Question

other factors affecting r - values - presence of gaps - result in contact points and increase heat transfer through conduction - proper calculation of insulation - 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, your 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 r - value(in) wood 0.71 drywall 0.45 brick 0.8 - 1.11 fiberglass (batts) 3.1 - 3.4 fiberglass blown (wall) 3.7 - 4.3 mineral wool (batts) 3.1 - 3.4 mineral wool blown (wall) 3.1 - 4.0 cellulose blown (wall) 3.6 - 3.9 polystyrene board 3.8 - 5.0 polyisocyanurate (foil - faced) 5.5 - 6.5

Explanation:

Step1: Identify R - values from the table

For wood, the R - value range is 0.71 - 1.41. For drywall, the R - value is 0.45. For brick, the R - value range is 0.8 - 1.11. For fiberglass (batt), the R - value range is 3.1 - 3.4. For fiberglass (blown), the R - value range is 3.1 - 3.4.

Step2: Calculate minimum R - value for thermos (Problem 1)

The minimum R - value occurs when we take the minimum values of each layer. The inner layer is regular fiberglass and outer layer is wood. Minimum R - value of fiberglass (batt) is 3.1 and minimum R - value of wood is 0.71. So minimum R - value for thermos = 3.1+0.71 = 3.81.

Step3: Calculate maximum R - value for thermos (Problem 1)

The maximum R - value occurs when we take the maximum values of each layer. Maximum R - value of fiberglass (batt) is 3.4 and maximum R - value of wood is 1.41. So maximum R - value for thermos = 3.4 + 1.41=4.81.

Step4: Calculate minimum R - value for house (Problem 2)

The house has two layers of drywall and an outer layer of brick. Minimum R - value of drywall is 0.45 and minimum R - value of brick is 0.8. There are two layers of drywall, so total R - value of drywall layers is 2×0.45 = 0.9. Minimum R - value for house = 0.9+0.8 = 1.7.

Step5: Calculate maximum R - value for house (Problem 2)

Maximum R - value of drywall is 0.45 and maximum R - value of brick is 1.11. Total R - value of two drywall layers is 2×0.45 = 0.9. Maximum R - value for house = 0.9+1.11 = 2.01.

Answer:

1. Minimum R - value for thermos: 3.81, Maximum R - value for thermos: 4.81
2. Minimum R - value for house: 1.7, Maximum R - value for house: 2.01