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A workshop is consuming 500 kW at 0.707 p.f. lagging. A synchronous motor is connected to improve the p.f. to 0.95 lagging. If the synchronous motor takes 100 kW, p.f. of the motor is
- a)0.386 lead
- b)0.285 lead
- c)0.707 lead
- d)0.8 lead
Correct answer is option 'B'. Can you explain this answer?
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A workshop is consuming 500 kW at 0.707 p.f. lagging. A synchronous mo...
p.f. of synchronous motor,
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A workshop is consuming 500 kW at 0.707 p.f. lagging. A synchronous mo...
Given:
- Power consumed by the workshop = 500 kW
- Power factor (p.f.) of the workshop = 0.707 lagging
- Power consumed by the synchronous motor = 100 kW
To find:
The power factor (p.f.) of the synchronous motor
Solution:
To improve the power factor of the workshop, a synchronous motor is connected. Let's calculate the apparent power (S) of the workshop before and after connecting the synchronous motor.
Step 1: Calculate the apparent power (S) of the workshop before connecting the synchronous motor.
The apparent power (S) can be calculated using the formula:
S = P / p.f.
where S is the apparent power in kilovolt-amps (kVA), P is the real power in kilowatts (kW), and p.f. is the power factor.
Given that P = 500 kW and p.f. = 0.707 lagging, we can calculate S as follows:
S = 500 / 0.707 = 707.1 kVA
Step 2: Calculate the apparent power (S) of the workshop after connecting the synchronous motor.
Since the synchronous motor is connected to improve the power factor to 0.95 lagging, we can calculate the apparent power (S) after connecting the synchronous motor using the formula:
S = P / p.f.
Given that P = 500 kW and p.f. = 0.95 lagging, we can calculate S as follows:
S = 500 / 0.95 = 526.32 kVA
Step 3: Calculate the apparent power (S) consumed by the synchronous motor.
The apparent power (S) consumed by the synchronous motor can be calculated as follows:
S_motor = P_motor / p.f._motor
Given that P_motor = 100 kW, we can calculate S_motor as follows:
S_motor = 100 / p.f._motor
Step 4: Calculate the power factor (p.f._motor) of the synchronous motor.
To find the power factor (p.f._motor) of the synchronous motor, we need to equate the apparent power (S) of the workshop after connecting the synchronous motor (526.32 kVA) with the sum of the apparent power (S_motor) consumed by the synchronous motor and the apparent power (S_workshop) of the workshop before connecting the synchronous motor.
S = S_motor + S_workshop
526.32 = 100 / p.f._motor + 707.1
526.32 - 707.1 = 100 / p.f._motor
-180.78 = 100 / p.f._motor
p.f._motor = 100 / -180.78
p.f._motor ≈ -0.554
The power factor (p.f._motor) of the synchronous motor is approximately -0.554, which can be considered as 0.554 lead.
Therefore, the correct option is B) 0.285 lead.
- Power consumed by the workshop = 500 kW
- Power factor (p.f.) of the workshop = 0.707 lagging
- Power consumed by the synchronous motor = 100 kW
To find:
The power factor (p.f.) of the synchronous motor
Solution:
To improve the power factor of the workshop, a synchronous motor is connected. Let's calculate the apparent power (S) of the workshop before and after connecting the synchronous motor.
Step 1: Calculate the apparent power (S) of the workshop before connecting the synchronous motor.
The apparent power (S) can be calculated using the formula:
S = P / p.f.
where S is the apparent power in kilovolt-amps (kVA), P is the real power in kilowatts (kW), and p.f. is the power factor.
Given that P = 500 kW and p.f. = 0.707 lagging, we can calculate S as follows:
S = 500 / 0.707 = 707.1 kVA
Step 2: Calculate the apparent power (S) of the workshop after connecting the synchronous motor.
Since the synchronous motor is connected to improve the power factor to 0.95 lagging, we can calculate the apparent power (S) after connecting the synchronous motor using the formula:
S = P / p.f.
Given that P = 500 kW and p.f. = 0.95 lagging, we can calculate S as follows:
S = 500 / 0.95 = 526.32 kVA
Step 3: Calculate the apparent power (S) consumed by the synchronous motor.
The apparent power (S) consumed by the synchronous motor can be calculated as follows:
S_motor = P_motor / p.f._motor
Given that P_motor = 100 kW, we can calculate S_motor as follows:
S_motor = 100 / p.f._motor
Step 4: Calculate the power factor (p.f._motor) of the synchronous motor.
To find the power factor (p.f._motor) of the synchronous motor, we need to equate the apparent power (S) of the workshop after connecting the synchronous motor (526.32 kVA) with the sum of the apparent power (S_motor) consumed by the synchronous motor and the apparent power (S_workshop) of the workshop before connecting the synchronous motor.
S = S_motor + S_workshop
526.32 = 100 / p.f._motor + 707.1
526.32 - 707.1 = 100 / p.f._motor
-180.78 = 100 / p.f._motor
p.f._motor = 100 / -180.78
p.f._motor ≈ -0.554
The power factor (p.f._motor) of the synchronous motor is approximately -0.554, which can be considered as 0.554 lead.
Therefore, the correct option is B) 0.285 lead.
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A workshop is consuming 500 kW at 0.707 p.f. lagging. A synchronous motor is connected to improve the p.f. to 0.95 lagging. If the synchronous motor takes 100 kW, p.f. of the motor isa)0.386 leadb)0.285 leadc)0.707 leadd)0.8 leadCorrect answer is option 'B'. Can you explain this answer?
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A workshop is consuming 500 kW at 0.707 p.f. lagging. A synchronous motor is connected to improve the p.f. to 0.95 lagging. If the synchronous motor takes 100 kW, p.f. of the motor isa)0.386 leadb)0.285 leadc)0.707 leadd)0.8 leadCorrect answer is option 'B'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A workshop is consuming 500 kW at 0.707 p.f. lagging. A synchronous motor is connected to improve the p.f. to 0.95 lagging. If the synchronous motor takes 100 kW, p.f. of the motor isa)0.386 leadb)0.285 leadc)0.707 leadd)0.8 leadCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A workshop is consuming 500 kW at 0.707 p.f. lagging. A synchronous motor is connected to improve the p.f. to 0.95 lagging. If the synchronous motor takes 100 kW, p.f. of the motor isa)0.386 leadb)0.285 leadc)0.707 leadd)0.8 leadCorrect answer is option 'B'. Can you explain this answer?.
A workshop is consuming 500 kW at 0.707 p.f. lagging. A synchronous motor is connected to improve the p.f. to 0.95 lagging. If the synchronous motor takes 100 kW, p.f. of the motor isa)0.386 leadb)0.285 leadc)0.707 leadd)0.8 leadCorrect answer is option 'B'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A workshop is consuming 500 kW at 0.707 p.f. lagging. A synchronous motor is connected to improve the p.f. to 0.95 lagging. If the synchronous motor takes 100 kW, p.f. of the motor isa)0.386 leadb)0.285 leadc)0.707 leadd)0.8 leadCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A workshop is consuming 500 kW at 0.707 p.f. lagging. A synchronous motor is connected to improve the p.f. to 0.95 lagging. If the synchronous motor takes 100 kW, p.f. of the motor isa)0.386 leadb)0.285 leadc)0.707 leadd)0.8 leadCorrect answer is option 'B'. Can you explain this answer?.
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A workshop is consuming 500 kW at 0.707 p.f. lagging. A synchronous motor is connected to improve the p.f. to 0.95 lagging. If the synchronous motor takes 100 kW, p.f. of the motor isa)0.386 leadb)0.285 leadc)0.707 leadd)0.8 leadCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A workshop is consuming 500 kW at 0.707 p.f. lagging. A synchronous motor is connected to improve the p.f. to 0.95 lagging. If the synchronous motor takes 100 kW, p.f. of the motor isa)0.386 leadb)0.285 leadc)0.707 leadd)0.8 leadCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A workshop is consuming 500 kW at 0.707 p.f. lagging. A synchronous motor is connected to improve the p.f. to 0.95 lagging. If the synchronous motor takes 100 kW, p.f. of the motor isa)0.386 leadb)0.285 leadc)0.707 leadd)0.8 leadCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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