Medium Transmission Lines

Demonstrative Video

Problem-1

\[\begin{aligned} A = D & = 0.936 + j 0.16 = 0.936 \angle{0.98}\\ B & = 33.5 +j138 = 142\angle{76.4^\circ}~\Omega \\ C & = \left(-5.18+j914\right) \times 10^{-6}~\Omega \end{aligned}\]

, 220 kV rated voltage, medium length TL are: The generalized circuit constants of a

Solution-1

Power received by load = 50 MW
\[\begin{aligned} & = \sqrt{3}V_rI_r\cos\phi \\ & = \dfrac{50\times 10^6}{\sqrt{3} \times 220 \times 10^3 \times 0.9}\\ & = 145.79~A \end{aligned}\]
\[I_r = 145.79\angle{25.84^\circ}\]
\(\cos\phi=0.9 \rightarrow \phi = 25.84^\circ\)Current at receiving end

\[\begin{aligned} \mathbf{V_s} & =\mathbf{A} \mathbf{V}_{\mathbf{r}}+\mathbf{B I}_{\mathbf{r}} \\ &=0.936 \angle 0.98^{\circ} \times(220 / \sqrt{3}) \times 10^{3} \\ &+14.2 \angle 76.4^{\circ} \times 145.79 \angle-25.84^{\circ} \\ &=133.246 \angle 7.77^{\circ} \mathrm{kV} \end{aligned}\]

\[\begin{aligned} \text { voltage } &=\sqrt{3} \times 133.246 \angle 7.77^{\circ} \mathrm{kV} \\ &=230.78 \angle 7.77^{\circ} \mathrm{kV} \end{aligned}\]

\[=230.78 \mathrm{kV}\]

Line to line sending end

Problem-2

\[\begin{array}{l} A=D=0.94 \angle 10^{\circ} \\ B=130 \angle 730^{\circ} \\ C=0.001 \angle 900^{\circ} \end{array}\]

\(\%\)\(240 ~\mathrm{kV},\) line are: constant of a The

Solution-2

\[\begin{aligned} &=\frac{\mathbf{V}_{\mathbf{R}(\mathrm{N} . \mathbf{L})}-\mathbf{V}_{\mathrm{R}(\mathrm{F} . \mathrm{L})}}{\mathrm{V}_{\mathrm{R}(\mathrm{F} . \mathrm{L})}} \times 100 \\ \mathrm{V}_{\mathrm{S}} &=\mathrm{AV}_{\mathrm{r}}+\mathrm{BI}_{\mathrm{r}} \end{aligned}\]

\[\begin{array}{l} =\dfrac{255.3-220}{220} \times 100 \\ =16 \% \end{array}\]

\(\therefore \%\)

\[\begin{array}{l} \therefore V_{R(N . L)}=\dfrac{V_{S}}{A}=\dfrac{240 \times 10^{3}}{0.94} =255.3 \mathrm{kV} \end{array}\]

\(\mathrm{I}_{\mathrm{R}}=0\)Voltage regulation

Problem-3

\[\begin{aligned} A &=D=0.9 \angle 0^{\circ} \\ B &=200 \angle 90^{\circ} \Omega \\ C &=0.95 \times 10^{-3} \angle 90^{\circ} \mathrm{S} \end{aligned}\]

The ABCD parameters of a 3-phase overhead transmission line are:

Solution-3

\[\begin{array}{l} \mathrm{A}=\mathrm{D}=0.9 \angle 0^{\circ} \\ \mathrm{B}=200 \angle 90^{\circ}\\ \mathrm{C}=0.95 \times 10^{-3} \angle 90^{\circ} \end{array}\]

Given:

\[\begin{array}{l} {\left[\begin{array}{ll} A^{1} & B^{1} \\ C^{1} & D^{1} \end{array}\right]=\left[\begin{array}{ll} A & B \\ C & D \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ Y & 1 \end{array}\right]} \end{array}\]

Problem-4

\[\begin{array}{l} A=D=0.95 \angle 2^{\circ} \\ B=60 \angle 70^{\circ} \Omega / \mathrm{ph} \\ C=0.6 \times 10^{-4} \angle 90^{\circ} \mathrm{S} / \mathrm{ph} \end{array}\]

\(110 ~\mathrm{kV}\) 3-phase TL as generalized circuit has constants as A

Solution-4

\[\begin{aligned} \mathrm{V}_{\mathrm{rph}}&=\dfrac{110 \mathrm{kV}}{\sqrt{3}} \angle 0^{\circ} \\ \mathrm{V}_{\mathrm{sph}}&=\mathrm{A.V}_{\mathrm{rph}}\\ &=0.95 \times \frac{110}{\sqrt{3}} \angle 2^{\circ} \end{aligned}\]

Problem-5

\[\begin{array}{l} A=D=0.938 \angle 1.2^{\circ} \\ B=131.2 \angle 72.3^{\circ} \Omega / \mathrm{ph} \\ C=0.001 \angle 90^{\circ} \mathrm{S} / \mathrm{ph} \end{array}\]

\(230 \mathrm{kV}\) transmission line are given as following: 3-phase The line constants of a

Solution-5

\[\mathrm{V_s}=\frac{230}{\sqrt{3}}=132.79 \mathrm{kV}\]
Sending-end phase voltage
Receiving end current will be zero, when the load is disconnected \(I_{R}=0\)
\[\mathrm{V}_{\mathrm{s}}=\mathrm{AV}_{\mathrm{R}}+\mathrm{BI}_{\mathrm{R}}=\mathrm{AV}_{\mathrm{R}}+\mathrm{B} \times 0\]
Sending-end phase voltage is given as

\[\mathrm{V}_{\mathrm{R}}=\frac{\mathrm{V}_{\mathrm{S}}}{\mathrm{A}}=\frac{132.79 \angle 0^{\circ}}{0.938 \angle 1.2^{\circ}}=141.5 \angle-12^{\circ} \mathrm{kV}\]
Receiving-end voltage,
\[\begin{aligned} \mathrm{I}_{C}=\mathrm{CV}_{\mathrm{R}} &=0.001 \angle 90^{\circ} \times 141.5 \times 10^{3} \angle-12^{\circ} \\ &=141.5 \angle 88.8^{\circ} \mathrm{A} \end{aligned}\]
Line charging current

Problem-6

A 3-phase, 50-Hz overhead TL 100 km long has the following constants:

\[\begin{aligned} \text{Resistance/km/phase} & =0.1 \Omega \\ \text{Inductive reactance/km/phase} & =0 \cdot 2 \Omega \\ \text{ Capacitive susceptance/km/phase} &=0.04 \times 10^{-4} ~\text{siemen} \end{aligned}\]

Determine (i) the sending end current (ii) sending end voltage (iii) sending end power factor and (iv) transmission efficiency when supplying a balanced load of \(10,000 ~\mathrm{kW}\) at \(66~ \mathrm{kV}, \mathrm{p}, f .0 \cdot 8\) lagging. Use nominal \(T\) method.

Solution-6

\[\begin{aligned} \text {Total resistance/phase}~R & =0.1 \times 100=10 ~\Omega \\ \text { Total reactance/phase } ~ X_{L} & =0.2 \times 100=20~ \Omega \\ \text { Capacitive susceptance }~ Y & =0.04 \times 10^{-4} \times 100=4 \times 10^{-4} \mathrm{S} \\ \text { Receiving end voltage/phase, } ~ V_{R} & =66,000 / \sqrt{3}=38105~ \mathrm{V} \\ \text{Load current}~I_{R}&=\frac{10,000 \times 10^{3}}{\sqrt{3} \times 66 \times 10^{3} \times 0.8}=109 ~\mathrm{A}\\ \cos \phi_{R} &=0.8 \\ \sin \phi_{R}&=0.6 \\ \text{Impedance per phase}~ \vec{Z} &=R+j X_{L}=10+j 20 \end{aligned}\]

Taking receiving end voltage as the reference phasor, we have,

\[\begin{aligned} \text{Receiving end voltage}~\overrightarrow{V_{R}} & =V_{R}+j 0=38,105 \mathrm{V}\\ \text{Load current}~\overrightarrow{I_{R}}&=I_{R}\left(\cos \phi_{R}-j \sin \phi_{R}\right)\\ &=109(0 \cdot 8-j 0 \cdot 6)\\ &=87 \cdot 2-j 65.4 \\ \text{Voltage across}~ C,~ \vec{V}_{1} & =\overrightarrow{V_{R}}+\overrightarrow{I_{R}} \cdot \vec{z} / 2\\ &=38,105+(87 \cdot 2-j 65-4)(5+j 10) \\ & =38,105+436+j 872-j 327+654\\ &=39,195+j 545 \end{aligned}\]

\[\begin{aligned} \text { Charging current, }~\overrightarrow{I_{C}} & =j Y \overrightarrow{V_{1}}=j 4 \times 10^{-4}(39,195+j 545)\\ &=-0 \cdot 218+j 15 \cdot 6\\ \text { Sending end current, }~ \quad \vec{I}_{S} &=\overrightarrow{I_{R}}+\overrightarrow{I_{C}}\\ &=(87 \cdot 2-j 65-4)+(-0 \cdot 218+j 15-6) \\ &=87 \cdot 0-j 49 \cdot 8=100 \angle-29^{\circ} 47^{\prime} \mathrm{A} \\ \therefore \quad \text { Sending end current } &=100 \mathrm{A} \end{aligned}\]

\[\begin{aligned} \text{Sending end voltage,} ~ \vec{V}_{S}&=\vec{V}_{1}+\overrightarrow{I_{S}} \vec{Z} / 2\\ &=(39,195+j 545)+\\ &(87 \cdot 0-j 49.8)(5+j 10) \\ &=39,195+j 545+434 \cdot 9+j 870-j 249+498 \\ &=40128+j 1170\\ &=40145 \angle 1^{\circ} 40^{\prime} \mathrm{V}\\ \text{sending end line voltage}~&=40145 \times \sqrt{3}=69533 \mathrm{V} \end{aligned}\]

\[\begin{aligned} \theta_{1} &=\text { angle between } \vec{V}_{R} \text { and } \vec{V}_{S}=1^{\circ} 40^{\prime} \\ \theta_{2} &=\text { angle between } \vec{V}_{R} \text { and } \vec{I}_{S}=29^{\circ} 47^{\prime} \\ \therefore \quad \phi_{S} &=\text { angle between } \vec{V}_{S} \text { and } \vec{I}_{S} \\ &=\theta_{1}+\theta_{2}=1^{\circ} 40^{\prime}+29^{\circ} 47^{\prime}=31^{\circ} 27^{\prime} \end{aligned}\]

\[\begin{aligned} \text { Sending end power factor, } ~\cos \phi_{\mathrm{S}}&=\cos 31^{\circ} 27^{\prime}=0.853 \mathrm{lag}\\ \text { Sending end power } &=3 V_{S} I_{S} \cos \phi_{S}\\ &=3 \times 40,145 \times 100 \times 0.853 \\ &=10273105 \mathrm{W}=10273 \cdot 105 \mathrm{kW} \\ \text { Power delivered } &=10,000 \mathrm{kW} \end{aligned}\]