Answer:
a. The magnitude of the line source voltage is
Vs = 4160 V
b. Total real and reactive power loss in the line is
Ploss = 12 kW
Qloss = j81 kvar
Sloss = 12 + j81 kVA
c. Real power and reactive power supplied at the sending end of the line
Ss = 540.046 + j476.95 kVA
Ps = 540.046 kW
Qs = j476.95 kvar
Step-by-step explanation:
a. The magnitude of the line voltage at the source end of the line.
The voltage at the source end of the line is given by
Vs = Vload + (Total current×Zline)
Complex power of first load:
S₁ = 560.1 < cos⁻¹(0.707)
S₁ = 560.1 < 45° kVA
Complex power of second load:
S₂ = P₂×1 (unity power factor)
S₂ = 132×1
S₂ = 132 kVA
S₂ = 132 < cos⁻¹(1)
S₂ = 132 < 0° kVA
Total Complex power of load is
S = S₁ + S₂
S = 560.1 < 45° + 132 < 0°
S = 660 < 36.87° kVA
Total current is
I = S*/(3×Vload) ( * represents conjugate)
The phase voltage of load is
Vload = 3810.5/√3
Vload = 2200 V
I = 660 < -36.87°/(3×2200)
I = 100 < -36.87° A
The phase source voltage is
Vs = Vload + (Total current×Zline)
Vs = 2200 + (100 < -36.87°)×(0.4 + j2.7)
Vs = 2401.7 < 4.58° V
The magnitude of the line source voltage is
Vs = 2401.7×√3
Vs = 4160 V
b. Total real and reactive power loss in the line.
The 3-phase real power loss is given by
Ploss = 3×R×I²
Where R is the resistance of the line.
Ploss = 3×0.4×100²
Ploss = 12000 W
Ploss = 12 kW
The 3-phase reactive power loss is given by
Qloss = 3×X×I²
Where X is the reactance of the line.
Qloss = 3×j2.7×100²
Qloss = j81000 var
Qloss = j81 kvar
Sloss = Ploss + Qloss
Sloss = 12 + j81 kVA
c. Real power and reactive power supplied at the sending end of the line
The complex power at sending end of the line is
Ss = 3×Vs×I*
Ss = 3×(2401.7 < 4.58)×(100 < 36.87°)
Ss = 540.046 + j476.95 kVA
So the sending end real power is
Ps = 540.046 kW
So the sending end reactive power is
Qs = j476.95 kvar