Electrical Engineering (EE) Exam > Electrical Engineering (EE) Questions > A supply system feeds the following loads(i) ...
Start Learning for Free
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?
| FREE This question is part of | Download PDF Attempt this Test |
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 = P2 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
Cosϕ5 = 0.789 leading
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
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
Attention Electrical Engineering (EE) Students!
To make sure you are not studying endlessly, EduRev has designed Electrical Engineering (EE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Electrical Engineering (EE).
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Similar Electrical Engineering (EE) Doubts
Top Courses for Electrical Engineering (EE)View all
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?
Question Description
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? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about 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? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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?.
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? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about 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? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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?.
Solutions for 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? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
Here you can find the meaning of 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? defined & explained in the simplest way possible. Besides giving the explanation of
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?, a detailed solution for 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? has been provided alongside types of 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? theory, EduRev gives you an
ample number of questions to practice 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? tests, examples and also practice Electrical Engineering (EE) tests.
|
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup