power factor of the synchronous motor so that station power factor may become unity.a)0.789 laggingb)0.789 leadingc)0.645 laggingd)0.645 leadingCorrect answer is option 'B'. Can you explain this answer?
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