Two different numbers are used to describe the insulating properties of construction materials: R-values and U-values. The R-value is a number that represents the thermal resistance of a material. The lvalue is a measurement of the thermal conductivity of a material. In the customary units, a lvalue of 1 means that 1 BTU/hr of heat will flow per square foot of area per degree difference in temperature on each side of the material. The term BTU stands for British Thermal Unit, and is the amount of heat required to raise the temperature of 1 lb of water 1 °F. In the SI unit system, the preferred unit for heat is kilojoules. A joule is equal to 9.478 E—4 BTU and is the amount of heat required to raise the temperature of 1 g of water by 1°C.

R-values are the inverse of the lvalue (R = 1 /U). Therefore, the insulating ability of a material increases as the R-value increases and as the lvalue decreases. The R-value of a nonhomogeneous building component is the sum of the R-values of the materials that make up the component. The nonhomogeneous material may be a construction component such as a concrete block, or a part of the building such as a wall or ceiling. The units for U are °FB™Xh or °CxJA xh, where Au is the unit area of the material through which the heat passes (generally in ft2 or m2). Since the units of R are the inverse of the units of U, the units of R are °Fg^xh or °CxAuXh. Appendix IX contains R-values for some common homogeneous and nonhomogeneous building materials.

The type of construction and the type and amount of insulating material used will determine the rate of heat flow (British Thermal Units/hr, customary, or Joules/hr, SI) for a particular building or building component. Common building components include walls, doors, windows, floors, ceilings and roofs. Knowledge of the relative insulating values of different construction materials and of how to put them together

^-values and ^-values 289

FIGURE 22.3. Typical wood frame wall construction viewed looking down.

^-values and ^-values 289

to estimate the overall insulating value of a building or a building component is a great help in selecting appropriate insulation materials for a particular application.

To determine the total thermal resistance of a wall, simply list and add together the thermal resistance of the individual parts. For homogeneous materials, the total thermal resistance value is determined by the thickness (inches) of the material used times the R-value per inch of thickness. For nonhomogeneous materials, select the thermal resistance value for the thickness of material specified.

Two additional factors are important in determining the total thermal resistance of a wall. The first is the thermal resistance associated with the layer of still air next to any surface, as well as any air space within the wall, floor, or ceiling. Still air is a good insulating material, and a thin air film clings to exterior and interior surfaces providing a measurable thermal resistance. Typical R-values for the inside and outside air film are found in Appendix IX. The second factor is an understanding of common construction methods. In wood frame construction, the walls usually are made from 2 x 4 or 2 x 6 inch lumber. Lumber is sold using nominal sizes; the actual size is less. A 2 x 4 is actually 1 and 1/2 inches by 3 and 1/2 inches, and a 2 x 6 is actually 1 and 1/2 inches by 5 and 1/2 inches. Frame walls usually are constructed with the long dimension of the board perpendicular to the wall, with different materials attached to both sides to form the wall. The cavity between the surfaces may be filled or partially filled with insulation, see Figure 22.3.

Problem: Determine the total thermal resistance (RT) ofa2 x 4 wood frame wall. The outside is covered with 1/2 inch medium-density particle board, vapor barrier is permeable felt and 2.5 inches of 100 lb per cubic foot fired brick. The inside surface is 1/2 inch plasterboard. The wall is filled with medium-density, loose-fill cellulose.

Solution: The R-value of the wall is the sum of the R-value of each component. For this type of problem, a diagram of the wall and a table of the information

Inside air film

Plasterboard (1/2 in) Cellulose (3.5 in x 3.70 R/in) Particle board (1/2 in) Felt

Wall R value

FIGURE 22.4. Solution to total thermal resistance problem.

will help prevent mistakes. Study Figure 22.4 for the solution. The R-values were obtained from Appendix IX.

For this particular wall the total thermal resistance (RT) is 15.34. Note: Because loose fill insulation is used, the thickness of the insulation is equal to the cavity in the wall, 3.5 inches. When a batt type of insulation is used, and it is not as thick as the wall, the wall will also have an air cavity or void, the R-value for two inside surfaces is used if the air cavity is less than 3/4 inch. For example, if the insulation in the wall in the pervious problem is 3.0 inch of low density batt insulation, then the Rt is 12.71, Table 22.1.

The type of same table and procedures can be used to determine the total R-value for any building component. Simply sum the R-values for each type of material and the inside and outside air films. Note: some sources of R-values include the air film values with the material.

Problem: What is the total R-value (RT) for a ceiling constructed of 1/2-inch plywood for the inside surface with 6.0 inches of low density batt insulation on top of the plywood?

Solution: This problem presents a different situation because neither surface is exposed to outside conditions. In buildings with very low attic ventilation the

TABLE 22.1. Solution for insulation problem.

Inside air film

Plaster board

9.00

Particleboard

Felt

Brick

Outside air film

Total

A A A A A A A A A A A A A A A A A A A A A A A A A <

Inside air film 0.68

Batt insulation 1 9.00

Inside air film 0.68

Ceiling total 20.98

FIGURE 22.5. Solution to ceiling total resistance problem.

R-value for inside air film should be used on both sides of the ceiling. If the attic is well ventilated, an outside air film R-value would be appropriate. For this problem the assumption is that the attic is not well ventilated.

In this example the total R-value of the ceiling is 20.98, see Figure 22.5. One area of a building that is more difficult to determine a total thermal resistance is the floor. A concrete slab floor on the ground does not have a uniform temperature gradient across the floor. The insulating properties of the ground will cause the temperatures between the inside space and the ground at the center of the floor to be different from the edges. The values for insulated and un-insulated floors in Appendix IX are estimates.

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