A 440 kV, 50 Hz, 500 km long symmetrical line is operated at the rated voltage. (a) What is the theoretical maximum power carried by the line? What is the midpoint voltage corresponding to this condition? (b) A series capacitor is connected at the midpoint of the line to double the power transmitted. What is its reactance? (c) A shunt capacitor of value 450 ohms is connected at the midpoint of the line. If the midpoint voltage is 0.97, compute the power flow in the line corresponding to this operating point. Data: L = I mH/km, c = 11.1 x IOA(-9) F/km.
Answers
Answer: Zn = √l/c = √10^6/11.1 = 300ohms
β = ω√lc = 2π · 50 √(1 × 11.1 × 10−12) x 180/π degrees = 0.06◦/km
θ = βd = 0.06 × 500 = 30◦
Pmax = Pn/sin θ = 440 x 440/300 sin(30◦) = 1290.7MW
Pmax occurs at δ = δmax = 90◦. The midpoint voltage corresponding to this
condition is:
Vm = (V cos δ/2)/ cos θ/2 = 440cos45◦/cos 15◦ = 322.1kV
(b) Since the power flow with a series capacitor is given by
P = Pn sin δ/ sin θ(1 − kse)
If power is to be doubled, kse = 0.5. The expression for kse (when the series capacitor is connected at the midpoint) is given by
kse = (Xc/2Zn) (cot θ/2)
Substituting the values for kse, θ and Zn, we get
Xc = 2kseZn tanθ/2 = 300 × tan 15◦ = 80.38ohms
(c) The midpoint voltage, with a shunt capacitor connected there, is
given by
Vm = (V cos δ/2)/ (cos θ/2 (1 − ksh)), ksh = (BcZn/2)( tanθ/2)
Hence,
cos δ/2 = 0.97 × cos 15◦ ( 1 - (300tan15◦ )/ (450 x 2)
= 0.823 ⇒ δ = 69.31◦
The power flow in the line is given by:
P = (V Vm sin δ/2)/ Zn sinθ/2 =( (1 × 0.97 × sin 34.65◦)/ sin 15◦) x Pn
= 6.90Pn = (6.90 x 450 x 450)/300 = 4657.5 MW
Explanation: