Question

Show the solution and determine the best answer

A short transmission line supplies power to a load of 30 MVA at 230 kV, 0.8 lagging power factor. Calculate the percent regulation for a line impedance of (12+j36) ohms.
a. 1.78%

b. 1.87%

c. 2.18%

d. 2.52%


The shunt capacitive reactance of a 40km 34.5 kv line is 90 Kohms per km. What is the total capacitive reactance of the line,

a. 3.5 M ohms

b. 2.25 Kohms

c. 6.75 Kohms

d. 3.6 Mohms

Answer 1

1) %regulation for short transmission line =

Power factor ,

Z=R +jX = 12+j36 ohms

Vr=230KV

By substituting. All value in the given formula. We get

Therefore percentage regulation is 1.76% which is near to option a so a is correct option .

2) given shunt capacitive reactance is 90k ohms per km

For 40km the reactance is 90k*40= 3600kohms =3.6 Mohms so option d is correct option .