Comment 29166

By JonC (registered) | Posted February 27, 2009 at 09:30:24

To be fair on comparing the R values...because insulation typically has a R value 10 times higher than that of windows, you need to replace 10 times as much area for a similar percentage increase in efficiency (ie 1% to 2% compared to 10% to 20%)

This looks messy without subscripts and the Greek alphabet, but bear with me.

Heat Loss (Q) = Area (A) * Change in temperature (DT) / R

Comparing your before (1) and after (2) Q1 - Q2 = (A1DT1)/R1 - (A2DT2)/R2 Assuming constant Area and temperature change DQ = ADT/R1 - ADT/R2 Cross multiplying DQ = R2ADT/(R1R2) - R1ADT/(R1R2) DQ = (R2ADT - R1ADT)/(R1R2) DQ = (R2-R1)ADT/(R1R2)

So if the R values increase by a factor of ten, the numerator increase is by ten, but the denominator increases by a hundred.

Your point is still entirely accurate, seals & insulation are a much more economical way to increase the efficiency, but not by the massive amount implied.

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