Question

Two units, rated at 100 MW and 150 MW, are enabled for economic load dispatch. When the overall incremental cost is 10,000 Rs./MWh, the units are dispatched to 50 MW and 80 MW respectively. At an overall incremental cost of 10,600 Rs./MWh, the power output of the units are 80 MW and 92 MW, respectively. The total plant MW-output (without overloading any unit) at an overall incremental cost of 11,800 Rs./MWh is ______________ (round off to the nearest integer). }

Hint:

In economic load dispatch, the power output of each unit depends linearly on the incremental cost \(\lambda\). Use the given data points to find these linear relations and apply unit limits to get the final dispatch.

Answer 1

Correct Answer: 216

We are given dispatch data at two incremental cost points:

At \( \lambda_1 = 10{,}000 \):
\( P_1 = 50 \) MW, \( P_2 = 80 \) MW

At \( \lambda_2 = 10{,}600 \):
\( P_1 = 80 \) MW, \( P_2 = 92 \) MW

We assume linear relationships for both units between \( P \) and \( \lambda \).

For Unit 1:
\[ P_1 = a_1 \lambda + b_1 \]
Using the two points:
\[ 50 = a_1 \cdot 10{,}000 + b_1 \quad (1) \]
\[ 80 = a_1 \cdot 10{,}600 + b_1 \quad (2) \]
Subtracting (1) from (2):
\[ 30 = a_1 (600) \Rightarrow a_1 = \frac{30}{600} = 0.05 \]
Substitute into (1):
\[ 50 = 0.05 \cdot 10{,}000 + b_1 \Rightarrow b_1 = 50 - 500 = -450 \]
So,
\[ P_1 = 0.05 \lambda - 450 \]

For Unit 2:
\[ P_2 = a_2 \lambda + b_2 \]
Using the two points:
\[ 80 = a_2 \cdot 10{,}000 + b_2 \quad (3) \]
\[ 92 = a_2 \cdot 10{,}600 + b_2 \quad (4) \]
Subtracting (3) from (4):
\[ 12 = a_2 \cdot 600 \Rightarrow a_2 = \frac{12}{600} = 0.02 \]
Substitute into (3):
\[ 80 = 0.02 \cdot 10{,}000 + b_2 \Rightarrow b_2 = 80 - 200 = -120 \]
So,
\[ P_2 = 0.02 \lambda - 120 \]

At \( \lambda = 11{,}800 \):
\[ P_1 = 0.05 \cdot 11{,}800 - 450 = 590 - 450 = 140~\text{MW} \]
\[ P_2 = 0.02 \cdot 11{,}800 - 120 = 236 - 120 = 116~\text{MW} \]

This exceeds generator limits. Given:
\[ P_1^{\max} = 100~\text{MW}, \quad P_2^{\max} = 150~\text{MW} \]

So we cap the outputs:
\[ P_1 = \min(140, 100) = 100~\text{MW} \]
\[ P_2 = \min(116, 150) = 116~\text{MW} \]

Final Total Output:
\[ P_{total} = 100 + 116 = \boxed{216~\text{MW}} \]