The complete and the correct format for the question is as follows:
A 60 Hz three-phase 115kV transmission line has a series impedance of 30+j150 ohms/phase. The shunt susceptance of this line is [tex]\dfrac{B_c}{2} = 4 \ mhos /phase[/tex]
The line to neutral voltages at buses p and q are and respectively. Compute , the three-phase real power losses in MW for this transmission circuit.
Answer:
0.183 MW
Explanation:
The diagrammatic illustration of the given information can be seen in the image attached below:
Thus: the current in the network can be computed as:
[tex]I =\dfrac{V_p-V_q}{Z +\dfrac{1}{Y}}[/tex]
[tex]I =\dfrac{63.51 \times 10^3 \angle 6^0-656.06 \times 10^3 \angle 0^0 }{30 + j 150 -j0.125}[/tex]
[tex]I = 45.173 \angle 27.28^0 \ A[/tex]
The [tex]P_{loss[/tex] can be computed as:
[tex]P_{loss} = 3 \times I^2 \times R[/tex]
[tex]P_{loss} = 3 \begin {bmatrix} 45.173 \end {bmatrix} ^2 \times 30[/tex]
Finally; the three-phase real power loss is:
[tex]P_{loss \ of \ sphere} = 0.183 \ MW[/tex]
