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A cylindrical rotor synchronous motor is used to improve the power factor of a load 5 MW at a power factor of 0.6 lag. The motor is required to meet an additional demand of 0.2 MW and raise the overall power factor to 0.8 (lag). If the efficiency of the motor is 75%. The power factor of the motor is – (up to two decimal places)
Correct answer is between '0.08,0.1'. Can you explain this answer?
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A cylindrical rotor synchronous motor is used to improve the power fac...
Rating of load is –
P1 = 5 MW, cos ϕ1 = 0.6
Q1 = P1 tan ϕ
= 5 × tan (cos-1 (0.6)) = 6.67 MVAR
Required motor output = 0.2 MW
Efficiency = 0.75
Total load = 5 + 0.267 = 5.267 MW
New power factor cos ϕ = 0.8 lag.
Reactive power, Q = P tan ϕ
= 5.267 × tan (cos-1 (0.8))
= 3.95 MVAR
MVA supplied by motor, Q2 = Q1 – Q = 6.67 – 3.95 = 2.72 MVAR
Q2 = P2 tan ϕ2
⇒ 2.72 = 0.267 tanϕ2
⇒ ϕ2 = 84.39°
Cosϕ2 = 0.098
P1 = 5 MW, cos ϕ1 = 0.6
Q1 = P1 tan ϕ
= 5 × tan (cos-1 (0.6)) = 6.67 MVAR
Required motor output = 0.2 MW
Efficiency = 0.75
Total load = 5 + 0.267 = 5.267 MW
New power factor cos ϕ = 0.8 lag.
Reactive power, Q = P tan ϕ
= 5.267 × tan (cos-1 (0.8))
= 3.95 MVAR
MVA supplied by motor, Q2 = Q1 – Q = 6.67 – 3.95 = 2.72 MVAR
Q2 = P2 tan ϕ2
⇒ 2.72 = 0.267 tanϕ2
⇒ ϕ2 = 84.39°
Cosϕ2 = 0.098
Most Upvoted Answer
A cylindrical rotor synchronous motor is used to improve the power fac...
Calculation of Power Factor of the Motor
- Given:
- Load power = 5 MW
- Initial power factor = 0.6 lag
- Additional demand = 0.2 MW
- Required power factor = 0.8 lag
- Motor efficiency = 75%
Calculating the Real Power and Reactive Power of the Load
- Real power (P) = Load power factor * Load power = 0.6 * 5 MW = 3 MW
- Reactive power (Q) = √(Load power^2 - Real power^2) = √(5^2 - 3^2) = √(25 - 9) = √16 = 4 MVAr
Calculating the New Real Power and Reactive Power with the Additional Demand
- New real power = Load power + Additional demand = 5 MW + 0.2 MW = 5.2 MW
- New reactive power = Q = 4 MVAr
Calculating the Apparent Power of the Load
- Apparent power (S) = √(Real power^2 + Reactive power^2) = √(5^2 + 4^2) = √(25 + 16) = √41 ≈ 6.40 MVA
Calculating the Apparent Power of the Motor
- Apparent power of the motor = New real power / Motor efficiency = 5.2 MW / 0.75 = 6.93 MVA
Calculating the Power Factor of the Motor
- Power factor of the motor = New real power / Apparent power of the motor = 5.2 MW / 6.93 MVA ≈ 0.75
Therefore, the power factor of the motor is approximately 0.75, which is not within the correct range of 0.08 to 0.1. It seems there might have been an error in the calculations or assumptions made.
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A cylindrical rotor synchronous motor is used to improve the power factor of a load 5 MW at a power factor of 0.6 lag. The motor is required to meet an additional demand of 0.2 MW and raise the overall power factor to 0.8 (lag). If the efficiency of the motor is 75%. The power factor of the motor is – (up to two decimal places)Correct answer is between '0.08,0.1'. Can you explain this answer?
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A cylindrical rotor synchronous motor is used to improve the power factor of a load 5 MW at a power factor of 0.6 lag. The motor is required to meet an additional demand of 0.2 MW and raise the overall power factor to 0.8 (lag). If the efficiency of the motor is 75%. The power factor of the motor is – (up to two decimal places)Correct answer is between '0.08,0.1'. 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 cylindrical rotor synchronous motor is used to improve the power factor of a load 5 MW at a power factor of 0.6 lag. The motor is required to meet an additional demand of 0.2 MW and raise the overall power factor to 0.8 (lag). If the efficiency of the motor is 75%. The power factor of the motor is – (up to two decimal places)Correct answer is between '0.08,0.1'. 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 cylindrical rotor synchronous motor is used to improve the power factor of a load 5 MW at a power factor of 0.6 lag. The motor is required to meet an additional demand of 0.2 MW and raise the overall power factor to 0.8 (lag). If the efficiency of the motor is 75%. The power factor of the motor is – (up to two decimal places)Correct answer is between '0.08,0.1'. Can you explain this answer?.
A cylindrical rotor synchronous motor is used to improve the power factor of a load 5 MW at a power factor of 0.6 lag. The motor is required to meet an additional demand of 0.2 MW and raise the overall power factor to 0.8 (lag). If the efficiency of the motor is 75%. The power factor of the motor is – (up to two decimal places)Correct answer is between '0.08,0.1'. 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 cylindrical rotor synchronous motor is used to improve the power factor of a load 5 MW at a power factor of 0.6 lag. The motor is required to meet an additional demand of 0.2 MW and raise the overall power factor to 0.8 (lag). If the efficiency of the motor is 75%. The power factor of the motor is – (up to two decimal places)Correct answer is between '0.08,0.1'. 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 cylindrical rotor synchronous motor is used to improve the power factor of a load 5 MW at a power factor of 0.6 lag. The motor is required to meet an additional demand of 0.2 MW and raise the overall power factor to 0.8 (lag). If the efficiency of the motor is 75%. The power factor of the motor is – (up to two decimal places)Correct answer is between '0.08,0.1'. Can you explain this answer?.
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