Question

A transmission line has the following parameters.
$A=D=0.96\angle {10}^o$, $B=100\angle {80}^o$

For a load of 50 MW at 0.8 p.f. lag ,110 kV,
sending end voltage is 120 kV. The maximum power that can be transmitted is

• A. 92.27 MW
• B. 121.74 MW
• C. 117.40 MW
• D. 98.43 MW

Answer 1

The correct option is A 92.27 MW

Given, $V_s=120kV$
$V_r=110kV$
$A=0.96$
$\alpha ={10}^o$
$B=100$
$\beta ={80}^o$

Maximum power transmitted is given by,

$P_{max}=\frac{V_s.V_r}{B}-\frac{A{V}_r^2}{B}cos(\beta -\alpha )$

$=\frac{110\times 120}{100}-\frac{0.96\times (110{)}^2}{100}cos({80}^o-{10}^o)$

$P_{max}=92.27$ MW