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

2.35 + j8.34 Ω

Step-by-step 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)` /
`√(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