Question

Three single-phase loads in parallel are supplied from a 1400V (RMS), 60 Hz supply. The loads are as follows: Load 1: Inductive load: 125 kVA, 0.28 power factor Load 2: Capacitive load: 10 kW, 40 kVAR Load 3: Resistive load: 15 kW Find the total kW, kVAR, kVA, and supply power factor. (5 points) Find the capacitive correction (in kVARs) required to improve the power factor to 0.8 and calculate the supply current with this correction in place. (10 points) What is the least current that can service these three loads and how much compensation would it require

Answer 1

The answer is below

Step-by-step explanation:

`\theta_1=cos^(-1)0.28=73.74^o\ lagging\\\\S_1=125\angle 73.74^o=35\ kW+j120\ kVAR\\\\S_2=10\ kW-j40\ kVAR\\\\S_3=15\ kW`

Total power = P = 35 kW + 10 kW + 15 kW = 60 kW

Total kVAR = 120 kVAR - 40 kVAR = 80 kVAR

`Total\ apparent \ power =S= S_1+S_2+S_3=(35+j120)+(10-j40)+(15)\\\\S=60\ kW+j80\ kVAR=100\angle 53.13^o\\\\Current(I)=(S^*)/(V^*) (100000\angle -53.13^o)/(1400\angle0)=71.43\angle-53.13^o\\ \\Power\ factor (PF)=cos(53.13)=0.6\ lagging\\\\The \ new\ power\ factor\ is\ to \ be\ 0.8[cos^(-1)0.8=36.87^o], hence\ since\ the\ total\ \\real\ power(P)= 60\ kW, the\ capacitor\ kVAR(Q_c)\ is:\\\\Q_c=60tan(53.13)-60tan(36.87)=80-45=35\ kVAR\\\\`

`C=(Q_c)/(wV^2) =(35000\ VAR)/(2\pi*60\ Hz*1400\ V)=47.38\ \mu f`

New current (I') =
`(S'^*)/(V^*)=(60000-j45000)/(1400)=53.57\angle-36.87^o`

Current reduce from 71.43 A to 53.57 A