Question

Heating Load Calculation
Given a heating-only garage (shown below) with the dimension of 20’ × 20’ × 10’ (L × W × H), and the other building information:

· A gate with the dimension of 15’ × 9’ and U-0.5 Btu/(hr.ft2.F)

· Wall construction – insulated precast concrete wall panels with 4” LW concrete (R- value of 0.625 (hr.ft2.F)/Btu), 3” rigid insulation in the middle (R-value of 15 (hr.ft2.F)/Btu), and another 4” LW concrete (R-value of 0.625 (hr.ft2.F)/Btu)

· Roof construction – 6” rigid insulation with R-5 (hr.ft2.F)/Btu per inch (others may be ignored due to their very low R-values)

· No windows

· Fully insulated slab (2” rigid insulation placed horizontally and vertically as shown below with R-5 per inch)

· Infiltration: 0.75 ACH

· Building location: Bismarck, ND

· Indoor design T: 60F

Questions:

9.             a) Determine the U-value of the wall construction (2 points)

10.          b) Determine the U-value of the roof construction (2 points)

11.          c) Determine the heat loss coefficient U’ of the slab (2 points)

12.          d) Determine the CFM of infiltration (2 points)

13.          e) Determine the total heat loss (heating load) of this garage in terms of Btu/hr (6 points)

2/3

Answer 1

U= 1/(R1+R2+R3)

9. a) U = 1/ (0.625+15+0.625)

= 0.061 Btu/hr.ft2.F

10. b) U = 1/5 = 0.2 Btu/hr.ft2.F

11. c) U = 1/5 =0.2 Btu/hr.ft2.F

12. d) ACH :Air capacity per hour

CFM: Cubic feet per minute

CFM = (ACH*volume)/60

=(0.75*(20*20*10)/60 = 50 CFM

13. e) Total heat loss = Q wall+Qroof

Let Toutside = 0 F

Tdiff = 60-0=60

Qwall = Uwall*A*Tdiff

= 0.061*(20*10+20*10+20*10+20*10)*60 = 2928Btu/hr

Qroof = Uroof*Ar*Tdiff

= 0.2*(20*20)*60 = 4800 Btu/hr

Heat load Q = 2928+4800 = 7728 Btu/hr