What is the correct way to manually calculate air infiltration heat loss?
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I am designing my own house and I would like to figure out the Heat Loss & Heat Gain numbers to verify if the subcontrator’s calculations are right. I think I’ve got the formulas for those, but I don’t under stand 2 numbers in the equation, if this is the correct equation?
Air Infiltration Heat Loss = Room Volume X Design Temperature Difference X AIR CHANGES PER HOUR X .O18.
My questions are: How are the number of Air Changes per Hour arrived at? and What is the .018?
Thankyou,
Ron Lay
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Replies
Rob,
0.18 is the factor required to multiply the other items on that side of the equation (room volume in cubic feet, design temperature difference in F degrees, and air changes per hour) to come up with a heat loss number in Btu/h.
The most accurate way to determine air changes per hour is to use a blower door. If the house isn't built yet, you have to guess. If you want to build a leaky house, you make this number high. If you want to build a very tight house, you make this number very low.
"0.18 is the factor..."
Order of magnitude erroe- try 0.018 BTU per cubic foot per degree-F.
Martin,
Let me see if I understand what you are saying. .018 is a factor in the equation, but is it a Constant that is always used in the equation for Air Infiltration Heat Loss, or is it variable? If it is a constant, what is it actually, as in what does the number represent?
Since this will be a new construction I will have to guess at paragraph two. But what is the average Range of Air Changes per Hour between a leaky or tight house?
Bill,
Thanks -- your answer is better than mine!
Bill,
I will have to read your recommendations more thoroughly to understand the ACH50 to ACH(nat) method.
But I wonder about your equation for ACH(actual). Is it just adding all of the possible ventilation in the house? As, HRV is the abbreviation for Heat Recovery Ventilator? What does (HRV eff.) mean, and is ACH(HRV) just the air changes that I would need to calculate for the ventilator? Would the ACH for Direct Vent Fans need to be added in since they are only used occasionally?
Rob,
The heat capacity of air at sea level is, on average, 0.018 Btu per cu. ft. per degree F. So the infiltration heat loss calculation is:
[delta-T] x [ACH(actual)] x [volume] x 0.018,
where ACH(actual) is the actual air exchange rate, in cu. ft. per hour, for your house. How to get this number? Calculate/Estimate it:
ACH(actual) = ACH(nat) + (1 - HRV eff.) x ACH(HRV) + ACH(direct vent fans) + ...
Here, ACH(nat) is the "natural" leakage of air into the house. Getting from the blower door test ACH50 to ACH(nat) is an interesting issue. Some people refer to this web site:
http://www.energystar.gov/ia/home_improvement/home_sealing/ES_HS_Spec_v1_0b.pdf
For ACH50 > 1, then ACH(nat) dominates the result.
Search the web using the terms “ach50” and “achnat” to gather more opinions on the subject!
This discussion has been very useful to me. We just built a high performance home in Gunnison, and are collecting data to be able to quantify our successes and failures and where we can improve and where the industry can improve.
We are at 77oo ft. If 0.018 is the heat capacity of average air at sea level, is there any information for dry thin air at high elevations?
Thanks for any direction you can give.
scaeascaea - we're designing for 7300' (Roaring Fk Valley) and I found this guidance to correct ASHP for altitude:
https://s3.amazonaws.com/greenbuildingadvisor.s3.tauntoncloud.com/app/uploads/2016/09/08060512/Mitsubishi%20M%20and%20P%20series%20Altitude%20correction.pdf
(in case the link is deleted, Google: "Application Note 1021:
M&P Series High Altitude Applications")
This doesn't directly address your question, but it was insightful to temper expectations for ASHP performance at altitude, which might be one of your questions. There seem to be a number of thin-air/high-altitude issues that are not immediately apparent.
Ron,
Correct.
ACH(actual) is the air leaking into the house per hour from all sources, unintentional leakage as well as due to all intentional "leakage".
In the above post:
HRV = Heat Recover Ventilator
HRV eff. = the efficiency of the heat recovery ventilator
For a rough approximation (that’s what it is anyway, even with more “precise” divisors) use ACH(nat) = ACH50 / 20. House with more stories or a climate with bigger delta T? Then you get a larger stack effect and more natural leakage, so you might divide by 12 rather than 20. I’d take all this with a grain of salt, but the general principles hold. Makes me wonder how (exactly) the PHPP software handles this...
Note that in the original post I had (HRV eff.) x ACH(HRV) but I should have written (1 - HRV eff.) x ACH(HRV). If the HRV is 75% efficient then HRV eff. = 0.75, if the HRV is 100 % efficient, then there is no heat loss! Sorry for the confusion.
If there are direct-vent fans on only part of the time, then multiply by the fan duty cycle (the fraction of the time it's on) to get an average heat loss for the entire day for that fan: a 100 cfm direct-vent bath-fan on one hour a day in an 18,000 cu. ft. house? Then ACH(bath fan) = (100/24) x 60 / 18,000 = 0.0139ACH. This seems small, but in a house with ACH50 = 1 it’s not an insignificant part of the total infiltration, but in terms a Btu/h of heat loss it’s not much unless it’s quite cold outside.
I hope all this helps.
House volume x ACH (nat) x 24 x hdd x .018 for the annual infiltration heat loss in Btu's. I believe Denver uses .015 but I can't say as to the altitude conversion factor.
Bill,
After reviewing your posts last night, I want to Thank You for the excellent information and very thorough explanations. You have made me understand so much that I just could not find otherwise.
I do have a couple of questions though:
I'm inferring that a Blower Door test actually needs to be done to arrive at a meaningfull ACH50 number? Although, after typing that, maybe that should have been obvious to me? What information is used to reach the ACH50, I'm thinking it might be the CFM50, Duct CFM50 ? & House Volumn. What would the formula be?
And about your example of the ACH(bath fan), I don't understand why you would divide the 100cfm by 24?
It seems like I would just take cfm X 60 (minutes) / house volume?
Ron,
Since air leakage is determined by the actual construction, the only way to know it is to measure it after you’re done. No formulas here. You have the same issue when using a R-value for the walls, ceilings, and floors: voids in the blown in insulation? fiberglass bats put in poorly? more lumber in the walls than originally anticipated? slightly different product actually used? After good design, quality control is king. Since you are designing/building the house, can you specify the tightness at the beginning of the process and require blower door test(s) to ensure you get approximately what you wanted?
Perhaps you are over thinking this. Remember that the formula for going from ACH50 to ACH(nat) is approximate (+/- 20% ??) - don't be lured into the illusion of absolute accuracy by the equals sign. I assume that your subcontractor is trying to calculate the heating system design capacity. If you design for ACH50 = 2 and you end up with a tighter house then no big deal (?). Push the numbers around and see how they change the total heating system design requirement. Seems like your contractor is on top of it.
Regarding picking a reasonable (or meaningful) leakage rate for the house, here's a good article:
http://www.buildingscience.com/documents/insights/bsi-053-just-right-and-airtight/files/BSI-053_Just_Right_and_Airtight.pdf
You know also that very tight houses with ACH50 < 0.6 are possible from reading here at GBA, but all that air sealing will cost more. That's house design!
As for dividing the bath fan cfm by 24, in the example above, the fan is on 1 hour a day; that is, it's on 1/24th of the time and the average flow is 4.2 cfm.
Old post, but should be corrected. ACH(nat) depends greatly on if you are talking about annual average or short term (relating to peak load calculation).
> the formula for going from ACH50 to ACH(nat) is approximate (+/- 20% ??)
On a short term basis, ACH(nat) could vary from near zero to ACH50 - depending on wind and stack effect.
Jon, you are correct. ACH(nat) is actually very hard to infer from ACH50. The blower door result doesn't really care about the location of the holes in your envelope or the wind and ACH(nat) does. If all the holes are at the neutral pressure plane, then ACH(nat) is much reduced. If you have a four story house, its the middle of a very cold winter and all your holes are split between the attic and basement, then your stack effect is larger forcing ACH(nat) to larger values. Calculating ACH(nat) is not to be taken terribly seriously, but you can get a crude estimate from the above and add that to your heat load for the house. I recommend assuming ACH50 of two in the design phase and then aiming for an ACH50 of one or less. But you have to tailor that to the type of construction, if it's new, or just measure ACH50 if it's an existing house. Upon reflection I'd put the accuracy at +/- 50% rather than +/- 20% just to emphasize the inherent unknowns, but that's a non scientific seat of the pants judgement on my part. I've only used the above to estimate peak heating load, not ventilation. I'm using an ERV for ventilation.
I UNDERSTAND. THANKS again. GREAT INFORMATION !
ACPH = {60Q/vol. of room or area}.
Where:
ACPH = number of air changes per hour; higher values correspond to better ventilation
Q = Volumetric flow rate of air in cubic feet per minute (cfm), if using Imperial units, or cubic meters per minute if using SI
Vol = Space volume L × W × H, in cubic feet if using Imperial units, or cubic meters if using SI
Ventilation rates are often expressed as a volume rate per person (CFM per person, L/s per person).
The conversion between air changes per hour and ventilation rate per person is as follows:
Rp= {ACPH*D*h/60.
Where:
Rp = ventilation rate per person (CFM per person, L/s per person)
ACPH = Air changes per hour
D = Occupant density (square feet per occupant, square meters per occupant)
h = Ceiling height (ft, meters)
AND,
0.018 is altitude conversion factor at sea level.
q = - k a (T2 -T1) where k = conductivity of the material/s (BtU/ hr * ft2 * F) or
(1 / Resistance eg 1 / R-value)
a = area (ft2)
T2 -T1 = temp out/inside (F)
q = heat loss/gain (BTU/hr)
Air infiltration can be determined with blower door test, otherwise it's a best guess.
Air exchange is usually considered in an hvac system and is different from heat loss/gain.
I did a quick heat loss calculation for btu/hr at a 10 degree design temp for the addition we're planning, Martha's Vineyard, zone 5, using coolcalc (tight house) and also manually. Total square footage is about 1,100, volume 11,360 (cathedral ceilings)
Coolcalc results in a total load of 18,000Btu/hr. of which about 2K is infiltration. Manually, I come up with about 14,500 in heat losses for walls, ceilings, etc, …That’s fairly close to the 16,000 cookcalc comes up with. But when I try to calculate the additional heat loss due to infiltration assuming only 1ACH, I'm coming up with about 12K. That seams totally out of whack, even given perhaps allowing an additional 25% for the additional volume of cathedral ceilings. It also seems like a disproportionate amount relative to the other heat losses. What am I doing wrong?
Here's my formula:
11,360cuft x 60 degrees temp delta x 1ACH x .018 = 12,269Btu/hr
That's correct if 1 ACH is occurring. But it will rarely be windy enough to cause that much infiltration.
On the other hand, ignore all talk about the long term average ACH(nat). You want ACH(design day) for design day load calculations. Possibly around 10% of ACH(50).
https://higherlogicdownload.s3.amazonaws.com/ACCA/c6b38bda-2e04-4f93-bd51-7a80525ad936/UploadedImages/Infiltration%20per%20Blower%20Door%20Test%20Oct2016.pdf
.018 is the specific heat of air. it is a constant that explains how many btu's are required to heat a cubic foot of air by one degree. to raise the temperature of one pound of water by one degree, the specific heat of water is 1, meaning it takes one btu to raise the temperature of 1 lb water by one degree. so you can see it takes a great deal more energy to heat water than to heat air. .018 is a constant, so when doing your infiltration calcs, use it. it doesn't change.
bob
scaeascaea,
re: comment 19, hello to fellow traveler. I'm at 8,000 just west of you and trying to get a handle on the losses of my next build. I stink at math, but the charts I found on Engineering Toolbox would lead me to believe that their is very, very little difference in the specific heat of dry air even at 8,000ft. If their was say a 20% difference, the the specific heat of dry air might be .0145. The graphing I found is far too coarse to show any usable results.
However, while rummaging about on their site I was shocked to find out that standard air pressure at about 7700 ft is 11 psi. More amazing is what hikers doing the fourteeners encounter. At 14,000 ft it is 8.5 psi. No wonder I get winded at even 10,000 ft.
A value I found for the volumetric change of "standard" air at different altitudes seems to indicate it takes 1.28 times more volume at our elevation to gather enough air mass to equal a standard volume at sea level. If I have not latched onto the wrong info, it suggests we would necessarily be losing less heat per ACH simply because our air is less dense. Conversely, we are stuck with needing to move more air through our furnaces to carry the same BTUs as our coastal friends.
The idea of worrying about the specific heat of dry air may moot since even we don't have truly dry air here. More relevant may be trying to determine specific heat content of the moisture at our altitude and relative humidities. Not sure if the psychrometric charts cover that. Hopefully, someone better schooled in this will chime in. My suspicion is that for practical purposes on could use the heat values for moist air at sea level and then actually reduce them by .78 if we accept that our cubic feet are the same size, but holding less mass.
A confounding issue comes up when trying to use air volume charts with fans. We have to de-rate fan volume related data in some cases because the air they are moving is less dense. Not sure how to deal with that yet.
Maybe Dana will bail us both out.