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What will be the heating load for 8' high by 600' long wall constructed as shown below. The design temperature difference is 59. Assume the wall is 80% insulated area, 20% framed area. Use table 3.2 to get R values. Assume horizontal airflow for interior air film value. - 4" (3-5/8") face brick. - 1" air space. - 3/4" plywood sheathing. - 2x6 wood studs 16" (1-1/2" x 5-1/2") on center with R13 batt insulation. - 1/2" drywall. a. 24,110 Btuh b. 32,240 Btuh c. 28,810 Btuhd. 19,510 Btuh

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To calculate the heating load for the wall, we need to determine the overall thermal resistance (R-value) of the wall and the surface area of the insulated and framed portions of the wall. We can then use the formula:

Heat Load = U x A x TD

where U is the overall heat transfer coefficient (the reciprocal of R-value), A is the surface area, and TD is the temperature difference.

Using Table 3.2, we can find the R-values for each layer of the wall:

Face brick: R = 0.44

Air space: R = 1.00

Plywood sheathing: R = 0.94

Wood studs with R13 batt insulation: R = 14.13 (80% insulated area)

Drywall: R = 0.45

The overall R-value for the wall is:

R = R_brick + R_air + R_plywood + (0.8 x R_studs) + R_drywall

R = 0.44 + 1.00 + 0.94 + (0.8 x 14.13) + 0.45

R = 23.57

So, the overall U-value is:

U = 1 / R

U = 0.0424 Btu/(h-ft^2-°F)

The surface area of the insulated portion of the wall is:

A_insulated = 8 ft x 600 ft x 0.8

A_insulated = 3,840 ft^2

The surface area of the framed portion of the wall is:

A_framed = 8 ft x 600 ft x 0.2

A_framed = 960 ft^2

The total surface area is:

A_total = A_insulated + A_framed

A_total = 4,800 ft^2

The temperature difference is given as 59°F.

Plugging in these values, we get:

Heat Load = U x A_total x TD

Heat Load = 0.0424 Btu/(h-ft^2-°F) x 4,800 ft^2 x 59°F

Heat Load = 11,975 Btu/h

Therefore, the heating load for the wall is approximately 12,000 Btu/h. None of the given answer choices match this result, so none of them are correct.

User Bobby Axe
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