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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

Consider the flat plate of Problem \(3.112\), but with the heat sinks at different temperatures, \(T(0)=T_{o}\) and \(T(L)=T_{L}\), and with the bottom surface no longer insulated. Convection heat transfer is now allowed to occur between this surface and a fluid at \(T_{\infty}\), with a convection coefficient \(h\). (a) Derive the differential equation that determines the steady-state temperature distribution \(T(x)\) in the plate. (b) Solve the foregoing equation for the temperature distribution, and obtain an expression for the rate of heat transfer from the plate to the heat sinks. (c) For \(q_{o}^{\prime \prime}=20,000 \mathrm{~W} / \mathrm{m}^{2}, T_{o}=100^{\circ} \mathrm{C}, T_{L}=35^{\circ} \mathrm{C}\), \(T_{\infty}=25^{\circ} \mathrm{C}, k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(L=100 \mathrm{~mm}, t=5 \mathrm{~mm}\), and a plate width of \(W=\) \(30 \mathrm{~mm}\), plot the temperature distribution and determine the sink heat rates, \(q_{x}(0)\) and \(q_{x}(L)\). On the same graph, plot three additional temperature distributions corresponding to changes in the following parameters, with the remaining parameters unchanged: (i) \(q_{o}^{\prime \prime}=30,000 \mathrm{~W} / \mathrm{m}^{2}\), (ii) \(h=200\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and (iii) the value of \(q_{o}^{\prime \prime}\) for which \(q_{x}(0)=0\) when \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A nuclear fuel element of thickness \(2 L\) is covered with a steel cladding of thickness \(b\). Heat generated within the nuclear fuel at a rate \(\dot{q}\) is removed by a fluid at \(T_{\infty}\), which adjoins one surface and is characterized by a convection coefficient \(h\). The other surface is well insulated, and the fuel and steel have thermal conductivities of \(k_{f}\) and \(k_{s}\), respectively. (a) Obtain an equation for the temperature distribution \(T(x)\) in the nuclear fuel. Express your results in terms of \(\dot{q}, k_{f}, L, b, k_{s}, h\), and \(T_{\infty}\). (b) Sketch the temperature distribution \(T(x)\) for the entire system.

Determine the percentage increase in heat transfer associated with attaching aluminum fins of rectangular profile to a plane wall. The fins are \(50 \mathrm{~mm}\) long, \(0.5 \mathrm{~mm}\) thick, and are equally spaced at a distance of \(4 \mathrm{~mm}\) ( 250 fins \(/ \mathrm{m})\). The convection coefficient associated with the bare wall is \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while that resulting from attachment of the fins is \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Measurements show that steady-state conduction through a plane wall without heat generation produced a convex temperature distribution such that the midpoint temperature was \(\Delta T_{o}\) higher than expected for a linear temperature distribution. Assuming that the thermal conductivity has a linear dependence on temperature, \(k=k_{o}(1+\alpha T)\), where \(\alpha\) is a constant, develop a relationship to evaluate \(\alpha\) in terms of \(\Delta T_{o}, T_{1}\), and \(T_{2}\).

One method that is used to grow nanowires (nanotubes with solid cores) is to initially deposit a small droplet of a liquid catalyst onto a flat surface. The surface and catalyst are heated and simultaneously exposed to a higher- temperature, low-pressure gas that contains a mixture of chemical species from which the nanowire is to be formed. The catalytic liquid slowly absorbs the species from the gas through its top surface and converts these to a solid material that is deposited onto the underlying liquid-solid interface, resulting in construction of the nanowire. The liquid catalyst remains suspended at the tip of the nanowire. Consider the growth of a 15 -nm-diameter silicon carbide nanowire onto a silicon carbide surface. The surface is maintained at a temperature of \(T_{s}=2400 \mathrm{~K}\), and the particular liquid catalyst that is used must be maintained in the range \(2400 \mathrm{~K} \leq T_{c} \leq 3000 \mathrm{~K}\) to perform its function. Determine the maximum length of a nanowire that may be grown for conditions characterized by \(h=10^{5} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=8000 \mathrm{~K}\). Assume properties of the nanowire are the same as for bulk silicon carbide.
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