How Thick Should Insulation Be for Optimal R-Value in Cold Climates?

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SUMMARY

The optimal insulation thickness for achieving an R-value of 30 in cold climates requires specific calculations for different materials. For polyurethane foam, with a thermal conductivity (k) of 0.024 W/mK, and for silver, with a thermal conductivity of 428 W/mK, the thickness can be derived using the formula R = L/k. The discussion highlights the importance of unit conversion and dimensional analysis in solving for thickness, particularly when dealing with mixed units such as feet, Fahrenheit, and Btu.

PREREQUISITES
  • Understanding of thermal resistance and R-value calculations
  • Familiarity with unit conversion between imperial and metric systems
  • Knowledge of thermal conductivity (k) values for different materials
  • Basic algebraic manipulation skills for solving equations
NEXT STEPS
  • Research the thermal conductivity values of various insulation materials
  • Learn about dimensional analysis in physics and engineering contexts
  • Explore the implications of R-value in energy efficiency standards
  • Study the differences between imperial and SI unit systems in engineering
USEFUL FOR

Homeowners, architects, insulation contractors, and engineers involved in building design and energy efficiency optimization in cold climates.

contrivance
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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)
k_p = 0.024
k_s = 428

Homework 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}}


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:

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

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):

<br /> {\frac{Wm^2sC^{\circ}}{mKJ}}<br />
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)
 
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contrivance said:
How was I to assume that the R-Value '30' was in US units instead of SI units
Just the way things are.
 

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