Question

What will be the U-factor for a wall constructed as follows? 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. Round value to closest 0.01 Btu/hr.ft2.°F. • 4" face brick. • 1" air space. • 1/2” plywood sheathing. • 2x4 wood studs 16" on center with R13 batt insulation. • 1/2" drywall.

Question 1 options: 0.05 Btu/hr.ft2.°F 0.06 Btu/hr.ft2.°F 0.07 Btu/hr.ft2.°F 0.08 Btu/hr.ft2.°F

Answer #1

Ans) We know,

U factor = 1/( R1 + R2.......R_{n})

R value of interior air film = 0.68

R value of 4" face brick = 0.44

R value for 1" air space = 1

R value for 1/2" plywood sheathing=0.62

R value for 2 x 4 wood struts 16" on center with R 13 batt insulation = 11.6

R value for 1/2" dry wall = 0.45

Total R value of insulated area = 0.80 x 12.44

= 9.952 hr ft^{2} F/BTU

Total R value of famed area = 0.20 x (0.68+0.44+1+0.62+0.45)

= 0.20 x 6.19

= 1.238 hr ft^{2} F/BTU

Therefore U value = 1/(9.952+1.238)

= 0.08 BTU/hr ft^{2} F

Hence , **U - value = 0.08 BTU/hr ft ^{2}
F**

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