# 1.6: Energy Cost and Battery Life

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As we have seen, knowing the voltage and current demands of a given device allows us to determine its power rating and energy consumption. The next steps are to determine the cost of operating a device and, if it's battery powered, how long the device will last before needing new batteries.

## Computing Energy Cost

Once we know the power drawn by the device and for how long it will be used, the cost to operate a device can be determined given the per unit cost of energy. In spite of the fact that many people refer to their local electricity supplier as “the power company”, we do not buy “power”, per se. Rather, we are billed for energy. Although it would be possible to determine the cost per joule (or more practically, per megajoule), suppliers normally bill based on kilowatt-hours, or kWh, the product of the power and time1. This unit is used because most residential and commercial devices and appliances are rated in terms of power consumption in watts. Multiplying the power consumption in watts by the length of time the device is used in hours yields a watt-hour value. This is then scaled by a factor of one thousand to arrive at kWh. For example, a 1500 watt toaster oven used for 30 minutes (i.e., 0.5 hours) yields 750 watt-hours, or 0.75 kWh. Finally, knowing the per kWh cost, a simple multiplication yields the cost of electricity. Thus, if the utility charges 10 cents per kWh, then the cost to run that toaster oven is 7.5 cents. If it is used for a full hour then it costs 15 cents, and so on.

To put usage in perspective, a typical home in the USA or Canada uses about 900 kWh per month while households in many countries in Europe might use one half to one quarter of that amount. Global electricity generation is around 25 million gigawatt hours per year. Typical electricity rates in the USA are between ten and twenty cents per kWh, depending on geographic region and sector (e.g., residential or commercial).

##### Example 2.6.1

A 100 watt incandescent light bulb is left on for 24 hours. If the cost of electricity is 15 cents per kWh, determine the cost to run the light.

$Cost = P \times t \times rate \nonumber$

$Cost = 100 W \times 24 hours \times 0.15 \/kWh \nonumber$

$Cost = \ 0.36 \nonumber$

At this point, it should be obvious that the more efficient a device is, the less expensive it will be to run. In fact, it is quite possible that in the long run, a device that is more expensive than a similar, though less efficient, device may be considerably less expensive to use over the course of its lifespan. A good example is a comparison between an ordinary incandescent light bub and an LED light. LED lighting can be an order of magnitude more efficient than incandescent lighting. Indeed, incandescent bulbs may convert less than five percent of their input into usable light output, the remaining 95 percent just turning into heat. While they are less expensive to purchase initially, their operating cost is much higher. A proper way to compare lights is to examine their light output in lumens, not their power consumption. For example, a 60 watt incandescent light produces about 800 lumens of illumination. That same level of illumination can be obtained with an LED drawing just 9 watts. Further, typically LED lights last ten to twenty times longer than incandescent bulbs.

This will be illustrated in the next example.

##### Example 2.6.2

A certain 14 watt LED light produces the same illumination as a 75 watt incandescent light bulb. Assume the LED costs $11 and has an expected life of 15,000 hours. The incandescent costs 50 cents each and has an expected life of 1000 hours. If the cost of electricity is 12 cents per kWh, determine the cost to run each version for 15,000 hours. First, it should be noted that 15 incandescent bulbs will be needed. At 50 cents each, that's$7.50 for the bulbs. The cost to run them is,

$Cost = P \times t \times rate \nonumber$

$Cost = 75 W \times 15000 hours \times 0.12 \/kWh \nonumber$

$Cost = \ 135.00 \nonumber$

The total is $142.50. Now for the single LED required: $Cost = P \times t \times rate \nonumber$ $Cost = 14 W \times 15000 hours \times 0.12 \/kWh \nonumber$ $Cost = \ 25.20 \nonumber$ The total is$36.20, a considerable savings, not to mention other positive factors including only having to change the light once instead of fifteen times; a considerable reduction in wasted energy, thus lowering demand and environmental impact; and finally, a reduction in burned out light bulbs for a further lowering of environmental impact (waste stream reduction).

## Batteries

A battery is a device used to store electrical energy, generally in the form of a chemical cell. Ideally, it presents a constant voltage, its current varying according to what it drives. In reality, as the battery is used, its voltage will begin to decrease. Eventually, the energy stored in the battery will be exhausted and its voltage will drop to zero. The storage capacity of a battery is measured in amp-hours, Ah (or milliamp-hours, mAh, for smaller batteries). All other factors being equal, the battery with the higher amp-hour rating will last longer before being depleted. This is true for the same size of battery using different compositions (e.g., zinc-carbon versus alkaline), as well as batteries of different physical sizes but having the same voltage. For example, all AAA, AA, C and D cell batteries have a nominal voltage of 1.5 volts. If they are all of the same type, such as alkaline, then the immediate practical difference is that the larger the physical size, the greater its energy storage, and thus the longer it will last.

Over a modest range of currents, the expected lifespan of a battery can be computed based on its amp-hour rating and the current drawn from it.

$\text{Battery life } \approx \text{ amp-hour rating / current draw} \label{2.9}$

This equation is best used as a rough guide. If the current draw is considerably higher than the current at which the battery was tested, the predicted lifespan will be overly optimistic. On the other hand, if the current draw is considerably lower, it is likely that the battery will last longer than predicted.

The maximum current output of a battery is likewise limited. If such were not the case, we might expect even very small batteries to produce phenomenally large currents for very shorts periods of time. This is not the case.

A graph of discharge curves at various load currents is shown in Figure 2.6.1 . Notice how the battery voltage begins at the rated 1.5 volts and then begins to fall. After a certain point, the rate of decrease accelerates and “drops off of a cliff”. The actual service life will also be application dependent in that some devices can tolerate a lower voltage than others. For example, an old-fashioned flashlight will still work with largely depleted batteries, it simply won't be that bright. If we were to consider 75% of rated voltage as the useful lower limit (a little over 1.1 volts), we see that at a 50 mA draw, the battery will last around 18 hours, achieving about 900 mAh. At 100 mA, it will last around 7 hours, yielding 700 mAh. If we used 1.0 volts as our lower usable limit, we would arrive at 1.05 Ah and 910 mAh, respectively.

Figure 2.6.1 : Battery discharge curves for various load currents at room temperature. Courtesy of Duracell

##### Example 2.6.3

A certain battery is rated at 10 Ah. Approximately how long will it last with a 0.5 amp draw?

$Lifespan \approx \frac{Ah}{I} \nonumber$

$Lifespan \approx \frac{10 Ah}{500mA} \nonumber$

$Lifespan \approx 20 hours \nonumber$

Remember, 20 hours is only an approximation. For a more accurate rendering, consider the following example.

##### Example 2.6.4

Using the graph of Figure 2.6.1 , determine the expected lifespan with a 100 milliamp draw for a lower voltage limit of 1.2 volts. Also determine the effective amp-hour rating at this point.

The 50 mA curve (purple) passes through 1.2 volts at approximately 5 hours. This is the expected lifespan. The corresponding amp-hour rating is

$Ah \approx I \times t \nonumber$

$Ah \approx 50mA \times 5 hours \nonumber$

$Ah \approx 250mAh \nonumber$

Batteries also tend to show decreased performance with reductions in temperature. This is shown in Figure 2.6.2 . For this particular battery, the capacity at freezing (0$$^{\circ}$$C) is roughly half of what it is at room temperature (21$$^{\circ}$$C). Peak current capacity may also be reduced with temperature. This is true of many kinds of batteries, including large 12 volt automotive batteries. This is a major reason why cars are often much more difficult to start on very cold days; their current capacity is reduced, thereby reducing the output of the starter motor.

Figure 2.6.2 : Battery discharge curves for different temperatures. Courtesy of Duracell

A table of typical amp hour ratings for common battery sizes is shown in Table 2.6.1 . These values are appropriate for good quality alkaline batteries. Rechargeable NiMH (nickel-metal hydride) would be around the same. Remember, these are just approximations.

Battery Size Capacity (mAh)
AAA 1000
AA 2500
C 5000
D 10000
9 Volt 500
Table 2.6.1 : Typical battery capacities.

## References

1Remember, power is the rate of energy usage per unit time, and thus multiplying power by time yields energy. One kWh is approximately equal to 3.6 megajoules.

This page titled 1.6: Energy Cost and Battery Life is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by James M. Fiore.