A 50 Hz three-phase transmission line is 300 km long. It has a total series impedance of 23+j75 Ω and a total shunt admittance of j500μS. It delivers 150 MW at 220 kV, with a power factor of 0.88 lagging.
c) Calculate the voltage regulation and efficiency of the line.
Answers
Answer:
Explanation:
approximation. (c)The long line equation. How accurate are the short and medium-length line approximations for this case? SOLUTION(a)In the short line approximation, the shunt admittance is ignored. The ABCDconstants for this line are:101ABZCD====(9-67) 1A=2375 78.473 BZj==+Ω =∠° Ω0 SC=1D=The receiving end line voltage is 220 kV, so the rated phase voltage is 220 kV / 3= 127 kV, and the current is()out150,000,000 W394 A33220,000 VLLLSIV===If the phase voltage at the receiving end is assumed to be at a phase angle of 0°, then the phase voltage at the receiving end will be 1270kVR=∠ °V, and the phase current at the receiving end will be 39428.4AR=∠ −°I. The current and voltage at the sending end of the transmission line are given by the following equations:SRRAB=+VVI( )()()()11270kV78.473 39428.4AS=∠ °+∠° Ω∠ −°V1518.2kVS=∠°VSRRCD=+IVI()()( )()0 S1330kV139428.4AS=∠ °+∠ −°I39428.4AS=∠ −°I(b)In the medium length line approximation, the shunt admittance divided into two equal pieces at either end of the line. The ABCDconstants for this line are:121142ZYABZZYZYCYD=+=§·=+=+¨¸©¹(9-73) ()()2375 0.0005 S110.98130.3422jjZYA+Ω=+=+=∠°2375 78.473 BZj==+Ω =∠° Ω()()()2375 0.0005 S10.0005 S144jjZYCYjªº+Ω§·=+=+¨¸«»©¹¬¼
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19044.9531090.2SC−=×∠°()()2375 0.0005 S110.98130.3422jjZYD+Ω=+=+=∠°The receiving end line voltage is 220 kV, so the rated phase voltage is 220 kV / 3= 127 kV, and the current is()out150,000,000 W394 A33220,000 VLLLSIV===If the phase voltage at the receiving end is assumed to be at a phase angle of 0°, then the phase voltage at the receiving end will be 1270kVR=∠ °V, and the phase current at the receiving end will be 39428.4AR=∠ −°I. The current and voltage at the sending end of the transmission line are given by the following equations:SRRAB=+VVI148.48.7kVS=∠°VSRRCD=+IVI36119.2AS=∠ −°I(c)In the long transmission line, the ABCDconstants are based on modified impedances and admittances: sinhdZZdγγ′ =(9-74) ()tanh/ 2/ 2dYYdγγ′ =(9-75) and the corresponding ABCDconstants are 121142Z YABZZ YZ YCYD′ ′′=+=′ ′′ ′§·′=+=+¨¸©¹(9-76) The propagation constant of this transmission line is yzγ=650010S2375 0.0006681.5300 km300 kmjjyzγ−§·×+Ω§·===∠°¨¸¨¸©¹©¹()()0.0006681.5300 km0.19881.5dγ=∠°=∠°The modified parameters are ()()sinh 0.19881.5sinh2375 77.973 0.19881.5dZZjdγγ∠°==+Ω=∠° Ω′∠°()4tanh/ 25.011089.9S/ 2dYYdγγ−==×∠°′and the ABCDconstants are
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19110.9830.332Z YA′ ′=+=∠°77.973 BZ==∠° Ω′14.9790.1S4Z YCY′ ′§·=+=∠°′¨¸©¹10.9830.332Z YD′ ′=+=∠°