Question

Given a single area with two generating units.

Unit	Rating	Speed drop R (per unit on unit base)
1	400	0.04
2	800	0.05

The units share a load of $P_1=200MW,P_2=500MW$. The units are operating in parallel to share a load of 700 MW at 50 Hz frequency. The load is increased by 130 MW with B = 0, The value of new frequency caused by steady state frequency deviation is _______ Hz.

• A. 49.75

Answer 1

The correct option is A 49.75
$\Delta f=\frac{-\Delta P}{\frac{1}{R_1}+\frac{1}{R_2}+B}$

On a common base of 1000 MVA

$R_1=\left(0.04\right)\left(\frac{1000}{400}\right)=0.1$

$R_2=\left(0.05\right)\left(\frac{1000}{800}\right)=0.0625$

$\Delta f=\frac{-130/1000}{\frac{1}{0.1}+\frac{1}{0.0625}}=-5\times {10}^{-3}p.u$

$\Delta f_{\left(actual\right)}=50\times \Delta fp.u=50\times -5\times {10}^{-3}$

$f=f_o+\Delta f_{\left(actual\right)}=50\times -(5\times {10}^{-3})\left(50\right)$

$=49.75Hz$