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A substation operating at its full load of 1000 kVA supplies a load of power factor 0.71 lagging. Calculate the permissible additional load at this power factor and the rating of an ideal synchronous condenser to raise the substation power factor to 0.87 lagging.?
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A substation operating at its full load of 1000 kVA supplies a load of...
Given data:
• Full load of substation = 1000 kVA
• Power factor of the load = 0.71 lagging
• Desired power factor = 0.87 lagging
Permissible Additional Load:
To calculate the permissible additional load, we first need to find the apparent power of the existing load.
Apparent power = Real power / power factor
Real power = Full load = 1000 kVA
Power factor = 0.71 lagging
Apparent power = 1000 / 0.71 = 1408.45 kVA
Now, we can calculate the permissible additional load as follows:
Permissible additional load = Apparent power x (desired power factor - existing power factor) / desired power factor
Permissible additional load = 1408.45 x (0.87 - 0.71) / 0.87 ≈ 450 kVA
Therefore, the permissible additional load at a power factor of 0.71 lagging is approximately 450 kVA.
Ideal Synchronous Condenser Rating:
To calculate the rating of the ideal synchronous condenser, we can use the following formula:
Rating of synchronous condenser = Full load x (tanδ1 - tanδ2) / (tanδ2 x √(1 - (cosφ2)^2))
Where,
• Full load = 1000 kVA (given)
• tanδ1 = tangent of the angle between voltage and current at existing power factor = tan⁻¹(0.71) ≈ 35.5°
• tanδ2 = tangent of the angle between voltage and current at desired power factor = tan⁻¹(0.87) ≈ 40.4°
• cosφ2 = desired power factor = 0.87
Rating of synchronous condenser = 1000 x (tan35.5° - tan40.4°) / (tan40.4° x √(1 - 0.87^2)) ≈ 358 kVAR
Therefore, the rating of the ideal synchronous condenser to raise the substation power factor to 0.87 lagging is approximately 358 kVAR.
• Full load of substation = 1000 kVA
• Power factor of the load = 0.71 lagging
• Desired power factor = 0.87 lagging
Permissible Additional Load:
To calculate the permissible additional load, we first need to find the apparent power of the existing load.
Apparent power = Real power / power factor
Real power = Full load = 1000 kVA
Power factor = 0.71 lagging
Apparent power = 1000 / 0.71 = 1408.45 kVA
Now, we can calculate the permissible additional load as follows:
Permissible additional load = Apparent power x (desired power factor - existing power factor) / desired power factor
Permissible additional load = 1408.45 x (0.87 - 0.71) / 0.87 ≈ 450 kVA
Therefore, the permissible additional load at a power factor of 0.71 lagging is approximately 450 kVA.
Ideal Synchronous Condenser Rating:
To calculate the rating of the ideal synchronous condenser, we can use the following formula:
Rating of synchronous condenser = Full load x (tanδ1 - tanδ2) / (tanδ2 x √(1 - (cosφ2)^2))
Where,
• Full load = 1000 kVA (given)
• tanδ1 = tangent of the angle between voltage and current at existing power factor = tan⁻¹(0.71) ≈ 35.5°
• tanδ2 = tangent of the angle between voltage and current at desired power factor = tan⁻¹(0.87) ≈ 40.4°
• cosφ2 = desired power factor = 0.87
Rating of synchronous condenser = 1000 x (tan35.5° - tan40.4°) / (tan40.4° x √(1 - 0.87^2)) ≈ 358 kVAR
Therefore, the rating of the ideal synchronous condenser to raise the substation power factor to 0.87 lagging is approximately 358 kVAR.
Community Answer
A substation operating at its full load of 1000 kVA supplies a load of...
A sub-station operating at full load of 1200 kVA supplies a load at 0.7 power factor lagging. Calculate the permissible additional load at this power factor and the rating of synchronous condenser to raise the substation power to 0.9 lagging.
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A substation operating at its full load of 1000 kVA supplies a load of power factor 0.71 lagging. Calculate the permissible additional load at this power factor and the rating of an ideal synchronous condenser to raise the substation power factor to 0.87 lagging.?
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A substation operating at its full load of 1000 kVA supplies a load of power factor 0.71 lagging. Calculate the permissible additional load at this power factor and the rating of an ideal synchronous condenser to raise the substation power factor to 0.87 lagging.? 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 substation operating at its full load of 1000 kVA supplies a load of power factor 0.71 lagging. Calculate the permissible additional load at this power factor and the rating of an ideal synchronous condenser to raise the substation power factor to 0.87 lagging.? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A substation operating at its full load of 1000 kVA supplies a load of power factor 0.71 lagging. Calculate the permissible additional load at this power factor and the rating of an ideal synchronous condenser to raise the substation power factor to 0.87 lagging.?.
A substation operating at its full load of 1000 kVA supplies a load of power factor 0.71 lagging. Calculate the permissible additional load at this power factor and the rating of an ideal synchronous condenser to raise the substation power factor to 0.87 lagging.? 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 substation operating at its full load of 1000 kVA supplies a load of power factor 0.71 lagging. Calculate the permissible additional load at this power factor and the rating of an ideal synchronous condenser to raise the substation power factor to 0.87 lagging.? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A substation operating at its full load of 1000 kVA supplies a load of power factor 0.71 lagging. Calculate the permissible additional load at this power factor and the rating of an ideal synchronous condenser to raise the substation power factor to 0.87 lagging.?.
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