Question

1) A frame wall of the bedroom of a house has the following specifications:

– Wall: 12 ft. by 8 ft., wood siding, wood sheathing, 2 in. of insulation with R-7 value, and inside finish

– Window: 3 ft. by 4 ft. 6 in., single glass, aluminum frame

The room temperature is 68º F and the outdoor temperature is 2º F. What is the heat transfer through the total assembly (wall and window combined)?

2) Consider a 1 ft × 1 ft × 1 ft box. Two of the sides are 1” thick extruded expanded polystyrene foam. The other four sides are 1” thick plywood. The inside of the box needs to be maintained at 40 °F. The air around the box is still and at 80 °F. How much cooling do you need?

Answer 1

Ans 1) We know,

Heat transfer, Q = U A

where, U = U value

A = Area

   = Temperature difference

Total heat transfer = Heat transfer through wall + Window

U-value of wall = 1 / R_wall = 1 / 7 = 0.1428 BTU / hr F ft²

=> Q_wall = 0.1428 x (12 x 8) (68 - 2)

=> Q_wall = 905.14 BTU/hr

R value of window with glass in aluminium frame = 0.91 + 0.61 = 1.52

=> U-value of window = 1 / 1.52 = 0.6578 BTU / hr F ft²

=> Q_wall = 0.6578 x (3 x 4) (68 - 2)

=> Q_wall = 521 BTU/hr

Total heat loss = 905.14 + 521 = 1426.14 BTU/hr

Ans 2) Cooling need = Heat loss

Also, Heat loss,Q = U A

U-value of sides with 1" thick extruded polystyrene foam = 1 / R = 1 / 5 = 0.20  BTU / hr F ft²

U- value of sides with 1" thick plywood = 1 / R = 1 / 1.125 = 0.888 BTU / hr F ft²

=> Heat loss from 2 sides with 1" thick extruded polystyrene foam = 2 x 0.20 x (1 x 1) x (80 - 40) = 16 BTU/hr

=> Heat loss from 4 sides with 1" thick plywood = 4 x 0.888 x (1 x 1) x (80 - 40) = 142 BTU/hr

Total cooling need = 16 + 142 = 158 BTU/hr