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fanpower needed to cool house overnight
Lacustral wrote:
I'd like to use an exhaust fan (fan from a ceiling grille to outside) to run overnight in the summer, with the windows open, so that my house is cool in the morning. I'm not sure how much CFM is needed. That depends on where you live and the house thermal mass and insulation and how cool you want it. Whole house fans can work almost as well as AC for all but a few days per year near Philadelphia. NREL says June has average daily min and max temps of 61.8 and 81.7 F. July is the warmest month, with 67.2 and 86.1 and humidity ratio w = 0.0133 pounds of water per pound of dry air. An indoor fan or a window AC for dehumidification can help ensure comfort in July. I don't want a big powerful whole house fan because I'd like it to be quiet. Just a small fan that keeps running overnight. But you may need a large fan. Lasko's 2155A 16" window box fan has a thermostat and sliding mounts to fit windows and it's efficient, moving 2470 cfm with 90 watts on the highest of 3 speeds, and it's fairly quiet. About $50 at ACE Hardware stores. A house with an open window or two upstairs is a natural cool air trap. At night, warm house air will rise up out of the window and cooler night air will slide in and replace it. During the day, air stops moving. More air and heat can flow with a fan that runs at night, eg an upstairs window with a 1-way damper outside (eg plastic film, hinged at the top edge) and an downstairs window with a damper inside and a differential thermostat or an outdoor 70 F thermostat that runs the fan when outdoor air is cooler. ... can you tell me how much CFM gets your house down to the temperature of the outside air It can never reach the daily min, but a larger fan can keep it closer. A Q cfm fan is like a 1/Q F-h/Btu thermal resistor or a Q Btu/h-F conductor between outdoor air and the house mass. This may be the bottleneck, if the exposed mass surface is large, with a 1.5 Btu/h-F-ft^2 airfilm conductance. (I could calculate the cubic feet of airspace in my house, divide by CFM of a fan, and come up with a guess, but I'm sure it's not that simple - the hot stuff in the house is heating up the air, fans aren't completely efficient about clearing out the inside air, etc.) If a house has infinite thermal mass (most don't :-) and G Btu/h-F of conductance and a Q cfm fan and a constant indoor temp Ti and 8-hour night and 16-hour day temps Tn and Td and the fan moves 8Q(Ti-Tn) Btu at night and the house gains 16(Td-Ti)G Btu/h during the day and these heatflows are equal, Ti = (QTn+2GTd)/(Q+2G), approximately. For example, G = 800 and Q = 100 and Ti = 80 and Tn = 70 makes Ti = (100x70+2x800x80)/(100+1600) = 79.4. G = 400 and Q = 2470 makes Ti = (2470x70+2x400x80)/(2470+800) = 72.4. We might model a house with C Btu/h of thermal mass and G Btu/h-F of conductance and a Q cfm fan and Tmin and Tmax indoor temps like this, viewed in a fixed font like Courier: 1/Q (closed at night) ---www--- \--- | | -------------- | 1/G | Tmax . . . Tn/Td ---------www--------*--- Tmin/Tmax . . . | Tmin . . | ------- ------- --- C night day --- | | - RCn = C/(Q+G) and RCd = C/G and Tmin=Tn+(Tmax-Tn)e^(-8/RCn) and Tmax = Td+(Tmin-Td)e^(-16/RCd), and Tmin = [Tn+(Td-Tn)e^(-8/RCn) -Tde^(-16/RCd-8/RCn)]/(1-e^(-16/RCd-8/RCn), if I did that right. For example, C = 6000 and G = 400 and Q = 2470 and Ti = 80 and Tn = 70 makes RCn = 2.1 hours and RCd = 15 hours, so... 20 C=6000'house thermal mass (Btu/F) 30 G=400'house conductance (Btu/h-F) 40 Q=2470'fan cfm 50 RCN=C/(G+Q)'night time constant (hours) 60 RCD=C/G'day time constant (hours) 70 TN=70'night temp (F) 80 TD=80'day temp (F) 90 TMIN=(TN+(TD-TN)*EXP(-8/RCN)-TD*EXP(-16/RCD-8/RCN))/(1-EXP(-16/RCD-8/RCN)) 100 TMAX=TD+(TMIN-TD)*EXP(-16/RCD) 110 PRINT TMIN,TMAX'min and max indoor temps (F) 70.14393 76.608 Nick |
fanpower needed to cool house overnight
No, I don't know, sorry. It's a 20-inch fan, the cheapo kind that goes
on sale at the hardware store for $12.97. Plastic housing, plastic blades. Nothing very special about it, but it works just great for exhausting the hot air and pulling in the cool air. Jo Ann Lacustral wrote: ) wrote: Just for the record -- I have ~1600 sq. ft. in two stories. At night, I put a box fan in one window, set on medium, blowing out. It matches the window size almost exactly, so I don't bother to block the sides. It is in a spare upstairs bedroom. By the time I walk back into my bedroom, also upstairs, I can already feel the difference in temp. I don't know how long it takes, since I just leave it on all night, but when I get up in the morning, the entire house is cool. Do you know how many CFM your box fan is? Thanks Laura |
fanpower needed to cool house overnight
A key point that seems to need underlining here is that there
are two elements. The air exhaust (via fan), and air *intake*. You can put in a humongous monster of a fan but if you don't open enough windows and interior doors throughout the house, the fan cannot get enough house air to exchange hence cannot cool the house. This is especially true in a newer, "buttoned up" heating/AC efficient house. Also, a good reason to install a whole house fan in an attic gable end (if you have one) is so that you have the choice of running the fan to cool just the attic (indirectly cooling the house, but with less perceptible noise and no draft), or to cool the whole house directly, by opening the attic door, selected windows, and interior doors (especially the bedroom doors). Catalogs (and websites) that specialize in farming equipment usually include a huge selection of ventilation fans. Una |
fanpower needed to cool house overnight
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