Question

Consider a house that is maintained at a constant temperature of \(22^{\circ} \mathrm{C}\). One of the walls of the house has three singlepane glass windows that are \(1.5 \mathrm{~m}\) high and \(1.8 \mathrm{~m}\) long. The glass \((k=0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(0.5 \mathrm{~cm}\) thick, and the heat transfer coefficient on the inner surface of the glass is \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Now winds at \(35 \mathrm{~km} / \mathrm{h}\) start to blow parallel to the surface of this wall. If the air temperature outside is \(-2^{\circ} \mathrm{C}\), determine the rate of heat loss through the windows of this wall. Assume radiation heat transfer to be negligible. Evaluate the air properties at a film temperature of \(5^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Short answer

#Answer# Based on the given information and calculations, the rate of heat loss through the windows is: $$ Q = \frac{24^{\circ}\mathrm{C}}{R_{total}} $$ Where $R_{total}$ is the combined thermal resistance found in Step 2. By plugging in the values and calculating, we can determine the rate of heat loss through the windows.

Step-by-step solution

Calculate the Area of the Windows

First, we will calculate the area of each window, which will be used later in our calculations. The dimensions given are 1.5 m high and 1.8 m long, so the area of one window is: $$ A = (1.5\,\mathrm{m})(1.8\,\mathrm{m}) = 2.7\,\mathrm{m}^2 $$ There are three windows, so the total area of the windows is: $$ A_{total} = 3 \times A = 3 \times 2.7\,\mathrm{m}^2 = 8.1\,\mathrm{m}^2 $$

Calculate Thermal Resistance of Glass and Convective Heat Transfer Coefficients

Now that we have the area of the windows, we will calculate the combined thermal resistance of conduction through the glass and convection at the inner surface of the glass. First, the thermal resistance of conduction through the glass can be expressed as: $$ R_{cond} = \frac{L}{kA} = \frac{0.005\,\mathrm{m}}{0.78\,\mathrm{W/m\,K}\times 8.1\,\mathrm{m}^2} $$ Next, compute the thermal resistance of convection at the inner surface of the glass: $$ R_{conv} = \frac{1}{hA} = \frac{1}{8\,\mathrm{W/m^2\,K} \times 8.1\,\mathrm{m}^2} $$ Then, find the total combined thermal resistance by adding these two values: $$ R_{total} = R_{cond} + R_{conv} $$

Calculate Temperature Difference

The temperature difference between the inside and outside of the house is needed to find the rate of heat loss. The inside temperature is 22°C, and the outside temperature is -2°C. The temperature difference is: $$ \Delta T = T_{inside} - T_{outside} = 22^{\circ}\mathrm{C} - (-2^{\circ}\mathrm{C}) = 24^{\circ}\mathrm{C} $$

Calculate Rate of Heat Loss

Now that we have the combined thermal resistance and temperature difference, we can determine the rate of heat loss through the windows using the following equation: $$ Q = \frac{\Delta T}{R_{total}} $$ Plug in the values calculated in previous steps and find the rate of heat loss through the windows.

Key concepts

Thermal Resistance

Understanding thermal resistance is key to evaluating how well a material can prevent heat flow through it. This concept is parallel to electrical resistance, where greater thermal resistance signifies less heat flow, similar to how greater electrical resistance signifies less current flow. Thermal resistance (\( R \) ) can be defined for different modes of heat transfer, such as conduction and convection. It is calculated by the formula:- For conduction: \[ R_{ ext{cond}} = \frac{L}{kA} \] - Here, \( L \) is the thickness of the material,\( k \) is the thermal conductivity, and \( A \) is the area.- For convection: \[ R_{ ext{conv}} = \frac{1}{hA} \] - \( h \) is the convective heat transfer coefficient, representing how well heat is transferred via convection.In the case of a wall with windows, like in the example, both conductive and convective resistances are involved, and their values are combined to find the total thermal resistance. This is done by simply adding the two:\[ R_{ ext{total}} = R_{ ext{cond}} + R_{ ext{conv}} \].Knowing the total thermal resistance lets us calculate the rate of heat loss from one side of a layer to the other, for instance, from inside the house to the outside.

Convection Heat Transfer

Convection heat transfer occurs when heat is transferred between a solid and a fluid in motion, such as air or water, across the surface of the solid. This process can be significantly influenced by the motion of the fluid, be it naturally occurring or mechanically induced wind. The strength of this transfer is often represented by the convective heat transfer coefficient (\( h \) ). A high \( h \) value indicates efficient heat transfer from the surface to the air. For example, an inside surface of a window in a heated room might have a \( h \) value of 8 \( ext{W/m}^2 ext{K} \) .Factors affecting convective heat transfer include:- **Fluid velocity:** Faster-moving fluids enhance the rate of convection.- **Surface geometry:** Affects how the fluid moves across a surface.- **Temperature difference:** Greater differences between surface and fluid temperatures increase convection rates.In practical terms, imagine a cold wind blowing across the windows of a warm house. This increases the rate of heat loss through convection, compounded by a high convective heat transfer coefficient, necessitating insulation.

Conduction through Glass

Conduction is the process by which heat energy is transferred through a material, due to a temperature difference. For windows, it is essential to understand how heat is transferred through the glass to know how energy-efficient a building is.The rate of conduction heat transfer through glass depends upon:- **Thickness (\( L \) )**: Thicker glass provides more resistance to heat flow.- **Thermal conductivity (\( k \) )**: A measure of how well the glass material conducts heat. High thermal conductivity translates to less insulation.- **Area (\( A \) )**: Larger areas can increase the overall heat transfer.The resistance to conduction through the glass is calculated using:\[ R_{ ext{cond}} = \frac{L}{kA} \]For practical application, consider glass with a conductivity of 0.78 \( ext{W/m⋅K} \) and a thickness of 0.5 cm; knowing these allows one to calculate the resistance the glass offers to heat flow. This calculation is essential, as lesser resistance translates into higher energy costs to maintain indoor temperature.