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Cost to cool building?
- Thread starter ctroyp
- Start date
"Typical" office cooling is about a ton of refrigeration for every 360 square feet. Finding tonnage and using the coefficient of performance of the AC system will get you compressor work, which will get you refrigeration cost knowing utility rates...
But that just gives demand cost for refrigeration. Then there's fan power that has pressure as a variable and there's no fixed relation between flow and operating cost. It's a function of duct design also.
We'd have to then add in water pumping with similar variables.
Each process or component would have to be adjusted for efficiency, then proportion all these requirements to load changes; mainly OA enthalpy.
There's maintenance costs, variable occupant loads and functions, etc.
I'd have to agree with Mint.
Best done by historical data - the US DOE has pretty decent book full of energy use data for different building regions/types/sizes.
OR, if there's an experienced building manager that's looked at this for his/her sampling of buildings - hopefully he/she would chime in...
But that just gives demand cost for refrigeration. Then there's fan power that has pressure as a variable and there's no fixed relation between flow and operating cost. It's a function of duct design also.
We'd have to then add in water pumping with similar variables.
Each process or component would have to be adjusted for efficiency, then proportion all these requirements to load changes; mainly OA enthalpy.
There's maintenance costs, variable occupant loads and functions, etc.
I'd have to agree with Mint.
Best done by historical data - the US DOE has pretty decent book full of energy use data for different building regions/types/sizes.
OR, if there's an experienced building manager that's looked at this for his/her sampling of buildings - hopefully he/she would chime in...
Assume or estimate cooling full load hours. You can estimate it by looking at a year record of electric bills & knowing the chiller Kw/ton.
Ton_hrs per year = (Kwh summer - Kwh winter)/(chiller kw/ton)
Full load hour per year = Ton_hrs / Total chiller tons installed
Estimated $ electric/year for cooling, pumps condenser fans but excluding $ for supply & return fan elect kwh = ($/kwh) x Total chiller tons x (Kw/ton chiller rating) x 1.2.
The factor 1.2 is based on about 0.1 kw/ton for condenser fans and 0.1 kw/ton for pumps.
Typical full load hours is 1000 to 2000 per year.
Ton_hrs per year = (Kwh summer - Kwh winter)/(chiller kw/ton)
Full load hour per year = Ton_hrs / Total chiller tons installed
Estimated $ electric/year for cooling, pumps condenser fans but excluding $ for supply & return fan elect kwh = ($/kwh) x Total chiller tons x (Kw/ton chiller rating) x 1.2.
The factor 1.2 is based on about 0.1 kw/ton for condenser fans and 0.1 kw/ton for pumps.
Typical full load hours is 1000 to 2000 per year.
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