Question

A double-glazed window is made of two glass panes of 6 mm thick with an air gap of 5mm between them. Taking the ambient and inside temperatures as 40 °C and 22 °C, outside and inside heat transfer coefficients as 9 W/m2 K and 15 W/m² K, thermal conductivity of glass and air as 1.40 W/m K and 0.025 W/m K, respectively, and assuming that the air layer is stagnant and only conduction is involved,

draw the thermal resistance schematics

Answer 1

The student's question requires calculations of thermal resistance and rate of heat conduction through a double-glazed window based on the thermal conductivity of glass and air, the thickness of materials, and the area of the window.

Step-by-step explanation:

The student's question involves calculating the thermal resistance and the rate of heat conduction through a double-glazed window, which is an application of the principles of thermodynamics in physics. To answer this question, we need to consider the separate thermal resistances offered by the air gap and the glass panes and then combine them to get the total resistance.

The formulas used in the calculations rely on the thermal conductivity of the materials, the thickness of the materials, and the area through which heat is being conducted. To make comparisons, similar calculations can be performed for windows of different thicknesses or materials and then the heat conduction rates can be contrasted to understand the insulation efficiency.

For example, if we need to calculate the thermal resistance for glass, we would use the formula R_glass = thickness / (area * conductivity) and similarly for air. Then, adding these thermal resistances together gives us the total resistance across the window. For the rate of heat conduction, we would apply Fourier's law of heat conduction Q = ΔT / R_total, where ΔT is the temperature difference across the window, and R_total is the total thermal resistance.