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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 information:
- 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
- Desired power factor after connecting the synchronous motor = 0.95 lagging
Step 1: Calculating the apparent power of the workshop
Apparent power (S) is the product of the power consumed (P) and the reciprocal of the power factor (p.f.).
S = P / p.f.
In this case, the power consumed by the workshop is 500 kW and the power factor is 0.707 lagging.
So, S = 500 / 0.707 = 707.1 kVA.
Step 2: Calculating the reactive power of the workshop
Reactive power (Q) is the product of the apparent power (S) and the sine of the angle between the apparent power and the real power.
Q = S * sin(angle)
In this case, the angle (θ) can be calculated using the power factor.
θ = cos^(-1)(p.f.)
θ = cos^(-1)(0.707)
θ ≈ 45.57 degrees
So, Q = 707.1 * sin(45.57)
Q ≈ 500 kVAR
Step 3: Calculating the apparent power of the synchronous motor
The power consumed by the synchronous motor is given as 100 kW.
Since the power factor is not given, let's assume it as x.
So, the apparent power of the synchronous motor is 100 / x kVA.
Step 4: Calculating the reactive power of the synchronous motor
The reactive power of the synchronous motor can be calculated using the apparent power and the power factor.
Q_m = √(S_m^2 - P_m^2)
In this case, the apparent power of the synchronous motor is 100 / x kVA and the power consumed is 100 kW.
So, Q_m = √((100 / x)^2 - 100^2) kVAR
Step 5: Calculating the new apparent power of the workshop
The new apparent power of the workshop can be calculated by adding the apparent powers of the workshop and the synchronous motor.
S_new = S + S_m
In this case, S_new = 707.1 + (100 / x) kVA.
Step 6: Calculating the new reactive power of the workshop
The new reactive power of the workshop can be calculated by adding the reactive powers of the workshop and the synchronous motor.
Q_new = Q + Q_m
In this case, Q_new = 500 + √((100 / x)^2 - 100^2) kVAR.
Step 7: Calculating the new power factor of the workshop
The new power factor (p.f.) of the workshop can be calculated using the new apparent power and the new reactive power.
p.f. = P_new / S_new
In this case, P_new = 500 kW and S_new = 707.1 + (100 / x) kVA.
So, p.f. = 500 / (707.1 + (100 / x)).
- 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
- Desired power factor after connecting the synchronous motor = 0.95 lagging
Step 1: Calculating the apparent power of the workshop
Apparent power (S) is the product of the power consumed (P) and the reciprocal of the power factor (p.f.).
S = P / p.f.
In this case, the power consumed by the workshop is 500 kW and the power factor is 0.707 lagging.
So, S = 500 / 0.707 = 707.1 kVA.
Step 2: Calculating the reactive power of the workshop
Reactive power (Q) is the product of the apparent power (S) and the sine of the angle between the apparent power and the real power.
Q = S * sin(angle)
In this case, the angle (θ) can be calculated using the power factor.
θ = cos^(-1)(p.f.)
θ = cos^(-1)(0.707)
θ ≈ 45.57 degrees
So, Q = 707.1 * sin(45.57)
Q ≈ 500 kVAR
Step 3: Calculating the apparent power of the synchronous motor
The power consumed by the synchronous motor is given as 100 kW.
Since the power factor is not given, let's assume it as x.
So, the apparent power of the synchronous motor is 100 / x kVA.
Step 4: Calculating the reactive power of the synchronous motor
The reactive power of the synchronous motor can be calculated using the apparent power and the power factor.
Q_m = √(S_m^2 - P_m^2)
In this case, the apparent power of the synchronous motor is 100 / x kVA and the power consumed is 100 kW.
So, Q_m = √((100 / x)^2 - 100^2) kVAR
Step 5: Calculating the new apparent power of the workshop
The new apparent power of the workshop can be calculated by adding the apparent powers of the workshop and the synchronous motor.
S_new = S + S_m
In this case, S_new = 707.1 + (100 / x) kVA.
Step 6: Calculating the new reactive power of the workshop
The new reactive power of the workshop can be calculated by adding the reactive powers of the workshop and the synchronous motor.
Q_new = Q + Q_m
In this case, Q_new = 500 + √((100 / x)^2 - 100^2) kVAR.
Step 7: Calculating the new power factor of the workshop
The new power factor (p.f.) of the workshop can be calculated using the new apparent power and the new reactive power.
p.f. = P_new / S_new
In this case, P_new = 500 kW and S_new = 707.1 + (100 / x) kVA.
So, p.f. = 500 / (707.1 + (100 / x)).
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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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