Corrected Electric-Input Ratio in eQuest

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To elaborate on this answer, From your calculations: For the first three rows of data: The left most grey column (1/"corrected EIR") is equal to the right most grey column (COP/PLR). For the last four rows of data: the "Assigned Load" (left most white column) is less than the input value used for "Minimum Part Load Ratio" of the equipment. Therefore your system is cycling and running for a fraction of the one hour timestep at the minimum allowed part load ratio. Thus, you will need to factor out the fractional run time in order to see the equality observed for the first three rows of data.

How do I then find capacity? I would think it would be nominal capacity times PLR, but that doesn't seem right. And how do I find CAPACITY-FT since eQuest doesn't output it? Obviously if I knew how to calculate CAPACITY I would then be able to calculate this from the equation you provided above for power.

And Molly- that is a really useful observation. Thank you!

The $CAPACITY$ variable above is the rated capacity input parameter. The operating capacity is an output variable labeled "Operating capacity at current conditions". It accounts for a chiller having more or less ability to cool depending on evaporator/condenser temperatures. The operating capacity is the $CAPACITY\text{-}FT$ multiplied by the rated capacity; so, you can get $CAPACITY\text{-}FT$ by dividing the "Operating capacity at current conditions" by the rated capacity. You can also evaluate it directly by using the curve coefficients and the evaporator/condenser temperatures.

Thanks! Somehow I missed that CAPACITY is an output variable

I'm running into this again and am having trouble with calculating COP at the times when the chiller is running for a fraction of the hour. To get actual hourly COP for the chiller when the PLR is 0.18 (as in the table above), Would I multiply the fraction of hour equipment operates by the assigned load at that hour and divide by the fraction of hour equipment operates times the fan power for that hour (converted to Btu/hr)? When I do this I don't get the exact COP I would expect, but it's closer.

1) Hmm - I don't follow your explanation above, but I'll try to help. This is how I would think about it: The PLR of the chiller has a minimum of value of 0.18 (this is a user input).For lower assigned loads the chiller will still run at PLR of 0.18, but it will run for a fraction of the hour to meet the assigned load. For Ex:if the assigned load divided by "operating capacity at current conditions" is 0.09 the chiller will operate at PLR of 0.18 for 1/2 hour.

2) So, to calculate the COP of the chiller (when it is operating) you want: "assigned load"/("electrical power"* run fraction). The term ("electrical power"* run fraction) gives the energy consumed. I can't remember if run fraction is an available report variable, but you should be able to calculate it as ("assigned load"/"OperCap at current hour")/ "report variable PLR" You will find that COP does not vary much at lows below the min cycling ratio (since PLR value entered in the perf. curves is fixed at the min val, efficiency will only vary with the two temp variables)

3) Using the formula I listed above, including the hourly run fraction, the "Calculated COP" values for the last four rows of your table above should be: 5.84; 5.84; 6.16; 6.14

(Also note you have inverted the COP formula listed in your "calculated COP" col. heading)

Thanks, Molly! I'll add this as a question as well, and if you can move your answer over to there I think it will be more helpful for others (here)

Anna, I looked at your other question. There aren't enough sig figs in the table values to verify if my above calculation is correct. But, the "fraction of hour equipment operates" reported by eQuest does not agree with what I think it should be - so I'm not sure what eQuest is doing there.