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\( W \) Work done on the fluid
\( Q_\text{in} \) Heat entering the air conditioned space
\( Q_\text{out} \) Heat removed by the air conditioner
\( Q_h \) Heat emitted to outside environment by air conditioner
dots \(\dot{Q}_\text{in}\;\dot{Q}_\text{out}\;\dot{W} \) Energy movement per time
\( T_h \) Temperature at the hot side
\( T_c \) Temperature on the cool side
\( \eta \) Coefficient of performance, reported as an EER or SEER rating on AC units
\( t \) AC expected lifespan

Wall units or central air?

Photo credit to Dan LeFebvre

Goals of this article

Because of the similarities in AC and refrigeration, this article will summarize concepts from the refrigeration thermodynamics article and apply them to home cooling to answer the question, am I better off with wall units or a central AC unit? The final section will answer the question, how much extra should I pay for an AC unit with a higher SEER rating?

What goes in and what goes out?

control volumes of a AC unit

[Caption] The energy balances on the house and air conditioner (heat pump) are the same as those on the refrigerator because the underlying physics are the same. Refrigerant in the air conditioner picks up heat on the cool side and drops it off on the hot side as it recirculates.

In cooling a room, we are concerned with the ratio between two things -- (1) how quickly is heat leaving? and (2) how quickly is heat entering? If the temperature of the room is not changing, these two quantities are equal. (See figure, air conditioned control volume)

$$ \dot{Q}_\text{in}=\dot{Q}_\text{out} $$

We control \(\dot{Q}_\text{out}\) with our AC unit. We control \(\dot{Q}_\text{in}\) with insulation, windows, limiting the opening of doors, and reducing the temperature difference between the inside and outside of the home (setting the AC hotter).

What makes an air conditioner?

In engineering parlance, the efficiency of a heat pump/refrigerator/AC unit is \(\eta\), the amount of heat moved (per time) per unit energy used (per time). This is the "coefficient of performance."

$$ \eta = \frac{\dot{Q}_\text{out}}{\dot{W}} $$

There exists a thermodynamic limit on what \(\eta\) can be and it is only a function of the temperatures of the hot and cold sides of the heat exchanger. For an AC unit cooling a house to 75°F in 95°F weather, the ratio between heat moved vs energy used cannot exceed 27. (Formula must use absolute units - Kelvin.) The derivation of this thermodynamic limit was explored in detail here.

$$ \eta_\text{max} < \frac{T_c}{T_h - T_c} = \frac{297}{308-297}=27 $$

This calculation is reported on air conditioners but instead of calling it the "coefficient of performance" as engineers do, it is called the energy efficiency ratio (EER). Unfortunately the EER, a number which is a unitless ratio, is defined as a ratio of units: BTU per watt hours. The conversion between \(\eta\) and EER is therefore:

$$ \text{EER} = \eta\times\left( 3.412 \frac{\text{BTU}}{\text{watt-hours}}\right) $$
$$ \text{EER}_\text{max} < \frac{T_c}{T_h - T_c}\times\left( 3.412 \frac{\text{BTU}}{\text{watt-hours}}\right) = \frac{297}{308-297}\times\left( 3.412 \frac{\text{BTU}}{\text{watt-hours}}\right) = 92 \frac{\text{BTU}}{\text{watt-hours}} $$

Obviously this ratio strongly depends on temperature for any heat pump but US standards dictate reporting at 95°F outside at 40% humidity and 75°F inside at 51% humidity. To make reporting more fair and to prevent manufacturers from heavily optimizing to testing conditions, the "seasonal energy efficiency ratio," SEER, was introduced. It's a more complicated formula to give a weighted average of performance across various reasonable operating conditions. Because the EER uses an extreme temperature ratio, the SEER is usually higher. SEER ratings are not necessarily bounded below 92 because the operating conditions that go into that average have lower temperature differences. Both measurements, however, are coefficients of performance with different operating conditions in strange units.

Putting it together

For a given set of conditions (interior temp, exterior temp, insulation, frequency of door openings, etc) we can solve for the electricity usage.

$$ \dot{W} = \frac{\dot{Q}_\text{in}}{ \eta } $$

For a wall, unit \(\eta\) is around 3-4 (SEER between 9-12); for a central unit, \(\eta \) is between 5-9 (SEER between 15-26). Air conditioners can be more efficient than refrigerators because the temperature difference between the inside and outside of the cooled volume is lower.

The wall unit air conditioner solution has two problems: (1) per joule of heat removed, the wall unit is 30% less efficient than a central AC system and (2) while the central AC is cooling a volume that is typical insulated on 6 sides (4 walls, roof, and floor), the wall unit usually cools a volume only insulated on 4 sides at most (2 walls if it's a corner room, floor if it's a bottom story room, and roof if it's a top story room). Despite these shortcomings, wall units can still sometimes come out ahead because they might cool 80% or less volume. If you have a 2,500 sq ft house but only need to cool one 250 sq ft bedroom at night, the wall unit can be much less efficient cooling a less insulated space and still use less energy than a central unit.

So what is the answer?

To answer this question properly requires a few empirical measurements of the central and wall unit ACs under operating conditions at your home. However, lots of people have done these measurements. The rule of thumb for the break even point in electricity use is between one and three bedrooms. If you only need to cool 10% of your house (1 bedroom), the wall unit will probably consume less energy. If you need to cool 30-50% of your house (3 bedrooms), central AC will probably consume less energy. Like all rules of thumb, accuracy will vary depending on the specifics of your situation.

How much extra should I pay for a higher SEER rating?

Given how much air conditioning seasonally affects our bills in the southern United States, how much a high SEER rating is worth is an important question. I live in a 1300 sq ft house and my AC removes approximately 50-100 GJ of heat per year. This translates to $500-$1000 per year in cooling costs.

We are going to set up this problem just like the fuel efficency problem for cars. With \(V\) as the cost of the unit, \(\rho\) as the cost of electricity, and \(t\) as the expected AC lifespan, the following condition must be met for unit 1 to be a better choice than unit 2.

$$\begin{align} V_1 +\rho\dot{W}_1t &< V_2 +\rho\dot{W}_2t\\ V_1 +\rho\frac{t\dot{Q}_\text{out}}{\eta_1} &< V_2 +\rho\frac{t\dot{Q}_\text{out}}{\eta_2} \\ V_1 - V_2 &< \rho t \dot{Q}_\text{out}\left(\frac{1}{\eta_2}-\frac{1}{\eta_1}\right)\\ V_1 - V_2 &< \rho t \dot{Q}_\text{out}\left(\frac{\eta_1 - \eta_2}{\eta_1\eta_2}\right)\\ \frac{V_1 - V_2}{\rho t \dot{Q}_\text{out}} &< \frac{\eta_1 - \eta_2}{\eta_1\eta_2}\\ \end{align}$$
graph of cost difference justified by higher SEER ratings

[Caption] Graph of cost difference justified by higher SEER ratings. SEER2 plotted from 8 to 28 by increments of 2. [Left axis] Non-dimensional cost in terms of rate of heat removed, AC lifespan, and electricity cost. [Right axis] Cost difference assuming \(t\) = 15 years, \(\dot{Q}_\text{out}\) = 100 GJ per year, and \(\rho\) = $0.15 per kWh.

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© MC Byington