(i) To determine the most economical division of load between the generators, we need to find the point where the total incremental cost of the two generators is minimized. This occurs when the incremental cost of each generator is equal. Therefore, we need to solve the following equation:
0.1P1 + 20 = 0.2P2 + 25
Simplifying this equation, we get:
0.1P1 - 0.2P2 = 5
Next, we use the fact that the total power output is constant at 30 MW:
P1 + P2 = 30
Solving these two equations simultaneously, we get:
P1 = 20 MW
P2 = 10 MW
Therefore, the most economical division of load between the generators is to have Generator 1 supply 20 MW and Generator 2 supply 10 MW.
(ii) To determine the saving in $/day thereby obtained compared to equal load sharing between the machines, we need to calculate the total cost of each scenario.
For equal load sharing between the machines, each generator would supply 15 MW. Therefore, the cost would be:
C = 0.1(15) + 20 + 0.2(15) + 25 = $28/h
For the most economical division of load between the generators, the cost would be:
C = 0.1(20) + 20 + 0.2(10) + 25 = $27/h
Therefore, the saving in $/day obtained by using the most economical division of load is:
$1/h x 24 h/day = $24/day