A three-phase. 50 Hz transmission line is built with two Blue jay conductors per
ID: 2081025 • Letter: A
Question
A three-phase. 50 Hz transmission line is built with two Blue jay conductors per bundle, as drawn in Figure 5.95. The distance between conductors in the bundle is 15 in. The separation distance between the center points of adjacent phases is 30 ft. The height of the line is 60 ft. The line has a length of 90 miles and delivers a maximum of power of 1000 MVA with a power factor of 0.85 lagging to the load. The voltage at the load terminals is 345 kV. Determine the total resistance, inductance, and capacitance of the transmission line, Draw the equivalent circuit, Find the required supply voltage and complex power, Plot the supply voltage and the voltage regulation versus the load, and determine the load value that causes a voltage regulation of 10%.Explanation / Answer
Radius of each blue jay sub-conductor , r = 0.1573 inches = 0.003995 m
GMR of each Sub-conductor = r` = 0.7788r = 0.7788*0.003995=0.003111306 m
Phase to Phase Seperation = D= 30 feet = 9.144 m
Spacing between sub-conductors of one phase = d= 15 inches = 0.381m
GMR, DD = Square root of (r`D) = Square root of (0.003111306*9.144) = 0.16867 m
Dab = Dbc = 4th square root of (dab dab`da`bda`b`) = 4th square root of (9.144*9.525*8.763*9.144) =9.1400 m
Dca = 4th square root of (dca dca`dc`adc`a`) = 4th square root of (18.288*17.907*18.669*18.288) =18.2860 m
Dm = cube root of (Dab Dbc Dca) = cube root of (9.1400*9.1400*18.2860) = 11.5169 m
inductance of the bundled conductor line per phase = L =0.2loge (Dm/DD) = 0.2loge (11.5169/1.6867) =0.3842mH/Km
Total inductance of 90 miles per phase = 144.84096 km * 0.3842 mH/km = 55.6478mH
inductive reactance = 2*pi*F*L = 2*3.14*50*55.6478*10^-3 =17.47341 ohms
capacitance of the bundled conductor line per phase 2*pi*e0 /(loge (Dm/DD)) =
= 2*3.14*8.854*10^-9 / (loge (11.5169/1.6867)) = 2.895955*10^-11 F/ Km
Total Capacitance of 90 miles per phase = 144.84096 km * 2.895955*10^-11 F/ Km = 4.1945* 10^-9 Farads
Capacitive reactance = 1/(2*pi*f*c) = 1/ (2*3.14*50*4.1945*10^-9) = 0.7588 Mega Ohms
line curret = 1000*10^6 / 345*10^3) = 2898.5 A
Sending Voltage = Vs = VR+ IXSintheta = 345000 + 2898.5(758782.5266)sin0.8 = 1578.04 MV
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