Air Conditioner Efficiency - Humidity & Precooling

[Cyphear]

I have two questions/ideas about the efficiency of A/C's that I'd like to bounce off you guys. I couldn't find anything on Google about either of these items being discussed before.

1) Precooling
The first thought I was wondering was whether it was more efficient to pre-cool your house in the morning, since an A/C will blow colder air when it is cooler outside. It may be blowing 50 degrees at the registers first thing in the morning, but 60 degrees when it's hot in the middle of the day. Obviously it's more energy efficient to get colder air coming out of the registers considering the energy use should be constant. I think the value of this technique would depend on how well your house is insulated, but with good enough insulation, I believe it could make sense to use this technique and avoid running the A/C in the heat of the day.

2) Humidity
My second idea I was able to hash out a little more, and it was in regard to humidity. It appears that it takes a descent amount of energy to remove humidity from the air (a byproduct of running the A/C). My calculations show that it takes 8.9kJ to dehumidify one kg of air from 75% RH to 50% RH at 25 degrees C, while it takes 1.012KJ to cool one KG of air 1 degree. With these assumptions, this means that if you let 75% RH air in your house, it will only be cost effective if it keeps you from cooling that air by 9 degrees.

This was fairly surprising to me, as I would previously have opened my windows if it's a little bit cooler outside in the morning (regardless of humidity). I'll be considering the humidity in the future before opening my windows.

Here are my math notes in case anyone is curious or wants to check my math:

25C = 298.15 Kelvin

Lwater(T) = - 0.0000614342T3 + 0.00158927T2 - 2.36418T + 2500.79
0.01831660673 + 0.4738408505 - 704.880267 + 2500.79
=1796.40189 kJ/kg (latent heat of vaporization @ 25C)

100% RH = 20g H2O per kg of air
75% RH = 15g H2O per kg of air
50% RH = 10g H2O per kg of air

To remove 5g of H2O:
1796.40189/1000*5 = 8.98200945 kJ to dehumidify one kg of air


1.012 J/g specific heat of air at 25C
To cool 1KG of air 1 degree: 1012J

 

[russ_watters]

The first thought I was wondering was whether it was more efficient to pre-cool your house in the morning, since an A/C will blow colder air when it is cooler outside.

Probably not, because:

It may be blowing 50 degrees at the registers first thing in the morning, but 60 degrees when it's hot in the middle of the day.

The loss in efficiency is nowhere near that high. It is probably more like 1-2F (5-15%). Also, at the same time the capacity goes down, the compressor energy goes up by a similar proportion.

I think the value of this technique would depend on how well your house is insulated, but with good enough insulation, I believe it could make sense to use this technique and avoid running the A/C in the heat of the day.

Yes. It would also depend on the temperature difference between day and night and how much efficiency your a/c loses with temperature.

It appears that it takes a descent amount of energy to remove humidity from the air (a byproduct of running the A/C). My calculations show that it takes 8.9kJ to dehumidify one kg of air from 75% RH to 50% RH at 25 degrees C, while it takes 1.012KJ to cool one KG of air 1 degree. With these assumptions, this means that if you let 75% RH air in your house, it will only be cost effective if it keeps you from cooling that air by 9 degrees.

Let me put it this way: If air is already saturated (100% humidity), a little less than 2/3 of the energy required to decrease its temperture by 1F goes toward dehumidification. So if you increase your airflow and thus increase your supply air temperature, you will increase the capacity of the unit (by blowing more air over the same coil) and make more of that energy go toward decreasing the room temperature than room humidity.

Of course, comfort depends on humidity too, so you'd never really want it to be 75% humidity in your house.

This was fairly surprising to me, as I would previously have opened my windows if it's a little bit cooler outside in the morning (regardless of humidity). I'll be considering the humidity in the future before opening my windows.

Yes, people don't always realize that. If you're going to be using your air conditioning in the afternoon, you don't want to have your windows open in the morning on a humid day (say, 65F and 90% humidity, a common condition in the northeast on summer nights).

 

[Cyphear]

Thanks for the reply Russ.

Probably not, because: The loss in efficiency is nowhere near that high. It is probably more like 1-2F (5-15%). Also, at the same time the capacity goes down, the compressor energy goes up by a similar proportion.

I'll do some more research on the temperature drop and see if I can find some concrete numbers. Can you define "capacity" and "compressor energy goes up" in the second sentence? I'm guessing the later means that the compressor will use more power, but what is capacity? If it's the amount of cooling the system can provide, you saying that the compressor uses more energy when it's hot out, correct?

Let me put it this way: If air is already saturated (100% humidity), a little less than 2/3 of the energy required to decrease its temperture by 1F goes toward dehumidification.

Is this using your math, or something you've heard, or are you just rehashing my data? I'm just curious if what your saying is restating my claim, or substantiating it.

So if you increase your airflow and thus increase your supply air temperature, you will increase the capacity of the unit (by blowing more air over the same coil) and make more of that energy go toward decreasing the room temperature than room humidity.

Good/Interesting point. I suppose you'd lose efficiency after a point (e.g. 30mph winds blowing out of your vents). Thanks again for the reply.

 

[russ_watters]

I'll do some more research on the temperature drop and see if I can find some concrete numbers. Can you define "capacity" and "compressor energy goes up" in the second sentence? I'm guessing the later means that the compressor will use more power, but what is capacity? If it's the amount of cooling the system can provide, you saying that the compressor uses more energy when it's hot out, correct?

Yes, capacity is the amount of cooling it provides. Here's some data on a typical residential air conditioner: http://www.docs.hvacpartners.com/idc/groups/public/documents/techlit/24acb3-3pd.pdf

On page 30, it has a performance chart. Picking some conditions, you can see that:

-At an outside air temperature of 75F and an indoor wet bulb temperature (a measure of the energy of the air by combining temperature and humidity) of 67F at a flow rate of 525CFM, it will have a total capacity of 18.79MBH and draw 1.22kW of power.

-If you increase the outside air temperature to 105F, you drop the total capacity to 16.12MBH, and raise the input power to 1.72kW.

That's a drop of 14% in capacity and a rise of 41% in input power. That's a bigger change in input power than I realized, though I did pick a very wide temperature difference between night and day. At 95F, the rise in input power is 25%.

-Now if you increase the airflow to 675CFM (29% increase), the total capacity goes to 19.93MBH and the input power goes to 1.27kW. That's an increase of 6% in capacity and an increase of 4% in input power. However, the input power doesn't include the fan and fan energy is maybe 10% of total system energy, but it rises as a cube function of airflow. So for example if it was .12kW, now it is .26kW. So in this case, your efficiency actually goes down a little.

Is this using your math, or something you've heard, or are you just rehashing my data? I'm just curious if what your saying is restating my claim, or substantiating it.

I was curious to know exactly what it was (it has come up before), so I calculated it. It was about 35%/65%. You calculate it with a psychrometric chart (I used a software version): https://www.nrc-cnrc.gc.ca/obj/irc/images/bsi/83-psy_E.gif

Good/Interesting point. I suppose you'd lose efficiency after a point (e.g. 30mph winds blowing out of your vents). Thanks again for the reply.

Yes, fan power can be a big factor, as it increases as a cube function of airflow if nothing else about the system changes (ie, twice the airflow means 8x the fan power).

 

[Cyphear]

That's some great information Russ. Those are also some pretty great links. I love me some thermodynamics!

Since we have the A/C side of the equation figured out, I wonder if there is a formula for how heat enters a house on an average day. Some heat that enters would remain static in a precooled house (sun energy), but more heat would enter a precooled house with any heat transfer (through windows, other surfaces, or openings/cracks). It'd be interesting do finitely conclude when precooling is more efficient.

By the way, to meet Energy Star requirements, fan energy has to be less than 2% of the total energy, so I think it'd be a safe guess that an average system may be 4-5%? Ironically, if they have to slow down the fan to meet Energy Star requirements, the the A/C could be less efficient than it would be at 3% of total energy. I suppose they could use more efficient technology (e.g. ball bearings) or a bigger fan (I have no idea, but I'm guessing power use is not cubed) to meet these requirements.

 

[Actuary_CFS]

Researching Air Conditioner Efficiency I came across this TWO YEAR OLD thread.

Might I question the following formula used by Cyphear (Russ appears under Engineering now):-

25C = 298.15 Kelvin

Lwater(T) = - 0.0000614342T3 + 0.00158927T2 - 2.36418T + 2500.79
0.01831660673 + 0.4738408505 - 704.880267 + 2500.79
=1796.40189 kJ/kg (latent heat of vaporization @ 25C)

What is the source of this formula?
- T, T2 & T3 appear identical. I wonder whether, in view of the increasing nature of the corresponding coefficients, T2 might not be T^2 (T squared) and T3 T^3 (T cubed)?
- The transposition of sign on the very small initial coefficient is not material to Cyphears calculated result.

 

[Cyphear]

Good point. I'm glad you found this thread and I'm glad I posted my math in the thread.

Wow, this thread has almost 5k views. If I built an online calculator, I wonder if anyone would use it.

I don't know where I got that formula either. In googling this, there are plenty of sources that don't square it (I may have used one of those), but in regoogling the latenent heat of water, wikipedia says it's squared, as you've suggested.

Reworking my math results in a bit more energy to dehumidify, further strengthening the logic behind this post.

25C = 298.15 Kelvin

Lwater(T) = - 0.0000614342T^3 + 0.00158927T^2 - 2.36418T + 2500.79
= (0.0000614342 * (25^3)) + (0.00158927 * (25^2)) - (2.36418 * 25) + 2500.79 (just type this into google and it'll solve it)
= 2443.6387 kJ/kg (latent heat of vaporization @ 25C)

100% RH = 20g H2O per kg of air
75% RH = 15g H2O per kg of air
50% RH = 10g H2O per kg of air

To remove 5g of H2O (figuring that's an average reduction):
2443.6387/1000*5 = 12.2181935 kJ to dehumidify one kg of air

1.012 J/g specific heat of air at 25C
To cool 1KG of air 1 degree: 1012J = 1.012kJ

In summary, dehumidifying air 25% takes equal energy as cooling it 12 degrees. Approximately a 2% humidity per degree.

 

[Actuary_CFS]

THANKS for your prompt reply Cyphear; BUT I found your original mixed use of F, C & K temperature scales a little confusing.

What result might you get if the Kelvin temperature was used in your formula, and the smallest coefficient signed negatively? EDIT - just found the following:
http://en.wikipedia.org/wiki/Latent_heat#Latent_heat_for_condensation_of_water
which uses degrees celsius, as you have.

I might initially assume a reduction of 12.5% in RH; from say 62.5% to 50%. I'm also wondering how sensitive this formula is to changes in temperature, and whether the warmer input temperature or the chilled output temperature should be used. Again if the A/C is largely working on recycled air any need to reduce RH might quickly taper off.

 

[Cyphear]

I found my use of Celsius (instead of Kelvin) a little confusing too. I'm not sure why I chose to define it at the top. You bring up a good point about my use of Fahrenheit. My numbers are all in Celsius, but I was thinking of them as Fahrenheit at the end. I'm going to edit my formula above to clarify all this. Thanks for carefully reviewing my math.

My intuition tells me that the formula should not be too sensitive to changes in temperature (I would think that evaporating water shouldn't take that much more energy at 18°C vs 25°C). Especially considering the output temperature of an A/C isn't THAT cold. The formula is "dominated" by the 2500.79 constant. I would guess that these numbers represent the reduced energy that it takes to make the phase change (which would be used to get it up to 100°C if this was a boiling scenario), while the 2500.79 is the phase change itself.

It's not letting me edit any post but this one, so I'm rewriting the formula here.

77°F = 25°C = 298.15 Kelvin (although Kelvin isn't used in this formula)

Lwater(T) = - 0.0000614342T^3 + 0.00158927T^2 - 2.36418T + 2500.79
= (0.0000614342 * (25^3)) + (0.00158927 * (25^2)) - (2.36418 * 25) + 2500.79 (just type this into google and it'll solve it)
= 2443.6387 kJ/kg (latent heat of vaporization @ 25C)

100% RH = 20g H2O per kg of air
75% RH = 15g H2O per kg of air
50% RH = 10g H2O per kg of air

To remove 5g of H2O (figuring that's an average reduction):
2443.6387/1000*5 = 12.2181935 kJ to dehumidify one kg of air

1.012 J/g specific heat of air at 25C
To cool 1KG of air 1 degree C: 1012J = 1.012kJ

In summary, dehumidifying air 25% takes equal energy as cooling it 12°Celsius (or 21.6°F). Approximately a 2% humidity per degree Celsius (or 1.15% (approximately 1%) humidity per degree Fahrenheit). So in layman's terms, the percent humidity change needs to be less than the degree change in Fahrenheit to justify opening windows.

Can this be right? Is letting in 30% more humid air really as inefficient as letting in 30 degree hotter air? In other words, if it's 70 degrees and 40% humidity inside, is letting in 70 degree air at 70% humidity as bad as letting in 100 degree air at 40% humidity? Maybe I'm mixing up my perception of the temperature of the air with the energy it takes to cool it. Thoughts? Maybe you can check my math again Actuary_CFS. :)

 

[Cyphear]

I haven't worked out the numbers, but one thing to think about is that relative humidity goes up with temperature drop, somewhere around 12% per 5C (9F). Absolute humidity is a better number to work with, but not one we normally reference.

I haven't had a chance to work it out, but maybe after adding a 1% per degree difference (roughly), the percent change in relative humidity must be no more than double the temperature drop in Fahrenheit? Thoughts?

 

[RottenMutt]

I've always wondered if programmable thermostats save money. I live in TX and the peak temperature is always at the time you would be coming home, thus poor efficiency because of the temperature differential (+100F out) and removing all the humidity in the house.

I would love to see the experiment ran with two identical houses.

 

[the_emi_guy]

Yes, capacity is the amount of cooling it provides. Here's some data on a typical residential air conditioner: http://www.docs.hvacpartners.com/idc/groups/public/documents/techlit/24acb3-3pd.pdf

On page 30, it has a performance chart. Picking some conditions, you can see that:

-At an outside air temperature of 75F and an indoor wet bulb temperature (a measure of the energy of the air by combining temperature and humidity) of 67F at a flow rate of 525CFM, it will have a total capacity of 18.79MBH and draw 1.22kW of power.

FYI, that Carrier data sheet seems to be off by 3 orders of magnitude in its capacity values. They should be KBtuH (meaning KBtu/h) not MBtuH.

12,000 Btu/h per ton (conversion)
400cfm per ton (conventional design)

 

[russ_watters]

FYI, that Carrier data sheet seems to be off by 3 orders of magnitude in its capacity values. They should be KBtuH (meaning KBtu/h) not MBtuH.

Americans use "M" instead of "k" for the prefix on BTUs.

 

[the_emi_guy]

Wow, thanks, I had no idea.

M as in the Roman numeral for 1000!

Clipped from Wikipedia page for "BTU":
The unit MBtu was defined as one thousand Btu, presumably from the Roman numeral system where "M" stands for one thousand (1,000). This is easily confused with the SI mega (M) prefix, which multiplies by a factor of one million (1,000,000). To avoid confusion many companies and engineers use MMBtu to represent one million Btu. Some companies also use BtuE6 in order to reduce confusion between a thousand Btu vs. a million Btu.