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 = z l = 14.85 + j 150.84 = 151.6 ∠ 84.38 ∘ Ω , Y = y l = j 1.541 × 10 − 3 S \begin{aligned}
z &= 0.033 + j(2\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 &= zl = 14.85 + j150.84 = 151.6\angle 84.38^\circ\ \Omega, \quad Y = yl = 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 = z l = 14.85 + j 150.84 = 151.6∠84.3 8 ∘ Ω , Y = y l = j 1.541 × 1 0 − 3 S Step 2: ABCD parameters
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 × 0.9 = 673.6 ∠ − 25.84 ∘ = 606.22 − j 293.60 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\times0.9} = 673.6\angle -25.84^\circ = 606.22 - j293.60\ \text{A}
\end{aligned} V R I R = 3 400 = 230.94∠ 0 ∘ kV = 3 × 400 × 1 0 3 × 0.9 420 × 1 0 6 = 673.6∠ − 25.8 4 ∘ = 606.22 − j 293.60 A Step 4: Sending-end voltage and current
V S = A V R + B I R = ( 204,610 + j 2541 ) + ( 50,925 + j 83,948 ) = 255,540 + j 86,489 = 269.78 ∠ 18.70 ∘ kV/phase V S ( L ) = 3 × 269.78 = 467.27 kV I S = C V R + D I R = ( − 1.33 + j 342.24 ) + ( 540.33 − j 253.46 ) = 539.01 + j 88.78 = 546.3 ∠ 9.35 ∘ A \begin{aligned}
V_S &= AV_R + BI_R = (204{,}610 + j2541) + (50{,}925 + j83{,}948) \\
&= 255{,}540 + j86{,}489 = 269.78\angle 18.70^\circ\ \text{kV/phase} \\
V_{S(L)} &= \sqrt3\times269.78 = 467.27\ \text{kV} \\
I_S &= CV_R + DI_R = (-1.33 + j342.24) + (540.33 - j253.46) \\
&= 539.01 + j88.78 = 546.3\angle 9.35^\circ\ \text{A}
\end{aligned} V S V S ( L ) I S = A V R + B I R = ( 204 , 610 + j 2541 ) + ( 50 , 925 + j 83 , 948 ) = 255 , 540 + j 86 , 489 = 269.78∠18.7 0 ∘ kV/phase = 3 × 269.78 = 467.27 kV = C V R + D I R = ( − 1.33 + j 342.24 ) + ( 540.33 − j 253.46 ) = 539.01 + j 88.78 = 546.3∠9.3 5 ∘ A Step 5: Sending-end power
p.f. = cos ( 18.70 ∘ − 9.35 ∘ ) = 0.9867 = \cos(18.70^\circ - 9.35^\circ) = 0.9867 = cos ( 18.7 0 ∘ − 9.3 5 ∘ ) = 0.9867
P S = 3 ∣ V S ∣ ∣ I S ∣ cos ϕ S = 3 × 269.78 × 10 3 × 546.3 × 0.9867 = 436.24 MW P_S = 3|V_S||I_S|\cos\phi_S = 3\times269.78\times10^3\times546.3\times0.9867 = 436.24\ \text{MW} P S = 3∣ V S ∣∣ I S ∣ cos ϕ S = 3 × 269.78 × 1 0 3 × 546.3 × 0.9867 = 436.24 MW Step 6: Voltage regulation
V R ( N L ) = ∣ V S ∣ ∣ A ∣ = 467.27 0.8861 = 527.35 kV (line) % VR = 527.35 − 400 400 × 100 = 31.84 % \begin{aligned}
V_{R(NL)} &= \frac{|V_S|}{|A|} = \frac{467.27}{0.8861} = 527.35\ \text{kV (line)} \\
\%\text{VR} &= \frac{527.35 - 400}{400}\times100 = 31.84\ \%
\end{aligned} V R ( N L ) % VR = ∣ A ∣ ∣ V S ∣ = 0.8861 467.27 = 527.35 kV (line) = 400 527.35 − 400 × 100 = 31.84 % Step 7: Efficiency
η = P R P S × 100 = 420 436.24 × 100 = 96.28 % \eta = \frac{P_R}{P_S}\times100 = \frac{420}{436.24}\times100 = 96.28\ \% η = P S P R × 100 = 436.24 420 × 100 = 96.28 % Answer: A = D = 0.8861 ∠ 0.71 ∘ A = D = 0.8861\angle 0.71^\circ A = D = 0.8861∠0.7 1 ∘ B = 145.8 ∠ 84.60 ∘ Ω B = 145.8\angle 84.60^\circ\ \Omega B = 145.8∠84.6 0 ∘ Ω C = 1.482 × 10 − 3 ∠ 90.22 ∘ C = 1.482\times10^{-3}\angle 90.22^\circ C = 1.482 × 1 0 − 3 ∠90.2 2 ∘ V S = 467.27 V_S = 467.27 V S = 467.27 I S = 546.3 I_S = 546.3 I S = 546.3 P S = 436.24 P_S = 436.24 P S = 436.24 = 31.84 % = 31.84\% = 31.84% = 96.28 % = 96.28\% = 96.28%