Air conditioners are rated by the number of British Thermal Units (BTU) of heat they can remove per hour. Another common rating term for air conditioning size is the "ton," which is 12,000 BTU per hour.

Each air conditioner has an energy-efficiency rating that lists how many BTUs per hour are removed or “pulled out” for each watt of power it draws.

- The efficiency rating for room conditioners is the
**Energy Efficiency Ratio**, or EER. - The efficiency rating for central air conditioners, is the
**Seasonal Energy Efficiency Ratio,**or SEER.

These ratings are posted on an Energy Guide Label, which must be conspicuously attached to all new air conditioners. Energy Star-labeled appliances mean that they have high EER and SEER ratings.

### Room Air Conditioners—EER

Energy Efficient Ratio (EER) measures how efficiently a room air conditioner will operate at a specific outdoor temperature. The higher the EER, the more efficient the system.

The EER can be calculated using this equation:

Remember that the EER energy-efficiency rating lists how many BTUs per hour are removed or “pulled out” for each watt of power it draws. Room air conditioners generally range from 5,500 BTU per hour to 14,000 BTU per hour.

National appliance standards require room air conditioners built after January 1, 1990, to have an EER of 8.0 or greater. A room air conditioner with an EER of at least 9.0 is recommended for milder climates, whereas in hotter climates an EER over 10 is preferred.

The Association of Home Appliance Manufacturers reports that the average EER of room air conditioners rose 47 percent from 1972 to 1991. If a 1970s-vintage room air conditioner with an EER of 5 is replaced with a new one with an EER of 10, air conditioning energy costs will be cut by 50 percent.

### Central Air Conditioners—SEER

Seasonal Energy Efficiency Ratio (SEER) measures how efficiently a central air conditioner will operate at a specific outdoor temperature. The higher the SEER, the more efficient the system.

The SEER can be calculated using this equation:

Again, the SEER energy-efficiency rating lists how many BTUs per hour are removed or “pulled out” for each watt of power it draws.

National minimum standards for central air conditioners require a SEER of 9.7 and 10.0, for single-package and split-systems, respectively. But you do not need to settle for the minimum standard—there is a wide selection of units with SEERs reaching nearly 17.

Before 1979, the SEERs of central air conditioners ranged from 4.5 to 8.0. Replacing a 1970s-era central air conditioner with a SEER of 6 with a new unit having a SEER of 12 will cut your air conditioning costs in half. Today's best air conditioners use 30% to 50% less energy to produce the same amount of cooling as air conditioners made in the mid 1970s. Even if your air conditioner is only 10 years old, you may save 20 to 40 percent of your cooling energy costs by replacing it with a newer, more efficient model.

In general, new air conditioners with higher EERs or SEERs have higher price tags. However, the higher initial cost of an energy-efficient model will be recovered several times during its lifespan. Some utility companies encourage the purchase of a more efficient air conditioner by offering incentives. Buy the most efficient air conditioner you can afford, especially if you use (or think you will use) an air conditioner frequently and/or if your electricity rates are high.

#### Example 1

**Calculate the power consumption of 5000 BTUs/h room air conditioner with an Energy Efficiency Ratio (EER) of 8.**

Solution: We know that

$$\text{EER}=\text{}\frac{\frac{\text{BTUs}}{\text{h}}\text{pulled out}}{\text{Watt}}$$

Given that the AC pulls out 5,000 BTUs per hour and its EER = 8, we have

$$\frac{\text{5000}\frac{\text{BTUs}}{\text{h}}}{\text{Wattt}}=\text{8 W}$$

Therefore, its wattage =

$$\frac{\text{5000}\frac{\text{BTUs}}{\text{h}}}{\text{8}}=\text{625 W}$$

#### Example 2

### Air Conditioner Efficiency Example 2

**An old room air conditioner with an EER 6 was replaced by a new air conditioner with an EER of 10.0. The power consumption with the old air conditioner was 1000 W. Calculate the power consumption of the new air conditioner.**

$$\text{EER=6}$$

$$\text{EER=}\frac{\text{BTUs}/\text{hr}}{\text{W}}$$

We have an old air conditioner with an EER of 6. EER is basically Energy Efficiency Ratio which is given by number of Btus the air conditioner is pulling out per hour divided by watts of power consumed.

$$\text{6=}\frac{\text{BTU}/\text{hr}}{\text{1,000}}$$

$$\text{X=6}\times \text{1,000=6,000}\text{BTU}/\text{hr}$$

And in this problem we are given the EER as 6 and we need to calculate the number of Btus it is capable of pulling out. We also know that it is consuming a thousand watts of power. So we need to calculate these Btus per hour that it is pulling out. So we can calculate the x, unknown, by multiplying thousand by 6 and we get six thousand Btus per hour.

$$\text{New=}\frac{\text{6,000}\text{BTU}/\text{hr}}{\text{Watts}}$$

The room size is not changing but we are just replacing the old air conditioner with the new one. The new EER is 10, the new air conditioner EER is 10 and it is still pulling 6000 Btus per hour out and the new one, how many watts of power does it consume?

$$\text{Power=}\frac{\text{6,000}\text{BTU}/\text{hr}}{\text{10}}\text{=600W}$$

To calculate the power, we have 6000 Btu/hour load and we know the EER, which is 10, so dividing by this we get the power which is 600 watts.

What we are doing here is, by replacing the old air conditioner which used to consume 1000 watts with this new air conditioner which has an EER of 10, we are reducing the power consumption to 600 watts.

#### Example 3

An old room air conditioner with an EER 6 was replaced by a new air conditioner with an EER of 10.0. The room requires 0.75 tons of air conditioning. Calculate the difference in power consumption between the old and new air conditioner.

Ok. An air conditioner, an old one, has an EER of 6. And this was replaced by an EER of 10 air conditioner. The room basically is required to pull out 0.75 or three quarters of a ton. You should remember that each ton, one ton of refrigeration or air conditioning is equal to, basically pulling out 12,000 Btus every hour. So it is pulling out ¾ of a ton, which happens to be 0.75 times 12,000 Btus per hour. That is 9000 Btus per hour.

$$\text{EER=6}$$

$$\text{EER=10}$$

$$\text{0}\text{.75ton}$$

$$\text{1ton=12,000}\text{BTU}/\text{hr}$$

$$\text{0}\text{.75}\times \text{12,000}\text{BTU}/\text{hr}$$

$$\text{=9,000}\text{BTU}/\text{hr}$$

So to pull out 9,000 Btus per hour with an air conditioner of EER equal to 6. So we are pulling out 9,000 Btus/hr and what is the wattage? Or watts? And wattage is equal to now, 9,000 divided by 6. That happens to be 1500 watts.

$$\text{6=}\frac{\text{9,000}\text{BTU}/\text{hr}}{\text{Wattage}}$$

$$\text{=}\frac{\text{9,000}}{\text{6}}\text{=1,500Watts}$$

Ok. Now if we were to replace this with an EER of 10 (air conditioner with 10). Now it still has to pull out 9000 Btus/hr and what would be the wattage? So watts equal to 9000 Btus/hr divided by 10, that would be 900 watts.

$$\text{EER10=}\frac{\text{9,000}\text{BTU}/\text{hr}}{\text{Watts}}$$

$$\text{W=}\frac{\text{9,000}}{10}\text{=900W}$$

So by replacing this air conditioner, which used to consume 1500 watts, by an energy efficient air conditioner with an EER of 10, we are able to bring down the power consumption to 900 watts. So that is a savings of 40% right there.

#### Example 4

**What is the annual cost for operating a 3 ton central air conditioner with an SEER of 10? Assume that the AC operates 2,000 hours in a year and the cost of electricity is 9.2 cents per kWh.**

Solution:

$$\text{SEER=}\frac{\frac{\text{BTUs}}{\text{h}}\text{pulledout}}{\text{Watt}}\text{=}\frac{\frac{\text{BTUs}}{\text{h}}\text{36,000}}{\text{Watt}}\text{=10}$$

$$\text{Watts=}\frac{\text{36,000}\frac{\text{BTUs}}{\text{h}}}{\text{10}}\text{=3,600W}$$

Recall that 1 ton =12,000 BTUs/h. Therefore, the cooling load is 3 x 12,000 BTUs/h = 36,000 BTUs/h

Recall also that 1,000 W = 1 kW. Therefore, power consumption = 3.6 kW.

Energy = Power x Time of Usage

= 3.6 kW x 2,000 h/year = 7,200 kWh/year.

Annual Cost = Units of energy x price per unit

$$\text{AnnualCost=7,200}\overline{)\text{kWh}}\text{}\times \text{}\frac{\text{\$0}\text{.092}}{\overline{)\text{kWh}}}\text{=\$662}\text{.40}$$#### Example 5

**Suppose you are comparing two air conditioners both of which last for 10 years. The least efficient air conditioner draws 775 W of power. The most efficient one uses 700 Watts. Assuming that the air conditioner operates 2,400 hours annually and that the local energy costs 0.08 per kWh, how much money and energy can you save with the energy efficient model? How much money are you willing to pay extra for the energy efficient model?**

Ok. Here we are trying to compare two air conditioners both of which are going to last ten years. One consumes less power and the other one high power. They both operate for 2,400 hours annually so we have the power and we have the time of usage data. So we need to calculate how much energy each of these will consume.

So life of one, our first air conditioner is 10 years and second one is 10 years. Power is 775, least efficient one, watts, and the other one is 700 watts. We have the time of usage. Time of usage is given as 2,400 hours and this one is also 2,400 hours.

Air conditioner 1 | Air conditioner 2 | |
---|---|---|

Life | 10 years | 10 years |

Power | 775 watt | 700 watt |

Time | 2,400 hours | 2,400 hours |

So the energy consumed is given by power multiplied by time of usage. So in this case it would be 775 watts multiplied by 2,400 hours and that will be 1,860,000 watt hours.

$$\text{Energy=P}\times \text{T}$$

$$\text{=775watts}\times \text{2,400hours}$$

$$\text{=1,860,000watt hours}$$

And for this air conditioner it would be 700 watts times 2,400 hours. That will be 1,680,000 watt hours.

$$\text{Energy=P}\times \text{T}$$

$$\text{=700watts}\times \text{2,400hours}$$

$$\text{=1,680,000watt hours}$$

So we need to convert this into kilowatt hours because we buy by kilowatt hours. So dividing this by a thousand and dividing here by a thousand, we get kilowatt hours. So this air conditioner will consume 1860 kilowatt hours per year. This air conditioner will consume 1680 kilowatt hours per year.

Air conditioner 1 | Air conditioner 2 |
---|---|

$$\frac{\text{1,860,000watt hours}}{\text{1,000}}$$ | $$\frac{\text{1,680,000watt hours}}{\text{1,000}}$$ |

$$\text{=1,860}\text{kWh}/\text{year}$$ | $$\text{=1,680}\text{kWh}/\text{year}$$ |

So in 10 years, the energy consumption is 18,600 kWh. This is simply multiplying yearly consumption by 10 years. And for the efficient one it would be 1680 kwh times 10 years which would be 16,800 kwh.

Air conditioner 1 | Air conditioner 2 | |
---|---|---|

In 10 years... | $$\text{18,600kWh}$$ | $$\text{16,800kWh}$$ |

We know the price of each of these. You know, each of the kilowatt hours. That is, it is sold at the rate of 0.08 dollars per kilowatt hour. So when you multiply that by 0.08 both sides, the cost to operate this one would be \$1,488 dollars and this one would be \$1,344 dollars. So the difference is, \$144 which means the least efficient one basically cost \$144 more and the best one would be, even if the best one is priced \$144 more, it would work out to be the same amount in the long run over 10 years. But we would be helping the environment by not burning the difference between these two kilowatt hours. You know a couple of thousand kilowatt hours over 10 years. That would be a help for the environment.

#### Example 6

**If the owner bought an air conditioner with an SEER of 15, which costs $500 more, instead of the model in the previous Illustration (7-3), what is the pay back period?**

The power consumption of this new model is $$\frac{\text{36,000}\frac{\text{BTUs}}{\text{h}}}{15}=\text{2,400 Watts}=\text{2.4 kW}$$

$$\text{Energy=2.4kWh}\times \text{}\text{2,000h}/\text{year}=\text{}\text{4,800kWh}/\text{year}$$

$$\text{AnnualCost=}\frac{\text{4,800kWh}}{\text{year}}\times \frac{\text{\$0.092}}{\text{kWh}}=\text{\$441.60}$$

$$\text{Savingsperyear}=\text{\$662.40}-\text{\$441.60}=\text{\$220.80}$$

$$\text{PaybackPeriod=}\frac{\text{AdditionalInvestment}}{\text{SavingsperYear}}\text{=}\frac{\text{\$500.00}}{\text{\$220.80}/\text{year}}\text{=2.3years}$$