Two generators have cost functions F₁ and F₂ . Their incremental-cost characteristics are

$\frac{d{F}_1}{d{P}_1}=40+0.2P_1$ and $\frac{d{F}_2}{d{P}_2}=32+0.4P_2$

They need to deliver a combined load of 260 MW. Ignoring the network losses, for economic operation, the generations P₁ and P₂ (in MW) are

The correct option is P₁ = 160, P₂ = 100.

Step-by-Step Explanation:

For the economic dispatch of multiple generating units when transmission losses are neglected, the fundamental condition for the most economical operation is that all operating generators must run at the same incremental fuel cost. That is:

$\frac{d{F}_1}{d{P}_1}=\frac{d{F}_2}{d{P}_2}$

We are given the incremental cost characteristics for the two generators as follows:
For Generator 1:
$\frac{d{F}_1}{d{P}_1}=40+0.2P_1$

For Generator 2:
$\frac{d{F}_2}{d{P}_2}=32+0.4P_2$

Equating the two incremental costs gives:
$40+0.2P_1=32+0.4P_2$

Rearranging the equation yields:
$0.2P_1-0.4P_2=-8$

Dividing the entire equation by 0.2 simplifies it to:
$P_1-2P_2=-40$ --- (Equation 1)

We are also given that the total combined load to be delivered is 260 MW:
$P_1+P_2=260$ --- (Equation 2)

Now, we can solve Equation 1 and Equation 2 simultaneously. Subtracting Equation 1 from Equation 2:
$(P_1+P_2)-(P_1-2P_2)=260-(-40)$

$3P_2=300$

$P_2=100\text{MW}$

Substituting the value of P₂ back into Equation 2:
$P_1+100=260$

$P_1=160\text{MW}$

Thus, for economic operation, the optimal generations are P₁ = 160 MW and P₂ = 100 MW.