Question

In a turbine generator set $R=0.04p.u.$ based on the generator rating of 100 MVA and 60 Hz. The generator frequency decreases by 0.02 Hz and the system adjusts to steady-state operation. The turbine output power increases by

• A. 1 MW
• B. 1.2 MW
• C. 0.83 MW
• D. 0.63 MW

Answer 1

The correct option is C 0.83 MW

Given R= 0.04 p.u.

$\Delta f_1=0.04\times 60=2.4Hz$

We know that $\frac{\Delta f_1}{\Delta P_1}=\frac{\Delta f_2}{\Delta P_2}$

$\frac{2.4}{100}=\frac{0.02}{\Delta P_2}$

$\Delta P_2=\frac{0.02\times 100}{2.4}=\frac{5}{6}=0.83MW$