A phase symmetric threephase 60Hz 230kV line of 200 miles ha

A phase symmetric three-phase 60-Hz, 230-kV line of 200 miles has the following parameters; z = j 0.8 Ohm/mi y = j5x10^-6 mho/mi = si/mi Full load at the receiving end of the line is 80 MW, 0.8 pf lagging, at the rated voltage of 230-kV. Find the voltage at the sending end by assuming; The short length line approximation, The medium length line approximation, The long length line approximation, Compare your results, note the difference, and state the reason for this difference.

Solution

z = j0.8/ mile

y= j5* 10^-6 s/mi

P= 80 MW

(a) In the short line approximation, the shunt admittance is ignored. The ABCD constants for this line are:

A = 1, B = Z

C = 0, D= 1

Thus, A =1, B= j0.8/ mile

C = 0, D = 1

Assuming that the receiving end voltage is at 0°, the receiving end phase voltage is

The receiving end line voltage is 230 kV, so the rated phase voltage is 230 kV / 3 = 132.8 kV.

Thus receiving end current is

I_{R =} P / 3 V cos

I_{R = 370 A}

The sending end voltage is given by

V_S = AV_R + BI_R

Thus, V_S = 1*132.8 + j0.8 *370

(b) In the medium length line approximation, the shunt admittance divided into two equal pieces at either end of the line. The ABCD constants for this line are:

A = ZY/2 +1 B = Z

C = Y(ZY/4 +1) D= (ZY/2) +1

The receiving end line voltage is 230 kV, so the rated phase voltage is 230 kV / 3 = 132.8 kV.

Thus receiving end current is

I_{R =} P / 3 V cos = 370 A

Thus, V_S = AV_R + BI_R

V_s = (ZY/2 +1 )*132.8 + Z*370

V_s = ((j0.8/ mile * j5* 10^-6 s/mi) /2 + 1 ) *132.8 + j0.8/ mile*370

(c) In the long transmission line, the ABCD constants are based on modified impedances and admittances:

Z^! = Z sinh d /d

Y! = Y tanh (d/2) /(d/2)

and the corresponding ABCD constants are

A = Z^!Y! /2 +1 B = Z!

C =Y!(Z!Y! /4 +1) D = Z!Y!/2 +1

Now we would apply he same procedure as above.