I overlooked a crucial consideration raised by denkenberger here that reduces the efficiency loss ~2x.
Thanks-it looks like you are referring to the net infiltration flow rate impact on the building. But there was also the consideration of humidity, and I did not see any humidity measurements in the data, so we are not able to resolve that one. Humidity sensors are fairly cheap, but notoriously unreliable. But one could actually measure the amount of water condensed pretty accurately to get an idea how much of the cooling of the air conditioner is going to condensing water versus cooling air (sensibly).
I didn’t know how to estimate this effect but I was guessing the total impact on the bottom line is much smaller than the factors of 2 from the other factors, at least in CA (though it’s definitely another factor I overlooked). I’m comfortable treating 85 to 70 in CA as a typical use case to benchmark efficiency for a portable AC.
That guess is coming from the rough sense that dehumidifiers use much less energy than air conditioners. I don’t know if that’s right and reflects that dehumidifying is pretty cheap (at least in CA), or if dehumidifiers are just normally used for relatively small humidity changes, or if I’m wrong about relative energy use. I also have a sense that when I run an AC it just doesn’t produce very much water (and that the energy cost is like ~0.6kWh per liter).
Actually this seems pretty non-trivial to estimate. Do you know reasonable ballpark figures?
If you want to geek out on this you can use a psychrometric chart. For instance, if outdoor air is 85F and 50% relative humidity (RH), that’s an enthalpy of about 35 BTU/lb of dry air. Typical exit air conditions on the cool side of an air conditioner are ~50F and 100% RH, so ~20 BTU/lb of dry air. The dehumidification portion would be going to 85F and ~30% RH or ~29 BTU/lb of dry air, so ~40% of the heat removed is in the form of condensing water (latent). This means you would take the sensible part and multiply by about 1.7 to get the total load on the air conditioner. If you were not drawing in outdoor air, the latent load would be much lower. So overall I think you’re right that in CA the humidity correction is not as big as the other factors.
I don’t think I can follow your calculation. My version would be:
You are intaking hot wet outside air (wet from both high RH and high temp). You need to cool it and condense a bunch of water out of it. There’s some ratio that’s fixed by the humidity and temperature of the outside vs inside air. I think that’s what you are saying is around 40%? I think actually the number you are giving isn’t what quite this calculation needs, but I’ll run with it anyway.
If all the heat was coming in from outside air (either before turning on AC or from infiltration), then you’d have a fixed ratio of latent to sensible heat removed, so the ratio wouldn’t depend on how much additional infiltration you caused, and we could just ignore humidity when thinking about the efficiency loss.
But in fact some of the heat is coming in from other channels. I guess the other big one is sunlight through windows. That heat doesn’t come with any more humidity. Extra infiltration from 2-hose AC increases how much latent heat you need to remove per unit of sensible heat, by increasing the relative importance of infiltrated air vs sunlight and other sources of heat. So if we just calculate how much extra sensible heat you have to remove, we’ll underestimate the efficiency loss.
The total extra infiltrated heat is about 25% of what the AC removes. At equilibrium, that’s 25% of all the heat gain in the house. If 13% of heat gain is normally from infiltration, then replacing that with 75% normal heat and 25% new infiltration would increase the fraction of heat from infiltration all the way to 35%. (I was super wrong about the 13% going in, I was expecting 25-50%!)
So per unit of heating, you are also increasing the fraction of heat coming from infiltrated air by 22%.
For the heat coming from infiltration, the extra cost of dehumidifying is about 2⁄3 of the sensible heat removed. So per unit of sensible heat removed, you need to remove an additional 15% of a unit of latent heat.
If the AC exhaust was more humid than the inside, then this would be lower, but my sense is that AC exhaust is basically as dry as indoor air?
So the net effect would be to take you from 25% efficiency loss (ignoring humidity) up to roughly 40% efficiency loss, which is pretty huge.
That was a super confusing calculation, definitely beyond my pay grade. I assume I got a ton of numbers/calculations and wrong, that there were much simpler ways to do it, and that this overall computation is likely to be conceptually confused in one or more ways. So I’d be pretty curious for your bottom line estimate or intuition about where it should have ended up.
(But I also understand if you want to stop talking about AC and put this thread to rest...)
I would say that is basically right. AC exhaust is about as humid as indoor air. The fraction of the heating load in the summer due to infiltration really does depend on how tight your building construction is. With the numbers Jeff was assuming for a very old house, infiltration would be a much larger percentage. There are some other sources of heat in a house that come with humidity, such as people and showers, but overall it is much less humidity than bringing in outdoor air (there is heat conduction through the walls, electricity use of lighting and appliances, etc.). So that might mean that it would take you from a 25% efficiency loss (ignoring humidity) up to a 35% efficiency loss, which is still a big deal. But I’m not sure if 85°F in California typically corresponds to 50% relative humidity.
Thanks-it looks like you are referring to the net infiltration flow rate impact on the building. But there was also the consideration of humidity, and I did not see any humidity measurements in the data, so we are not able to resolve that one. Humidity sensors are fairly cheap, but notoriously unreliable. But one could actually measure the amount of water condensed pretty accurately to get an idea how much of the cooling of the air conditioner is going to condensing water versus cooling air (sensibly).
I didn’t know how to estimate this effect but I was guessing the total impact on the bottom line is much smaller than the factors of 2 from the other factors, at least in CA (though it’s definitely another factor I overlooked). I’m comfortable treating 85 to 70 in CA as a typical use case to benchmark efficiency for a portable AC.
That guess is coming from the rough sense that dehumidifiers use much less energy than air conditioners. I don’t know if that’s right and reflects that dehumidifying is pretty cheap (at least in CA), or if dehumidifiers are just normally used for relatively small humidity changes, or if I’m wrong about relative energy use. I also have a sense that when I run an AC it just doesn’t produce very much water (and that the energy cost is like ~0.6kWh per liter).
Actually this seems pretty non-trivial to estimate. Do you know reasonable ballpark figures?
If you want to geek out on this you can use a psychrometric chart. For instance, if outdoor air is 85F and 50% relative humidity (RH), that’s an enthalpy of about 35 BTU/lb of dry air. Typical exit air conditions on the cool side of an air conditioner are ~50F and 100% RH, so ~20 BTU/lb of dry air. The dehumidification portion would be going to 85F and ~30% RH or ~29 BTU/lb of dry air, so ~40% of the heat removed is in the form of condensing water (latent). This means you would take the sensible part and multiply by about 1.7 to get the total load on the air conditioner. If you were not drawing in outdoor air, the latent load would be much lower. So overall I think you’re right that in CA the humidity correction is not as big as the other factors.
I don’t think I can follow your calculation. My version would be:
You are intaking hot wet outside air (wet from both high RH and high temp). You need to cool it and condense a bunch of water out of it. There’s some ratio that’s fixed by the humidity and temperature of the outside vs inside air. I think that’s what you are saying is around 40%? I think actually the number you are giving isn’t what quite this calculation needs, but I’ll run with it anyway.
If all the heat was coming in from outside air (either before turning on AC or from infiltration), then you’d have a fixed ratio of latent to sensible heat removed, so the ratio wouldn’t depend on how much additional infiltration you caused, and we could just ignore humidity when thinking about the efficiency loss.
But in fact some of the heat is coming in from other channels. I guess the other big one is sunlight through windows. That heat doesn’t come with any more humidity. Extra infiltration from 2-hose AC increases how much latent heat you need to remove per unit of sensible heat, by increasing the relative importance of infiltrated air vs sunlight and other sources of heat. So if we just calculate how much extra sensible heat you have to remove, we’ll underestimate the efficiency loss.
The total extra infiltrated heat is about 25% of what the AC removes. At equilibrium, that’s 25% of all the heat gain in the house. If 13% of heat gain is normally from infiltration, then replacing that with 75% normal heat and 25% new infiltration would increase the fraction of heat from infiltration all the way to 35%. (I was super wrong about the 13% going in, I was expecting 25-50%!)
So per unit of heating, you are also increasing the fraction of heat coming from infiltrated air by 22%.
For the heat coming from infiltration, the extra cost of dehumidifying is about 2⁄3 of the sensible heat removed. So per unit of sensible heat removed, you need to remove an additional 15% of a unit of latent heat.
If the AC exhaust was more humid than the inside, then this would be lower, but my sense is that AC exhaust is basically as dry as indoor air?
So the net effect would be to take you from 25% efficiency loss (ignoring humidity) up to roughly 40% efficiency loss, which is pretty huge.
That was a super confusing calculation, definitely beyond my pay grade. I assume I got a ton of numbers/calculations and wrong, that there were much simpler ways to do it, and that this overall computation is likely to be conceptually confused in one or more ways. So I’d be pretty curious for your bottom line estimate or intuition about where it should have ended up.
(But I also understand if you want to stop talking about AC and put this thread to rest...)
I would say that is basically right. AC exhaust is about as humid as indoor air. The fraction of the heating load in the summer due to infiltration really does depend on how tight your building construction is. With the numbers Jeff was assuming for a very old house, infiltration would be a much larger percentage. There are some other sources of heat in a house that come with humidity, such as people and showers, but overall it is much less humidity than bringing in outdoor air (there is heat conduction through the walls, electricity use of lighting and appliances, etc.). So that might mean that it would take you from a 25% efficiency loss (ignoring humidity) up to a 35% efficiency loss, which is still a big deal. But I’m not sure if 85°F in California typically corresponds to 50% relative humidity.