# Long Transmission Line

Lines having length more than 200 km are called long transmission line. As the line constants of the transmission line are uniformly distributed over the entire length of the line . However ,reasonable accuracy can be obtained in line calculations for short and medium transmission lines by considering these constants as lumped. If such assumptions of lumped constants is applied to long transmission lines there will be the introduction of serious errors in the performances calculations. In order to obtain fair degree of accuracy in the performance calculations of long lines, the line constants are considered as uniformly distributed throughout the length of the line. Rigorous mathematical treatment is required for the solution of such lines. The figure above shows the equivalent circuit of a 3-phase long transmission line on a phase-neutral basis. The whole line length is divided into n sections, each section having line constants (1/n )th of those for the whole line.

• The line constants are uniformly distributed over the entire length of line.
• The resistance and inductive reactance are the series elements.
• The leakage susceptance(B) and leakage conductance(G) are shunt elements. The leakage susceptance is due to the capacitance exists between line and neutral .The leakage conductance is due to energy losses occurring through leakage over the insulators or due to corona effect between conductors.
• The leakage current through shunt admittance is maximum at the sending end of the line and decreases continuously as the receiving end of the circuit is approached at which point its value is zero.

Let us consider an infinitely small length dx of the line at a distance x from the receiving end. where,

r=Resistance per unit of length of line

x=Reactance per unit length of line

b=Susceptance per unit length of line

g=conductance per unit length of line V=voltage per phase at the end of element towards receiving end

V+dV=Voltage per phase at the end of element towards sending end Series impedance of element dx of the line=Z dx

Shunt admittance of element dx of the line= y dx

The rise of voltage over the element length in the direction of increasing x

dV=Iz dx, voltage drop in element dx The difference of current entering the element and leaving the element is,

dI=Vydx ,the current drawn by this element Now, Differentiating equation (i) w.r.t x ,we get, The solution of above differential equation is A1 and A2 are unknown constants.

Now, Differentiating equation (iV) w.r.t x , we get, From equation (i) and (V) we get, The value of V and I can be known from equation (iv) and (vi) in the form of unknown constant A1 and A 2 .Where A1 and A2 can be calculated by applying receiving end conditions,

At receiving end , Now applying the value of V and I in equation (iv) and (vi) we get,  Form equation (vii) and (viii) we have, Now the expression for V and I become, From equation (ix) and (x) we get, The sending end voltage Vs and sending end current Is can be obtained by substituting x=l in equation (xi) and (xii) we get, Z is the total impedance of the line and  Y is the total admittance of the line. It is helpful to expand hyperbolic sine and cosine in terms of their power series. comparing the equation(xiii) and (xiv) with the general voltage and current equations of the line , we get, 