Question

A 1.0 m × 1.5 m double-pane window consists of two 20-mm-thick layers of glass (k = 0.78 W/m·K) that are separated by a 13-mm air gap (kair = 0.025 W/m·K). The heat flow through the air gap is assumed to be by conduction. The inside and outside air temperatures are 20°C and −20°C, respectively, and the inside and outside heat transfer coefficients are 40 W/m2·K and 20 W/m2·K. Determine the daily rate of heat loss through the window in steady operation. (You must provide an answer before moving on to the next part.)

Answer 1

Answer:

$\dot Q=105.042\ W$ is the heat loss by conduction.

Explanation:

Given:

• area of window pane,$A=1.5\ m^2$

• thickness of the glass pane, $x=0.02\ mm$

• thickness of the air gap, $x'=0.013\ m$

• thermal conductivity of the glass, $k=0.78\ W.m^{-1}.K^{-1}$

• thermal conductivity of the air, $k_a=0.025\ W.m^{-1}.K^{-1}$

• inside temperature, $T_i=20^{\circ}C$

• outside temperature, $T_o=-20^{\circ}C$

From the Fourier's law of conduction we have the rate of heat transfer as:

$\dot Q=k.A.\frac{dT}{dx}$

$\dot Q=dT\div \frac{dx}{k.A}$

where:

k = thermal conductivity of the material

A = area subjected to the conduction

$dT =$ temperature difference across the two surfaces

dx = thickness of the surface

$\frac{dx}{k.A} =$ regarded as thermal resistance on electrical analogy

According to question here the  heat transfer occurs due to conduction of the air.

Here we have two surfaces with air sandwiched between them. So we find an equivalent resistance:

$R_e=2\times \frac{x}{k.A}+ \frac{x'}{k_a.A}$

$R_e=2\times \frac{0.02}{0.78\times 1.5} +\frac{0.013}{0.025\times 1.5}$

$R_e=0.3808\ K.W^{-1}$

Therefore:

$\dot Q=40\div 0.3808$

$\dot Q=105.042\ W$