On mechanical ventilation in attics – GreenBuildingAdvisor

I’ve been thinking about this question since June, because at the start of the summer there’s always a lot of people complaining that the upstairs of their house is hot. They often want to ventilate their attic to help cool off the upstairs. But I was busy in the summer; things have slowed down for me a bit now so I’ve had time to jot down some thoughts.

I know that the standard advice is not to do it, that any sort of mechanical ventilation will just pull air out of the living space. But you could argue that if the boundary between the attic and the living space is properly sealed that’s not a concern. And I was surprised to read Joe Lstiburek say this: “In hot climates, the primary purpose of attic or roof ventilation is to expel solar heated hot air from the attic to lessen the building cooling load.” (see https://buildingscience.com/documents/digests/bsd-102-understanding-attic-ventilation ). Et tu, Joe[1]? So let’s assume that we have a house that is properly built with a ventilated attic that is well air-sealed to the interior. What happens if you put mechanical ventilation in the attic?

Imagine a house with a footprint of 1,000 square feet. It has a ventilated attic. In the summer, when it’s 90F out and the sun is out, the attic heats up to 140F. Below the attic is occupied space that is cooled to 75F, and between the attic and the occupied space is R40 insulation. Assume also that between the roof and gables, the attic has 2,000 square feet of surface area exposed to the outside, and that the surface area has an average effective R-value[2] of 2.0.

Let’s say you want to mechanically ventilate that attic so that instead of being at 140F, it’s at 110F. First, let’s look at the impact on air conditioning use. With R40 between the attic and occupied space, 140F in the attic and 75F in the occupied space, and 1000 square feet, you’d see heat transfer[3] of (140-75)*1000/40= 1625 BTU/hour.  With the attic at 110F, the heat difference is now 35F and you’d get 35*1000/40= 875 BTU/hour, a difference of 750 BTU/hr. So you’d save 750 BTU/hr of air conditioning.

The next question is, how much ventilation would you need to do that? The sun is beating down on the attic, heating it up. Since the attic is hotter than the surrounding environment, it is losing heat to the environment. When the attic is at 140F and it’s 90F outside, that’s a 50F difference, with 2,000 square feet of exposed area and R-2, the heat transfer is 50*2,000/2= 50,000 BTU/hr. That’s how much heat the sun is putting into the attic.

When the attic is at 110F, the difference is 20F, and the heat transfer is 20*2000/2= 20,000 BTU/hr. In order to maintain that temperature when the sun is pouring in 50,000 BTU/hr, the ventilation has to be removing 30,000 BTU/hr.

The ventilation is sucking out attic air at 110F, and replacing it  with outside air at 90F. The heat transfer of an airflow is given by the temperature difference times the airflow in CFM, times 1.08. To move 30,000 BTU/hr with a 20F temperature difference, you need 1388 CFM of airflow.[4]

So what does an attic fan that can move that kind of flow look like? It looks like this:  [link removed by spam filter]

That’s the first one I found on Google.

The specifications say it requires 4.5 Amps at 120V, which would be 540 Watts[5].

How does that compare with the 750 BTU/hr in air conditioning that you’re saving? Let’s say the air conditioner can manage a COP of 3.0, so that’s 250 BTU/hr of electricity usage, which is about 75 Watts.

So the ventilating fan uses about seven times as much electricity to “help” the air conditioning as the air conditioning would use without the help.

NOTES:

[1] To be fair, Joe goes on to say, “The amount of cooling provided by a well ventilated roof exposed to the sun is very small. Field monitoring of numerous attics has confirmed that the temperature of the roof sheathing of a unvented roof will rise by a few to no more than 10 F more than a well ventilated attic.”

[2] “Effective R-value” is a slight of hand. Regular R-value just measures heat through conduction and ignores radiation and convection. But sometimes it’s useful to lump all the heat transfer together. That’s what effective R-value does.

[3] For all these calculations I’m using the formula for heat transfer through an assembly, which is temperature difference times area in square feet divided by R-value gives heat flow in BTU/hr.

[4] You could argue that the reduced heat loss to below should be added in as well, but this is such a rough approximation that such precision wouldn’t add anything to the conclusion.

[5] I know you can’t just multiply Volts by Amps, you have to include a power factor. And that might be peak Amps. So yeah, the average Watts is probably lower. But how much, and does it change the picture?