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ProblemsAndSolutions_Ch-9

# ProblemsAndSolutions_Ch-9 - (c The voltage regulation of...

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181 (c) The voltage regulation of the transmission line is 10000 8000 VR 100% 100% 25% 8000 S R R V V V = × = × = Problems 9-8 through 9-10 refer to a single phase, 8 kV, 50-Hz, 50 km-long underground cable consisting of two aluminum conductors with a 3 cm diameter separated by a spacing of 15 cm. 9-8. The single-phase transmission line referred to in Problems 9-3 through 9-7 is to be replaced by an underground cable. The cable consists of two aluminum conductors with a 3 cm diameter, separated by a center-to-center spacing of 15 cm. As before, assume that the 50 Hz ac resistance of the line is 5% greater than its dc resistance, and calculate the series impedance and shunt admittance of the line in ohms per km and siemens per km. Also, calculate the total impedance and admittance for the entire line. S OLUTION The series inductance per meter of this transmission line is given by Equation (9-22). 1 ln H/m 4 D l r µ π § · = + ¨ ¸ © ¹ (9-22) where 7 0 4 10 H/m µ µ π = = × . 7 6 0 1 0.15 m 4 10 H/m 1 0.15 m ln ln 1.021 10 H/m 4 0.015 m 4 0.015 m l µ π π π × § · § · = + = + = × ¨ ¸ ¨ ¸ © ¹ © ¹ Therefore the inductance of this transmission line will be ( ) ( ) 6 1.021 10 H/m 50,000 m 0.0511 H L = × = The inductive reactance of this transmission line is ( )( ) 2 2 50 Hz 0.0511 H 16.05 X j L j fL j j ω π π = = = = The resistance of this transmission line is the same as for the overhead transmission line calculated previously: AC 2.1 R = . The total series impedance of this entire line would be 2.1 16.05 Z j = + , so the impedance per kilometer would be ( ) ( ) 2.1 16.05 / 50 km 0.042 0.321 /km Z j j = + = + The shunt capacitance per meter of this transmission line is given by Equation (9-41). ln c D r πε = § · ¨ ¸ © ¹ (9-41) ( ) 12 11 8.854 10 F/m 1.21 10 F/m 0.15 ln 0.015 c π × = = × § · ¨ ¸ © ¹ Therefore the capacitance per kilometer will be 8 1.21 10 F/km c = × The shunt admittance of this transmission line per kilometer will be ( ) ( ) 8 6 2 2 50 Hz 1.21 10 F/km 3.80 10 S/km sh y j fc j j π π = = × = × Therefore the total shunt admittance will be ( ) ( ) 6 4 3.80 10 S/km 50 km 1.90 10 S sh Y j j = × = × 9-9. The underground cable is operating with the receiving side of the line open-circuited.

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