Economic Dispatch Problem Explained

This problem asks us to find the optimal generation amounts, $P_1$ and $P_2$, for two generators to meet a total load of 260 MW while minimizing the total cost. The key principle for economic operation, ignoring network losses, is that the incremental cost (the cost to produce one additional MW) must be the same for both generators.

Incremental Cost Functions

We are given the incremental cost functions for the two generators:

• Generator 1: $\frac{{d{F_1}}}{{d{P_1}}} = 40 + 0.2{P_1}$
• Generator 2: $\frac{{d{F_2}}}{{d{P_2}}} = 32 + 0.4{P_2}$

Here, $\frac{{d{F}}}{{d{P}}}$ represents the marginal cost of generating power $P$ for a specific generator.

Condition for Economic Operation

For the most economical operation, the marginal costs of both generators must be equal:

$$ \frac{{d{F_1}}}{{d{P_1}}} = \frac{{d{F_2}}}{{d{P_2}}} $$

Calculating Optimal Generations

We set the two incremental cost functions equal to each other:

$$ 40 + 0.2{P_1} = 32 + 0.4{P_2} $$

Rearranging this equation:

$$ 0.2{P_1} - 0.4{P_2} = 32 - 40 $$

$$ 0.2{P_1} - 0.4{P_2} = -8 $$

To simplify, we can multiply the entire equation by 10:

$$ 2{P_1} - 4{P_2} = -80 $$

Dividing by 2:

$$ P_1 - 2{P_2} = -40 \quad (*) $$

We also know the total load requirement:

$$ P_1 + P_2 = 260 \, \text{MW} $$

From this total load equation, we can express $P_1$ in terms of $P_2$:

$$ P_1 = 260 - P_2 $$

Now, substitute this expression for $P_1$ into the simplified incremental cost equation ($*$):

$$ (260 - P_2) - 2{P_2} = -40 $$

Combine the $P_2$ terms:

$$ 260 - 3{P_2} = -40 $$

Solve for $P_2$:

$$ 3{P_2} = 260 + 40 $$

$$ 3{P_2} = 300 $$

$$ P_2 = \frac{300}{3} $$

$$ P_2 = 100 \, \text{MW} $$

Finally, substitute the value of $P_2$ back into the equation for $P_1$:

$$ P_1 = 260 - P_2 $$

$$ P_1 = 260 - 100 $$

$$ P_1 = 160 \, \text{MW} $$

Verification

Let's check if these values satisfy the conditions:

• Total Load: $P_1 + P_2 = 160 + 100 = 260$ MW (Matches the requirement).
• Equal Incremental Costs:
  • Generator 1: $40 + 0.2(160) = 40 + 32 = 72$
  • Generator 2: $32 + 0.4(100) = 32 + 40 = 72$
  The incremental costs are equal ($72$).

Therefore, the economic generation schedule is $P_1 = 160$ MW and $P_2 = 100$ MW.

Conclusion

The generation amounts that ensure economic operation for the combined load of 260 MW are $P_1 = 160$ MW and $P_2 = 100$ MW.