Given: 50 Hz, 400 kV, l = 450 l = 450 l = 450 r = 0.033 Ω r = 0.033\ \Omega r = 0.033 Ω L = 1.067 L = 1.067 L = 1.067 C = 0.0109 μ C = 0.0109\ \mu C = 0.0109 μ
Step 1: Line constants
z = 0.033 + j 2 π × 50 × 1.067 × 10 − 3 = 0.033 + j 0.3352 Ω / km y = j 2 π × 50 × 0.0109 × 10 − 6 = j 3.424 × 10 − 6 S/km Z = 14.85 + j 150.84 = 151.6 ∠ 84.38 ∘ Ω , Y = j 1.541 × 10 − 3 S \begin{aligned}
z &= 0.033 + j2\pi\times50\times1.067\times10^{-3} = 0.033 + j0.3352\ \Omega/\text{km} \\
y &= j2\pi\times50\times0.0109\times10^{-6} = j3.424\times10^{-6}\ \text{S/km} \\
Z &= 14.85 + j150.84 = 151.6\angle 84.38^\circ\ \Omega, \quad Y = j1.541\times10^{-3}\ \text{S}
\end{aligned} z y Z = 0.033 + j 2 π × 50 × 1.067 × 1 0 − 3 = 0.033 + j 0.3352 Ω/ km = j 2 π × 50 × 0.0109 × 1 0 − 6 = j 3.424 × 1 0 − 6 S/km = 14.85 + j 150.84 = 151.6∠84.3 8 ∘ Ω , Y = j 1.541 × 1 0 − 3 S Step 2: γ l \gamma l γ l Z c Z_c Z c
γ l = Z Y = 0.4833 ∠ 87.19 ∘ = 0.02370 + j 0.48271 Z c = Z / Y = 313.6 ∠ − 2.81 ∘ Ω A = D = cosh γ l = 0.88599 + j 0.01100 = 0.8861 ∠ 0.71 ∘ B = Z c sinh γ l = 145.8 ∠ 84.60 ∘ Ω C = sinh γ l Z c = 1.482 × 10 − 3 ∠ 90.22 ∘ S \begin{aligned}
\gamma l &= \sqrt{ZY} = 0.4833\angle 87.19^\circ = 0.02370 + j0.48271 \\
Z_c &= \sqrt{Z/Y} = 313.6\angle -2.81^\circ\ \Omega \\
A = D &= \cosh\gamma l = 0.88599 + j0.01100 = 0.8861\angle 0.71^\circ \\
B &= Z_c\sinh\gamma l = 145.8\angle 84.60^\circ\ \Omega \\
C &= \frac{\sinh\gamma l}{Z_c} = 1.482\times10^{-3}\angle 90.22^\circ\ \text{S}
\end{aligned} γ l Z c A = D B C = Z Y = 0.4833∠87.1 9 ∘ = 0.02370 + j 0.48271 = Z / Y = 313.6∠ − 2.8 1 ∘ Ω = cosh γ l = 0.88599 + j 0.01100 = 0.8861∠0.7 1 ∘ = Z c sinh γ l = 145.8∠84.6 0 ∘ Ω = Z c sinh γ l = 1.482 × 1 0 − 3 ∠90.2 2 ∘ S Step 3: Receiving-end values
V R = 400 3 = 230.94 ∠ 0 ∘ kV I R = 420 × 10 6 3 × 400 × 10 3 × 0.95 = 638.1 ∠ − 18.19 ∘ = 606.22 − j 199.25 A \begin{aligned}
V_R &= \frac{400}{\sqrt3} = 230.94\angle 0^\circ\ \text{kV} \\
I_R &= \frac{420\times10^6}{\sqrt3\times400\times10^3\times0.95} = 638.1\angle -18.19^\circ = 606.22 - j199.25\ \text{A}
\end{aligned} V R I R = 3 400 = 230.94∠ 0 ∘ kV = 3 × 400 × 1 0 3 × 0.95 420 × 1 0 6 = 638.1∠ − 18.1 9 ∘ = 606.22 − j 199.25 A Step 4: Sending-end voltage and current
V S = A V R + B I R = ( 204,610 + j 2541 ) + ( 37,233 + j 85,242 ) = 241,840 + j 87,783 = 257.28 ∠ 19.95 ∘ kV/phase V S ( L ) = 3 × 257.28 = 445.63 kV I S = C V R + D I R = ( − 1.33 + j 342.24 ) + ( 539.30 − j 169.87 ) = 537.97 + j 172.37 = 564.9 ∠ 17.77 ∘ A \begin{aligned}
V_S &= AV_R + BI_R = (204{,}610 + j2541) + (37{,}233 + j85{,}242) \\
&= 241{,}840 + j87{,}783 = 257.28\angle 19.95^\circ\ \text{kV/phase} \\
V_{S(L)} &= \sqrt3\times257.28 = 445.63\ \text{kV} \\
I_S &= CV_R + DI_R = (-1.33 + j342.24) + (539.30 - j169.87) \\
&= 537.97 + j172.37 = 564.9\angle 17.77^\circ\ \text{A}
\end{aligned} V S V S ( L ) I S = A V R + B I R = ( 204 , 610 + j 2541 ) + ( 37 , 233 + j 85 , 242 ) = 241 , 840 + j 87 , 783 = 257.28∠19.9 5 ∘ kV/phase = 3 × 257.28 = 445.63 kV = C V R + D I R = ( − 1.33 + j 342.24 ) + ( 539.30 − j 169.87 ) = 537.97 + j 172.37 = 564.9∠17.7 7 ∘ A Step 5: Sending-end power and power factor
Angle between V S V_S V S I S I_S I S = 19.95 ∘ − 17.77 ∘ = 2.18 ∘ = 19.95^\circ - 17.77^\circ = 2.18^\circ = 19.9 5 ∘ − 17.7 7 ∘ = 2.1 8 ∘
p.f. = cos 2.18 ∘ = 0.9993 lagging P S = 3 ∣ V S ∣ ∣ I S ∣ cos ϕ S = 435.71 MW , Q S = 16.61 Mvar \begin{aligned}
\text{p.f.} &= \cos 2.18^\circ = 0.9993\ \text{lagging} \\
P_S &= 3|V_S||I_S|\cos\phi_S = 435.71\ \text{MW}, \quad Q_S = 16.61\ \text{Mvar}
\end{aligned} p.f. P S = cos 2.1 8 ∘ = 0.9993 lagging = 3∣ V S ∣∣ I S ∣ cos ϕ S = 435.71 MW , Q S = 16.61 Mvar Line loss = 435.71 − 420 = 15.71 = 435.71 - 420 = 15.71 = 435.71 − 420 = 15.71 = 96.39 % = 96.39\% = 96.39%
Answer: V S = 445.63 V_S = 445.63 V S = 445.63 ∠ 19.95 ∘ \angle 19.95^\circ ∠19.9 5 ∘ I S = 564.9 ∠ 17.77 ∘ I_S = 564.9\angle 17.77^\circ I S = 564.9∠17.7 7 ∘ P S = 435.71 P_S = 435.71 P S = 435.71 = 0.9993 = 0.9993 = 0.9993