Question

Consider the economic dispatch problem for a power plant having two generating units. The fuel costs in Rs/MWh along with the generation limits for the two units are given below:

\(\begin{array}{l} {C_1}\left( {{P_1}} \right) = 0.01P_1^2 + 30{P_1} + 10;100MW \le {P_1} \le 150MW\\ {C_2}\left( {{P_2}} \right) = 0.05P_2^2 + 10{P_2} + 10;100MW \le {P_2} \le 180MW \end{array}\)

The incremental cost (in Rs/MWh) of the power plant when it supplies 200 MW is ___________

Answer 1

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 the most economical division of load between plants is that all the unit is must operate at the same incremental fuel cost.

Calculation:

\(\begin{array}{l} \frac{{d{C_1}}}{{d{P_1}}} = 2 \times 0.01{P_1} + 30 = 0.02{P_1} + 30\\ \frac{{d{C_2}}}{{d{P_2}}} = 2 \times 0.05{P_2} + 10 = 0.10{P_2} + 10\\ \frac{{d{C_1}}}{{d{P_1}}} = \frac{{d{C_2}}}{{d{P_2}}}\\ ⇒ {P_2} = {\rm{ }}200,{\rm{ }}{P_1} = 0\end{array}\)

But P₂  upper limit is 180 MW.

⇒ P₂ = 180 MW, P₁ = 200 - 180 = 20 MW

Now, P₁ lower limit is 100 MW

Therefore,

⇒ P₁ = 100 MW, P₂ = 200 - 100 = 100 MW

Incremental fuel cost of both the power plant 1

\(\frac{{d{C_1}}}{{d{P_1}}}= 0.02{P_1} + 30=32 Rs/MWh\)

Incremental fuel cost of both the power plant 2
\(\frac{{d{C_2}}}{{d{P_2}}}= 0.10{P_2} + 10=20 Rs/MWhr\)

Power plant 2 has more control over power generation as the increase in the demand can be supported only by power plant 2

Therefore, the incremental cost (in Rs/MWh) of the power plant when it supplies 200 MW is 20 Rs/MWh