Answer :
Answer:
Source voltage (L-L) = 3479.50<2.13 V (in polar form)
Source voltage (L-L) = 3477.1 + j129.57 V (rectangular form)
Given Information:
A 3 phase source is feeding three loads connected in parallel.
Load voltage (L-N) = VLoad = 2000 V
Impedance of line = ZLine = 0.2 + j1.0 Ω
Load 1 = S1 = P + jQ = 150 + j120 kVA
Load 2 = S2 = Delta connected with Z2 = 150 - j48 Ω
Load 3 = S3 = 120 KVA at PF = 0.6 leading
Source voltage (L-L) = ?
Explanation:
The source voltage is = VLoad + total current*(ZLine)
Where total current is = I1 + I2 + I3
Lets first find current flowing in each of the loads
Load 1:
3 phase apparent power is given S1 = 150 + j120 kVA
Convert into per phase by diving by 3
S1 = (150 + j120)/3
S1 = 50 + j40 kVA
As we know, S = V1* (where * is the conjugate)
I1 = S1*/VLoad
I1 = (50,000 - j40,000)/2000 (notice minus sign due to conjugate)
I1 = 25 - j20 A
Load 2:
first convert delta impedance into wye by the relation
Zy = Zdelta/3
Zy = (150 - j48)/3
Zy = 50 - j16 Ω
I2 = V/Zy
I2 = 2000/(50 - j16)
I2 = 36.29 + j11.61 A
Load 3:
apparent power = 120 KVA
PF = cos(θ) = 0.6 leading
As we know, P = S*cos(θ) and Q = S*sin(θ)
θ = cos⁻¹(PF) = cos⁻¹(0.6) = 53.13°
P3 = 120*cos(53.13°) = 72 kW
Q3 = 120*sin(53.13°) = 96 kVAR
S3 = 72 + j96 kVA
Convert into per phase by diving by 3
S3 = (72 - 96)/3 (minus sign due to leading PF)
S3 = 24 - j32 kVA
I3 = S3*/VLoad
I3 = (24,000 + j32,000)/2000
I3 = 12 + j16 A
Total current = I1 + I2 + I3
Total current = (25 - j20) + (36.29) + j11.61) + (12 + j16)
Total current = 73.29 + j7.61
Source voltage (L-N) = VLoad + total current*(ZLine)
Source voltage (L-N) = 2000 + (73.29 + j7.61)*(0.2 + j1.0)
Source voltage (L-N) = 2007.05 + j74.81
Source voltage (L-L) = [tex]\sqrt{3}[/tex]*Source voltage (L-N)
Source voltage (L-L) = [tex]\sqrt{3}[/tex] (2007.05 + j74.81)
Source voltage (L-L) = 3477.1 + j129.57 V
Source voltage (L-L) = 3479.50<2.13 V (in polar form)