Question

Consider a \(1.5\)-m-high and 2 -m-wide triple pane window. The thickness of each glass layer \((k=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(0.5 \mathrm{~cm}\), and the thickness of each air space \((k=0.025 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(1 \mathrm{~cm}\). If the inner and outer surface temperatures of the window are \(10^{\circ} \mathrm{C}\) and \(0^{\circ} \mathrm{C}\), respectively, the rate of heat loss through the window is (a) \(75 \mathrm{~W}\) (b) \(12 \mathrm{~W}\) (c) \(46 \mathrm{~W}\) (d) \(25 \mathrm{~W}\) (e) \(37 \mathrm{~W}\)

Short answer

Answer: The rate of heat loss through the triple-pane window is approximately 37 W.

Step-by-step solution

Calculate the total area of the window

The total area of the window can be found by multiplying its height by its width: \(A = 1.5\,\mathrm{m} \cdot 2\,\mathrm{m} = 3\, \mathrm{m}^2\)

Calculate the resistance for each layer

The thermal resistance for each layer can be calculated using the formula: \(R = \frac{d}{kA}\), where \(d\) is the thickness of the material, \(k\) is the thermal conductivity, and \(A\) is the area. For each glass layer, the resistance is: \(R_\text{glass} = \frac{0.005\,\mathrm{m}}{0.80\,\mathrm{W/m\,K} \cdot 3\,\mathrm{m}^2} = 0.00208\, \mathrm{K/W}\) For each air space layer, the resistance is: \(R_\text{air} = \frac{0.01\,\mathrm{m}}{0.025\,\mathrm{W/m\,K} \cdot 3\,\mathrm{m}^2} = 0.13333\, \mathrm{K/W}\)

Calculate the total thermal resistance

As there are three glass layers and two air layers, the total resistance can be found by adding the individual resistances: \(R_\text{total} = 3R_\text{glass} + 2R_\text{air} = 3(0.00208\, \mathrm{K/W}) + 2(0.13333\, \mathrm{K/W}) = 0.00624\, \mathrm{K/W} + 0.26667\, \mathrm{K/W} = 0.27291\, \mathrm{K/W}\)

Calculate the rate of heat loss

The rate of heat loss through the window can be calculated using the formula: \(Q = \frac{\Delta T}{R_\text{total}}\), where \(\Delta T\) is the temperature difference between the inner and outer surfaces of the window. Here, \(\Delta T = 10^{\circ} \mathrm{C} - 0^{\circ} \mathrm{C} = 10\, \mathrm{K}\) \(Q = \frac{10\,\mathrm{K}}{0.27291\, \mathrm{K/W}} \approx 36.6\, \mathrm{W}\) The rate of heat loss is closest to option (e) \(37\, \mathrm{W}\). So the correct answer is (e).

Key concepts

Thermal Resistance

Understanding thermal resistance is crucial when delving into heat loss calculations, as it represents the measure of a material's ability to resist the flow of heat. In simpler terms, it's like a measure of insulation effectiveness, but instead of keeping warmth in, it emphasizes hindering heat from escaping through materials.
Think of it as a form of 'thermal obstruction', similar to how thicker clothes can keep you warmer by offering more resistance to the cold air trying to creep in.

Importance in Heat Loss Calculation

When we analyze a structure like a triple pane window, the thermal resistance of each layer - whether glass or air - influences the overall insulating property of the window. The formula for calculating thermal resistance is:
\[ R = \frac{d}{kA} \]
where \( d \) is the material's thickness, \( k \) the thermal conductivity, and \( A \) the cross-sectional area. More significant thermal resistance value indicates better insulation, hence less heat loss. For multi-layer structures, we simply sum up the resistances of individual layers to get the total thermal resistance, which is used to determine heat transfer across the whole assembly.

Thermal Conductivity

In the context of heat loss through windows or any barrier, thermal conductivity is a fundamental property. It is defined as the material's ability to conduct heat and is denoted by \( k \) in heat transfer equations. Highly conductive materials, such as metals, allow heat to pass through quickly, while low conductivity materials, like air or glass wool, slow down the heat transfer process.

Role in Insulating Materials

It's the thermal conductivity that dictates the efficiency of insulating layers within a window. Lower thermal conductivity means a material is a better insulator. This is why windows may have air gaps – air has a very low thermal conductivity, providing a good layer of insulation.
Materials like glass, with higher conductivity than air but lower than metals, strike a balance between transparency and insulation. It explains why multiple layers of glass separated by air gaps can significantly enhance a window's insulating properties without compromising its transparency.

Heat Transfer Through Windows

Windows are critical elements in a building's design, particularly concerning heat transfer. They can be a major source of heat loss in colder climates or unwanted heat gain in warmer climates. The way heat moves through windows can be broken down into conduction, convection, and radiation.

Controlling Heat Transfer

In practice, understanding heat transfer through windows allows for smarter design choices to improve energy efficiency. For instance, a triple pane window setup used in our exercise employs multiple layers of glass and air gaps to minimize conductive heat loss.
The thickness and thermal properties of the glass, the size and number of air spaces, and the overall area of the window dictate how much heat is lost. By calculating each parameter, including the thermal resistance and conductivity discussed earlier, we can optimize a window's design to reduce energy consumption and maintain comfortable indoor temperatures, which is crucial for sustainable building practices.