Question

other factors affecting r - values: presence of improper laps of insulation result in contact points and increase heat transfer through conduction. to calculate the total r - value of a structure, you need to add the r - values of each layer. this would include the r - value for bricks, the insulation, the drywall, air spaces, etc. for example: r - value(brick)+r - value(insulation)+r - value(plasterboard)=r - value(total). remember, the higher your r - value, the better insulated your house is. 1) a thermos is made out of an inner layer of regular fiberglass and an outer layer of wood. calculate the minimum and maximum r - values for the thermos. 2) a building has two layers of drywall and an outer layer of brick. calculate the minimum and maximum r - values for the house. r - value of materials and depths
material\tr - value
wood\t0.71 - 1.41
drywall\t0.45
brick\t0.8 - 1.11
fiberglass (bat)\t3.1 - 3.4
fiberglass blown (wall)\t3.7 - 4.3
mineral wool (bat)\t3.1 - 3.4
mineral wool blown (wall)\t3.1 - 4.0
cellulose blown (wall)\t3.8 - 3.9
polystyrene board\t3.8 - 5.0
polyurethane board\t5.5 - 6.5
polysocyanurate (foil - faced)\t6.0 - 6.5

Explanation:

Step1: Identify R - values for materials

From the table, for wood, the R - value range is 0.71 - 1.41. For regular fiberglass (bat), the R - value range is 3.1 - 3.4. For brick, the R - value range is 0.45 - 1.41. For drywall, assume a common value of around 0.45.

Step2: Calculate minimum and maximum R - values for thermos (question 1)

Minimum R - value:

Use the minimum R - values of each material. If the inner layer is fiberglass (min R = 3.1) and outer layer is wood (min R = 0.71), the minimum R - value of the thermos is $0.71+3.1 = 3.81$.

Maximum R - value:

Use the maximum R - values of each material. If the inner layer is fiberglass (max R = 3.4) and outer layer is wood (max R = 1.41), the maximum R - value of the thermos is $1.41 + 3.4=4.81$.

Step3: Calculate minimum and maximum R - values for house (question 2)

Minimum R - value:

Use the minimum R - values of each material. Two layers of drywall (min R = 0.45 each) and outer layer of brick (min R = 0.45). The total minimum R - value is $0.45+0.45 + 0.45=1.35$.

Maximum R - value:

Use the maximum R - values of each material. Two layers of drywall (assume max R = 0.45 each) and outer layer of brick (max R = 1.41). The total maximum R - value is $0.45+0.45+1.41 = 2.31$.

Answer:

1. Minimum R - value for thermos: 3.81, Maximum R - value for thermos: 4.81
2. Minimum R - value for house: 1.35, Maximum R - value for house: 2.31