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.