1) Determine the watts. Read directly or multiply amps by the volts (115 VAC or 230 VAC as appropriate).

2) A watt is a measure of the rate at which energy is consumed. (Power is the rate at which energy is consumed so a watt is a unit of electrical power.) To calculate energy consumed in kwh multiply the kilowatts (watts ¸ 1000) by the hours used. The following calculations are based on a cost of 6.5 cents per kwh.

1) For electric resistive heaters one kilowatt hour consumed produces one kilowatt hour of heat. (To convert from BTU's per hour to watts, multiply by 0.293 or from watts to BTU's per hour multiply by 3.41.) To supply 14,000 BTU's per hour requires an average of 4100 watts continuously which costs 26.6 cents per hour or $6.40 per day or $192 per month using electrical heating methods other than heat pumps.

2) Since rate of energy loss depends upon the temperature difference, a 'heating degree-day' has been devised as a measure of the average temperature difference between inside and outside temperatures. Degree-days per winter may be determined by taking the average temperature difference between the inside and outside during the winter for each day (or approximately by using the average of the daily high and low temperature) and summing this for the entire winter. Because individual home owners use different thermostat settings, if degree-days are to be used for comparison of heating demands in various locations and for comparison of one winter season with another, it is necessary to use a fixed inside reference temperature. The reference temperature currently in use is 65 ^OF.  Using this reference temperature the average winter in the Clarksville, Tennessee area produces about 3600 heating degree-days from November through March. For example, the average temperature in January for this area is 30 ^OF.  Thus an average January here will produce [(65 - 30) degrees x 31 days] or 1085 heating degree-day units.  If the inside temperature is actually kept at 70 ^OF, then the actual relative heat loss will be the equivalent of 155 degree-days (5 degrees x 31 days) higher for a typical January.  For the entire winter this would be the equivalent of 750 degree-days (5 degrees x 150 days) higher.  Therefore, for an inside temperature of 70 ^OF the actual equivalent heating degree-days per winter would be about 4350 rather than 3600.

For the example house above the $192 for the month is for 1200 actual heating degree-days [(70 - 30) x 30].  The cost per winter is therefore about 3.6 times this, which is about $700.

Note: As will be seen shortly, heat is actually derived not only from the heating system but also from other sources including electrical appliances.  Therefore the 3600 heating degree-day units may be thought of as roughly representing the heating demand placed upon the heating system for a typical Clarksville winter.  Thus the $700 estimate includes the cost of operating appliances and an electric heating system.  (Heat pump systems will be covered separately later.)

3) It is very important to realize that all the energy used to operate electrical appliances except a hot water heater and clothes dryer ends up as heat in the house. (There are some exceptions to this, since cooking food does change the chemical bonding and evaporating water does require energy as well, but nevertheless much of the energy used by a stove ends up as heat in the house.) This is a consequence of the First Law of Thermodynamics which states that energy can be converted from one form to another but it can not be destroyed. For example, operating a 70 watt TV contributes 70 watts of heat to the house. If one used about 59 of these 70 watt TV's, they would supply the 4100 watts to heat the house mentioned above. Or you could turn on 41 lamps that had 100 watt light bulbs. Or you could use some combination of lights and TV's to supply all or part of the 4100 watts. The consequence of this for the example house is that the total energy produced by an electric heater plus the energy 'consumed' by appliances must add up to 4100 watts. Or looking at this another way, appliances cost nothing to operate during cold weather! Reading by a small light with the rest of the house dark during the winter simply means that the electric furnace comes on more often than had all the lights and TV's been on in the house. (In the summer, however, you pay dearly for having all the lights and TV's on in an air conditioned house.)

Example: If the appliance operating cost is $50 per month, then the electric furnace cost for our example house will be $192 - $50 = $142. If by using appliances frugally, the appliance operating cost is reduced to $25, then the electric furnace cost will be $192 - $25 = $167. The total energy cost is the same --- $192. (It is easy to become discouraged about conserving energy if you do not go about it correctly.)

1) Heat pumps use electrical energy to transport thermal energy (heat) from outside the house to the inside. It is more efficient to use the electrical energy to transport thermal energy than to change the electric energy directly into heat as is done with resistive electric furnaces. Heat pump operation may be listed as BTU input per hour rather than watts. The BTU output from the heat pump depends upon the temperature difference between the inside and outside temperatures. The greater the temperature difference the more difficult it is to transport heat. The ratio of the BTU output over the BTU input is called the coefficient of performance, COP. Heat pump COP's are listed at 47 ^OF and 17 ^OF. For example, a heat pump COP at 47 ^OF might be 2.8 but only 1.9 at 17 ^OF. A COP of 2.8 means that for every BTU put in, 2.8 BTU's are produced.

2) To derive all the heating needs from a heat pump during the coldest winter night might require the use of a heat pump four times as large as one that would be adequate at 30 ^OF. To reduce heat pump purchase price, units may be designed to use the heat pump down to 30 ^OF and then use auxiliary resistive electric heaters to help out the heat pump at lower temperatures. These auxiliary heaters like all resistive heaters produce one BTU of heat for each BTU of energy input and are therefore more expensive to operate. (Note: Auxiliary heaters may also come on if the thermostat setting is suddenly increased significantly. The auxiliary heaters are used in order to help get the temperature of the house up to the new setting rapidly which may please the occupants until they get their heating bill.)

3) If the example house used a heat pump with an average COP of 2, the heating costs for operating the heat pump would be half that of operating a resistive heating electric furnace. Appliance operation will reduce the heat pump operation time but now only half the cost of operating the appliances may be subtracted from the cost of operating the more efficient heat pump. If the example house used $50 for appliance operation, the heat pump operating cost would be reduced $25. The cost to heat the house with a heat pump with no contribution from appliances would be $192 ¸ 2 = $86. $50 worth of energy from operating appliances would reduce the heat pump cost from $86 to $61 (i.e. 86 - 25). The total electric bill would then be heat pump plus appliance operating cost or $61 + $50 = $111. If a heat pump is used, then using appliances conservatively will reduce electrical energy cost.

Note: For houses heated with gas much of the above discussion applies. Electrical appliance operation will reduce the amount of gas used to operate the gas furnace. Cost of gas and electricity must be compared to see how much, if any, money is saved by conservative use of electrical appliances.

A. Formula: BTU loss per hour = (area in square feet ¸ R value) ´ temp difference in ^OF

B. Calculations are for a house which is 24' x 62.5' (1500 sq ft) with 8' ceilings having wallboard ceiling and interior walls, polystyrene sheathing (R3), brick exterior, carpeted floors, and 8% of the wall area window or door glass. All glass is double pane or single pane with storm windows. Inside temperature 70 ^OF and outside temperature including attic 30 ^OF and basement or crawl space 50 ^OF. R is the inherent resistance to heat flow. R values are additive. For example, 3 inches of fiberglass has an R of 10 and 6 inches an R of 20.

C. Calculated heat loss for the above house with no fiberglass (or equivalent) insulation. (Percentage is the % of BTU lost by that particular part of the house.)

A. Cost per gallon of bath or shower water

The cost per gallon of warm water used depends on a temperature difference. If the cold water input is 60 ^OF and the shower temperature is 105 ^OF, the cost depends on the difference in temperature which in this case is 45 ^OF. (It does not matter how hot the hot water is provided it exceeds 105 ^OF.) One gallon of water weighs 8.34 pounds and requires 8.34 BTU to raise the temperature one degree. In this case, the energy required to heat one gallon of water is 8.34 ´ 45 = 375 BTU which is 0.110 kwh or 0.72 cents per gallon -- less than one cent per gallon.

B. Bathing costs

C. Laundry and dishwashing hot water costs are quite variable but can be calculated in a similar fashion.

D. Other hot water costs include loss of heat from hot water left in the water line after the hot water tap is shut off depending on the location of the water line and the season of the year. The hot water tank also continuously loses heat to its surroundings.