Currently stuck at this part of the question Recieving end Voltage 0.96pu, rated

Question

Currently stuck at this part of the question

Recieving end Voltage 0.96pu, rated full load current = 1.9kA, series impedence 0.0166+0.301j. transmission line voltage = 750kV

Answers are i) 751.64 kV, 22.369 degrees ii) 796.17 kV , 22.183 degrees iii) 10.579 precent

How do i obtain the above answers, been trying for awhile. if you need any more values just ask me. ive done the other parts.

Part 4: The full load current per-phase for the transmission line is I. The transmission line is required to supply the full load current to a load at a receiving end voltage of V pu. The new value for V is to be obtained from the excel spread sheet. FL i. Calculate the sending end line to line voltage and power angle during full-load operation at unity power factor. Specify your answer in kV and degrees. The load is switched off. The load current is zero. Assuming the sending end voltage magnitude is regulated at the same value as in part 4-(i), calculate the no- load receiving end voltage. Specify your answer in kV and degrees. ii. ii. Calculate the voltage regulation for the transmission line as a percentage. 41-rhlx1 VEL Voltage regulation

Explanation / Answer

The sending end voltage(Vs) = 1.02*750=765v

SIL=2199.8MW

Angle ()=36

SIL in MVA= SIL in MW/COS

= 2199.8/cos 36

=2719.102 MVA

SIL=(Vs) 2/Zbase*ZcP.U

Zbase*ZcP.U=(Vs) 2/SIL= 765*765/2719.102=215.2272

Receivind end voltage (Vr)=0.98*750=735v

SIL in receiving end voltage side= (Vr)2/ Zbase*ZcP.U=735*735/215.2272=2510.02MVA

SILin receiving in MW= 2510.02*C0S 36=2030.65MW

practical line loadability = SIL in sending end side - SIL in receiving end side

= 2199.8-2030.65

=169.14

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