The problem I need solved is problem 3. the 2.52 is there just for reference. Ch

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The problem I need solved is problem 3. the 2.52 is there just for reference.  Chegg has solutions to 2.52 but not to the specific questions asked in my "Question 3" Please help

Refer to system of prob. 2.52 in text, page 88/89. Calculate the following for two cases: case 1 - with system as described in the problem; case 2 - add a 3 phi shunt capacitance, wye grounded, at the load with a capacitance of 230 microfarads per phase. using VJn at the source as the reference voltage, find the load voltage for phase a [magnitude and angle] Phase a line current [magnitude and angle] P3 phi and Q3 phi from the source A balanced three-phase load is connected to a 4.16-kV, three-phase, four-wire, grounded-wye dedicated distribution feeder. The load can be modeled by an impedance of ZL = (4.7 + y9) ohm /phase, wye-connected. The impedance of the phase conductors is (0.3 + ji1) ohm. Determine the following by using the phase A to neutral voltage as a reference and assume positive phase sequence: Line currents for phases A, B, and C. Line-to-ncutral voltages for all three phases at the load. Apparent, active, and reactive power dissipated per phase, and for all three phases in the load. Active power losses per phase and for all three phases in the phase conductors.

Explanation / Answer

Z of the line is 0.3+j0.8. Z of the load is 5+j13.

Based on the above values the solutions look like this:

a) Since the system is balanced we proceed by solving it on a single phase basis and as assuming the other phase values will be displaced by +/-120 degrees. Solve for total Z by adding the line and load Z. Divide this into the line to ground voltage for "a" phase to get the "a" pahse current.

total Z = 5.0.3 +j (13 + 0.8) = 5.3 + j 13.8 = 14.8 ang 69

Ia = Va/Z = 2402 ang 0 / 14.8 ang 69 = 162.5 ang -69
Ib= 162.5 ang 171
Ic= 162.5 ang 51

b) Here you must subtract the line voltage drop from the source voltage. The voltage dropped on the line would be, V line drop = Ia * Z line = (162.5 ang -69)*(0.9 ang 69.4) = 138.8 ang 0.5 volts.

'a"phase voltage at the load = V source - V line drop = (2402 ang 0) - (138.8 ang 0.5) = (2401.8 + j 0) - (138.8 + j 1.1) = 2263 -j 1.1 = 2263 ang -0.03
Va at the load = 2263 ang -0.03
Vb = 2263 ang (240-0.03)
Vc = 2263 ang (120-0.03)

c) Apparent power = EI* = (2263 ang -0.03)*(162.5 ang 69) = 367.7 ang 69.02 for a phase. Now add the 120 and 240 degree angles to this to get the C and B phase Apparent power. To get kW and kVAR take 367.7 ang 69.02 and convert it to rectangular values to find 131.7 kW and 343.3 kVAR for "a" phase. Follow similar work for "b" and "c".

d) Line losses = E(line drop)I(current in the line)* = 22.6 kVA ang 68.54, 8.3 kW, 21 kVAR

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