A concrete building slab, the temperature is normally on the upper surface of the slab (the inside), andit is-on the lower surface, with a linear temperature profilein between. efficiency of 66%, how much energy must be utilized to maintain the basement temperaturefor 90daysassuming that nearly all heat losses occur through the slab

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?

a) Rate of heat loss,

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:

b) Energy,

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

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explanation: