R-value calculator

[Jillian Burgess]

I’m having trouble understanding the R-value calculator.  I believe, to get the universally accepted R-value, I will need to model the wall at the FTC established values (50F – 100F).  I tried this out on a simple model of 1” mineral wool insulation, which given the conductivity at 0.24 BTU.in/hr.ft2.F, the published R-value is 4.2 (http://www.thermafiber.com/Portals/0/PDF/RainBarrier%20Data%20Sheet.pdf).  However, the model returns a value of 5.25.  Can you help me understand what other variables the model is taking in, that the published data would not

[Christian Kohler]

Jillian,

"Universally accepted R-value" in this case would refer to the R-value used in the insulation industry. That R-value does not include film coefficients (also called heat transfer coefficients). It is purely the conductivity through the material. When THERM gives you a U-factor or R-value, it is always for a complete assembly, which includes interior and exterior film coefficients. The R-values is simply the inverse of the U-factor (R=1/U). So when people talk about an R-10 window, they talk about R-values with film coefficients, but if they talk about R-10 insulation it is for the material only, without film coefficients. Very confusing, I agree.

But you can get the 'insulation style' R-value out of THERM, but it takes a little bit of math.

• Your example had a conductivity in BTU.in/hr.ft2.F, which is not the default in THERM. Make sure THERM is setup to use these units under Options, Preferences, Conductivity Units (look for Btu-in/ instead of Btu/)
• You mentioned 50 and 100 F, but we also need to use a film coefficient in THERM, for now lets use 1. This is fine for solids like mineral wool, but if you model something with frame cavities in it, it should be the actual film coefficient used in a test.
• Create a new boundary condition: Libraries, Boundary Conditions, New - pick model Simplified and use temperature 50F and Film Coefficient=1
• Create another boundary condition with 100F and Film Coefficient=1 (you can pick a name for both of these boundary conditions)
• Draw the 1" thick insulation layer and give it some height (5" or 10 " for example)
• Put the boundaries you just created on either side
• Put a U-factor tag on one side
• Calculate the example
• Note the R-value or U-factor (U=1/R)
• Calculate the file
• To calculate the R-values without the film coefficient, you have to subtract them. The way you do this is by subtracting R-values (the inverse of a film coefficient is the R-value). R_material=R_overall - 1/exterior_film_coefficient - 1/interior_film_coefficient.
• For your example of 1" mineral wool (cond=0.24 btu-in/hr-ft2-f). Roverall=6.166 (reported by THERM, see attached sample file). exterior and interior film coefficients are 1, so 6.166- 1/1 - 1/2=4.166
• More simply, the THERM reported R-value - 2. Remember this works in this case because we set the interior and exterior film coefficients to 1.

Christian

[Jérémy L.]

Christian,

I was trying out the R-Value capability of THERM when I ran into a problem similar to Jillian's. What I wanted to do was to manually calculate the conductivity of a insulation material, input it in THERM, run it and check the R-Value, however the results were not what I expected.

The insulation material has a R-Value of 21. It's thickness is 3.5 in so around 0.29 ft. So it's conductivity k should be its thickness / its R-Value, ie, 0.29 / 21 which is about 0.0138 Btu/h.ft.F. I use the model that you attached to your previous message and edited the conductivity of the material, ran the calculation and obtained a R-Value of about 7.95 h.ft2.F/Btu (even without the film coefficients) which is different than the R-Value of 21.

What am I missing?

Thank you for your help!

[Dragan Vidanovic]

You need to make model thickness to be 3.5 in. Example is 1 in thick and your calculations are for 3.5 in.

Simon

[Jérémy L.]

Of course  ... Thank you!