Question

A 60 Hz three-phase 115kV transmission line has a series impedance of 30+j150 ohms/phase. The shunt susceptance of this line is phasemhosBc/42. The line to neutral voltages at buses p and q are kVVp0.651.63and kVVq006.65, respectively. Compute lossP, the three phase real power losses in MW for this transmission circuit.

Answer 1

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
`(B_c)/(2) = 4 \ mhos /phase`

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

Step-by-step 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:

`I =(V_p-V_q)/(Z +(1)/(Y))`

`I =(63.51 * 10^3 \angle 6^0-656.06 * 10^3 \angle 0^0 )/(30 + j 150 -j0.125)`

`I = 45.173 \angle 27.28^0 \ A`

The
`P_{loss` can be computed as:

`P_(loss) = 3 * I^2 * R`

`P_(loss) = 3 \begin {bmatrix} 45.173 \end {bmatrix} ^2 * 30`

Finally; the three-phase real power loss is:

`P_(loss \ of \ sphere) = 0.183 \ MW`