# Archived R-Value Dimensional Analysis

1. Dec 8, 2013

### contrivance

1. The problem statement, all variables and given/known data

The ceiling of a single-family dwelling in a cold climate should have an R-value of 30. To give such insulation, how thick would a layer of (a) polyurethane foam and (b) silver have to be?
per my book (W/mK)
$$k_p = 0.024$$
$$k_s = 428$$
2. Relevant equations

This is Thermal Resistance to Conduction (R-Value) and the equation is:
$$R={\frac{L}{k}}$$
The 'R-Value' units per my book are
$${\frac {ft^2 F^{\circ} h}{Btu}}$$

3. The attempt at a solution

I assumed that I would just solve for L and find the thickness of each material, although I ran into the problem with mixed units (F, Btu, Ft). The solutions manual simply lays down a dimensional analysis, although I'm having trouble following it and getting the same resulting unit.

For Poly:

$$({\frac{0.024W}{mK}}) ({\frac{30ft^2 F^{\circ} h}{Btu}}) ({\frac{m}{3.281ft}})^2 ({\frac{5C^{\circ}}{9F^{\circ}}}) ({\frac{3600s}{h}}) ({\frac{Btu}{1055J}})$$

EDIT
Heh, as I was writing this, I figured it out. I wasn't thinking very carefully about which units cancel. So i'll finish anyways:

After cancelling most of the obvious stuff, we end up with (taking out numbers for analysis):

$${\frac{Wm^2sC^{\circ}}{mKJ}}$$
Which is where I was stuck, so there are two things here:

1. Rewrite C in K, and then they cancel
2. Rewrite J as W=J/s and then cancel.

We end up with m, which is the thickness (L) for the original question.

Now, my question is: How was I to assume that the R-Value '30' was in US units instead of SI units as my book has no mention of this value in SI units? Wouldn't it just be 1 h·ft²·°F/Btu = 0.176110 K·m²/W? (per wikipedia)

2. Sep 2, 2016

### Bystander

Just the way things are.