Question

The air exchange coefficient for my home has been determined to be 0.35 hr-1. My house volume is 13,757 ft3. Assume a constant house temperature of 67 F.
For the West Face Wall, Area = 693 ft^2, total R-value (or U-value) = 5.53 ft^2- F- hr / Btu.

Calculate:

[15] 1b the heat loss through the west wall for December 17 (average daily temperature was 1 °F). Note
that the HDD value needs to be multiplied by 24 (24 hrs / day) to get the daily heat flux value
since R has units in hours.

[20]1c. The energy needed to heat air for Dec 17 (average daily temperature was 1 °F).

Answer 1

Ans) We know,

Heat Loss (Q) = U A T

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

A = Area of wall

T = Temperature difference

=> Q = (1 / 5.53) x 693 x (67 - 1)

=> Q = 8270.88 BTU/hr

Heat loss due to infilteration,Q' = 0.018 x ACH x V x T

where, V = volume

=> Q' = 0.018 x 0.35 x 13757 x (67 -1)

=> Q' = 5720.16 BTU/hr

Total heat loss = 8270.88 + 5720.16 = 13991 BTU/hr

=> Heat loss = 13991 BTU/hr x 24 hr/day = 335784 BTU

Ans) We know,

Energy required to heat air = specific heat capacity of air x weight x temperature difference

Specific heat capacity of air = 0.241 BTU/lb F

Density of air = 0.086 lb/ft^3

=> Weight of 13757 ft^3 air = density x volume = 0.086 lb/ft^3 x 13757 ft^3 = 1183.1 lbs

=> Energy required = 0.241 BTU/lb F x 1183.1 lb x (67 - 1) F

=> Energy required to heat air = 18818.38 BTU or 19854.44 kJ