Engineering, 04.10.2020 06:01 jgrable5175
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
Answers: 3
![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)=\frac{S^*}{V^*} \frac{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\\\\](/tpl/images/0787/4366/829b7.png)
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Three single-phase loads in parallel are supplied from a 1400V (RMS), 60 Hz supply. The loads are as...
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