答复

1. 
   jklingel|Oct 16, 2011 01:41am|#1

   Is interior poly required in your area? Many folks are avoiding it, preferring to do a bang-up job of air sealing. If you will ever use an air conditioner, poly inside can be problematic. Are you familiar w/ alternative methods of "sealing" your walls and NOT using poly? BTW: I'm not familiar w/ that dew point calc method. Do you use the R value of the insulation, or the whole wall? With 2x6, those figures will be a bit different.

2. GBA Editor
   马丁·霍拉迪（Martin Holladay）|Oct 16, 2011 05:57am|#2

   麦克风，
   Here's an article that discusses these calculations:Calculating the Minimum Thickness of Rigid Foam Sheathing。我不确定您是在气候区域6还是7。如果您在气候区域6，则需要R-11.25的最低泡沫R值。如果您在气候区7中，则需要R-15的最低泡沫R值。无论哪种情况，我都认为2英寸的Polyiso都可以使用。（取决于您如何计算Polyiso的R值，2英寸的Poyiso可能只有R-13 - 但我认为这足够接近以使您摆脱困境。）

   我会跳过内部聚。我知道，在加拿大，一些建筑检查员坚持内部poly。如果您的建筑检查员是这样的，我将投资膜。因为它看起来像poly，所以它们通常会使用。

3. 
   user-1040028|Oct 16, 2011 09:44am|#3

   约翰：室内poly仍然是我所在地区的除法标准，我的“老学校”建筑商似乎认为是墙壁建筑中的一种。我最近与当地的首席建筑检查员聊天，他说他们看不到许多建筑商脱离它（因为他们最常处理最低码的建筑商），但是如果我能证明露点计算工作并有关于护套有能力充当蒸气屏障的能力的文档，他们会接受它。计算是基于2x6壁（具有R-22 Roxul腔绝缘）进行的。

   Martin: I believe I fall smack dab on the line of zone 6 and 7, however even though average winter temps are about 7F (-13C), it is not rare to get a cold snap of -20F. With this said, I'd rather play it safe. Typically for polyiso, I stick with the ajusted value of around R6.5 per inch. But even at R-15, I fail to see how this would work given the dew point calculation.

4. 
   jklingel|Oct 17, 2011 12:40am|#4

   Dick: Yes. What I meant was, for that particular calc, would one enter the R of the insulation, or the R of the whole wall, taken as an assembly. I would assume one would use the R of the wall, "averaged" for studs, but was not sure if that concept was sort of built into the equation.... or fudged in, generally ignoring the studs. Thanks for the thorough run-down of the method, though. Cheers. j

5. 
   DickRussell|Oct 17, 2011 10:11am|#5

   约翰，没有任何魔术来计算墙壁组件中的稳态温度曲线。当然，稳态永远不会真正存在，因为外部温度总是在变化，但是出于墙壁分析的目的，稳态的假设是有用的。鉴于此，通过壁的每一层的传热速率必须相同，或者将增加或降低热含量，从而在不同层中温度。由于该传热率（Q，每个单位面积）由以下方式给出：

   q = u*dt = dt/r

   where U is the heat transfer coefficient, and equal to 1/R, and dT is the temperature difference across that layer having that R value, then first the heat loss rate Q is calculated from the overall temperature difference and the overall R of the assembly. We are talking about a piece of area where the R is uniform perpendicular to the direction of heat flow, so that would be center of cavity; more on that below.

   由于总r是层的R值的总和，并且Q对于所有层都是恒定的，因此，对于一层，从以下方面发现温度差：

   Q = dTL/RL = dTO/RO

   其中“ l”和“ o”分别是指层和整体。解决DTL我们得到：

   DTL = DTO *（RL/RO）

   meaning, simply, that the fraction of the overall temperature difference that is across the layer is the ratio of layer R value to overall R value. Thus, to use Mike's numbers, with R22 inside and R10 more outside, the dT across the inner layer would be 22/32 = 0.6875 of the overall dT (63 degrees), or 43 degrees, giving 70-43 = 27 degrees at the boundary.

   通过木头，木材护套的温度剖面，最后可以以相同的方式计算外部泡沫，尽管该轮廓将与两个绝缘层不同。整个木螺柱上的温度下降将比整个绝缘层较低，因为该木材的R是该片片组件的总R的小部分。

   但是，该分析忽略了平行于内部和外部表面的传热。在框架的附近，由于框架和内部绝缘层的温度曲线不同，因此将有一些从绝缘到木材的热量。在大多数情况下，可以忽略这是为了计算护套层处于最差的情况，因为在内部绝缘层的外部比在更有导电木螺柱的外边缘的外部更冷。

6. 
   user-1040028|2011年10月18日下午1:27|#6

   马丁，

   I'm bringing this back up because I re-read the article that you linked to in your original response. This just leaves me a little bit more confused on what to do... if I base myself solely on the dewpoint calculation, R-15 on exterior walls isn't close to satisfying the test for a class III vapour retarder. However, if it's based on the table found in the the IRC supplement, it says I'm fine. Which one should I trust and follow?

7. GBA Editor
   马丁·霍拉迪（Martin Holladay）|Oct 18, 2011 01:55pm|#7

   麦克风，
   I'm confident that the values given in the table in my article will keep you out of trouble. Those values are based on dew-point calculations.

   我只能假设您在露点计算中犯了数学错误，但是我现在没有时间检查您的数学。