The incremental fuel cost two generating units are given by

IC_{1} = 10 + 0.2 PG_{1}

IC_{2} = 16 + 0.2 PG_{2}

Where PG_{1} and PG_{2} are real power generated by the units. Find the economic allocation for a total load of 170 MW.

This question was previously asked in

DMRC AM Electrical Official Paper 2020

Option 3 : PG_{1} = 100 MW, PG_{2} = 70 MW

ST 1: Logical Reasoning

1253

20 Questions
20 Marks
20 Mins

Concept:

The condition for optimum operation and economic dispatch is

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

Where,

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

The above equation shows that the criterion for a most economical division of load between a plants is that all the unit is must operate at the same incremental fuel cost.

Calculation:

Given, Incremental fuel costs

\(\frac{{d{F_1}}}{{d{P_1}}} = 0.2{PG_1} + 10\)

\(\frac{{d{F_2}}}{{d{P_2}}} = 0.2{PG_2} + 16\)

For optimum operation

\(\frac{{d{F_1}}}{{d{P_1}}} = \frac{{d{F_2}}}{{d{P_2}}}\)

⇒ 0.2PG1 + 10 = 0.2PG2 + 16

⇒ 0.2PG1 - 0.2PG2 = 6

Total load shared by the two generating plant is

PG1 + PG2 = 170

On solving above equations, we get

PG1 = 100 MW, PG2 = 70 MW