Question

What is the new steady state frequency, of a 10 MW load increase occurs in an area with frequency 50 Hz, governor regulation R = 0.1 pu., and frequency dependent component of load change D = 0.8 p.u. on 100 MVA base?
1. 49.889 Hz
2. 49.537 Hz
3. 49.074 Hz
4. 48.554 Hz

Answer 1

Correct Answer - Option 2 : 49.537 Hz

Steady-state frequency deviation (∆ f) for an Area Control system:

The steady-state frequency deviation is

∆ f = -∆ P_D / (D +\(\frac{1}{R}\))

∆ P_D = Load changes

∆ P_D is positive for load demand or load increase

∆ P_D is negative for load loss

D = (% Change in Power x Rated power) / (% Change in frequency x Rated frequency) in MW / Hz

R = Speed regulation constant in Hz / MW

∆ f = f_n - f_i 

f_n = New steday-state frequency

f_i = Intial steday-state frequency

Calculation:

∆ P_D = 10 MW (positive because load incresases)

Base MVA = 100 MVA

Frequency f = f_i = 50 Hz

D = 0.8 pu

⇒D = 0.8 x Base MVA / Frequency

⇒D = 0.8 x 100 / 50

∴ D = 1.6 MW / Hz

R = 0.1 pu

1/ R = 1/ 0.1 = 10 pu

⇒1 / R = 10 x Base MVA /Frequency

⇒ 1 / R = 10 x 100 / 50

∴ 1/ R = 20 MW / Hz

Applying all in the below equation 

∆ f  = fn - fi  = -∆ PD / (D +\(\frac{1}{R}\))

⇒ f_n - 50 = - 10 / (1.6 + 20)

⇒ f_n = -0.463 + 50

∴ New steady-state frequency = f_n = 49.537 Hz.