10.6 Air Conditioner Efficiency

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

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

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

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

Therefore, its wattage =

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

Example 2

Air Conditioner Efficiency. 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.

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

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

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.

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

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?

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

To calculate the power, we have 6,000 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 1,000 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 13.0. The room requires 0.75 tons of air conditioning. Calculate the difference in power consumption between the old and new air conditioners.

The old AC unit has an EER of 6. And this was replaced by one with an EER of 13. 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 9,000 Btus per hour.

$$\text{EER = 6}$$

$$\text{EER = 13}$$

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

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

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

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

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. This is 1,500 watts.

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

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

Ok. Now if we were to replace this with an EER of 13. Now it still has to pull out 9,000 Btus/hr and what would be the wattage? So watts equal to 9,000 Btus/hr divided by 13, that would be 900 watts.

$$\text{EER 13 = }\frac{\text{9,000 BTU}/\text{hr}}{\text{Watts}}$$

$$\text{W = }\frac{\text{9,000}}{13}\text{ = 692 W}$$

So by replacing this air conditioner, which used to consume 1,500 watts, by an energy efficient air conditioner with an EER of 12, we are able to bring down the power consumption to 692 watts. So that is a savings of over 53% 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 19.2 cents per kWh.

Solution:

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

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

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

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

Energy = Power × Time of Usage

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

Annual Cost = Units of energy × price per unit

$$\text{Annual Cost = 7,200}\cancel{\text{kWh}}\times \frac{\text{\$0}\text{.192}}{\cancel{\text{kWh}}}\text{ = \$1382.40}$$

Example 5

Suppose you are comparing two air conditioners, both of which are expected to last for 10 years. The least efficient air conditioner draws 775 W of power. The most efficient one uses 600 Watts. Assuming that the air conditioner operates 2,400 hours annually and that the local energy costs 0.128 per kWh, how much money and energy can you save with the energy-efficient model? 

Air Conditioner #1

$$\begin{gathered} \text{Energy}=\text{Power}\times \text{Time} \\ =\text{775} \text{Watts}\times \text{2,400} \text{hours} \\ =\text{1,860,000} \text{Wh} \\ =\text{1,860} \text{kWh}/\text{year} \end{gathered}$$

Air Conditioner #2

$$\begin{gathered} \text{Energy = Power ×Time} \\ = \text{600 Watts × 2,400 hours} \\ \text{= 1,440,000 Wh } \\ = \text{1,440 kWh/year} \end{gathered}$$

Electricity price is estimated at $0.128/kWh. Over the ten years you will pay $2,380.80 in electricity to run AC #1, while you would only pay $1,848.32 for AC #2. This assumes electricity prices stay the same for those ten years. You would save $532.48 over 10 years by purchasing unit #2.