- 14

- 1

**1. Homework Statement**

*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)

[tex] k_p = 0.024[/tex]

[tex] k_s = 428[/tex]

**2. Homework Equations**

This is

*Thermal Resistance to Conduction (R-Value)*and the equation is:

[tex] R={\frac{L}{k}} [/tex]

The 'R-Value' units per my book are

[tex] {\frac {ft^2 F^{\circ} h}{Btu}}[/tex]

**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:

[tex]

({\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}})

[/tex]

**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):

[tex]

{\frac{Wm^2sC^{\circ}}{mKJ}}

[/tex]

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)