COP of VAV in calculating energy while using thermostat

asked 2016-02-15 05:57:44 -0600

I have used a thermostat (Dual Setpoint,18ºC and 25 ºC) for my model considering a HVAC template (Ideal Load Air System). The calculated annual heating and cooling energies is through considering a Variable-air-volume (VAV) terminal unit in Energyplus. I checked the heating and cooling energies through end-uses district cooling and district heating in results, please correct me if I am wrong. Moreover, How can I see (and change) the coefficient of performance (COP) of this VAV?

Thanks a lot in advance!

answered 2016-02-15 11:09:57 -0600

Aaron Boranian

14155 ●83 ●24 http://bigladdersoftware.com/

The Ideal Load Air system template object is similar to a black box that is attached directly to a zone. This black box can provide heating and cooling in order to meet the zone's demand. To answer your questions:

The heating and cooling required are reported as district heating and cooling use, since the "black box" isn't connected to any plant loops
The heating and cooling is provided at PERFECT EFFICIENCY, so you can't change the COP

This object is intended to be used for sizing purposes, so that the modeler can see required capacity and other parameters in order to select "real" equipment to simulate that will meet heating and cooling demands. If you want to specify COP, you'll need to simulate a VAV system using other EnergyPlus objects.

You can find more info about the Ideal Load Air system in EnergyPlus here and here.

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Thanks a lot for your detailed answer Aaron!! just one more doubt: Does the "perfect efficiency" for these reported district cooling and district heating energies mean that the considered COP was 1? and if so, does it make sense that for the COP of 4 (for example), I divide the required energy by 4? thanks again!

Pouya ( 2016-02-17 10:49:32 -0600 )edit

Correct. You can think of the energy use results in district cooling and heating outputs as "energy output required" to meet cooling and heating demand, respectively. If you know the efficiency of your "real" system, then divide the "energy output required" by your system efficiency to get the "energy input required".

Aaron Boranian ( 2016-02-17 11:18:57 -0600 )edit

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