Question

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

Answer 1

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, $\dot{Q} = 875 W$

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:

$\dot{Q} = KA \frac{dT}{dx} \\\dot{Q} = 1.4 * 15 \frac{291 - 266}{0.6}\\\dot{Q} = 1.4 * 15 \frac{25}{0.6}\\\dot{Q} = 875 W$

b) Energy, $E = \dot{Q}t$

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