You are absolutely correct in that VC may have just thrown their hands up and refused to test with an R-Rated hearth assembly. It doesn't make any sense at all- but it is possible. In other words, why limit the options of your customers...or, worse yet, have a fire caused because you didn't include the heat shield standard and almost no existing floor would do without it. And, keep in mind that VC is the only company in the entire industry that has this "bare earth" thing. If nothing else, we may get to the bottom of that.
In the case of HearthStone, it makes sense to assume they actually did the testing....and the requirement of 20 layers of Durarock makes it clear in my mind that the floor temps hit MANY hundreds of degrees (on top of hearth)....in order to still hit 160 under all those layers.
I'm getting a lot of info, but the final piece of the puzzle would be an actual test report on one of these stoves.....exactly what the thermocouples under the stoves were reading. I'm sure anyone with decent heat transfer knowledge could work it out backwards....like this.....
Assume wood floor was at 150 degrees (passing temps)...now assume R 6.6 layer - what would the temp have to be on the top of the layer to have 150 on the bottom? This should be simply algebra on the R-Values, but I dropped out of that class! In simpler terms, what is the temp % reduction achieved by R 6.6 ?
Anyone?
In the case of HearthStone, it makes sense to assume they actually did the testing....and the requirement of 20 layers of Durarock makes it clear in my mind that the floor temps hit MANY hundreds of degrees (on top of hearth)....in order to still hit 160 under all those layers.
I'm getting a lot of info, but the final piece of the puzzle would be an actual test report on one of these stoves.....exactly what the thermocouples under the stoves were reading. I'm sure anyone with decent heat transfer knowledge could work it out backwards....like this.....
Assume wood floor was at 150 degrees (passing temps)...now assume R 6.6 layer - what would the temp have to be on the top of the layer to have 150 on the bottom? This should be simply algebra on the R-Values, but I dropped out of that class! In simpler terms, what is the temp % reduction achieved by R 6.6 ?
Anyone?