Medium Transmission Line

The transmission line having a length between 80km to 240km and voltage level is between 20kv to 100kV is considered as a medium transmission line.

In case of medium transmission line, we cannot neglect the effect of capacitance like short transmission line.

The capacitance of the line is distributed uniformly over the entire length of a line. But, we can consider capacitor as a lumped parameter. And need not to make a distribution model like long transmission line.

While modeling of medium transmission line, according to the placement of capacitor, it is classified in three types;

  • End Condenser method
  • Nominal-T method
  • Nominal-π method

End Condenser Method

In this method, the capacitor is assumed to be placed at the end of the transmission line. This capacitor is assumed to a lumped throughout the line. But in actual cases, it is distributed over the line.

Due medium length of the line, we can assume the capacitor as a lumped parameter. The graphical diagram of the end condenser method is shown below figure.

End Condenser Method
End Condenser Method

IR = receiving end current
IS = sending end current
R = line resistance
XL = series reactance
VS = sending end voltage
VR = receiving end voltage
IC = capacitor current

The phasor diagram of the End Condenser method is shown in the below figure.

Phasor Diagram of End Condenser Method
Phasor Diagram of End Condenser Method

In the phasor diagram, the receiving end voltage is taken as a reference vector.

Therefore, Receiving end voltage;

    \[ \vec{V_R} = V_R + j 0 \]

Receiving end current;

    \[ \vec{I_R} = I_R (cos \phi_R - j sin \phi_R) \]

Current passing through the capacitor is;

    \[ \vec{I_C} = j \omega C \vec{V_R}  \]

From the circuit diagram of End condenser method, we can see that the sending end current is a phasor summation of capacitor current and receiving end current.

So, sending end current is;

    \[ \vec{I_S} = \vec{I_R} + \vec{I_C} \]

    \[ \vec{I_S} = I_R (cos \phi_R - j sin \phi_R) + j \omega C \vec{V_R}  \]

Sending end voltage;

    \[ \vec{V_S} = \vec{V_R} + \vec{I_S} (R + jX_L) \]

Where,

    \[ \vec{I_S} (R + jX_L) = Voltage \; drop \; per \; phase \]

Limitations

Representation of this method is simple. But the results are not accurate as we have assumed the capacitor as a lumped parameter. This error may increase as increases the length. Therefore, we cannot exactly pretend the value of capacitance in this method.

Nominal-T Method

In this method, the capacitance is connected at the center of the line. And diagram looks like the letter “T”. Therefore, the name of this method is the nominal T method.

Here, the line resistance and reactance are divided into equal parts. And a capacitance is placed between both parts.

The equivalent circuit of the nominal T method is shown below figure.

Nominal-T Method
Nominal-T Method

VC=Capacitor voltage
IC=Capacitor current

The phasor diagram of the nominal T method is shown in the below figure.

Phasor Diagram of Nominal-T Method
Phasor Diagram of Nominal-T Method

Receiving end voltage is;

    \[\vec{V_R}=V_R+j0\]

Receiving end current is;

    \[\vec{I_R}=I_R (cos\phi_R-jsin\phi_R)\]

Receiving end voltage VR is taken as a reference vector (OP). The receiving end current IS lag behind VR by angle фR.

In the transmission line; resistive drop (PQ) and reactive drop (QR) occurs. Here, PQ is in the direction of IR and QR leads IR by 90˚.

The voltage across capacitor VC was noted as OR. And the capacitor current IC leads VC by 90˚.

Sending end current IS is the phasor summation of receiving end current IR and capacitor current IC.

Similarly, for the second part of a line, the resistive drop RS is in the direction of sending end current and reactive drop ST leads sending end current IS by 90˚.

Therefore,

Resistive drop PQ = IR(R/2) and RS = IS(R/2)

Reactive drop QR = IR(XL/2) and ST = IS(XL/2)

Voltage across capacitor is;

    \[ \vec{V_C}=\vec{V_R} + \vec{I_R} \frac{Z}{2} \]

    \[ \vec{V_C}=\vec{V_R} + I_R (cos \phi_R -j sin \phi_R) \left( \frac{R}{2} + j \frac{X_L}{2} \right) \]

Capacitor current is;

    \[ \vec{I_C}=j \omega C\vec{V_1} \]

Sending end current is;

    \[ \vec{I_S}=\vec{I_R}+\vec{I_C} \]

Sending end voltage is;

    \[ \vec{V_S}=\vec{V_C}+\vec{I_S}\frac{Z}{2} \]

Nominal π Method

In this method, capacitance is divided into two parts. One part is placed near the sending end side and the other part is placed near the receiving end side.

The equivalent circuit of the nominal π method is as shown below figure.

Nominal-π Method
Nominal-π Method

IC1 and IC2 = current across capacitor placed near to receiving end and sending end respectively

I1 = Line current

The phasor diagram of this system is as shown below figure.

Phasor Diagram of Nominal-π Method
Phasor Diagram of Nominal-π Method

In the phasor diagram, the receiving end voltage VR is a reference vector. The receiving end current IR lags VR by фR.

The capacitor current IC1 leads VR by 90˚. Line current I1 is a phasor summation of IR and IC1.

Resistive drop due to line current is PQ. And it is in phase with the line current. Reactive drop due to line current is QR. And it leads line current by 90˚.

Vector OR represents the sending end voltage VS. The capacitor current IC2 leads VS by 90˚. Hence, the sending end current is a phasor summation of IC2 and I1.

Sending end power factor derived from the angle фS.

Receiving end voltage is;

    \[ \vec{V_R} = V_R + j0 \]

Receiving end current is;

    \[ \vec{I_R} = I_R (cos \phi_R - j sin \phi_R) \]

Capacitor current at receiving end;

    \[ \vec{I_{C1}} = j \omega \frac{C}{2} \vec{V_R} \]

Line current is;

    \[ \vec{I_L} = \vec{I_R} + \vec{I_{c1}} \]

Sending end voltage is;

    \[ \vec{V_S} = \vec{V_R} + \vec{I_L} Z \]

Capacitor current at sending end is;

    \[ \vec{I_{C2}} = j \omega \frac{C}{2} \vec{V_S} \]

Sending end current is;

    \[ \vec{I_S} = \vec{I_L} + \vec{I_{C2}} \]

ABCD Parameter for Medium Transmission Line

To find the ABCD parameters for medium transmission line, we need to find the equations of the transmission line equivalent to the equation of ABCD parameters.

And we know the equations of ABCD parameter is;

(1)   \begin{equation*} V_S = AV_R + BI_R \end{equation*}

(2)   \begin{equation*} I_S = CV_R + DI_R \end{equation*}

ABCD Parameters for End Condenser Method

From the circuit diagram of end condenser method,

    \[ I_S = I_C + I_R \]

    \[ I_C = YV_R \]

(3)   \begin{equation*} I_S = YV_R + I_R \end{equation*}

Compare above equation with eq-2;

    \[ C = Y \quad and \quad D = 1 \]

Now, from KVL,

    \[ V_S = V_R + Z I_S \]

    \[ V_S = V_R + Z (YV_R + I_R) \]

(4)   \begin{equation*} V_S = V_R (1 + ZY) + Z I_R \end{equation*}

Compare above equation with eq-1;

    \[ A = 1+ZY \quad and \quad B = Z \]

ABCD Parameters for Nominal-T Method

From the circuit diagram of nominal-T method;

Source Voltage VS

    \[ V_S = V_1 + I_S \frac{Z}{2} \]

Voltage across capacitor VC is;

    \[ V_C = V_R + I_R \frac{Z}{2} \]

Sending end current IS is;

    \[ I_S = I_R + I_C \]

    \[ I_S = I_R + YV_C \]

    \[ I_S = I_R + Y \left( V_R + I_R \frac{Z}{2} \right) \]

    \[ I_S = I_R + YV_R + I_R \frac{ZY}{2} \]

(5)   \begin{equation*} I_S = YV_R + I_R \left( 1 + \frac{ZY}{2} \right) \end{equation*}

Compare this equation with eq-2;

    \[ C = Y \quad and \quad D = 1 + \frac{ZY}{2} \]

    \[ V_S = V_R + I_R \frac{Z}{2} + I_S \frac{Z}{2} \]

    \[ V_S = V_R + I_R \frac{Z}{2} + \frac{Z}{2} \left[  YV_R + I_R \left( 1 + \frac{ZY}{2} \right) \right] \]

    \[ V_S = V_R + I_R \frac{Z}{2} + V_R \frac{ZY}{2} + I_R \frac{Z}{2} + I_R \frac{YZ^2}{4} \]

    \[ V_S = \left( 1 + \frac{YZ}{2} \right) V_R + \left( Z + I_R \frac{YZ^2}{4} \right) I_R \]

Compare this equation with eq-1;

    \[ A = 1 + \frac{ZY}{2} \quad and \quad Z + I_R \frac{YZ^2}{4} \]

ABCD Parameters for Nominal-π Method

From the circuit diagram of nominal- π method,

Current passing through the transmission line;

    \[ I_1 = I_R + I_{C1} \]

Sending end current;

    \[ I_S = I_1 + I_{C2} \]

Where;

    \[ I_{C1} = \frac{Y}{2} V_R \quad and \quad I_{C2} = \frac{Y}{2} V_S \]

Sending end voltage;

    \[V_S = V_R + I_1 Z \]

    \[ V_S = V_R + Z \left( I_R + I_{C1}  \right) \]

    \[ V_S = V_R + Z \left( I_R + \frac{Y}{2} V_R \right) \]

    \[ V_S = V_R + I_R Z + V_R \frac{ZY}{2} \]

(6)   \begin{equation*} V_S = V_R \left( 1 + \frac{ZY}{2} \right) + I_R Z \end{equation*}

Compare the above equation with eq-1;

    \[ A = 1 + \frac{ZY}{2} \quad and \quad B = Z \]

    \[ I_S = I_1 + I_{C2} \]

    \[ I_S = I_R + I_{C1} + I_{C2} \]

    \[ I_S = I_R + \frac{Y}{2} V_R + \frac{Y}{2} V_S \]

    \[ I_S = I_R + \frac{Y}{2} V_R + \frac{Y}{2} \left[ V_R \left( 1 + \frac{ZY}{2} \right) + I_R Z \right] \]

    \[ I_S = I_R + \frac{Y}{2} V_R + \frac{Y}{2} \left[ V_R + \frac{ZY}{2} V_R + I_R Z \right] \]

    \[ I_S = I_R + \frac{Y}{2} V_R + \frac{Y}{2} V_R + \frac{ZY^2}{4} V_R + \frac{ZY}{2} I_R \]

(7)   \begin{equation*} I_S = I_R \left[ 1 + \frac{ZY}{2} \right] + V_R \left[ Y + \frac{ZY^2}{4} \right] \end{equation*}

Compare the above equation with ex-2;

    \[ C = Y + \frac{ZY^2}{4} \quad and \quad D = 1 + \frac{ZY}{2} \]

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