Question

Consider a flat ceiling that is built around \(38-\mathrm{mm} \times\) \(90-\mathrm{mm}\) wood studs with a center-to-center distance of \(400 \mathrm{~mm}\). The lower part of the ceiling is finished with 13-mm gypsum wallboard, while the upper part consists of a wood subfloor \(\left(R=0.166 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\), a \(13-\mathrm{mm}\) plywood, a layer of felt \(\left(R=0.011 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\), and linoleum \(\left(R=0.009 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\). Both sides of the ceiling are exposed to still air. The air space constitutes 82 percent of the heat transmission area, while the studs and headers constitute 18 percent. Determine the winter \(R\)-value and the \(U\)-factor of the ceiling assuming the 90 -mm-wide air space between the studs ( \(a\) ) does not have any reflective surface, (b) has a reflective surface with \(\varepsilon=0.05\) on one side, and ( ) has reflective surfaces with \(\varepsilon=0.05\) on both sides. Assume a mean temperature of \(10^{\circ} \mathrm{C}\) and a temperature difference of \(5.6^{\circ} \mathrm{C}\) for the air space.

Short answer

Short Answer: To determine the R-value and U-factor for each configuration, follow these steps: 1) Calculate the R-value for the air space and the rest of the assembly based on their configuration. 2) Determine the total R-value of the assembly by accounting for the 82% heat transmission through the air space and 18% through other components. 3) Calculate the U-factor by taking the reciprocal of the total R-value for each configuration. Compare the performance of each configuration by evaluating its R-value and U-factor.

Step-by-step solution

Calculate the R-value for Air Space and the Assembly

First, we need to find the R-value for the air space and the rest of the assembly. For the air space, the formula to use depends on the configuration: a) No reflective surface: \(R_\text{air\_space} = \frac{d}{k_\text{air}}\) b) One side with a reflective surface: \(R_\text{air\_space} = \frac{1}{h_c + \frac{1 - \varepsilon}{\varepsilon} \cdot h_r}\) c) Both sides with reflective surfaces: \(R_\text{air\_space} = \frac{1}{2\cdot\left( h_c + \frac{1 - \varepsilon}{\varepsilon} \cdot h_r\right)}\) where \(d\) is the distance between the studs (90 mm), \(k_\text{air}\) is the thermal conductivity of air, \(h_c\) is the convective heat transfer coefficient, and \(h_r\) is the radiative heat transfer coefficient. For the rest of the assembly, sum up the R-values of the individual components: \(R_\text{rest} = R_\text{subfloor} + R_\text{plywood} + R_\text{felt} + R_\text{linoleum}\)

Determine the R-value for the Assembly

With the R-values for components and the percentage of heat transmission being 82%, calculate the total R-value of the assembly for each configuration: \(R_\text{total} = \frac{0.82}{R_\text{air\_space}} + \frac{0.18}{R_\text{rest}}\)

Calculate the U-factor

To find the U-factor, take the reciprocal of the R-value for each configuration: \(U = \frac{1}{R_\text{total}}\) Following these steps, calculate the R-value and U-factor for each configuration described in the problem, and then compare their performance to evaluate the effectiveness of the different arrangements of air space and reflective surfaces.

Key concepts

Heat Transmission in Buildings

Understanding how heat is transmitted in buildings is crucial to maintaining energy efficiency and comfort. Heat can travel through building materials and air spaces by conduction, convection, and radiation. Materials have an R-value, which is a measure of thermal resistance. The higher the R-value, the better the material insulates. Conversely, the U-factor measures overall heat transfer and is the reciprocal of the R-value.

In a building, heat flows from warmer to cooler areas. During the winter, this typically means heat moves from the heated inside of a building through the walls and roof to the colder outside environment. The exercise provided demonstrates how different material layers and airspaces in a ceiling affect the R-value and U-factor, indicating the insulation's effectiveness and the building's potential energy use for heating.

Thermal Conductivity

Thermal conductivity is a property that tells us how well a material conducts heat. It is represented by the symbol \( k \) and measured in watts per meter-Kelvin \( (W/m\cdot K) \). Materials with high thermal conductivity, such as metals, are good conductors of heat, meaning they allow heat to pass through them easily. Insulators like wood or foam have low thermal conductivity and thus resist the flow of heat.

In the context of our problem, understanding the thermal conductivity of air \( k_\text{air} \) is essential, as it helps us to calculate the R-value for the air space within the ceiling structure. Thermal resistance \( R \) of a material or an air space is calculated by the thickness of the layer divided by its thermal conductivity, \( R = \frac{d}{k} \). This relationship indicates that increasing the thickness or reducing the thermal conductivity enhances an assembly's insulating properties.

Convective Heat Transfer Coefficient

The convective heat transfer coefficient, denoted by \( h_c \), is a measure of the convective heat transfer between a solid surface and a fluid, such as air, in contact with it. It is expressed in units of watts per square meter-Kelvin \( (W/m^2\cdot K) \). A higher value of \( h_c \) means that the material is more effective at transferring heat through convection.

Natural convection occurs as a result of fluid movement caused by temperature differences within the fluid. In buildings, this happens when warm air rises and cool air falls, creating a circulating flow. For the ceiling in the problem, \( h_c \) plays a role in determining the R-value when an airspace is present, as convective currents within the air can contribute significantly to the overall heat transfer.

Radiative Heat Transfer Coefficient

Radiative heat transfer is the process of heat exchange through electromagnetic waves, such as infrared radiation, without needing a medium. This form of heat transfer can occur even across a vacuum. The radiative heat transfer coefficient \( h_r \) quantifies how much heat is transferred by radiation per unit area and temperature difference, using the same units as the convective heat transfer coefficient.

In the given exercise, the presence of reflective surfaces modifies the radiative heat transfer. The emissivity \( \varepsilon \) of a surface can greatly affect \( h_r \), with lower emissivity materials reflecting more radiant heat, and thus reducing the radiative heat transfer coefficient. The formulas provided in the solution apply different methods to calculate \( h_r \) depending on whether there are zero, one, or two reflective surfaces.

U-factor Determination

The U-factor, also known as the overall heat transfer coefficient, represents the heat transfer per unit area and temperature difference. It is critical in quantifying how well an assembly, such as a wall or ceiling, prevents heat from passing through. The U-factor is the inverse of the total R-value, \( U = \frac{1}{R_\text{total}} \), indicating that lower U-factors signify better thermal performance.

To determine the U-factor of the ceiling in our problem, we first calculated the R-values for each configuration of air space and materials, taking into account the percentage of the ceiling covered by various materials. By understanding the R-values and how they combine to form the overall thermal resistance, we can better assess how changes to the structure, such as adding reflective surfaces, can improve energy efficiency and reduce heat transmission.