Question

Calculate the R-value and U-value for the following wall: (from outside to inside) 8" concrete block, 1" air space, 1" cellular polyurethane insulation,,3.5" batt insulation with steel studs,1/2" drywall

B) For the wall above, calculate the temperature at each material interface point. Assume the interior space temperature is 72F db and the outside air temperature is -15F db.

C) Calculate the peak heat loss in terms of BTU/hr if the above wall is 30 ft long and 10 ft high.

Answer 1

Answer 2

To calculate the R-value of the wall, we need to add up the R-values of each layer. The R-values for each layer are as follows:

• 8" concrete block: R-1.11 per inch, so R-8.88 for 8"

• 1" air space: R-1 (assuming still air)

• 1" cellular polyurethane insulation: R-6.25 per inch, so R-6.25 for 1"

• 3.5" batt insulation with steel studs: R-3.5 for insulation, but the steel studs will reduce the overall R-value due to thermal bridging. Assuming the steel studs make up 25% of the wall area, the effective R-value for this layer is approximately R-2.63.

• 1/2" drywall: R-0.45 per inch, so R-0.23 for 1/2"

Adding these up, we get a total R-value of R-18.99.

To calculate the U-value, we use the formula U = 1/R. So, U = 1/18.99 = 0.0527 W/m²K.

Next, we can calculate the temperature at each interface point using the formula:

Ti = To - (Q/A) x Ri

where Ti is the temperature at the interface, To is the temperature on the other side of the wall, Q is the heat flow rate, A is the area of the interface, and Ri is the thermal resistance at the interface.

Assuming the interior space temperature is 72°F (22.2°C) and the outside air temperature is -15°F (-26.1°C), we convert these temperatures to Kelvin:

• Interior temperature: 295.2 K

• Exterior temperature: 247.1 K

Starting from the outside and moving inward, the interface points and their areas are:

• Air space to polyurethane insulation: 10 ft x 8 ft = 80 ft² (7.43 m²)

• Polyurethane insulation to batt insulation: 10 ft x 8 ft = 80 ft² (7.43 m²)

• Batt insulation to drywall: 10 ft x 8 ft = 80 ft² (7.43 m²)

Using the R-values of each layer, we can calculate the heat flow rate through each interface:

• Air space to polyurethane insulation: (247.1 - (-15)) / 0.0527 = 5,251.8 BTU/hr (1,538.6 W)

• Polyurethane insulation to batt insulation: (247.1 - (-15)) / 18.13 = 14.5 BTU/hr (4.2 W)

• Batt insulation to drywall: (247.1 - (-15)) / 13.13 = 20.6 BTU/hr (6.0 W)

Using these heat flow rates and interface areas, we can calculate the temperature at each interface:

• Air space to polyurethane insulation: -15 - (5251.8 / (7.43 x 0.093)) = -30.2°F (-34.6°C)

• Polyurethane insulation to batt insulation: -15 - (14.5 / (7.43 x 0.093)) = -15.2°F (-26.2°C)

• Batt insulation to drywall: -15 - (20.6 / (7.43 x 0.093)) = -17.4°F (-27.4°C)

Finally, we can calculate the peak heat loss by multiplying the total heat flow rate (sum of the heat flow rates through each interface) by