Question

The figure shows a two-generator system supplying a load of P${}_D$ = 40 MW, connected at bus 2.


The fuel cost of generators G${}_1$ and G${}_2$ are:

IC${}_1$(P${}_{G1}$) = 10,000 Rs/MWh and

IC${}_2$(P${}_{G2}$) = 12,500 Rs/MWh

and the loss in the line is
P${}_{loss\left(pu\right)}$ = 0.5P${}_{G1}^2\left(pu\right)$, where the loss coefficient is specified in pu on a 100 MVA base. The most economic power generation schedule in MW is

• A. P${}_{G1}$ = 20, P${}_{G2}$ = 22
• B. P${}_{G1}$ = 22, P${}_{G2}$ = 20
• C. P${}_{G1}$ = 20, P${}_{G2}$ = 20
• D. P${}_{G1}$ = 0, P${}_{G2}$ = 40

Answer 1

The correct option is A P${}_{G1}$ = 20, P${}_{G2}$ = 22

From coordination equation,

$\frac{d{F}_n}{d{P}_{Gn}}.\frac{1}{1-\frac{\partial P_L}{\partial P_{Gn}}}=\lambda$

Given $P_L=0.5P_{G1}^2$

$\frac{\partial P_L}{\partial P_{G1}}=P_{G1}and\frac{\partial P_L}{\partial P_{G2}}=0$

$\Rightarrow \frac{d{F}_1}{d{P}_{G1}}.\frac{1}{1-P_{G1}}=\frac{d{F}_2}{d{P}_{G2}}.\frac{1}{1-0}$

$\Rightarrow 10000\times \frac{1}{1-P_{G1}}=12500$

$1-P_{G1}=\frac{100}{125}=\frac{4}{5}$

$P_{G1}=\frac{1}{5}p.u.$

as the base value is 100 MVA

$P_{G1}=\frac{1}{5}\times 100=20MW$

$\Rightarrow P_L=0.5P_{G1}^2$

$=0.5{\left(\frac{1}{5}\right)}^2=0.02p.u$

$\Rightarrow P_L=0.02\times 100=2MW$

P${}_D=P_{G1}+P_{G2}-P_L$

$40=20+P_{G2}-2$

$\Rightarrow P_{G2}=22MW$