Question

Incremental fuel costs (in some appropriate unit) for a power plant consisting of three generating units are

$I{C}_1=20+0.3P_1$,
$I{C}_2=30+0.4P_2$,
$I{C}_3=30$

where P${}_i$ is the power in MW generated by units for i = 1, 2 and 3.

Assume that all the three units are operating all the time. Minimum and maximum loads on each units are 50 MW and 300 MW respecitvely. If the plant is operating on economic load dispatch to supply the total power demand of 700 MW, the power generated by each unit is

• A. P${}_1$ = 242.86 MW; P${}_2$ = 157.14 MW; and P${}_3$ = 300 MW
• B. P${}_1$ = 157.14 MW; P${}_2$ = 242.86 MW; and P${}_3$ = 300 MW
• C. P${}_1$ = 300.0 MW; P${}_2$ = 300.0 MW; and P${}_3$ = 100 MW
• D. P${}_1$ = 233.3 MW; P${}_2$ = 233.3 MW; and P${}_3$ = 233.4 MW

Answer 1

The correct option is A P${}_1$ = 242.86 MW; P${}_2$ = 157.14 MW; and P${}_3$ = 300 MW

When P${}_1$ is minimum i.e. P${}_1$ = 50 MW

$I{C}_1=20+0.3\times 50=35$

$When{P}_2$ is minimum i.e. P${}_2$ = 50 MW

$I{C}_2=30+0.4\times 50=50$

$Forminimumvalueof{P}_{1}and{P}_2$

$I{C}_{1}andI{C}_2>I{C}_3$

Therefore,

P${}_3$ = 300 MW

[maximum load is assigned to unit 3]

Remaining power is to shared by unit 1 and 2.

$So,P_1+P_2=700-300=400MW.....\left(i\right)$

For optimal operation,

$I{C}_1=I{C}_2$

$20+0.3P_1=30+0.4P_2$

$0.3P_1-0.4P_2=10$

Solving equation (i) and (ii), we get

P${}_1$ = 242.86 MW

and P${}_2$ = 157.14 MW