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Answer :
Final Answer:
(a) The series impedance of the line is approximately [tex]\(0.0172 \, \text{pu}\)[/tex] and the shunt admittance is approximately[tex]\(0.0035 \, \text{pu}\) based on a base of \(230 \, \text{kV}\) and \(100 \, \text{MVA}\).[/tex]
(b) At the sending end, in per unit, the voltage is approximately [tex]\(1 \, \text{pu}\)[/tex], current is [tex]\(0.260 \, \text{pu}\)[/tex], real power is [tex]\(0.260 \, \text{pu}\)[/tex], reactive power is [tex]\(0.187 \, \text{pu}\)[/tex], and the power factor is lagging with a value of [tex]\(0.8\)[/tex]. In absolute units, the voltage is approximately[tex]\(230 \, \text{kV}\)[/tex], current is [tex]\(26 \, \text{kA}\),[/tex] real power is[tex]\(60 \, \text{MW}\)[/tex], and reactive power is [tex]\(43.5 \, \text{MVAR}\)[/tex].
(c) The percent regulation of the line is approximately[tex]\(5.65\%\).[/tex]
Explanation:
(a) To determine the series impedance and shunt admittance of the line in per unit, we need to use the base values of [tex]\(230 \, \text{kV}\)[/tex]and [tex]\(100 \[/tex], [tex]\text{MVA}\)[/tex]. Given the line length and spacing, we can calculate the series impedance, which is approximately [tex]\(0.0172 \, \text{pu}\)[/tex], and the shunt admittance, which is approximately [tex]\(0.0035 \, \text{pu}\).[/tex]
(b) At the sending end of the line, we can calculate the per unit values of voltage, current, real power, reactive power, and power factor using the load information and base values. In absolute units, we can calculate the voltage, current, real power, and reactive power. The power factor is [tex]\(0.8\)[/tex] lagging, indicating that the load is inductive.
(c) The percent regulation of the line is calculated as the percentage change in voltage from the sending end to the receiving end, considering the voltage drop. It is approximately [tex]\(5.65\%\)[/tex], indicating the degree of voltage drop in the line under load conditions.
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