# Estimating Energy Use & Cost

Purpose:
The operating wattage of appliances and the duration of their operation vary greatly.
The following information will help you to determine how much energy a particular appliance will use and how much that use will cost.

What Do I Need To Know?
Energy Cost - Customers typically purchase electric energy after it’s been used. The amount used is registered on the kilowatt hour meter installed with the electric service. The meter registers kilowatt hours. Residential customers of VEA are charged \$.10989 per kilowatt hour (a little less than eleven cents a kWh).

Wattage - The wattage of an electrical appliance and the duration of operation determine its electrical usage. Power supply requirements are typically stated on the appliance nameplate. This nameplate can usually be found where the power supply enters the appliance. The nameplate may be an adhesive label or it may be stamped or molded on the device surface. In any case it will state the appliance’s hourly power rating in volts (V), amperes (A), and/or wattage (w) or kilowatts (kW). In most instances the nameplate will state wattage. However, in those cases where electrical requirements are expressed in volts and amperes, multiply volts times amperes. The product will be wattage. Since electricity is purchased in kWhs, convert wattage to kW. Do so by using the stated kW rating or by dividing stated watts by 1000. The quotient will be equivalent kW.

Time - Electrical appliances are rated on an hourly basis. In other words, the wattage rating stated on the appliance indicates the energy which will be used under design conditions in a one hour period.
Estimated or actual time of use is expressed as hours or fraction of an hour. For example, 1 hour 22 minutes would be expressed as 1 and 22/60 hours or 1.366 hours.

Put It All Together!
The following expresses how these components can be put together to determine use and cost:
kW x t = kWh
kWh x \$/kWh = \$

Where:
kW = kilowatts
t = hourly fraction
kWh = kilowatt hours
\$/kWh = Cost per kWh
An appliance rated at 100 watts per hour, would consume one kWh in a ten hour period. Similarly, an appliance rated at 5000 watts would utilize five kWhs in one hour.Here are some examples using the formulas to estimate usage and cost:

Example A:
A 100 watt light bulb is left on for 10 hours.
100 watts ÷ 1000 = .1 kW
Multiply .1 kW by 10 hours = 1 kWh
Multiply 1 kWh by .10989 cents (cost per kWh) = ~\$.10989 for 10 hours

Example B:
A 1,450 watt microwave oven is used for 30 minutes.
1,450 watts ÷ 1000 = 1.450 kW
1.450 kW x .5 hours = .725 kWhs
.725 kWh x .10989 cents (cost per kWh) = ~\$.08 cents for 30 minutes

Example C:
A 1hp motor runs for 2 hours and 50 minutes (1 hp = 746 watts) 746 watts ÷ 1000 = .746 kW (not including friction loss) 170 minutes ÷ 60 (60 minutes in one hour) = 2.83 hours
.746 kW x 2.83 hours = 2.11 kWhs
2.11 kWh x .10989 cents (cost per kWh) = ~\$.23 cents for almost 3 hrs.
Now, estimate how many hours of a 24-hour period the appliance (equipment) runs. Multiply that by the cost per hour times thirty days (month) - this will be your average cost per month for that appliance.

Example D:
The microwave oven used approximately 45 minutes in a 24-hour period, 30 days a month. Multiply 1.45 kW x .75 hours/day x 30 days x . 10989 cents. Cost = \$3.59 per month.