Connections to the Voltmeter and Ammeter for Measuring the Input to an Electric Heater

The efficiency of any machine is defined as being the ratio of the output to the input expressed as a percentage, and both quantities must be measured in the same units. For example, the output of a motor is 10 hp. when the power taken by the motor from the electric circuit to which it is connected is 9,325 watts. What is the efficiency? Since the output must be expressed in the same units it is necessary to change the 10 horsepower to watts or the 9,325 watts to horsepower. There are 746 watts in each horsepower. The 9,325 watts are equivalent to 9,325 ÷ 746 or 12.5 hp. The efficiency is then equal to 10 ÷ 12.5 or .8; that is, the output is .8 of the input or, when multiplied by 100 to change it to percentage, 80 percent.

By way of an example, consider the efficiency of an electric heater, like the one shown in the illustration, which is immersed in water placed in a suitable vessel. The energy input to this heater in a given time may be easily determined by measuring the current passing through the heater circuit and the difference in pressure between the terminals of the heater. These measurements may be made, in case the heater is operated on a direct circuit, by means of any ammeter and voltmeter of suitable capacity, connected as shown. If the heater is operated on an alternating-current circuit, only alternating-current instruments can be used, as certain types of instruments will not operate when connected to such a circuit. In either case, the product of the ammeter reading in amperes and the voltmeter reading in volts will give the power taken by the heater in watts, assuming the heater winding to be noninductive. If the heater winding is not noninductive, then the current and the electrical pressure will no longer be in phase when the device is operating on an alternating-current circuit, and a wattmeter must be used. Practically all heating elements are wound noninductively so that the power may be measured by means of an ammeter and voltmeter.

The energy taken by a heater in a given time will be equal to the product of the average power and the time. For example, if the heater takes 300 watts for 30 minutes—¹⁄₂ hour—then the energy consumed is equal to 300 times ¹⁄₂ or 150 watt-hours, which is equal to .15 kilowatt-hour.

To determine the output of the heater is a little more difficult, but it may be approximated as follows: Since the object of the device is to convert electrical energy into heat energy the output must be measured in heat units. The unit of heat most commonly employed is the calorie, which is the heat required to raise the temperature of one gram of water one degree centigrade. Hence, if a certain weight of water has its temperature increased a definite number of degrees centigrade by the electric heater, then the total heat imparted to the water in calories will equal the weight of the water in grams multiplied by the change in temperature in degrees centigrade. Of course, the heat generated by the heater exceeds that obtained by the above calculation, due to the fact that some heat is imparted to the vessel containing the water and to the supports for the vessel, but it is only the heat imparted to the water that must be considered, as the other heat is not useful.

When the temperature of the water is raised to the boiling point and a part of the water is evaporated, the foregoing method of calculating the heat imparted to the water no longer holds good, and the following method must be used. Weigh the water before and immediately after the test to determine the amount of evaporation. For each gram of water evaporated there will be required approximately 536 calories, and the heat in calories imparted to the water to raise its temperature to the boiling point will be equal to the difference between 100 and the initial temperature of the water multiplied by the weight of the water at the start. To determine the efficiency, the input to the heating element in electrical units must be changed to heat units which may be done by multiplying the power in watts by the time in seconds and this product in turn by .24, giving the result in calories. The following example may serve as a help in performing such an experiment or test.

Weight of water at the start500.0grams.
Weight of water at the end of test474.5grams.
Temperature of water at the start25 deg. C.
Average current taken by the heater6.5amperes.
Average pressure at the heater terminals110 volts.
Time heater is connected5¹⁄₂minutes.
Change in temperature of the water75 deg. C.
Heat developed in heater: 6.5 × 110 × 5¹⁄₂ × 60 × .24 =56,628 calories.
Heat absorbed by water in coming to boiling point: 500 × 75 =37,500 calories.
Heat used in evaporating 25.5 grams of water: 536 × 25.5 =13,668 calories.
Total heat absorbed by water51,168 calories.
Efficiency of heater: 51,16856,628 × 100 =90.4per cent.

This value of efficiency may be increased by insulating the vessel with a nonconductor of heat and providing a covering for it, thus decreasing the losses to the air and surrounding objects.

The efficiency of an electric stove or electric iron, or, in fact, any electrically heated device, may be determined in a manner similar to the water heater. In the case of a stove, place a vessel filled with water on it and measure the heat imparted to the water in a given time, also the input to the heating element in the same time, from which data the efficiency may be calculated. In the case of an electric iron, dampened cloths may be ironed and the actual water evaporated by the iron, determined by weighing the cloths before and after the ironing, together with the increase in weight of the cloth on the ironing board, the time the iron is in use and the temperature of the cloths. The actual water evaporated is the difference in the weight of the cloths before and after ironing, minus the increase in weight of the cloth on the ironing board, which takes up some of the moisture from the cloths being ironed.