# Thermal Resistance

1. Jan 23, 2008

### Math Jeans

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

The wall of a house consists of three layers: A wooden outer wall with an R factor of 1.0, a 3 inch layer of fiberglass with an R factor of 11, and a gypsium-board inner wall with an R factor of 0.33. (All R factors are in h*ft^2*F/BTU). When the temperature is 68 degrees Fahrenheit inside, and -4.0 outside, what is the rate of heat loss through an 8ft by 15ft section of this wall.

2. Relevant equations

3. The attempt at a solution

I was a little confused as to what they means by "R factor", and basically tried to add of the R factors as the regular resistance, and calculate the heat loss through:

rate of heat flow= (delta)T/R_eq

However (not suprisingly), I was wrong. I know I am missing something with this problem, but I cannot figure it out.

2. Jan 23, 2008

### blochwave

From what I've read, you just sum them, so Rwood+Rfiberglass+Rgypsum=Rtotal and that's what you use.

Make sure you take into account the surface area that was given, a quick check of your units in the equation you end up using should show an extra ft^2 that shouldn't be there

3. Jan 23, 2008

### Math Jeans

So are you saying that the R Factor is the resistance per square foot?

4. Jan 23, 2008

### blochwave

Well the R-factor's units wouldn't make that make sense.

If you look at your equation, which is correct but not quite finished, you have said

rate of heat flow = deltaT/R

rate of heat flow is power. R is in units of temp*area/power, deltaT is temp

so you've said rate of heat flow = power/area, which isn't true

What that equation gives you is power per unit area. Now you need to multiply by the actual area to find the net power

5. Jan 23, 2008

### Math Jeans

Ok. I'll try again. So I need to take the original answer I got (basically taking the R-factors and treating them like the actual resistance of each part of the wall), and multiplying that answer by the area?