Demonstrative Video
Long Transmission Lines
SECTION 01
Revision of Important Formulas
\[\begin{aligned}
V_{x}&=\left(\dfrac{V_{R}+I_{R} Z_{C}}{2}\right) e^{\gamma
x}+\left(\dfrac{V_{R}-I_{R} Z_{c}}{2}\right) e^{-\gamma x} \\
I_{x}&=\left(\dfrac{V_{R} / Z_{c}+I_{R}}{2}\right) e^{\gamma
x}+\left(\dfrac{V_{R} / Z_{c}-I_{R}}{2}\right) e^{-\gamma x}
\end{aligned}\]
\[\begin{aligned}
V_{x}&=V_{R}\left(\frac{e^{\gamma x}+e^{-\gamma x}}{2}\right)+I_{R}
Z_{c}\left(\frac{e^{\gamma x}-e^{-\gamma x}}{2}\right)\\
&=V_{R} \cosh (\gamma x)+I_{R} Z_{c} \sinh (\gamma x) \\
I_{x}&=V_{R} \frac{1}{Z_{c}}\left(\frac{e^{\gamma x}-e^{-\gamma
x}}{2}\right) e^{\gamma x}+I_{R}\left(\frac{e^{\gamma x}+e^{-\gamma
x}}{2}\right)\\
&=I_{R} \cosh (\gamma x)+V_{R} \frac{1}{Z_{c}} \sinh (\gamma x)
\end{aligned}\]
SECTION 02
when \(x = l, \quad V_{x} = V_{s}, \quad I_{x} = I_{s}\)
\[
\boxed{
\left[\begin{array}{c}
V_S \\
I_S
\end{array}\right]=\left[\begin{array}{cc}
\cosh (\gamma l) & Z_c \sinh (\gamma l) \\
\frac{1}{Z_c} \sinh (\gamma l) & \cosh (\gamma l)
\end{array}\right]\left[\begin{array}{c}
V_R \\
I_R
\end{array}\right]
}
\]
\[\begin{aligned}
V_{x}&=\left|\frac{V_{R}+I_{R} Z_{c}}{2}\right| e^{\alpha x}
e^{j\left(\beta x+\phi_{1}\right)}+\left|\frac{V_{R}-I_{R}
Z_{c}}{2}\right| e^{-\alpha x} e^{-j\left(\beta x-\phi_{2}\right)}
\end{aligned}\]
\[\begin{aligned}
&\boxed{Z_{c}=\sqrt{\left(\dfrac{z}{y}\right)}} \\
&\boxed{\gamma=\sqrt{y z}} \\
\gamma&=\sqrt{y z}=\boxed{\alpha+j \beta}
\end{aligned}\]
\[\begin{aligned}
\boxed{\cosh (\alpha l+j \beta l)} & =\cosh (\alpha l) \cos
(\beta l)+j \sinh (\alpha l) \sin (\beta l) \\
\boxed{\sinh (\alpha l+j \beta l)}& =\sinh (\alpha l) \cos
(\beta l)+j \cosh (\alpha l) \sin (\beta l)
\end{aligned}\]
SECTION 03
Problem
\[R =
0.2~ \Omega, ~ L = 1.3~\text{mH}, ~ \text{and}~ C = 0.01~\mu F\]
The incident voltage to neutral at the receiving end (reference)
The reflected voltage to neutral at the receiving end
The incident and reflected voltage to neutral at 120 km from the receiving end
Also,
Determine the efficiency of the line, if the line is 120 km long and delivers 40 MW at 132 kV and 0.8 pf lagging
Determine the ABCD parameters of the line
Determine the sending end voltage and efficiency using nominal-\(\pi\) and nominal-T methods.
SECTION 04
Solution
- The series impedance per unit length of the line\[\begin{aligned} z &=r+j x=\left(0.2+j 1.3 \times 314 \times 10^{-3}\right)=(0.2+j 0.408) \\ &=0.454 \angle 63.88^{\circ} \end{aligned}\]
- The shunt admittance\[=3.14 \times 10^{-6} \angle+90^{\circ}\]
The characteristic impedance \(Z_{c}=\sqrt{\dfrac{z}{y}}=\sqrt{\dfrac{0.454}{0.314} \times 10^{5} \angle 63.88-90^{\circ}} =380 \angle-13.06^{\circ}\)
- Propagation constant:\[\begin{array}{l} \gamma=\sqrt{y z}=\sqrt{0.314 \times 0.454 \times 10^{-6}} \angle(90+63.88)^{\circ} / 2 \\ =(0.2714+j 1.169) \times 10^{-3} \\ =(\alpha+j \beta) \end{array}\]
- The receiving end line to neutral voltage\[=\frac{132 \times 1000}{\sqrt{3}}=76200 \text { volts }\]
- The receiving end current under open circuited condition\[I_{r}=0\]
- \[=\frac{V_{r}+I_{r} Z_{c}}{2}\]\(I_{r}=0\) Incident voltage to neutral at the receiving end\[\text { Incident voltage }=\frac{V_{r}}{2}=\frac{76200}{2}=38100 \text { volts }\]
- Reflected voltage to neutral at the receiving end\[\frac{V_{r}-I_{r} Z_{c}}{2}=\frac{V_{r}}{2}=38100 \mathrm{volts}\]
- from the receiving end At a distance of\[\begin{aligned} V_{r}^{+} &=V_{r} \exp (\alpha x) \exp (j \beta x) \\ &=76.2 \exp \left(0.2714 \times 120 \times 10^{-9}\right) \exp \left(j 1.169 \times 120 \times 10^{-3}\right) \\ &=78.7 \angle 8.02^{\circ} \\ V_{r}^{-} &=76.2 \exp (-\alpha x) \exp (-j \beta x)\\ &=76.2 \exp (-0.0325) \exp (-j 0.140) \\ &=73.76 \angle-8.02^{\circ} \end{aligned}\]
- from the receiving end Reflected voltage at a distance of\[=\frac{73.76}{2} \angle-8.02^{\circ}=36.88 \angle-8.02^{\circ} \mathrm{kV}\]
- from the receiving end Incident voltage at a distance of\[=\frac{78.7}{2} \angle 8.02^{\circ}=39.35 \angle 8.02^{\circ}\]
- Receiving end current\[\begin{aligned} I_{r} &=\frac{40 \times 1000}{\sqrt{3} \times 132 \times 0.8} \\ &=218.7 \text { amps. } \end{aligned}\]
- Characteristics impedance\[Z_{c}=380 \angle-13.06^{\circ}\]
- length of line, and For\[\begin{aligned} e^{\alpha x} e^{j \beta x} &=1.033 \angle 8.02^{\circ} \\ e^{-\alpha x} e^{-j \beta x} &=0.968 \angle-8.02^{\circ} \end{aligned}\]
\[\begin{aligned}
I_{r} &=218.7 \angle-36.8^{\circ} \\
V_{s}^{+} &=\frac{V_{r}+I_{r} Z_{c}}{2} e^{\alpha x} e^{j \beta x}
\\
&=\frac{76200+380 \times 218.7 \angle-13.06 \angle-36.8}{2} \times
1.033 \angle 8.02^{\circ} \\
&=74.63 \angle-18^{\circ} \\
V_{s}^{-} &=\frac{V_{r}-I_{r} Z_{c}}{2} e^{-\alpha x} e^{-j \beta x}
\\
&=\frac{76200-380 \times 218.7 \angle-49.86}{2} \times 0.968
\angle-8.02^{\circ} \\
&=32.619 \angle 62.37 \mathrm{kV} \\
V_{s} &=V_{s}^{+}+V_{s}^{-}=74.63 \angle-18^{\circ}+32.619 \angle
62.37^{\circ} \\
&=86077+j 5751=86.26 \angle 3.82^{\circ}
\end{aligned}\]
\[\begin{aligned}
I_{s} &=\frac{V_{r} / Z_{c}+I_{r}}{2} e^{\alpha x} e^{j \beta
x}-\frac{V_{r} / Z_{c}-I_{r}}{2} e^{-\alpha x} e^{-j \beta x} \\
&=\frac{V_{s}^{+}}{Z_{c}}-\frac{V_{s}^{-}}{Z_{c}}\\
&=\left(\frac{74.63 \angle-18^{\circ}}{380
\angle-13.06^{\circ}}-\frac{32.619 \angle 62.37^{\circ}}{380
\angle-13.06^{\circ}}\right) \mathrm{kA} \\
&=200.39 \angle-29.9^{\circ}
\end{aligned}\]
SECTION 05
Efficiency
- Power at the sending end\[\begin{aligned} &=3 \times\left|V_{s}\right|\left|I_{s}\right| \cos \phi_{s} \\ &=3 \times 86.26 \times 200.39 \mathrm{cos} 33.72 \\ &=43.132 ~\mathrm{MW} \end{aligned}\]
- Efficiency\[\% \eta =\dfrac{40}{43.132} \times 100=92.7 \%\]
SECTION 06
ABCD parameters
\[\begin{aligned}
& \gamma l=(0.2714+j 1.169) 120 \times 10^{-3}\\
&=0.03254+j 0.1402
\end{aligned}\]
\[\begin{aligned}
A = D &=\cosh \gamma l\\
&=\cosh (0.03254+j 0.1402) \\
&=\cosh 0.03254 \cos 0.1402+j \sinh 0.03254 \sin 0.1402 \\
&=0.99+j 0.004435=0.99 \angle 0.26^{\circ}
\end{aligned}\]
\[\begin{aligned}
\sinh \gamma l &=\sinh \alpha l \cos \beta l+j \cosh \alpha l \sin
\beta l \\
&=\sinh 0.03254 \cos 0.1402+j \cosh 0.03254 \sin 0.1402 \\
&=0.031958+j 0.1386 \\
&=0.1422 \angle 77^{\circ}
\end{aligned}\]
\[\begin{aligned}
B &=Z_{c} \cdot \sinh \gamma l=380 \angle-13.06^{\circ} \times
0.1422 \angle 77^{\circ} \\
&=54.03 \angle 64^{\circ}
\end{aligned}\]
\[\begin{aligned}
C &=\dfrac{1}{Z_{c}} \cdot \sinh \gamma l\\
& =\dfrac{1}{380 \angle-13.06^{\circ}} \times 0.1422 \angle
77^{\circ} \\
&=3.74\times 10^{-4} \angle 90.06^{\circ}
\end{aligned}\]
\[\begin{aligned}
V_{s} &=A V_{r}+B I_{r} \\
&=0.99 \angle 0.26 \times 76200+54.03 \angle 64^{\circ} \times 218.7
\angle-36.8^{\circ} \\
&=75438+11772 \angle 27.2^{\circ} \\
&=85908+j 5380 \\
&=86.07 \angle 3.588^{\circ}
\end{aligned}\]
SECTION 07
Nominal-\(\pi\)
The resistance of the line \(=0.2 \times 120=24\) ohms
The inductive reactance \(=1.3 \times 10^{-3} \times 120 \times 314=48.98 \Omega\)
The capacitance \(\quad=0.01 \times 10^{-6} \times 120=1.2 \mu \mathrm{F}\)
The nominal- \(\pi\) circuit will be
\[\begin{aligned}
I_{r} &=218.7(0.8-j 0.6)=174.96-j 131.22 \\
I_{c_{1}} &=j 314 \times 0.6 \times 10^{-6} \times 76200=j 14.356
\mathrm{amp} \\
I_{l} &=I_{c_{1}}+I_{r}=174.96-j 116.86=210.39 \angle 33.73^{\circ}\\
V_{s} &=76200+(174.96-j 116.86)(24+j 48.38) \\
&=76200+4199+j 8596-j 2804+5723 \\
&=86122+j 5765 \\
&=86314 \angle 3.82^{\circ} \mathrm{volts} \\
\text { The loss } &=3 \times 210.39^{2} \times 24=3.187
\mathrm{MW} \\
\% \eta &=\frac{40 \times 100}{43.187}=92.69 \%
\end{aligned}\]
SECTION 08
Nominal-T
\[\begin{aligned}
V_{c} &=76200(0.8+j 0.6)+218.7(12+j 24.49) \\
&=60960+j 45720+2624+j 5356 \\
&=63584+j 51076 \\
I_{c} &=j 314 \times 1.2 \times 10^{-6}(63584+j 51076) \\
&=j 23.95-19.24 \\
I_{s} &=218.7+j 23.95-19.24=199.46+j 23.95 \\
&=200.89 \angle 6.8^{\circ} \\
V_{s} &=63584+j 51076+(199.46+j 23.95)(12+j 24.49) \\
&=63584+j 51076+2393+j 4884.7+j 287.4-586.5 \\
&=65390+j 56248 \\
&=86.25 \angle 40.70^{\circ} \\
\text { The loss }&=3 \times 12\left(200.89^{2}+218.7^{2}\right)\\
&=3.174~ \mathrm{MW} \\
\% \eta & =\frac{40}{43.174} \times 100\\
&=92.64 \%
\end{aligned}\]
Taking receiving end current
as reference