All that follows has admitted simplification, for easy hand-calculation. Rigor can be added, to a point of paralysis. Anyone, please, treat this as a wiki. I will edit in attributed contributions of better analysis. And know, wherever my simplification comes with numbers specific to my climate and heating appliance, your numbers can readily be imagined, and substituted. Costs Associated With Heat Loss By Convection and Conduction: In the absence of heat loss by draft airflow, the relationship between total (system) R-value of a wall, ceiling, or floor, and the dollar cost of the heat lost through that surface, is: Cost = Fuel Cost * HDD * 24 / Eff * (Area/R) Fuel Cost is $/Btu, and HDD is Heating Degree Days, Eff = Furnace Fuel Efficiency Fuel Cost (natural gas) = $.00002/BTU ($2 per added therm. This is a reasonably high number consistent with enforcement of conservation. I see no reason to use a current number, such as my most recent bill, at $1.27 per added therm. With a trend to replace oil or electric heat, apply math only for natural gas heat, at this mildly inflated number. I wish we could call it, and pay it, at $4 per therm with a carbon tax, promoting conservation. Assume an average modern natural gas furnace with Eff = 0.88. Potentially as adjustments, consider better numbers of efficiency, about 0.8 for a non-condensing gas forced-air furnace, up to 0.95 for a condensing gas furnace. HDD = 4400, for Portland Oregon. Find HDD values for other locations in the United States, at degreedays.net, linked from discussion at Wikipedia. Area = Area of heat transfer, measured in sq ft R = System R-value of airtight boundary and insulation = 3 + Ri Ri = Insulation or component inherent R-value The number 3 of the formula R = 3 + Ri is a reasonable allowance for heat transfer resistance at surfaces of contained insulation and framing, a summation of values L/k for conduction through solid materials, and heat transfer coefficients, h, for convection at air barriers. With the "reasonable" number, 3, costs math can be brought to completion as in dollar savings with fixing or adding insulation. Better numbers in lieu of 3, will range from about 2 for a flat roof, to more than 4 for well-built walls lacking insulation. A number smaller than 3 will give prohibited over-estimate of savings due to added insulation, in a sturdily-built home. Anyone may later adjust the math that has been carried to completion with known simplification. There will be reasons for questioning and adjusting insulation math, far more more significant than small surface R-value uncertainties, where an insulated face of a structure acts against other than outdoor conditions in a crawl space or attic. Find satisfaction with a supportive 1983 summary by Randy L. Martin at Colorado Energy. Randy demonstrates one calculation of the effective R-value of a complex wall, applying a 0.25 framing factor for 2x4 @16" oc framing. Assembly R-value = 1 / (Assembly U-value) = 1 / (U-studs x % + U-cavity x %) http://www.coloradoenergy.org/procorner/stuff/r-values.htm Find resistance values collected by prestigious University of Washington in R-Value Table, Common Building Materials . Find resistance values collected by Gary Reysa, at Build-It-Solar: http://www.builditsolar.com/References/MILRvalues/MILRvalues.htm Numbers add up as increments of L/k (thickness divided by thermal conductivity), and h (convective heat transfer coefficient at each air barrier). Air barriers are very significant, with each contributing a summed h value of about 1. It is easy for numbers to add up to three or more with multiple air barriers, and hard in careless construction with pass-through air flow. My 1955 home has high quality Celotex ship-lap sheathing 3/4" thick, with R-value of about 2 , as an air-tight barrier. My old-growth cedar siding complements the significant structural strength of the Celotex and is one more air barrier. We must not overlook value of 1950s construction materials and methods vs. awful OSB leaky sheathing and floppy and extremely leaky fiber-cement siding in justifying a monster new home, that replaces a foolishly-demolished 1950s home. I accuse there is criminal purpose in analysis by US DOE Home Energy Score, that credits the older home (not so old) and its replacement, with only Convection and the conduction resistances of air-tight cladding add to a total ranging from more than 2, to about 3, and a highest number gives wished 1/(R where f ## Annual Cost of Heat = (.00002 * 4400 * 24 / .88) * (Area/R) = 2.4* (Area/R), with cost in dollars, and sq ft area.Multiply by: (Your HDD)/4400 and by 0.88/ (Your Furnace efficiency, perhaps 0.95 with a condensing gas furnace). Be mindful that incentive to insulate is first in maximizing attic insulation in any climate, for summertime comfort without cost and complications of air conditioning. Be mindful with weatherization decisions, that easy energy for home comfort is scarce and will always rise in cost, faster than general inflation. This properly inspires that we do all we can for energy independence, now not later. Energy independence is wealth that must be offered equally to every person. What if I have an electric heat pump?Dollar consequences are complex. You might spend less per year for comfort with given insulation and tightness in your home. Maybe not. There is great virtue in using a single dollar conversion regardless of heating/ cooling method, in making or motivating weatherization decisions. If you might replace or supplement a gas furnace with a heat pump someday, that should not affect anything you do in weatherization. Always place the maximum amount of tightly-placed insulation that you can, and do not fail to first do thorough sealing of the attic floor and to prevent wind-washing of insulation in your walls. Let us all apply the same motivational math.Moderation in an Attic or Crawl Space:The above is for heating conditions only, and does not account for moderating by sky and ground effects. The temperature above an attic floor has daytime rise above ambient, and may fall below ambient at night. For an attic, I stick with the formula. For a moderately-ventilated crawl space, ground heat keeps temperature above ambient. Should I assume a 25% reduction of HDD. Annual Cost of Heat = 1.8 * (Area/R), for Portland, Oregon? I invite critique of all postings here; this on crawl spaces, especially, is just a blind step in the right direction, and is not heart-felt. I will continue to apply the same dollar conversion for R and area, for all home surfaces.Heated Surfaces (Hydronic, Radiant Heat): Now, what if the surface losing heat to ambient is HEATED, is the source of building heat? That is the case with ceiling hydronic or electric radiant heat. I think the acknowledgement may be in increase of the HDD number. How about by 20%? With this adjustment, with surface radiant heat, Annual Cost of Heat, Portland Oregon numbers, is 2.88 * (Area/R). With justice, we disdain electric resistance heating. There is 100% efficiency in conversion, but dollars per therm are high. $2.93 per therm at ten cents per KWH. At a common 15 cents per KWH, motivate extreme insulation over electric-resistance hydronic heat, with a number much larger than the 2.88 factor above. Extreme insulation is in fact all that will fit, and can not compensate for less-efficient or wasteful heating.
True cost of energy considers federal expenditures for military adventure to protect fuel imports, and criminal ruin of land and water never recoverable, in domestic fuel fracking. Consider responsibility for global warming including effects of exponential release of accelerant methane. How about making weatherization decisions based on energy cost of Carbon Footprint: The cost of replacement heat as carbon footprint, is 1.2 * (Area/R), in therms of natural gas. Where a therm of natural gas is 100,000 BTU, and is approximately the energy equivalent of burning 100 cubic feet Ccf, of natural gas. A therm corresponds to emission of about 12 pounds of CO2. As pounds of CO2 emitted for this locale and furnace efficiency, the round number is 14.4 * (Area/R). Usage of Electricity: The work product of Bonneville Power Administration/ Regional Technical Forum, as it presumes to be a director of weatherization, is an Excel spreadsheet, last noted March 2011, as RESWXSF FY10v2.3, stating electricity savings for insulation measures, and more. We need a lot less hydroelectric or nuclear power in heating homes. Still, the numbers for Utility management purposes, are there for comparison with other predictions. Savings are expressed in KWH/sq ft/ year. One therm is 29.3 KWH. With different matters of efficiency, it is messy to relate hydro power and gas-furnace heat costs. I apply KWH numbers converted to therms, and to dollars at $2 per therm. The review is published at this site, here. ## Insulation With Parallel Paths:Use inverse summation to calculate effective Ri with parallel paths, over areas of unequal resistance. For lumber of framing and for applied insulation, find unbiased statement of materials R-value at Wikipedia . For 2x joists 16” on-center: 1/Reff = (1.5/16)/ Rjoists + (14.5/16)/ Rinsulation where Rjoists is taken as 0.94 per inch (douglas fir), 2.4 for 2x3, 3.3 for 2x4, 5.2 for 2x6, 6.8 for 2x8, 8.7 for 2x10. Above joists, add crossing insulation Ri for all area. For ordinary fiberglass insulation batts, apply "thickness" as claimed R-value, divided by R3.5 per inch. For high density fiberglass insulation batts, apply "thickness" as claimed R-value, divided by R4.2 per inch. Find then that R19 batts have thickness 5.4" and are not sufficient to fill 2x6 framing. R15 HD batts have thickness 3.6" and do fill 2x4 framing, just barely. Prefer R19 batts in 2x4 walls. Yes, use R19 in 2x4 walls, but not in 2x6 walls. R19 compressed to 3.5" is then R15. The applied cost of R19 in 2x4 walls is about 25% below that of R15. Likewise prefer R25 batts in a 2x6 wall, compressed to R21, for less applied cost than R21. Here do math in an attic floor, where batts compress in tight packing and in bearing covering batts. Compression to "thickness" is without penalty vs. free thickness. For all batts on an attic floor, apply bag-claimed R-value. Example 1: 2x6 attic floor, joists 16” oc, with R19 fit between joists (partially filling to depth perhaps 4 1/2"). 1/Reff = 1.5/16/5.2 + 14.5/16/19. Reff = 15.2 Example 2: 2x4 attic floor, joists 16” oc, with R15 fit between joists, and R25 crossing batts. 1/Reff = 1.5/16/3.3 + 14.5/16/15. Reff = 11.3. Ri = 11.3 + 25 = 36.3. This is fairly close to the intended R40. Example 3: 2x8 attic floor joists at 16” oc, with R38 batts fit between joists, no covering insulation on joists. Batt insulation would slightly over-fill the 7.25" joists. Such framing is rare in older homes, and in newer homes would likely be found with loose-fill covering joists. 1/Reff = 1.5/16/6.8 + 14.5/16/38. Reff = Ri = 26.6 Here the joist thermal shorting is painful. To have >R38 in this example, starting with R26.6, I would add R19 crossing over joists, for total R45. I would not add thinner batts, as R19 is especially affordable, and comes in convenient sizes. I mainly use Johns Manville AU333 here, nice postage stamps for the puzzle, 24" by 48", and AU395, 16" by 96", often cut in half. It is important to not fight the roof-joist spacing. It is important to not have long strips to pick up and stow, for service access. With 2x8 framing, learn to accept total insulation beyond an R38 target, when using batt insulation. Example 3a: 2x6 attic floor joists at 16” oc, with R38 batts fit between joists, no covering insulation on joists. Batt insulation would slightly over-fill the 7.25" joists. Such framing is rare in older homes, and in newer homes would likely be found with loose-fill covering joists. 1/Reff = 1.5/16/5.2 + 14.5/16/38. Reff = Ri = 23.9 Example 3b: 2x10 attic floor joists at 24” oc, with R38 high-density batts fit between joists, no covering insulation on joists. Batt insulation just fills the 9.25" joists. This is uncommon framing, but is another case in the topic of uncovered framing thermal shorts. 1/Reff = 1.5/24/8.7 + 22.5/24/38. Reff = Ri = 31.4 The wider joist spacing, and greater joist thickness, help. Getting to true R38 requires some crossing insulation. Example 3c: 2x10 attic floor joists at 24” oc. Batt insulation just fills the 9.25" joists. Add 2x4 decking 24" oc, with contained R19 unfaced batts. Call the base layer R38 between joists. Call the top layer R15 with over-compression, though covering plywood adds compensating points. Base layer: 1/Reff = 1.5/24/8.7 + 22.5/24/38. Reff = Ri = 31.4. Top layer: 1/Reff = 1.5/24/3.3 + 22.5/24/15. Reff = Ri = 7.3. Total insulation R38.7. Add R3 as system R value. And do math for wall insulation: Example 4: 2x4 wall joists at 16” oc, with R15 batts fit between joists. R15 is any insulation completely filling the wall volume, stilling air circulation in bypass of insulation. 1/Reff = 1.5/16/3.3 + 14.5/16/15. Reff = 11.3 Example 4a: 2x4 wall joists at 16” oc, with batts not completely filling the wall volume. The insulation may have no value. The least penalty may be with R13 nearly filling the air space assuming blocked air circulation within the wall. 1/Reff = 1.5/16/3.3 + 14.5/16/13. Reff = 10.2. This drop by 1.1 in effective R-value is with very large opportunity cost. The drop could be avoided with negligible effort. Example 5: 2x4 knee wall joists at 16” oc, with R15 batts fit between joists, with crossing 2x3 supports and a second layer of R15 outboard (on attic side). 1/Reff, inner = 1.5/16/3.3 + 14.5/16/15. Reff, inner = 11.3 1/Reff, outer = 1.5/16/2.4 + 14.5/16/15. Reff, outer = 10.0 Ri for the doubled R15, is 21.3, with simplified math. Think how this compares to a result if placement were not crossing, joists R = 2.4 + 3.3 + 5.7. R30 insulation. 1/Reff = 1.5/16/5.7 + 14.5/16/30. Reff = 21.4 I think the crossing should give a large improvement, and know the simplified math is not rigorous. Annual Heating Cost Calculations for These Examples: My definition, System R-value of airtight boundary and insulation = 3 + Ri, is intended to not over-estimate found heating cost, and thus be high in prediction of savings. The "three" is system convection and R-value of the ceiling, elsewhere taken as as little as 1.2. With "three", I, too, avoid the impossible area-averaging of resistance of floor joists, with the "zero" between. The choice matters especially for found conditions with no insulation. Here consider starting conditions with no insulation. Annual heating cost for 1000 sf is 2.4*1000/3 = $800. The heating cost for Example 1, is 2.4*1000/18.2 = $132. Savings vs. no insulation are $668. The heating cost for Example 2, is 2.4*1000/39.3 = $61. Savings vs. no insulation are $739. The heating cost for Example 3, is 2.4*1000/29.6 = $81. Savings vs. no insulation are $719. The heating cost for Example 4, is 2.4*1000/14.3 = $168. Savings vs. no insulation are $632. The heating cost for Example 4a, is 2.4*1000/13.2 = $182. Savings vs. no insulation are $618. The heating cost for Example 5, is 2.4*1000/24.4 = $98. Savings vs. no insulation are $702.
For Example 1, I would place about 910 sf of R19. I would use only unfaced insulation, and would ask about $560. Simple payback is in less than one year, with simple return on investment of more than 100%. At January, 2012, with addition of Example 6 above, and a link to critique of the math of Bonneville Power Administration, study "Line 12" information that corresponds to Example 1. The critique, and numbers here, are consistent. Payback in less than a year. For Example 2, I would place about 910 sf of R15, and 1000 sf of R25. I would use only unfaced insulation, and would ask about $1760. Simple payback is in less than three years, with simple return on investment of about 42%. For Example 3, I might place two layers of R19 within framing, 1820 sf. I would use only unfaced insulation, and would ask about $1120. Simple payback is in less than two years, with simple return on investment of about 64%. For Example 4, I would place one layer of R15 within framing, 910 sf. I would use only unfaced insulation, and would ask about $910. Simple payback is in less than two years, with simple return on investment of about 70%. For Example 5, For insulation only, I would ask about $1820. Simple payback is in less than three years, with simple return on investment of about 39%.
Consider a 1000 sf attic, with joists covered. Add R25 crossing batts. The found Annual heating cost is 2.4*1000/21 = $114/ yr. The improved heating cost at R43 is 2.4*1000/46 = $52/ yr. Savings $62/ yr. I would ask about $1000. Simple payback is in about sixteen years, with simple return on investment of about 6%. This is near the boundary at which rule-makers authorize incentives. Example 6: 2x8 crawl space joists at 16” oc, with R30 batts fit between joists. 1/Reff = 1.5/16/6.8 + 14.5/16/30. Reff = 22.7 If batts don't fill joists, the joist thermal shorting is greater, and batts claimed 10" thick at R30, might be of lesser value. I find framing revealed at least 1" on average, where batts are pushed tightly to the floor sheathing. At maximum, think of containment in joists R5.8. 1/Reff = 1.5/16/5.8 + 14.5/16/30. Reff = 21.6 I will apply Reff = 21 in a critique of the math of Bonneville Power Administration, at "line 10." This form of presentation is more interesting, and is consistent with payback study numbers above, for Examples 1 through 5. Example 7: One baffled can light in every 100 sf of an "R38" attic floor. Say, cellulose twelve to fifteen inches depth. Real R38 insulation except for holes. Baffle 12" dia., area 0.8 sf uninsulated or worse, with driven leakage. 100/Reff = 99.2/38 + 0.8/3. Reff = 34.8. As if insulation were less deep by an inch or two. Why would one "waste" all that insulation, to avoid replacing the ugly can light? The baffle, with a cover in a deep insulation bed, must cost as much as the light replacement, say $60 for a T91 LED plate light in a junction box? Correct actions are achieved by prohibiting rebates if wounds and dangers in a house are not fixed. With math for my home, the uninsulated area at each can light loses $2.4*0.8/3 = $0.64 per year. Bulb local heat is all lost in winter. Countable pennies are lost in heat-driven excess of house fresh air exchange through any can gaps. As Applied for Attic Ladders and Portal Doors (Attic Hatches): At December 2015, please know that wooden ladders I work with, Fakro and Calvert, are rapidly moving toward offering of very high insulation value, at higher cost that may hurt sales. The Fakro LWT has a 3 1/8" door thickness, called R12.5, and costs $495. The LWP with 1.25" door thickness costs $340. This page has had math now in italics, as the only offered form.
Now repeat this math of parallel paths in a panel door in a manner that gives higher door R-value, where paths aren't fully separate, but have heat flow division also affected by common resistances in convection at surfaces and conduction through door skins. Reality is somewhere in-between, and is probably nearer the more-favorable result below. The newer and favored analysis is again not in italics. 1/(R With 1" EPS, this is evaluated as: 1/(R However you do the math, save a bit more than $2 per year at system R12, vs. system R5. This $2 difference is with very large increase of ladder cost perhaps never repaid, but with strength and buyer satisfaction. The more-complicated ladder had better be more durable. A compromise "high-end" ladder might be R7.5, a bit thicker than 2", and with not more than a dollar-per-year to gain or lose, with other options. Please see feeding of r5portals better-code-proposal , from math here. Ladders at more than R12 are not on offer, with thickness impractical. Achievement of actual R12 in a 3" door should present structural compromise. Manufacturer-claimed R-values correspond to System R, 1/U, not panel Reff. Note that R values can not be measured or computed with great accuracy. Wikipedia lists a higher R-value for foil-faced polyisocyanurate rigid panels, pentane expanded, aged five to ten years. Manufacturers stick with expanded polystyrene. Expect that R12 claims for a ladder might not be defensible. ## Heat Cost Due to Drafts:Methods in Energy Center of Wisconsin Report 208-1, (at Equation (5), on Page 8), and in the Energy Conservatory Blower Door Operation Manual (at pdf page 45, labeled page 35), are applied to assess the virtue and cost-effectiveness of making a ladder installation air tight (fully gasketed and with sealing of the rough opening). The Equation 5 relationship is: Qh = 1.06 · Ah · ( Ph) where: Qh = rate of air flow through the hole (cfm) Ah = hole area (square inches) Ph = pressure drop across the hole (Pa) With Ph = 50 Pa, Qh = CFM50 = 7.5 * Ah. From Energy Conservatory: Annual 26 x HDD x Fuel Price x CFM50 Heating = ------------------------------------ x 0.6 Cost N x Seasonal Efficiency - HDD is the annual base 65 F heating degree days for the building location (4400). - The Fuel Price is the cost of fuel in dollars per Btu ($.00002). - N is the Energy Climate Factor from the Climate Information Screen (adjusted for wind shielding and building height). (21 for Portland, Oregon). - Seasonal Efficiency is the AFUE rating of the heating system, taken as 0.88 for a common value. The formula is further explained by The Energy Conservatory, in an old edition of its Blower Door Manual: The "26" factor is derived from multiplying the heat capacity of air (0.018 BTU/cubic ft/Degree Fahrenheit) by 60 min/hr and 24 hr/day. The "0.6" value is a correction factor which is used to adjust the theoretically calculated energy savings for the following interactions: - The theoretical model assumes that homes need heat at an outside air temperature of 65 degrees. In fact, many homes do not need heat until the outside air temperature is below 65 degrees F.
- Some air that leaks into the house may not get heated to the inside temperature before it leaks out again.
- Air leakage through small cracks reduces conduction losses by tempering surrounding surfaces. This process is referred to as the dynamic insulation effect.
The 0.6 multiplier has been described in error at this page, citing the same factors as for the "26" number, attributed to The Energy Conservatory. At March, 2013, the discussion is deleted. Annual Heating Cost = 0.074 * CFM50 (There is virtue in reducing math to a simple formula, without further statement and defense of details, independent of the economy in better heating systems.) CFM50 is the air flow artificially outward through all leakage paths, under blower door test conditions, with a pressure differential of 50 Pascal (0.0073 psi, 0.2 inches water column). CFM50 is computed from leakage path area, under these conditions, as: CFM50 = 7.5 * (Path Area, sq in), if the path is a common hole. Flow is higher where area is distributed in rectangular gaps. Annual Heating Cost = $.555 * (Path Area, sq in) The cost for a 1/4" annular gap about a 4" ceiling junction box is $0.555 * 3.14 /4 * (4.5² - 4²) = $1.85 per year. The area is 3.3 square inches. The cost for a 1/4" annular gap about a 7" can light is $0.555 * 3.14 /4 * (7.5² - 7²) = $3.16 per year. The area is 5.7 square inches.
I should wish to have 0.35 to 0.5 actual air changes in my home every hour, 0.35 to 0.5 ACHnat. This is accepted as necessary for good health. Corresponding blower door numbers are dependent on the air volume in my home. Say in round numbers for my modest 1955 ranch home, 1000 sq ft living space, eight foot ceilings, that volume is 8000 cubic feet. 0.35 changes per hour converts to CFM at natural conditions, as 0.35 Air Changes/hr*hr/60 min*8000 ft Each ACH50 for my home is 8000/60, 133 CFM50, and corresponds to makeup heat cost of $0.074*133 = $10 per year. The size of a hole that could carry this flow is 9.86/0.555 = 17.8 sq in, a circle 4.8" diameter. Every ACH50 I may covet, could come through its 5" hole. 5" holes! No big deal. Driven by natural pressure differential without cost of electric power and with measurable increment of good health. Think about this please. Math is for fuel cost bumped up to truer cost at $2 per therm. It's a 1000 sf ft home in a moderate climate formerly 4400 65° HDD and already down to 4000 HDD with global warming. If in silliness I have tightened my home to 3 ACH50, from 0.5 ACHnat, I am paying less for cost of heating the air from ambient to 65°F, to the tune of $10 * (10-3) = $70. Now I need to pay for the driven fresh air. Say it is 25 watts, continuous, six months per year. I will be erratic in regulating the system while mindful of opening windows more in mild weather. 25 watts *24 hr * 365/2 * KW/1000 watts *$0.11/KWH = $12 per year. There are maintenance costs. I am not saving enough, for the bother. I admit though, I have an unmentionable reason to be on this path. I have a radon problem largely solved by letting ample fresh air in at bedside windows at night, and will be happy to relax that some if I achieve a plan to exhaust stale air through-roof, via a sealed, conditioned crawl space. With this solution, I don't visibly and audibly display the radon problem. I think I have answers and not many problems. Silly people with blower doors know none of this. They We should not allow careless leakage, as at can lights in the building envelope. In this, consider 2015 International Energy Conservation Code (Residential), Section R402.4.5:
How much can light leakage area is allowed? Scale down first to 50 Pa differential, as 2.0 cfm * (50/75) 0.2 = 0.7854 *(7 I think that the blower door people know
A gasketed ladder might replace an old ladder 22" by 54" with an average gap tabulated here. Path Area = Perimeter*Average Gap. Perimeter = 152". Assume distributed-gap air flow is increased by times 1.37, vs. a simple hole. CFM50 = 1.37*7.5*152*Average Gap. Divide CFM50 by twenty, to note leakage at natural conditions, Q Average Gap Path Area, Sq. In. CFM50 Q 0.01" 1.52 15.6 1 $1 $6.60 0.06" 9.12 94 5 $7 $6.60 0.12" 18.2 187 9 $14 $6.60 0.25" 38 390 20 $29 $6.60 This considers only the gap between the door and the ladder frame, probably less than 0.06" on average, in any not-decrepit ladder. In addition, consider gaps between the ladder frame and its rough opening. A 1/8" average gap is common here, where molding against a rough ceiling may only hide the problem from view. Gaps about the ladder frame must be plastered or caulked. I use no molding,
Where Energy Star standards give an allowed maximum leakage for an exterior swinging door, Q <= 0.5 CFM/ft CFM50 = Q50 = (50/75) To achieve this, gasketing must be indefinitely resilient, and have large compression, say at least 1/8". Say more-reasonable leakage past gasketing is the allowed, times ten. What does that cost? CFM Know though that a healthy home has 5 ACH50 fresh air exchange, with nearly unavoidable cost to heat incoming air. For my 1000 sf home with 8-ft ceilings, this is Q
With a leaky, uninsulated attic ladder, the cost due to leakage may exceed the cost of convection and conduction through the door. Often, leakage paths around a ladder frame are larger than those between the door face and the frame. Yet, leakage at a most-awful attic ladder will not rival the 80 cfm Q Leakage beyond the Energy Star standard must be prohibited, and that is entirely a matter of reforming installation methods. It is achievable at no cost, that there shall be
Imagine the cumulative effect of blocking a few "holes": Item Path Area, Sq. In. CFM50 Annual Heating Cost, Leakage** Closet door not gasketed* 9 68 $5 Closet open to attic, if other doors open* 40 300 $22 Hole in outside wall, draped by panel* 40 300 $22 * - These are chosen for discussion from actual experiences. Numbers have consistent math, but have quite uncertain bases. The gasket consequence is not worth debating. We would like to fix it. The door impact is best inferred from cycling the door, while running a blower door test. How would one guess at the path area? We can estimate the area of an obstructed hole, or infer it from before/ after blower door tests. We should offer modest rewards for air sealing, with action driven more by the rewards of comfort and freedom to not isolate rooms. ** - I am distressed by these small numbers. I have wanted to believe there are larger penalties for such lack of care, and that rebates offered in Oregon for air sealing achievement, at $1 per cfm, are justified. Imagine a modest single-story house of 1000 sq ft, 8000 cubic ft volume. If it measures1333 CFM50 (at minus 50 pascal test conditions), that is 1333 *60/ 8000 = 10.0 ACH50 (ten changes per hour of the house air volume). That house has more infiltration than is needed for a healthy home, say 7 ACH50. The ACH50 reduction from 10, to 7 is a blower door test reduction of 400 CFM50. The math above says the involved cost saving is $30 per year, and a $1 per cfm rebate is $400. Do these numbers make sense? If my math is in error, I ask: someone please contact me. I have cared deeply to understand this spectrum of conservation opportunity. Sealing measurable by a blower door tester sometimes includes furnace air duct leakage to an attic or crawl space. Leakage pressurized by a furnace blower has far greater significance, than other boundary leakage. I have experienced more than 200 CFM50 benefit from reattaching separated ducts, where the detachment had been hidden by crummy insulation wrap. Such repairs require only diligence, the conditions fixed are a result of negligence, and repairs cost almost nothing. I am sure that savings far exceed $15 per year, yet don't see justification in much of a rebate incentive. I don't think measured air sealing achievement should ever be a basis for incentive rebates. When air sealing achievement is rewarded, there is incentive to game the readings. Test results do wander. A large threshold for sealing-achievement rebates, 200 cfm, is a measure of test uncertainty. Conflict of interest problems are real, and should be avoided by a shift in concepts for "home performance testing." Plenty of this testing should be going on, but always with impartiality, and with inclusion of big-picture concepts including moisture control. Fewer providers of repair services should be in the business. What happens when they are not the best choice for a particular repair? The home owner becomes unfairly locked into employing the tester.
The amplified heat loss under a big opening in the attic floor, to walls below, is typically larger than the loss through involved attic floor area, by an order of magnitude. I have blog posts on this. It is clear that big openings, like a pit to the ceiling of a stairwell, are - - pits. They are not "bypasses." They are almost entirely a convection opportunity. It is less clear what is going on at a far end of opening size to space below, in common wall header gaps, that might average 1/16" combined on both sides of a wall. In my house I expect that not more than 300 CFM50 of sealing to be achieved is in wall header gaps. The small expectation is consistent with a disrespect for such sealing in most weatherization, where change of blower door readings is so slight, and 400 CFM50 reduction targets are impossibly elusive. I can only allege that respect Interior Wall Header Gaps As Attic Floor Pits: In my own home, nearing completion of energy improvements in 2015, I find that the length of all wall headers is 107 ft, vs heated area now 1080 sf. The area of the gaps is 107*12*.062 = 80 sq in. That's a lot of area if it were fully engaged in infiltration. CFM50 = 7.5 * Area = 600, high by at least times two. The gaps are active and visible for upward flow by dirt deposition in insulation, but I believe there is a churning, flow both up and down, where cold air from the attic must fall, to feed the convection engine, tending to equalize in-wall and attic temperature. Gaps can be much greater than 1/16" both sides, raising the need to direct action for other than change of infiltration. I propose the math on-average , as part of the invention of "bare area equivalents:" The effect of sealing wall headers may also be estimated in equivalent bare area fully insulated. Say the area both sides of interior walls is the same as attic floor area, and stilling leakage/ floor pit bathing, is 10% of that which would occur if interior walls were directly open to the attic. In this, use further simplification, that interior wall area, two sides, is same as attic floor area. This is very unscientific, and none of the assumptions deserve sharpening for specific home details. Bare area equivalent as 10% of floor area gives an appropriate degree of importance. For my home with 107 ft of headers and eight-foot ceilings, the wall area involved with header throttled pits is 1700 sf. 170 sf not-insulated would be added to attic floor heat loss, if wall headers were not sealed. The cost is $2.4*170/3 = $136 per year, compared to infiltration cost for 300 cfm, of $0.074*300 = $22. I prefer the bigger number, despising an area 13 ft square left bare. Does it really seem less significant than that? The smaller number wrongly inspires inaction. |

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