Answer: To find the total real, reactive, and apparent power used by the facility, we need to calculate the power consumption of each individual load first.
Explanation:
a. The power consumed by the 50hp motor is given by:
P = (50 hp) / (0.86 × 0.70) = 83.63 kW
The reactive power consumed by the motor is given by:
Q = P × tan(cos⁻¹(0.70)) = 57.63 kVAR
b. The power consumed by the 100hp motor is given by:
P = (100 hp × 0.82) / 0.89 = 91.01 kW
The reactive power consumed by the motor is given by:
Q = P × tan(cos⁻¹(0.80)) = 54.72 kVAR
c. The power consumed by each of the two 20hp motors is given by:
P = (20 hp) / (0.92 × 0.85) = 25.08 kW
The reactive power consumed by each motor is given by:
Q = P × tan(cos⁻¹(0.85)) = 14.07 kVAR
d. The power consumed by the 300hp motor is given by:
P = (300 hp × 0.75) / 0.92 = 245.11 kW
The reactive power consumed by the motor is given by:
Q = P × tan(cos⁻¹(0.84)) = 160.89 kVAR
e. The power consumed by the incandescent lighting is given by:
P = 50 kW
The reactive power consumed by the lighting is zero, since it is a resistive load.
Now we can find the total real, reactive, and apparent power used by the facility:
Total real power = 83.63 kW + 91.01 kW + 2 × 25.08 kW + 245.11 kW + 50 kW = 529.91 kW
Total reactive power = 57.63 kVAR + 54.72 kVAR + 2 × 14.07 kVAR + 160.89 kVAR + 0 kVAR = 301.98 kVAR
Total apparent power = √(529.91² + 301.98²) = 609.57 kVA
The total power factor is given by:
cos(θ) = 529.91 kW / 609.57 kVA = 0.8691
θ = cos⁻¹(0.8691) = 29.59 degrees
Therefore, the total power factor is 0.869 lagging. The real power used by the facility is 529.91 kW, the reactive power used is 301.98 kVAR, and the apparent power is 609.57 kVA.
SPJ11