Re: Heat Loss calculation for Insulation thickness?

On 31 Jan 2006 05:41:44 -0800, "gtslabs" <gts@xxxxxx> wrote:

>Thank you. Can you explain how you got 3.5 inches from R = 5/0.28=18?

U is the metric kindly unit for thermal conductance in
watts per degreeC of temp drop across a square meter of surface

R is the American-customary-friendly unit for thermal resistance
in degreesF of temp drop across a square foot of surface
per BTU per hr.

So 1 R is 1 X 5/9 (degC/degF) x 12/39.37 X 12/39.37 (sq m/ sq ft)

-----------------------------------------------------------------
0.293 (watts/BTU/hr)

= 1/5.68 ( U/R)

But the one is the reciprocal of t'other, and I used 5 as a
multiplier - shoulda been 5.68
i.e I started with U = 0.28
took the reciprocal 1/U = 3.57
and multipied 3.57 times 5 = R18

But more accurately,
R = 1/U X 5.68
So actually R = 3.57 X 5.68 = 20 is closer. but I approximated in
various places without worrying too much...

You specified insulation at R = 5 per inch. That's a thickness of
four inches for R = 20

Brian Whatcott Altus OK
.