Question

People radiate energy at a net rate of roughly 100W;this is essentially waste heat from various chemical processes needed for bodily functioning.

Consider 4 humans in a roughly square hut measuring 5m × 5m,with a flat roof at 3m height. The exterior walls of the hut are maintained at 0 ◦C by the external environment. For each of the following construction materials for the walls and ceiling,

(i) compute the R-value (T/q) for the material

(ii) compute the equilibrium temperature in the hut when the four people are in it:

(a) 0.3m (1 foot) concrete walls/ceiling;

(b) 10 cm (4") softwood walls/ceiling; and

(c) 2.5 cm (1") softwood, with 0.09m (3.5") fiberglass insulation on interior of walls and ceiling.

Take thermal conductivities and/or R-values from the literature.

Answer 1

Conductivity of

Soft wood K_w = 0.146 W/m-K

Concrete K_c = 1.36 W/m-K

Fiber Glass K_g = 0.046 W/m-K

totla area of the walls and the roof A= 4*(5*3) + 3*3 = 69 sq.m

Thermal resistance R = x/KA , x is the length/thickness of the material

0.3m Concrete R = 0.3/1.36*69 = 0.0032 K/W

10 cm wood R = 0.1/0.146*69 = 0.01 K/W

2.5 cm wood R =  0.025/0.146*69 = 0.0025 K/W

0.09m glass R = 0.09/0.046*69 = 0.028 K/W

2.5cm wood + 0.09m fiber glass will add up = 0.028+0.0025 = 0.0305 K/W

2) with 4 people the heat produced = 4*100 = 400W

at thermal equilibrium heat transfer outside is equal to inside

400 = T/R

tmeprature diff T = 400R

0.3m concrete wall   T = 400*0.0032 = 1.28 K

outside temp = 0 C

inside temp = 1.28 C

10cm wood  T = 400*0.01 = 4 K

inside temp = 4 C

2.5cm wood + 0.09m fiber glass  T = 400*0.0305 = 12.2 K

inside temp = 12.2 C