The most economic division of load between the generators is 140.909 MW and 159.091 MW, and the saving obtained compared to equal load sharing is Rs 5.127 per day.
A constant load of 300 MW is supplied by two 200 MW generators, for which incremental fuel costs are:
dC₁/dPG₁ = 0.10PG₁ + 20.0
dC₂/dPG₂ = 0.12 PG₂ + 15
where PG is in MW and C in Rs/hr.
We can solve this problem by setting up a system of equations and using substitution to solve for x and y.
Step 1: Set up the system of equations
We know that the total load is 300 MW, so we can write the following equation: x + y = 300
We also know that the incremental fuel costs for each generator are equal at the most economic division of load.
So we can set the two incremental fuel cost equations equal to each other: 0.10x + 20.0 = 0.12y + 15
Step 2: Solve the system of equations
We can solve the system of equations by substitution. First, solve the first equation for x: x = 300 - y
Then, substitute this expression for x into the second equation:
0.10(300 - y) + 20.0 = 0.12y + 15
Simplifying this equation, we get:
30 - 0.10y + 20.0 = 0.12y + 15
Combining like terms, we get:
-0.22y = -5
Solving for y, we get:
y = 159.090909
Now that we know y, we can substitute it back into the equation x = 300 - y to solve for x:
x = 300 - 159.090909
Solving for x, we get:
x = 140.909091
Step 3: Calculate the saving
The saving obtained compared to equal load sharing is the difference between the total cost of supplying the load with equal load sharing and the total cost of supplying the load with the most economic division of load.
The total cost of supplying the load with equal load sharing is:
2(0.10 * 150 + 20.0) = 60
The total cost of supplying the load with the most economic division of load is:
0.10 * 140.909091 + 20.0 + 0.12 * 159.090909 + 15 = 54.872727
Therefore, the saving obtained is:
60 - 54.872727 = 5.127273