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8 km line receiving 15000 kVA at 33 kV

Short answer

Sending end voltage = 35.17 kV, sending end power factor = 0.828 lagging, voltage regulation = 6.58%, efficiency = 96.38%.

15000 kVA is received at 33 kV at 0.85 power factor lagging over an 8 km three-phase overhead transmission line. Each line has R = 0.29 ohm/km and X = 0.65 ohm/km. Calculate: (i) voltage at the sending end (ii) power factor at the sending end (iii) voltage regulation of the line (iv) efficiency of the transmission line.

Short line (8 km): shunt capacitance neglected.

Data

R=0.29×8=2.32 Ω,X=0.65×8=5.2 ΩVR=33 000/3=19 052.6 VI=15 000×1033×33 000=262.43 Acos⁡ϕR=0.85, sin⁡ϕR=0.527\begin{aligned} R &= 0.29\times8 = 2.32\ \Omega,\quad X = 0.65\times8 = 5.2\ \Omega \\ V_R &= 33\,000/\sqrt3 = 19\,052.6\ \text{V} \\ I &= \frac{15\,000\times10^3}{\sqrt3\times33\,000} = 262.43\ \text{A} \\ \cos\phi_R &= 0.85,\ \sin\phi_R = 0.527 \end{aligned}

(i) Sending-end voltage

VS=VR+I∠−31.79∘ (2.32+j5.2)=19 052.6+1494.3∠34.17∘=20 288.6+j839.3=20 306.3∠2.37∘ V\begin{aligned} V_S &= V_R + I\angle-31.79^\circ\,(2.32 + j5.2) \\ &= 19\,052.6 + 1494.3\angle34.17^\circ \\ &= 20\,288.6 + j839.3 = 20\,306.3\angle2.37^\circ\ \text{V} \end{aligned}

Line value =3×20 306.3=35.17= \sqrt3\times20\,306.3 = 35.17 kV.

(ii) Sending-end pf

Angle between VSV_S and II =2.37∘+31.79∘=34.16∘= 2.37^\circ + 31.79^\circ = 34.16^\circ.

cos⁡ϕS=cos⁡34.16∘=0.828 lagging\cos\phi_S = \cos34.16^\circ = 0.828\ \text{lagging}

(iii) Voltage regulation

%VR=35.17−3333×100=6.58%\%VR = \frac{35.17 - 33}{33}\times100 = 6.58\%

(iv) Efficiency

PR=15 000×0.85=12 750 kWPloss=3I2R=3(262.43)2(2.32)=479.3 kWη=12 75012 750+479.3×100=96.38%\begin{aligned} P_R &= 15\,000\times0.85 = 12\,750\ \text{kW} \\ P_{loss} &= 3I^2R = 3(262.43)^2(2.32) = 479.3\ \text{kW} \\ \eta &= \frac{12\,750}{12\,750 + 479.3}\times100 = 96.38\% \end{aligned}

Answer: (i) 35.17 kV, (ii) 0.828 lagging, (iii) 6.58 %, (iv) 96.38 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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