3 phase delta connection load?
anyone knows how to solve this?
A balanced delta-connected load with the per-phase impedance of 30+j15 Ω is connected to the main supply through three identical conductors each has an impedance of 0.6+j0.8 Ω. The voltage at the load terminals is measured as 400 V
a)Draw the schematic diagram of the circuit
b)Calculate the line current,
c) Calculate the line voltage across the source terminals and draw the phasor diagram.
d)Calculate the active and reactive powers delivered by the source
e)Calculate the active and reactive power losses in the conductors.
Answers & Comments
Verified answer
Z load = √Rl^2+Xl^2 = √30^2+15^2 = 33.54 ohm
Rt = Rl+Rc = 30+0.6 = 30.6 ohm
Xt = Xl+Xc = 15+0.8 = 15.8 ohm
Zt = √Rt^2+Xt^2 = √30.6^2+15.8^2 = 34.44 ohm
Line current I = Vl/(√3*Zl) = 400/58,1 = 6.89 A
Line to line voltage = I*Zt*√3 = 410.7 V
Ps = 3*I^2*Rt = 4352.0 watt
Qs = 3*I^2*Xt = 2247.1 war
Pc = 3*I^2*Rc = 85.0 watt
Qc = 3*I^2*Xc = 113.7 war
The voltage around the line AB could be evaluate Van (the place n is impartial) - Vbn. Van = 240 attitude 0 Vbn = 240 attitude -a hundred and twenty Vab = 240 @ 0 deg - 240 @-120deg Vab = 240 - 240 cos(-a hundred and twenty) + j240 sin (-a hundred and twenty) Vab = 240( a million - cos(-a hundred and twenty) + jsin(-a hundred and twenty)) Vab = 240 ( 3/2 + j sqrt(3)/2) Taking the exponential type of this offers you with the angles you're searching for. Extrapolate for the different lines and attitude mixtures. Technically, your answer is actual for the magnitude, in spite of the shown fact that doesn't hold the phasor information.