Question

The speed of a 200 MW, 50 Hz alternator decreases by 4% from no load to full load. Its rated capacity is 200 MW. The speed regulation of alternator in Hz/MW will be

• A. 0.43 Hz/MW
• B. 0.65 Hz/MW
• C. 0.01 Hz/MW
• D. 0.25 Hz/MW

Answer 1

The correct option is C 0.01 Hz/MW

$P_{rated}=200MW$

$\Delta f=\left(\frac{1}{B_{p.u.}+\frac{1}{R}}\right)(\Delta P_C-\Delta P_D)$

Here, $\frac{1}{R}$ is in $\frac{P.u.-MW}{Hz};$

So R is in $\frac{Hz}{P.u.-MW}$

$R=\frac{4}{100}\times 50=2Hz/p.u.-MW$

$R\left(\frac{Hz}{MW}\right)=\frac{2}{P_{rated}}=\frac{2}{200}=0.01Hz/MW$