33 kV, 15 MW line: efficiency and regulation

Short line: $Z = 5 + j18\ \Omega$/phase, shunt capacitance neglected.

Current

$$\begin{aligned} V_R &= 33\,000/\sqrt3 = 19\,052.6\angle0^\circ\ \text{V} \\ I &= \frac{15\times10^6}{\sqrt3\times33\,000\times0.9} = 291.59\ \text{A},\quad \phi_R = 25.84^\circ\ \text{lag} \end{aligned}$$

Efficiency

$$\begin{aligned} P_{loss} &= 3I^2R = 3(291.59)^2(5) = 1.275\ \text{MW} \\ \eta &= \frac{15}{15 + 1.275}\times100 = 92.16\% \end{aligned}$$

Voltage regulation

$$\begin{aligned} IR &= 1458.0\angle-25.84^\circ\ \text{V},\quad IX = 5248.6\angle64.16^\circ\ \text{V} \\ V_S &= V_R + I(R + jX) = 19\,052.6 + 5447.4\angle48.63^\circ \\ &= 23\,018.5\angle10.23^\circ\ \text{V} \end{aligned}$$

Line value $= \sqrt3\times23.019 = 39.87$ kV.

$$\%VR = \frac{23\,018.5 - 19\,052.6}{19\,052.6}\times100 = 20.82\%$$

Phasor diagram (lagging load)

                         Vs (23.02 kV, 10.23 deg)
                        /|
                       / | I*X (5.25 kV, leads I by 90)
                      /  |
                     /   |
     O-------------------+--> I*R (1.46 kV, along I)
      \        Vr (19.05 kV, reference)
       \
        \  phi_R = 25.84 deg
         v
          I (291.6 A)

• $V_R$ is the reference.
• $I$ lags $V_R$ by $\phi_R = 25.84^\circ$.
• $IR$ is drawn from the tip of $V_R$ parallel to $I$; $IX$ is drawn from its tip, perpendicular to $I$ (leading).
• $V_S$ joins $O$ to the end of $IX$ and leads $V_R$ by $\delta = 10.23^\circ$; $I$ lags $V_S$ by $36.07^\circ$ (sending pf 0.808 lag).

Answer: Efficiency ≈ 92.16 %, regulation ≈ 20.82 % ($V_S$ = 39.87 kV).