Question

A balanced three phase source with vL=240 V rms is supplying 8 kVA at 0.6 powder factor lagging to two wye connected parallel loads. If one load draws 3kW at unity powder factor , calculate impedence per phase of the second load.

Answer 1

Answer:

2.35 + j8.34 Ω

Explanation:

Voltage = V$_{L}$ = 240 V rms

supplying power = S$_{s}$ = 8 kVA

power factor = pf$_{s}$ = 0.6

Let P₁ represents one load draws 3kW at unity powder factor

The power angle is:

θ$_{s}$ = cos⁻¹  pf$_{s}$ = cos⁻¹  0.6 = 53.13°

Complex power supplied source is:

S$_{s}$ =  S$_{s}$ < θ$_{s}$ = 8<53.13° kVA

Complex power for first load:

S₁ = P₁ = 3kVA

Since the power angle of first load is  θ₁ = 0°

According to principle of conservation of AC power, the power of second load is:

S₂ =  S$_{s}$ - S₁

    = 8<53.13° - 3

    = 6.65<74.29° kVA

Since the second load is a Y connected load the phase voltage:

V$_{p}$ =  V$_{L}$ / $\sqrt{3}$

    = 240/1.732051

    = 138.564

    = 138.56 V

Complex power of second load:

S₂ = 3 V$_{p}$² / Z$_{p}$

impedance per phase of the second load:

Z$_{p}$ =  3 V$_{p}$² / S₂

   = 3 (138.56)² /  6.65<74.29°

   = 3(19198.8736) / 6.65<74.29°

   = 57596.6208 / 6.65<74.29°

Z$_{p}$ = 2.35 + j8.34Ω