Rigorous Method The figure below shows a phase and neutral connection of a three-phase line with uniformly distributed impedance and shunt admittance. Before looking at the analysis of long transmission lines in a rigorous manner, first look at the line constant effects of long transmission lines, otherwise, you may have difficulty understanding the subject. Think,
The minor component of the full length of the line is dx and its distance from the receiving end is x.

Besides,

Z = series impedance per unit length of line

Y = shunt admittance per unit length of line

V = Voltage at the receiving end of the component

V + dV = voltage from the end of the element to the sending end

I + dI = dx is the incoming voltage of the element

I = dx current leaving the component.

Here for the minor component dx

zdx = series impedance

Obviously;

dV = Izdx

or, \frac{dV}{dx}=\ Iz ………… (i)

Now, the current is entering the element at rate I+dI and the current is leaving the element at the rate I. This current difference is caused by the supply shunt admittance.

dI = shunt admittance caused by element current.

Or, dI = Vy dx

Or, \frac{dI}{dx}=Vy ……….. (ii)

Differentiating equation (ii) with respect to x we get,

\frac{d^2V}{dx^2}=z\ \frac{dI}{dx}

=z\ \left(Vy\right)\ \ \ \ \ \ \left[\frac{dI}{dx}=Vy\right]

or, \frac{d^2V}{dx^2}=\ yzV ………… (iii)

(iii) Solving the equation, we get,

V=K_1​\cosh(x\sqrt{yz}​)+K_2​\sinh(x\sqrt{yz}​) ………… (iv)

(iv) By differentiating the equation with respect to, we get,

\frac{dx}{dx}​=K_1​\sqrt{yz}​\sinh(x\sqrt{yz})+K_2​\sqrt{yz}​\cosh(x\sqrt{yz}​)

But, \frac{dV}{dx}=IZ

So, I_z=K_1\sqrt{yz}\sinh\ \left(x\sqrt{yz}\right)+K_2​\sqrt{yz}​\cosh(x\sqrt{yz}​)

or, I=\sqrt{\frac{y}{z}}​​[K_1​\sinh(x\sqrt{yz}​)+K_2\sqrt{yz}​​\cosh(x\sqrt{yz}​)] ………….. (v)

Now, if at the receiving end, X = 0, Y = Vr and I = Ir, then substituting (iv) no. and (v) no., we get,

V_R​=K_1​          That is, K_1​ = V_R​

I_R​=\sqrt{yz}​​K_2​          That is, K_2​=\sqrt{\frac{y}{z}}I_R​

Now, substituting the values of K_1 and K_2 in equations (iv) and (v), we get,

V=V_R​\cosh(x\sqrt{yz}​)+\sqrt{yz}​I_R​​\sinh(x\sqrt{yz}​) ……….. (vi) And

I=\sqrt{zy}​​V_R​\sinh(x\sqrt{yz}​)+I_R​\cosh(x\sqrt{yz}​) ……….. (vii)

Now, the distance from receiving end to transmitting end is l, so putting x=l in equation (vi) no and (vii) no gives voltage V_S and current I_S at transmitting end.

So, V_S=\ V_R\cosh\left(l\sqrt{yz}\right)+\sqrt{\frac{z}{y}}I_R\ \sinh\ \left(l\sqrt{yz}\right)

And, I_S=\ \sqrt{\frac{y}{z}}V_R\ \sinh\left(l\sqrt{yz}\right)+\ I_R\ \cosh\ \left(l\sqrt{yz}\right)

Now, \left(l\sqrt{yz}\right)=\sqrt{ly.\ lz}=\sqrt{YZ}

And, \sqrt{\frac{y}{z}}=\sqrt{\frac{yl}{zl}}=\sqrt{\frac{Y}{Z}}

When, Y = total shunt admittance of the line

Z = total series impedance of the line

Thus V_S and I_S of a long transmission line in the Rigorous method will be,

V_S​=V_R​\cosh\sqrt{YZ}​+I_R​\sqrt{\frac{Y}{Z}}​​\sinh\sqrt{YZ}​

I_S​=V_R​\sqrt{\frac{Y}{Z}}\sinh\sqrt{YZ}​+I_R​\cosh\sqrt{YZ}​

To facilitate hyperbolic sine and cosine analysis in power series

\cosh\sqrt{YZ}​=(1+\frac{YZ}{2}​+\frac{Z^2Y^2}{24}+\dots\dots\dots)

\sinh\sqrt{YZ}​=(\sqrt{YZ}+\frac{\left(YZ\right)^{\frac{3}{2}}}{6}​\dots\dots\dots)

A rigorous method analysis of long transmission lines is shown, as well as the equivalent circuit of a phase and neutral connection of a three-phase line with uniformly distributed impedance and shunt admittance.