1. FACP FAN AIRFLOW CONTROL POINT
Represents the desired fan airflow.
2. CFA COOLING FAN AIRFLOW
The set point (0-100%) that corresponds to the fan airflow desired when the cooling demand (CLG) is 100%.
3. CLG COOLING DEMAND
4. DFA DEADBAND FAN AIRFLOW
The set point (0-100%) that corresponds to the fan airflow desired during deadband mode when the cooling demand is 0% and the heating demand is 0%. (DFA is always set below the minimum of CFA and HFA. This is a satisfied condition in the room.)
5. HFA HEATING FAN AIRFLOW
The set point (0-100%) that corresponds to the fan airflow desired when the heating demand (HTG) is 100%.
6. HTG HEATING DEMAND
During the cooling mode, FACP = DFA + CLG (CFA – DFA)
Cooling demand (CLG) is expressed as a controller calculation from 0-100%.
During the heating mode, FACP = DFA + HTG (HFA – DFA)
Heating demand (HTG) is expressed as a controller calculation from 0-100%.
During the deadband mode, FACP = DFA.
Assume a zone where the cooling fan volume is 1400 cfm. Also assume that this zone has a heating requirement of 3 kW.
CFM = (kW)(3160)/(T2-T1)
Cooling cfm = 1400 cfm
Second Stage Heating cfm = 1740 cfm
First Stage Heating cfm = 1200 cfm
Deadband cfm = 700 cfm
For the size 5 unit with the a/c motor, .46 Watts per cfm @ 1740 cfm. .8 kW,
For the size 5 unit with the d/c motor, .31 Watts per cfm @ 1740 cfm, 0.54 kW,
.22 Watts per cfm @ 1400 cfm, 0.308 kW,
.17 Watts per cfm @ 1200 cfm, .204 kW,
.14 Watts per cfm @ 1050 cfm, .147 kW, and
.08 Watts per cfm @ 700 cfm, 0.056 kW.
Energy savings at 1740 cfm = 33%,
Energy savings at 1400 cfm = 52%
Energy savings at 1200 cfm = 63%
Energy savings at 1050 cfm = 70%, and
Energy savings at 700 cfm = 83%.
Total energy savings will be the weighted sum of these figures along with the varying amount as the fan speed changes with the modulating valve. These values came from comparing the values of the Watts per CFM at the different values shown above with the Watts per CFM as shown for a size 5 series fan powered unit with an a/c motor.
Actual dollar savings can then be calculated by multiplying the run time by the KW by the $/kWh. Obviously, this is going to be a long and tedious job if you intend to look at every zone on a building. There are some shortcuts that can be used. For instance, estimate the % of days that the unit will run at full cooling, second stage heating, first stage heating, deadband, and medial cooling. We made the following assumptions:
5% @ full cooling @ .52 savings plus
65% @ medial cooling @ .70 savings plus
5% @ deadband @ .83 savings plus
20% @ 1st stage heating @ .63 savings plus
5% @ 2nd stage heating @ .33 savings.
Then multiply the total savings by the total energy that would be expended with a constant volume unit using an AC induction motor per hour running time. See the attached spreadsheet for the calculations.
If the heating and cooling seasons vary, then similar percentages could be substituted. Once these values are selected, a spreadsheet could be developed that would allow the values to be picked rather quickly from the charts on the front of THE ECM MOTOR STORY flier (as was done in the example shown above) for each unit as laid out in the schedule. Average heater capacities would also have to be chosen, but this should get you close to what you are trying to predict.
Source: Nailor Product Bulletin