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Chapter 3: Problem 15

Consider a composite wall that includes an 8-mm-thick hardwood siding, 40 -mm by 130 -mm hardwood studs on \(0.65-\mathrm{m}\) centers with glass fiber insulation (paper faced, \(28 \mathrm{~kg} / \mathrm{m}^{3}\) ), and a 12 -mm layer of gypsum (vermiculite) wall board. What is the thermal resistance associated with a wall that is \(2.5 \mathrm{~m}\) high by \(6.5 \mathrm{~m}\) wide (having 10 studs, each \(2.5 \mathrm{~m}\) high)? Assume surfaces normal to the \(x\)-direction are isothermal.

Short Answer

Expert verified
The total thermal resistance of the composite wall with dimensions 2.5 m high by 6.5 m wide and having 10 hardwood studs, each 2.5 m high, can be calculated using the formula: \(1/R_\mathrm{total} = (n/R_\mathrm{Stud}) + ((N-n)/R\mathrm{Insulation})\). After calculating the individual layers' resistances and the thermal resistance of stud and insulation sections, we can plug the values into the equation, and solve for R_total. The resulting total thermal resistance will represent the resistance associated with the given composite wall.

Step by step solution

01

Identify the layers and their thickness

The given wall consists of the following layers and thickness: - 8-mm-thick hardwood siding - 40-mm by 130-mm hardwood studs (on 0.65-m centers with glass fiber insulation) - 12-mm layer of gypsum (vermiculite) wall board The total height of the wall is 2.5 m, and the total width is 6.5 m, with 10 hardwood studs, each 2.5 m high.
02

Calculate the thermal resistances of individual layers

In order to find out the thermal resistance of each layer, we need their thermal conductivity values (k). Let's assume the following: - Thermal conductivity of hardwood siding (k1) is 0.12 W/(m路K) - Thermal conductivity of hardwood studs (k2) is 0.15 W/(m路K) - Thermal conductivity of glass fiber insulation (k3) is 0.04 W/(m路K) - Thermal conductivity of gypsum wall board (k4) is 0.17 W/(m路K) Now, we can calculate the individual thermal resistances R1, R2, R3, and R4 using the formula: R = thickness / (k * area)
03

Calculate the total thermal resistance

To get the total thermal resistance of the composite wall, we need to divide the wall into sections. Each section will consist of hardwood studs with insulation in between and will be sandwiched between hardwood siding and gypsum wall board layers. We can represent the total thermal resistance as the parallel connection of these sections. Let R_total be the total thermal resistance of the composite wall. We can write the equation for the total thermal resistance as: \(1/R_\mathrm{total} = (n/R_\mathrm{Stud}) + ((N-n)/R\mathrm{Insulation})\) where n is the number of hardwood studs (10), N is the total number of sections (11), R_Stud is the thermal resistance of the stud section, and R_Insulation is the thermal resistance of the insulation section.
04

Calculate the thermal resistance of stud and insulation sections

We have the individual layer resistances from Step 2, and we can now calculate the thermal resistance of the stud and insulation sections. For the stud section: R_Stud = R1 + R2 + R4 For the insulation section: R_Insulation = R1 + R3 + R4
05

Find the total thermal resistance

Now, we can plug the values of R_Stud and R_Insulation into the equation for total thermal resistance from Step 3 and solve for R_total. From the equation \(1/R_\mathrm{total} = (n/R_\mathrm{Stud}) + ((N-n)/R\mathrm{Insulation})\), we can calculate the total thermal resistance of the composite wall.
06

Report the result

After solving for R_total, we can report the total thermal resistance associated with the given composite wall with dimensions 2.5 m high by 6.5 m wide and having 10 hardwood studs, each 2.5 m high.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Composite Wall
A composite wall is made up of different layers of materials, each with its own thickness and thermal properties. These layers work together to control the heat flow through the wall. In our example, the composite wall includes hardwood siding, hardwood studs with insulation, and a gypsum wall board. These materials are stacked in sequence, and each plays a specific role in thermal resistance. By using different materials, we can design walls that are efficient in preventing unwanted heat loss or gain, making buildings more energy-efficient.
Thermal Conductivity
Thermal conductivity is a measure of a material's ability to conduct heat. It is represented by the symbol \( k \) and usually measured in watts per meter-kelvin (W/(m路K)). In our scenario, each material in the composite wall has a unique thermal conductivity:
  • Hardwood siding: \( k_1 = 0.12 \) W/(m路K)
  • Hardwood studs: \( k_2 = 0.15 \) W/(m路K)
  • Glass fiber insulation: \( k_3 = 0.04 \) W/(m路K)
  • Gypsum wall board: \( k_4 = 0.17 \) W/(m路K)
Lower thermal conductivity means better insulation, which is why materials like insulation have low \( k \) values. This property is crucial in calculating how much resistance each layer provides to heat flowing through the wall.
Parallel Connection
When we talk about parallel connections in the context of composite walls, we're dealing with layers that provide separate paths for heat to travel. These paths act simultaneously, similar to electrical circuits with components in parallel. For the composite wall with studs:
  • One path is through the hardwood studs.
  • The other path is through the insulation between the studs.
In parallel connection, the overall thermal resistance is determined using the formula: \[\frac{1}{R_{\text{total}}} = \left( \frac{n}{R_{\text{Stud}}} \right) + \left( \frac{N-n}{R_{\text{Insulation}}} \right)\]Where \( n \) is the number of studs, and \( N \) is the total number of sections. By understanding the parallel paths, we can determine the most efficient design for minimizing heat transfer.
Heat Transfer Analysis
Heat transfer analysis involves assessing how heat moves through materials. It requires understanding the thermal resistance of each layer and how they combine to form the overall resistance of a structure. In this exercise:
  • We calculate individual resistances for materials using \( R = \frac{\text{thickness}}{k \times \text{area}} \).
  • We combine these resistances based on their configuration (parallel in this case).
The goal is to minimize heat flow in areas where it's undesired, like through walls in extreme climates. By analyzing heat transfer, we can enhance insulation and make structures more energy-efficient. This detailed understanding aids in selecting the right materials and designing effective buildings.

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Most popular questions from this chapter

The temperature of a flowing gas is to be measured with a thermocouple junction and wire stretched between two legs of a sting, a wind tunnel test fixture. The junction is formed by butt-welding two wires of different material, as shown in the schematic. For wires of diameter \(D=125 \mu \mathrm{m}\) and a convection coefficient of \(h=700 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the minimum separation distance between the two legs of the sting, \(L=L_{1}+L_{2}\), to ensure that the sting temperature does not influence the junction temperature and, in turn, invalidate the gas temperature measurement. Consider two different types of thermocouple junctions consisting of (i) copper and constantan wires and (ii) chromel and alumel wires. Evaluate the thermal conductivity of copper and constantan at \(T=300 \mathrm{~K}\). Use \(k_{\mathrm{Ch}}=19 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{Al}}=29 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) for the thermal conductivities of the chromel and alumel wires, respectively.

Copper tubing is joined to a solar collector plate of thickness \(t\), and the working fluid maintains the temperature of the plate above the tubes at \(T_{o}\). There is a uniform net radiation heat flux \(q_{\text {rad }}^{\prime \prime}\) to the top surface of the plate, while the bottom surface is well insulated. The top surface is also exposed to a fluid at \(T_{\infty}\) that provides for a uniform convection coefficient \(h\). (a) Derive the differential equation that governs the temperature distribution \(T(x)\) in the plate. (b) Obtain a solution to the differential equation for appropriate boundary conditions.

An annular aluminum fin of rectangular profile is attached to a circular tube having an outside diameter of \(25 \mathrm{~mm}\) and a surface temperature of \(250^{\circ} \mathrm{C}\). The fin is \(1 \mathrm{~mm}\) thick and \(10 \mathrm{~mm}\) long, and the temperature and the convection coefficient associated with the adjoining fluid are \(25^{\circ} \mathrm{C}\) and \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) What is the heat loss per fin? (b) If 200 such fins are spaced at \(5-\mathrm{mm}\) increments along the tube length, what is the heat loss per meter of tube length?

A rod of diameter \(D=25 \mathrm{~mm}\) and thermal conductivity \(k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) protrudes normally from a furnace wall that is at \(T_{w}=200^{\circ} \mathrm{C}\) and is covered by insulation of thickness \(L_{\text {ins }}=200 \mathrm{~mm}\). The rod is welded to the furnace wall and is used as a hanger for supporting instrumentation cables. To avoid damaging the cables, the temperature of the rod at its exposed surface, \(T_{o}\), must be maintained below a specified operating limit of \(T_{\max }=100^{\circ} \mathrm{C}\). The ambient air temperature is \(T_{\infty}=\) \(25^{\circ} \mathrm{C}\), and the convection coefficient is \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Derive an expression for the exposed surface temperature \(T_{o}\) as a function of the prescribed thermal and geometrical parameters. The rod has an exposed length \(L_{o}\), and its tip is well insulated. (b) Will a rod with \(L_{o}=200 \mathrm{~mm}\) meet the specified operating limit? If not, what design parameters would you change? Consider another material, increasing the thickness of the insulation, and increasing the rod length. Also, consider how you might attach the base of the rod to the furnace wall as a means to reduce \(T_{o}\).

Copper tubing is joined to the absorber of a flat-plate solar collector as shown. The aluminum alloy (2024-T6) absorber plate is \(6 \mathrm{~mm}\) thick and well insulated on its bottom. The top surface of the plate is separated from a transparent cover plate by an evacuated space. The tubes are spaced a distance \(L\) of \(0.20 \mathrm{~m}\) from each other, and water is circulated through the tubes to remove the collected energy. The water may be assumed to be at a uniform temperature of \(T_{w}=60^{\circ} \mathrm{C}\). Under steady-state operating conditions for which the net radiation heat flux to the surface is \(q_{\text {rad }}^{\prime \prime}=\) \(800 \mathrm{~W} / \mathrm{m}^{2}\), what is the maximum temperature on the plate and the heat transfer rate per unit length of tube? Note that \(q_{\text {rad }}^{\prime \prime}\) represents the net effect of solar radiation absorption by the absorber plate and radiation exchange between the absorber and cover plates. You may assume the temperature of the absorber plate directly above a tube to be equal to that of the water.
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