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A 208V star connected 3ph synchronous motor has a synchronous reactance of 4 ohm per phase and armature reactance is negligible. At a certain load the Motor takes 7.2 kw at 0.8 pf lag. If power develop by motor remains same while excitation Voltage is increase by 50 percent raising field excitation. Determine 1. new pf
2. new armature current?
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A 208V star connected 3ph synchronous motor has a synchronous reactanc...
Given Data
- Line Voltage (V) = 208V
- Synchronous Reactance (Xs) = 4Ω/phase
- Power (P) = 7.2 kW
- Power Factor (pf) = 0.8 lag
1. New Power Factor
- The motor operates at a certain load with a lagging power factor. When the excitation voltage is increased by 50%, the motor will draw more reactive power, which will improve the power factor.
- New Excitation Voltage (Ef) = 1.5 * Ef (initial)
- The power developed by the motor remains constant (7.2 kW). The new power factor can be calculated using:
New pf = Real Power / Apparent Power
- Real Power (P) = 7.2 kW, Apparent Power (S) = V * I (where I is the armature current).
- With increased excitation, the motor draws more current but with a higher voltage, which decreases the reactive power drawn, enhancing the power factor.
2. New Armature Current
- Initially, the motor's apparent power can be determined:
S = P / pf = 7.2 kW / 0.8 = 9 kVA
- The initial current (I) can be calculated as:
I = S / (√3 * V) = 9 kVA / (√3 * 208V)
- With increased excitation, the armature current increases due to the increased reactive power demand.
- New Apparent Power (considering improved pf) can be calculated:
Assuming a new power factor of approximately 0.95:
New S = P / New pf = 7.2 kW / 0.95 = ~7.58 kVA
- New Armature Current:
New I = New S / (√3 * V) = 7.58 kVA / (√3 * 208V)
Conclusion
- The new power factor improves to around 0.95.
- The new armature current can be calculated accordingly, showing a slight increase due to enhanced excitation.
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A 208V star connected 3ph synchronous motor has a synchronous reactance of 4 ohm per phase and armature reactance is negligible. At a certain load the Motor takes 7.2 kw at 0.8 pf lag. If power develop by motor remains same while excitation Voltage is increase by 50 percent raising field excitation. Determine 1. new pf 2. new armature current?
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A 208V star connected 3ph synchronous motor has a synchronous reactance of 4 ohm per phase and armature reactance is negligible. At a certain load the Motor takes 7.2 kw at 0.8 pf lag. If power develop by motor remains same while excitation Voltage is increase by 50 percent raising field excitation. Determine 1. new pf 2. new armature current? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A 208V star connected 3ph synchronous motor has a synchronous reactance of 4 ohm per phase and armature reactance is negligible. At a certain load the Motor takes 7.2 kw at 0.8 pf lag. If power develop by motor remains same while excitation Voltage is increase by 50 percent raising field excitation. Determine 1. new pf 2. new armature current? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 208V star connected 3ph synchronous motor has a synchronous reactance of 4 ohm per phase and armature reactance is negligible. At a certain load the Motor takes 7.2 kw at 0.8 pf lag. If power develop by motor remains same while excitation Voltage is increase by 50 percent raising field excitation. Determine 1. new pf 2. new armature current?.
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