Question

Three parallel three-phase loads are supplied from a 480V (line-line RMS), 60 Hz three-phase supply. The loads are as follows: Load 1: A 20HP motor operating at full load, 90% efficiency and a 0.8 lagging power factor. Load 2: A balanced resistive load that draws a total of 20kW. Load 3: A Y-connected capacitor bank with a total rating of 20kVAr. What is the total system kW ?

Answer 1

Answer:

The total system active power P = P_1 + P_2 + P_3 = 34.91 KW

Explanation:

Load 1: Active power P_1 = 20 HP = 14.91kW;

Reactive power $Q1 = P tan(\phi)$

                               $= 14.91\times tan(cos^{-}0.8) = 11.18 kvar$

Load 2: Active power P_2 = 20 kW;

Reactive power Q2 = 0 since the load is purely resistive.

Load 3: Active power $P_3$ = 0 due to purely capacitiveload

           Reactive power $Q_3$ = -20 Var

a) since all three loads are connected in parallel therefore

    The total system active power P = P_1 + P_2 + P_3 = 34.91 KW

Total system reactive power Q = Q_1 + Q_2 + Q_3 = 11.18 + 0 -20 = -8.82 kVar

Since Q = 0, the power factor is unity.

Supply current per phase is given by

$I = \frac{P}{\sqrt{3}V_{L}}$

$= \frac{34910}{\sqrt{3}\times 480} = 41.99 A$