Long Transmission Line

Long Transmission Line

The transmission line having a length of more than 250km is considered a long transmission line. The line parameters are distributed over the entire length of a transmission line.

Similar to short transmission line and medium transmission line, you cannot consider a lumped model for long transmission lines.

The equivalent circuit of a long transmission line is as shown below figure.

Distributed Model of Long Transmission Line
Distributed Model of Long Transmission Line

The entire length of a line is divided into several parts (n). Each part has a line constant (1/n) times the total line.

While modeling of long transmission line, we assume that the line constants are uniformly distributed over the line. And the resistance (R) and reactance (X) are series of lines.

Capacitance susceptance (B) and leakage conductance (G) are shunt elements of the line. The leakage current passes through the shunt admittance is maximum at the sending end side. And it decreases continuously towards the receiving end. At receiving end, the leakage current is zero.

Rigorous Method (Exact Solution of Long Transmission Line)

The exact solution of a long transmission line is given by the rigorous method. The equivalent circuit diagram of a small part of a transmission line is shown in the below figure.

Rigorous Method of Long Transmission Line
Rigorous Method of Long Transmission Line

Consider the length of this small part of the line is dx. And it is at X distance from the receiving end.

Where;

V = voltage at the end of element towards receiving end
V+dV = voltage at the end of element towards sending end
I = leaving current from the element dx
I+dI = current entering from the element dx
z = series impedance of line per unit length
y = shunt admittance of line per unit length

For small part dx;

Series impedance = zdx
Shunt admittance = ydx

In the direction of increasing x, rise in voltage over the element is;

    \[ dV = I \, zdx \]

(1)   \begin{equation*} \frac{dV}{dx} = Iz \end{equation*}

Similarly, current entering to the element is I+dI and leaving from the element is I.

Current drawn by element is;

    \[ dI = V ydx \]

(2)   \begin{equation*} \frac{dI}{dx} = Vy \end{equation*}

Differentiating equation-1 with respect to x.

    \[ \frac{d^2V}{dx^2} = z \frac{dI}{dx} = z (Vy) \]

    \[ \frac{d^2V}{dx^2} = yzV \]

The solution of this differential equation is;

(3)   \begin{equation*} V = k_1 cosh(x\sqrt{yz}) + K_2 sinh(x\sqrt{yz}) \end{equation*}

Differentiate above equation with respect to x. Therefore;

    \[ \frac{dV}{dx} = k_1 \sqrt{yz} sinh (x\sqrt{yz}) + k_2 \sqrt{yz} cosh(x\sqrt{yz}) \]

    \[ Iz = k_1 \sqrt{yz} sinh (x\sqrt{yz}) + k_2 \sqrt{yz} cosh(x\sqrt{yz}) \]

(4)   \begin{equation*} I = \sqrt{\frac{y}{z}} \left[ k_1 sinh(x\sqrt{yz}) + k_2 cosh(x\sqrt{yz}) \right] \end{equation*}

Equation-3 and 4 gives the value of voltage and current in the form of unknown constants k1 and k2.

By applying end conditions; we can find the values for constants k1 and k2.

At receiving end;

    \[ x=0; \quad V=V_R, \quad I=I_R \]

Now, put these values in equation-3;

    \[ V_R = k_1 cosh(0) + k_2 sinh(0) \]

    \[ V_R = k_1 \]

    \[ I_R = \sqrt{\frac{y}{z}} \left[ k_1 sinh(0) + k_2 cosh(0) \right] \]

    \[ I_R = \sqrt{\frac{y}{z}} k_2 \]

    \[ k_2 = \sqrt{\frac{z}{y}} I_R \]

We have values of both unknowns. So, we can find the equations for voltage and current.

(5)   \begin{equation*} V = V_R cosh(x\sqrt{yz}) + \sqrt{\frac{z}{y}} I_R sinh(x\sqrt{yz}) \end{equation*}

(6)   \begin{equation*} I = \sqrt{\frac{y}{z}}  V_R sinh(x\sqrt{yz}) + I_R cosh(x\sqrt{yz})  \end{equation*}

The entire length of transmission line is L. Therefore, if we consider the entire transmission line, we need to put;

    \[ V= V_S, \quad I=I_S, \quad x=L \]

(7)   \begin{equation*} V_S = V_R cosh(x\sqrt{yz}) + \sqrt{\frac{z}{y}} I_R sinh(x\sqrt{yz}) \end{equation*}

(8)   \begin{equation*} I_S = \sqrt{\frac{y}{z}}  V_R sinh(x\sqrt{yz}) + I_R cosh(x\sqrt{yz})  \end{equation*}

Now,

    \[ Z_C = \sqrt{\frac{z}{y}} \quad and \quad \gamma=\sqrt{yz} \]

Where;

ZC = surge impedance or characteristic impedance
ϒ = propagation constant

Therefore, the equation of long transmission line is;

(9)   \begin{equation*} V_S = V_R cosh(\gamma L) + I_R Z_C sinh(\gamma L) \end{equation*}

(10)   \begin{equation*} I_S = \frac{V_R}{Z_C} sinh(\gamma L) + I_R cosh(\gamma L) \end{equation*}

Compare these equations with equations of ABCD parameters.

    \[ A = cosh \gamma L \]

    \[ B = Z_C sinh \gamma L \]

    \[ C = \frac{1}{Z_C} sinh \gamma L \]

    \[ D = cosh \gamma L \]

Equivalent Network of Long Transmission Line

The long transmission line network can be converted into two types of equivalent network;

  • Equivalent-T network
  • Equivalent-π network

Let’s derive equations for both types of networks;

Equivalent-T Network

Consider the circuit diagram similar to the Nominal-T method of a medium transmission line. Here, Z and Y are replaced by Z’ and Y’ respectively.

Equivalent-T Model of Long Transmission Line
Equivalent-T Model of Long Transmission Line

From the above figure;

    \[ I_S = I_R + I_Y \]

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

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

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

(11)   \begin{equation*} I_S = Y'V_R + \left( 1 + \frac{Y'Z'}{2} \right) I_R \end{equation*}

Apply KVL to the loop;

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

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

    \[ V_S = V_R + I_R \frac{Z'}{2} + \frac{Y'Z'}{2}V_R + \frac{Z'}{2} I_R + \frac{Y'Z'^2}{4} I_R \]

(12)   \begin{equation*} V_S = V_R \left( 1+ \frac{Y'Z'}{2} \right) + Z' \left( 1 + \frac{Y'Z'}{4} \right) I_R \end{equation*}

Compare above equations with the equations of ABCD parameters;

    \[ A = D = 1+ \frac{Y'Z'}{2} \]

    \[ B = Z' \left( 1 +\frac{Y'Z'}{4} \right) \]

    \[ C = Y' \]

These ABCD parameters are the same as the ABCD parameter of the Nominal-T Method.

From the exact solution of long transmission line by Rigorous method;

    \[ V_S = V_R cosh(\gamma L) + I_R Z_C sinh(\gamma L) \]

    \[ I_S = \frac{V_R}{Z_C} sinh(\gamma L) + I_R cosh(\gamma L) \]

Now, compare the equation of equivalent-T method with the equations of rigorous method;

    \[ 1+\frac{Y'Z'}{2} = cosh \gamma L \]

    \[ Z' \left( 1 +\frac{Y'Z'}{4} \right) = Z_C sinh \gamma L \]

    \[ Y' = \frac{1}{Z_C} sinh \gamma L \]

Find Shunt Branch of Equivalent-T Method

    \[ Y' = \frac{1}{Z_C} sinh \gamma L \]

    \[ Y' = \sqrt{\frac{Y}{Z}} sinh \gamma L \]

Multiply and divide by ϒ L;

    \[ Y' = \sqrt{\frac{Y}{Z}} sinh \gamma L \times \frac{\gamma L}{\gamma L} \]

    \[ Y' = \sqrt{\frac{Y}{Z}} sinh \gamma L \times \frac{\sqrt{YZ} L}{\gamma L} \]

    \[ Y' = YL \frac{sinh \gamma L}{\gamma L} \]

    \[ Y' = Y'' \frac{sinh \gamma L}{\gamma L} \]

Where;

    \[ Y'' = YL \]

Find Series Impedance of the Equivalent-T Method

    \[ 1+\frac{Y'Z'}{2} = cosh \gamma L \]

Now, put the value of Y’ from above equation;

    \[ 1+\frac{Z'}{2} \left( \frac{1}{Z_C} sinh \gamma L \right)= cosh \gamma L \]

    \[ \frac{Z'}{2} \left( \frac{1}{Z_C} sinh \gamma L \right) = cosh \gamma L -1 \]

    \[ \frac{Z'}{2} \frac{2 sinh(\frac{\gamma L}{2}) cosh(\frac{\gamma L}{2})}{Z_C} = 2sinh^2(\frac{\gamma L}{2}) \]

    \[ \frac{Z'}{2} = Z_C \frac{sinh(\frac{\gamma L}{2})}{cosh(\frac{\gamma L}{2})} \]

    \[ \frac{Z'}{2} = Z_C tanh(\frac{\gamma L}{2}) \]

    \[ \frac{Z'}{2} = \sqrt{\frac{Z}{Y}} tanh(\frac{\gamma L}{2}) \times \frac{\frac{\gamma L}{2}}{\frac{\gamma L}{2}} \]

    \[ \frac{Z'}{2} = \frac{ZL}{2} tanh (\frac{\gamma L}{2}) \frac{1}{\frac{\gamma L}{2}} \]

    \[ Z' = Z'' tanh (\frac{\gamma L}{2}) \frac{1}{\frac{\gamma L}{2}} \]

The series branch of equivalent T circuit multiply by the above factor.

Equivalent-π Method

The circuit diagram of equivalent-π method is as shown in below figure.

Equivalent-π Model of Long Transmission Line
Equivalent-π Model of Long Transmission Line

Sending end voltage;

    \[ V_S = V_R + IZ' \]

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

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

(13)   \begin{equation*} V_S = V_R (1 + \frac{Y'Z'}{2}) + I_RZ' \end{equation*}

Sending end current;

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

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

(14)   \begin{equation*} I_S = V_R \left( 1+\frac{Y'Z'}{4} \right) Y' + I_R \left( 1+\frac{Y'Z'}{2} \right) \end{equation*}

Compare above equation with the equation of ABCD parameters;

    \[ A = D = 1 + \frac{Y'Z'}{2} \]

    \[ B = Z' \]

    \[ C = Y' (1+\frac{Y'Z'}{4}) \]

These ABCD parameters are same as the ABCD parameter of Nominal-π Method.

Compare above equations with the equations derived from the rigorous method;

    \[ 1 + \frac{Y'Z'}{2} = cosh \gamma L \]

    \[ Z' = Z_C sinh \gamma L \]

    \[ Y' (1+\frac{Y'Z'}{4}) = \frac{sinh \gamma L}{Z_C} \]

Find Series Impedance of Equivalent-π Method.

    \[ Z' = Z_C sinh \gamma L \]

    \[ Z' = \sqrt{\frac{Z}{Y}} sinh \gamma L \times \frac{\gamma L}{\gamma L} \]

    \[ Z' = Z L \frac{sinh \gamma L}{\gamma L} \]

    \[ Z' = Z \frac{sinh \gamma L}{\gamma L} \]

Find Shunt Admittance of Equivalent-π Method.

    \[ 1 + \frac{Y'Z'}{2} = cosh \gamma L \]

    \[ 1 + \frac{Y'}{2} Z_C sinh \gamma L = cosh \gamma L \]

    \[ \frac{Y'}{2} Z_C sinh \gamma L = cosh \gamma L -1 \]

    \[ \frac{Y'}{2} Z_C sinh \gamma L = 2sinh^2 \frac{\gamma L}{2} \]

    \[ \frac{Y'}{2} Z_C \left( 2 sinh \frac{\gamma L}{2} cosh \frac{\gamma L}{2} \right) = 2sinh^2 \frac{\gamma L}{2} \]

    \[ \frac{Y'}{2} = \frac{1}{Z_C} tanh \frac{\gamma L}{2} \]

    \[ \frac{Y'}{2} = \frac{YL}{2} \frac{tanh \frac{\gamma L}{2}}{\frac{\gamma L}{2}} \]

    \[ Y' = Y'' \frac{tanh \frac{\gamma L}{2}}{\frac{\gamma L}{2}} \]

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