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 answer is 250, 300.

Step-by-Step Explanation:

For the economic load dispatch and optimal sharing of load between power plants, the incremental cost of each operating plant must be equal to the system incremental cost (λ), provided no generator limits are violated.

We are given the following parameters:
• Total power demand, P_D = 550 MW
• Incremental cost of power, λ = 8 rupees/MWh
• Fuel cost function of Plant 1: C₁ = 350 + 6P₁ + 0.004P₁²

First, we write down the load dispatch equation that relates the generation of the two plants to the total demand:

$$P_1+P_2=550$$

Next, we determine the incremental fuel cost of Plant 1 (IC₁) by taking the derivative of its cost function C₁ with respect to P₁:

$$I{C}_1=\frac{d{C}_1}{d{P}_1}=\frac{d}{d{P}_1}\left(350+6P_1+0.004P_1^2\right)$$

$$I{C}_1=6+0.008P_1$$

Under optimal operation conditions, the incremental cost of Plant 1 must be equal to the system incremental cost, λ:

$$I{C}_1=\lambda$$

$$6+0.008P_1=8$$

Subtracting 6 from both sides of the equation:

$$0.008P_1=2$$

Solving for P₁:

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

Now, substituting the value of P₁ back into the load dispatch demand equation to solve for P₂:

$$250+P_2=550$$

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

Both generations (250 MW and 300 MW) are within the limits of the 600 MW capacity of each power plant. Thus, the optimal generation values for plant 1 and plant 2 are 250 MW and 300 MW, respectively.