Question

QUESTION 2 Calculate heat loss through square foot of construction assembly consisting of 4 in brick veneer, vapor barrier, ½ in plywood, insulation with 2x4 @16” stud spacing (R- value of 10) and ½” drywall DB difference in temperature of 20°F. Round off R of wall to whole number.

Answer 1

Ans) We know,

Heat loss, q = U A T

where, U = U-factor of wall = 1 / R-value

A = Area of wall

  T = Temperature difference

R-value = R₁ + R₂..................R_n

R₁ = R-value of 4" brick veneer = 0.80 F ft² hr/BTU

R₂ = R-value of vapor barrier = 0

R₃ = R-value of plywood = 0.62 F ft² hr/BTU

R₄ = R-value of insulation = 10 F ft² hr/BTU

R₅ = R-value of 1/2" drywall = 0.45 F ft² hr/BTU

=> Total R -value = 0.80 + 0.62 + 10 + 0.45 = 11.87 F ft² hr/BTU

Hence, U-value = 1/11.87 = 0.0842 BTU/ F ft² hr

Putting values,

=> q = 0.0842 BTU/ F ft² hr x 1 ft² x 20 F

=> q = 1.684 BTU/hr per ft²

Hence, heat loss per square ft of given assembly is 1.684 BTU/hr