Q6Power System - I

Question

Q.6. A synchronous motor, having a power consumption of 50kW, is connected in parallel with a load of 200kW, 0.8 p.f. lagging. The excitation of the motor is adjusted until combined power factor becomes 0.9 lagging. Determine the kVA input and power factor of synchronous motor.

Answer 1

Given: Load: P_load = 200 kW, cosφ_load = 0.8 lagging. Synchronous motor: P_motor = 50 kW (power consumption). Combined (load+motor) power factor = 0.9 lagging. Find: kVA input and power factor of the synchronous motor.

Step 1 — reactive power of the load:

Q_{l o a d} = P_{l o a d} tan (cos^{- 1} 0.8) = 200 \times 0.75 = 150 kVAR (lagging)

Step 2 — total combined real and reactive power: total real power P_total = P_load+P_motor = 200+50 = 250 kW. Since the combined power factor is 0.9 lagging:

Q_{t o t a l} = P_{t o t a l} tan (cos^{- 1} 0.9) = 250 \times 0.4843 = 121.08 kVAR (lagging)

Step 3 — reactive power of the synchronous motor: since Qtotal = Qload + Qmotor,:

Q_{m o t or} = Q_{t o t a l} - Q_{l o a d} = 121.08 - 150 = - 28.92 kVAR

The negative sign indicates the synchronous motor is actually supplying (generating) reactive power — i.e., it is operating overexcited and drawing a leading reactive component, which is exactly why the combined power factor (0.9) is closer to unity than the load's original power factor (0.8): the motor's leading VARs partially cancel the load's lagging VARs, an application of a synchronous motor as a synchronous condenser for power-factor correction while simultaneously performing useful mechanical work.

Step 4 — kVA input and power factor of the motor:

S_{m o t or} = P_{m o t or}^{2} + Q_{m o t or}^{2} = 5 0^{2} + 28.9 2^{2} = 2500 + 836.3 = 57.76 kVA

cos ϕ_{m o t or} = \frac{P _{m o t or}}{S _{m o t or}} = \frac{50}{57.76} = 0.8656 (leading)

Result: the synchronous motor draws a kVA input of approximately 57.76 kVA at a power factor of about 0.866 leading. This result illustrates the practical application of over-excited synchronous motors as combined mechanical-load-driving and power-factor-correcting devices in industrial installations, avoiding the need for a separate, purely reactive synchronous condenser or capacitor bank to achieve the desired improved overall plant power factor.

Cross-check: combined complex power S_total = P_total + jQ_total = 250 + j121.08 kVA, so |S_total| = √(250²+121.08²) = √(62,500+14,660) = √77,160 = 277.78 kVA, and cosφ_total = 250/277.78 = 0.900, exactly matching the given combined power factor of 0.9 lagging, confirming the internal consistency of the computed Qmotor value. This calculation demonstrates the general principle that any over-excited synchronous machine (motor or dedicated synchronous condenser) can be used to improve an installation's overall power factor by supplying leading VARs to offset the lagging VARs drawn by induction motors and other inductive loads, with the added benefit — unlike a static capacitor bank — of also performing useful mechanical work while doing so, and of being continuously adjustable in real time simply by varying the field excitation current.

Q6Power System - I

Question

Answer