Question

A constant load of 300 MW is supplied by two 200 MW generators, for which incremental fuel costs are:
$\frac{d{C}_1}{d{P}_{G1}}=0.10P_{G1}+20.0;\frac{d{C}_2}{d{P}_{G2}}=0.12P_{G2}+15$
[Where $P_G$ is in MW and C in Rs/hr]
If the most economic division of load between the generators is X and Y MW, then saving in Rs/day obtained compared to equal load sharing is Rs, _____ per day.

• A. 218.16

Answer 1

The correct option is A 218.16

The most economic division of load is

$0.10P_{G1}+20=0.12P_{G2}+15$

$0.10P_{G1}-0.12P_{G2}=-5$

$P_{G1}+P_{G2}=300$

$P_{G1}=140.9MW$

$P_{G2}=159.1MW$

Equal load sharing $P_{G1}=P_{G2}=150MW$

Increase in cost for unit-1;
$=\underset{140.9}{\overset{150}{\int }}(0.1P_{G1}+20)d{P}_{G1}$

$=0.1{\frac{{\left(P_{G1}\right)}^2}{2}|}_{140.9}^{150}+20\left(P_{G1}\right){|}_{140.9}^{150}=314.36Rs/hr$

Increase in cost for unit-2

$\underset{159.1}{\overset{150}{\int }}(0.12P_{G2}+15)d{P}_{G2}=-305.27Rs/hr$

Net saving = $(314.36-305.27)\times 24$

$=Rs218.16/day$