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(minute by minute) and considers what effect using an alternative stepping algorithm may have had. For our

alternative stepping algorithm, we are again going to make certain assumptions. We know the common return

temperature, and we will assume that will also be the temperature that the chilled water will reach each of

the chillers. We also know the common supply temperature. If we are using both chillers, then we will assume

that both chillers will achieve that common supply temperature. That may not precisely happen in practice,

but we know that the system is configured so that both chillers have the same temperature set point. The set

point is the temperature that each chiller is striving to achieve. We also know at each point in time, the total

flow (t) of chilled water in L/S. The only question is, how is this flow divided between the two chillers? This is

therefore the decision that we will seek to optimize. We have three options:

1. Use only chiller 1 to handle all t L/S of flow (and switch off chiller 2)

2.

Use only chiller 2 to handle all t L/S of flow (and switch off chiller 1)

3. Use both chillers, with a L/S sent to chiller 1 and b L/S sent to chiller 2 (where a+b=t)

Our algorithm will choose the option that produces the lowest estimated KWE. The KWE for options 1 and 2

can be computed using the functions that we created above to estimate kWE for each chiller. For option 3, if

we knew a and b then we could again use the functions created above for each chiller and simply sum the

electricity consumed by each chiller. To find the optimal values of a and b, we will use a loop to consider all

possible values of a (from 0 tot in 1 litre increments) and will use the value that leads to the lowest overall

electricity consumed by both chillers. Create a Python function that implements this algorithm.