Incremental fuel costs in Rs/MWh for a plant consisting of two generating units are given by dF1/dP1 = 0.4P1 +400 and dF2/dP2 =  0.48P2 +320. The allocation of loads P1 and P2 between generating units 1 and 2, respectively, for the minimum cost of generation to serve a total load of 900 MW, neglecting losses, is

Incremental fuel costs in Rs/MWh for a plant consisting of two generating units are given by dF1/dP1 = 0.4P1 +400 and dF2/dP2 =  0.48P2 +320. The allocation of loads P1 and P2 between generating units 1 and 2, respectively, for the minimum cost of generation to serve a total load of 900 MW, neglecting losses, is

Right Answer is: 400 MW and 500 MW

SOLUTION

$\frac{{d{F_1}}}{{d{P_1}}} = \frac{{d{F_2}}}{{d{P_2}}} = \frac{{d{F_3}}}{{d{P_3}}} \ldots \ldots \ldots \ldots = \frac{{d{F_n}}}{{d{P_n}}} = \lambda$

Where

$\frac{{dF}}{{dP}}$ is incremental fuel costs in Rs/MWh for a power plant.

Calculation:

The incremental fuel cost is Rs/MWh are:

$\frac{{dF}}{{dP}}$ = 0.4P1 +400

$\frac{{dF}}{{dP}}$ =  0.48P2 +320

Where

The total load shared by both generating units is:

P1 + P2 = 900 ———(1)

Equating both incremental cost equations, we get

0.4P1 +400 = 0.48P2 +320

0.48P2 =0.4P1 + 80

Substituting the value in equation 1

P1 = 400 MW

P2 = 500 MW

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