The fuel cost functions in rupees/hour for two 600 MW thermal power plants are given by

Plant 1 : C₁ = 350 + 6P₁ + 0.004P₁²

Plant 2 : C₂ = 450 + aP₁ + 0.003P₁²

where P₁ and P₂ are power generated by plant 1 and plant 2, respectively, in MW and a is constant. The incremental cost of power ( λ) is 8 rupees per MWh. The two thermal power plants together meet a total power demand of 550 MW. The optimal generation of plant 1 and plant 2 in MW, respectively, are

The correct option is 250, 300.

Let's analyze the given problem step-by-step to find the optimal generation of Plant 1 and Plant 2.

We are given the fuel cost function for Plant 1 in rupees per hour as:

$C_1=350+6P_1+0.004P_1^2$

For economic load dispatch, the incremental fuel cost of Plant 1 (${\mathrm{IC}}_1$) is equal to the system incremental cost of power ($\lambda$). The incremental cost is obtained by taking the derivative of the cost function with respect to the power generated by that plant:

${\mathrm{IC}}_1=\frac{d{C}_1}{d{P}_1}$

Differentiating $C_1$ with respect to $P_1$:

${\mathrm{IC}}_1=6+2·0.004P_1=6+0.008P_1$

We are given that the incremental cost of power ($\lambda$) is 8 rupees per MWh. Setting the incremental cost of Plant 1 to $\lambda$:

$6+0.008P_1=8$

Subtracting 6 from both sides:

$0.008P_1=2$

Solving for $P_1$:

$P_1=\frac{2}{0.008}=250\text{MW}$

The two thermal power plants together meet a total power demand of 550 MW. Therefore, the load balance equation is:

$P_1+P_2=550\text{MW}$

Substituting the value of $P_1$ into the equation:

$250+P_2=550$

Solving for $P_2$:

$P_2=550-250=300\text{MW}$

Thus, the optimal generation of plant 1 and plant 2 are 250 MW and 300 MW, respectively.