Placards with Electrical Data

Placards containing electrical usage data are found on most all appliances. These labels denote the voltage and current ratings used by the device along with other electrical data.  Although they vary in format, they can provide a reasonable estimate of power used in lieu of metering the load.

Electrical data must be expressed in units of power (watts or kilowatts) and energy (watt-hours or kilowatt-hours) for an accurate home energy audit using our Power Panel Profiler.  The best way to determine wattage is to keep a couple of simple electric formulas in mind. The first is the calculation of single phase power:

Power (watts) = Volts x Amps x Power Factor*

Power (kilowatts) = Power (watts) / 1000

*Power factor is the ratio between real power in kilowatts and apparent power in kilovolt-amperes (KVA).  Click here for a detailed explanation of power factor.  For estimating purposes in the Profiler, use 80% for inductive loads (motors) and 100% or unity for all other loads unless your metering equipment provides actual power factor measurements.

Trash Compactor

Let's look at a simple electrical data placard from a trash compactor to see how to determine the amount of power the appliance uses.

The value of 0.80 kilowatts would be entered under the appropriate circuit breaker if using our Power Panel Profiler.  The power factor would be entered as 80 percent or 0.8 since a trash compactor is purely an inductive or motor load.  Assuming the compactor is cycled twice a day for a total run time of one minute in 24 hours, the total energy consumed in kilowatt-hours will be very small.

Pool Pump Example

When working with motors another useful formula to remember is the conversion of horsepower to kilowatts.  Since there are 746 watts in one horsepower the equation can be expressed as follows:

Power (kilowatts) = Horsepower(hp) x 0.746

Looking at the electrical data placard for a pool pump we see that both horsepower (HP) and kilowatts (KW) are listed:

If motors ran at 100 percent of load and lost no energy in the conversion of electrical to mechanical power, 1.10 kilowatts would be the actual load. However, motors typically run at 70 to 80 percent of rated load to increase longevity and are only 90 to 95 percent efficient in converting electrical energy to mechanical energy.

Mathematically, this translates into a decrease in power used when motors run at less than rated load.  Inversely, energy lost in the conversion of electrical to mechanical energy increases the amount of power required to produce a given horsepower.  These adjustments to rated load can be expressed in the following formula:

Actual Load (kilowatts) = Rated load (kilowatts) x  % Loaded
                                       Motor Efficiency (%)  

If you know these values they can be applied to adjust the rated load before entering it into the Profiler.  If you do not know these values, use 85 percent as a rule of thumb for adjusting the rated load of a motor to its actual load.  A good way to verify this is to net meter the device using your home energy monitor system or use a plug-in meter if it is a smaller 120 volt load.

In the pool pump example estimating run time is very straight forward if it is set up to run on a daily timer.  However, since a pool pump is a seasonal load that is significant, you may wish to create a separate summer and winter home energy audit baseline.  Simply set the daily run time to zero in the winter baseline.

HVAC System Placards

When it comes to estimating the electrical load of your HVAC system using placards, things get a bit more complicated.  First, there are a minimum of three different electrical loads; the compressor, the condenser fan and the air handler fan.  If your unit has more than one stage there will be a separate load for each compressor.  If the system is all-electric, auxiliary heating strips will add an additional load during the cold weather.

Let's begin by looking at the placard on the condenser unit that sits outside:

The condenser unit placard contains electrical data about the compressor and the condenser fan.  Locate the R.L.A (Run Load Amps) for the compressor and the F.L.A (Full Load Amps) for the condenser fan.   Total Amps on this placard adds these two values together.   Multiply the voltage (230 V. in this case) by the total amps (19.8) to get total watts of 4,554.  Divide by 1000 to get 4.554 kilowatts for the total load of the outdoor unit.  Apply the 85 percent rule of thumb discussed above to estimate the actual load to be 3.871 kilowatts.

Next, let's look at the placard for the air handling unit.  The air handler is a fan motor that circulates air across the heat exchanger or evaporator and the auxiliary heat strips.  It is sized to provide adequate air flow to distribute this conditioned air throughout the home and return it to the heat exchanger to repeat the process.

This air handler placard shows the fan motor to be rated at 3/4 horsepower.   Using the conversion formula shown in the diagram we can calculate the power usage to be 0.56 kilowatts at full load.  Since this is a variable speed air handler it will not be possible to determine the amount of energy used over time from the placard alone.

The third component in this system is the auxiliary heat strip.  This is an electric heating element, similar to the one in a clothes dryer, that resides in the air handler's air stream.  Typically, activation of this heat strip requires outside temperatures to be below 40 degrees F.  Heat strips are used because a heat pump's heating capacity looses efficiency quickly at colder temperatures.

As you can see from the placard below the heat strip consumes more power than all other elements in the system combined.  

Due to this high level of energy consumption it is important to understand what conditions will turn the heat strip on. Outside air temperature and the amount of temperature change requested by the thermostat are the primary determinants.  The addition of an outdoor thermostat can assist in managing heat strip loads.

Likewise, it is important to understand the benefit heat strips provide.   In weather below 20 degrees F. a heat pump struggles just to maintain the desired temperature.  If the thermostat is bumped up 3 to 4 degrees, it may take several hours for a heat pump to satisfy that request under these cold conditions.

All of that extra compressor run time translates into spent kilowatt hours. Activating the heat strips can typically satisfy a three to four degree thermostat request in ten to fifteen minutes.  This reduces compressor run time significantly saving overall kilowatt-hours.

If you are using our Power Panel Profiler, apply the 85 percent rule of thumb to the motor loads, ie. compressor and fans, and assign a power factor of 80 percent if you do not have an actual measurement.  Enter the heat strip loads at 100 percent of their rated value and use a power factor of unity as this load is purely resistive.

Electrical placard data from the components of an HVAC system can give you a fairly accurate estimate of the amount of power it consumes.  However, these placards will not provide any information about the system's run time.  Given the frequent on-off cycling of the compressor, air handler and heat strips in winter, sub-metering is the most accurate method of measuring how much energy it really takes to heat and cool your home.