Respuesta :
Complete Question:
A concrete building slab, on a basement floor is 5 m long, 3 m wide and 0.6 m thick. During the winter, the temperature is normally 18°C on the upper surface of the slab (the inside), and it is -7°C on the lower surface, with a linear temperature profile in between. If the concrete has a thermal conductivity of 1.4 W/m-K, what is the rate of heat loss through the slab? If the basement is heated by a gas furnace operated with an efficiency of 66%, how much energy must be utilized to maintain the basement temperature for 90 days assuming that nearly all heat losses occur through the slab?
Answer:
a) Rate of heat loss, [tex]\dot{Q} = 875 W[/tex]
b) Energy that must be utilized to maintain the basement temperature,
Em = 10309 MJ
Explanation:
Length of the slab, l = 5 m
Width of the slab, w = 3 m
Thickness of the slab, t = 0.6 m
Cross Sectional Area of the slab, A = l x b
A = 5 x 3
A = 15 m²
Upper Surface Temperature, T₁ = 18°C = 18 + 273 = 291 K
Lower Surface Temperature, T₂ = -7°C = -7 + 273 = 266 K
a) The rate of heat loss is given by the formula:
[tex]\dot{Q} = KA \frac{dT}{dx} \\\dot{Q} = 1.4 * 15 \frac{291 - 266}{0.6}\\\dot{Q} = 1.4 * 15 \frac{25}{0.6}\\\dot{Q} = 875 W[/tex]
b) Energy, [tex]E = \dot{Q}t[/tex]
t = 90 days = 90 * 24 * 3600 = 7776000 s
E = 875 * 7776000
E = 6804000000J
E = 6804 MJ
If Efficiency = 66%, Energy that must be utilized to maintain the basement temperature.
6804 = 66% * Em
6804 = 0.66 * Em
Em = 6804/0.66
Em = 10309 MJ
