Final answer:
The question pertains to constructing power triangles for an induction motor, a synchronous motor, and their combined load when connected to a three-phase power source, taking into account the power factor, real power, and reactive power of each.
Step-by-step explanation:
The question involves deriving the power triangles for two three-phase motors and their combined load when connected to a three-phase source. Specifically, the question supplied given details about an induction motor with a power draw of 300 kW at a power factor of 0.8 lagging, and a synchronous motor with a draw of 150 kVA at a power factor of 0.95 leading. To draw the power triangles, we would first need to calculate the apparent power (S), real power (P), and reactive power (Q) for each motor.
For the induction motor (Power factor lagging):
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- Real Power (P) = 300 kW
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- Power Factor (pf) = 0.8 lagging
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- Apparent Power (S) = P / pf = 300 kW / 0.8 = 375 kVA
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- Reactive Power (Q) = S × √(1 - pf^2) = 375 kVA × √(1 - 0.8^2) ≈ 225 kVAr (lagging)
For the synchronous motor (Power factor leading):
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- Apparent Power (S) = 150 kVA
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- Power Factor (pf) = 0.95 leading
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- Real Power (P) = S × pf = 150 kVA × 0.95 = 142.5 kW
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- Reactive Power (Q) = S × √(1 - pf^2) = 150 kVA × √(1 - 0.95^2) ≈ 48.97 kVAr (leading)
The power triangle for each motor includes a horizontal component for the real power and a vertical component for the reactive power, which points upward for a leading power factor and downward for a lagging power factor. The combined power triangle will be constructed by vectorially adding the real and reactive powers of the two motors.