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Chapter 5: Problem 159

Design a fire-resistant safety box whose outer dimensions are \(0.5 \mathrm{~m} \times 0.5 \mathrm{~m} \times 0.5 \mathrm{~m}\) that will protect its combustible contents from fire which may last up to \(2 \mathrm{~h}\). Assume the box will be exposed to an environment at an average temperature of \(700^{\circ} \mathrm{C}\) with a combined heat transfer coefficient of \(70 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the temperature inside the box must be below \(150^{\circ} \mathrm{C}\) at the end of \(2 \mathrm{~h}\). The cavity of the box must be as large as possible while meeting the design constraints, and the insulation material selected must withstand the high temperatures to which it will be exposed. Cost, durability, and strength are also important considerations in the selection of insulation materials.

Short Answer

Expert verified
Short Answer: To design a fire-resistant safety box with outer dimensions of 0.5 m x 0.5 m x 0.5 m, capable of withstanding a fire lasting 2 hours at 700掳C without the internal temperature exceeding 150掳C, we can use ceramic fiber insulation with a thickness of 1.24 mm and inner dimensions of approximately 0.49752 m x 0.49752 m x 0.49752 m.

Step by step solution

01

Calculate the total heat transfer

First, we need to calculate the total heat transfer allowed over the 2-hour period. From the given data, we have the combined heat transfer coefficient, \(h=70\,\frac{W}{m^2\cdot K}\). The surface area of the box, \(A\), can be calculated as: \(A=6(0.5\,m\times0.5\,m)=1.5\,m^2\), since it has 6 faces. The difference in temperature between the box and the environment, \(\Delta T = T_{env} - T_{box} = 700掳C - 150掳C = 550\,K\). Using the formula for heat transfer, \(Q=h\,A\,\Delta T\,t\), where \(t=2\,h=7200\,s\), we can calculate the heat transfer: \(Q=70\,\frac{W}{m^2\cdot K}(1.5\, m^2)(550\,K)(7200\,s)=39,690,000\,J\)
02

Choose an insulation material

The next step is selecting an appropriate insulation material that meets the requirements of high temperature resistance, cost-effectiveness, durability, and strength. Some materials to consider are: 1. Ceramic fiber insulation 2. Mineral fiber insulation 3. Calcium silicate insulation Each insulation material will have a unique thermal conductivity (\(k\)) value and maximum temperature resistances. For example, ceramic fiber insulation has a thermal conductivity of \(0.1\,\frac{W}{m\cdot K}\), a maximum temperature resistance of \(1260^{\circ} \mathrm{C}\), and is relatively light and inexpensive. Note: You can explore the available insulation materials suitable for this specific application and choose one for the safety box.
03

Calculate the insulation thickness

After selecting an appropriate insulation material, we need to calculate the thickness of the insulation layer needed to keep the interior temperature below 150掳C under the given circumstances. Using the formula for heat transfer through a solid, \(Q=\frac{k\,A\,\Delta T}{d}\), where \(d\) is the insulation thickness: \(d=\frac{k\,A\,\Delta T}{Q}\) For ceramic fiber insulation: \(d= \frac{(0.1 \,\frac{W}{m\cdot K})(1.5 \, m^2)(550 \,K)}{39,690,000\,J} = 0.00124 \,m = 1.24 \,mm\)
04

Determine the inner dimensions of the safety box

Now we know the thickness of the insulation material, we can determine the inner dimensions of the safety box. Since we want the cavity to be as large as possible while meeting design constraints, we need to subtract twice the insulation thickness from each outer dimension. Inner dimensions: \((0.5 - 2(1.24 \,mm))\, m \times (0.5 - 2(1.24 \,mm))\, m \times (0.5 - 2(1.24 \,mm))\, m = (0.49752\, m) \times (0.49752\,m) \times (0.49752\,m)\) In conclusion, to design a fire-resistant safety box that meets the given requirements, we will need to use an insulation material such as ceramic fiber insulation with a thickness of 1.24 mm. The inner dimensions of the box will be approximately 0.49752 m x 0.49752 m x 0.49752 m.

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Key Concepts

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

Heat Transfer Calculation
Understanding how heat transfers helps us design effective thermal insulation. The goal is to limit the heat entering the box so internal temperatures remain below critical levels. To calculate this, let's examine the heat transfer formula:

\[Q = h \times A \times \Delta T \times t\]
- **\(h\)**: Combined heat transfer coefficient, given as \(70 \frac{W}{m^2 \cdot K}\).- **\(A\)**: Surface area of the box. Calculating this, one finds \(1.5 m^2\) (since it has 6 equal size faces).- **\(\Delta T\)**: Temperature difference between the internal box and external environment, which is \(700^{\circ} C - 150^{\circ} C = 550 K\).- **\(t\)**: Time duration in seconds, as given, \(2 \text{ hours} = 7200 \text{ seconds}\).

Inserting these values, the total heat transferred over 2 hours is computed to be \(39,690,000 J\). Understanding this principle ensures the insulation keeps heat infiltration at a manageable level.
Fire Resistance
Fire resistance is crucial in ensuring safety and protecting contents from high temperatures over a significant duration. This concept refers to the ability of a material or assembly to withstand fire without losing its integral properties. For our safety box:

- **Box Exposure**: The outer environment may be up to \(700^{\circ} C\) for 2 hours.- **Internal Safety Goal**: Maintain temperatures below \(150^{\circ} C\).

Selecting materials that inherently resist burning and insulate heat effectively is key to maintaining such resistance. Materials used must not degrade or allow heat to permeate through them quickly. They must also be tested for performance over time to ensure they maintain their protective qualities throughout exposure. This guarantees safety boxes maintain contents intact even during intense fire scenarios.
Insulation Material Selection
Choosing the right insulation is critical to effective thermal protection. To ensure maximal protection:

- **Ceramic Fiber Insulation**: Offers high-temperature resistance \(1260^{\circ} C\), and thermal conductivity of \(0.1 \frac{W}{m \cdot K}\).- **Mineral Fiber Insulation**: Known for cost-effectiveness and good thermal performance.- **Calcium Silicate Insulation**: Durable with commendable thermal resistance.

Factors in selection include:- Heat resistance: Must endure high temperatures.- Cost-effectiveness: Viable for commercial applications.- Durability and strength: Long-lasting performance under stress conditions.
Selecting the best insulation ensures that safety boxes remain robust, effective, and economically viable.
Thermal Conductivity
Thermal conductivity stands at the heart of insulation effectiveness. It measures how well heat is transmitted through a material. Lower values signify better insulating properties. For instance:

- **Ceramic Fiber Insulation**: \(0.1 \frac{W}{m \cdot K}\) indicates excellent insulation for high temperatures.

The thermal conductivity of a material influences the required thickness and, subsequently, the overall volume available inside the box:
  • It determines how much material is needed to attain desired insulation levels.
  • Impacts the cost and structural considerations of the safety box design.
Selecting materials with appropriate thermal conductivities ensures the box is effective and minimizes unnecessary thickness. Calculations derived from conductivity values guide decision-making during the design phase, ensuring optimal safety and material efficiency.

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Most popular questions from this chapter

Consider a large plane wall of thickness \(L=0.4 \mathrm{~m}\), thermal conductivity \(k=2.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=20 \mathrm{~m}^{2}\). The left side of the wall is maintained at a constant temperature of \(95^{\circ} \mathrm{C}\), while the right side loses heat by convection to the surrounding air at \(T_{\infty}=15^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming steady one-dimensional heat transfer and taking the nodal spacing to be \(10 \mathrm{~cm},(a)\) obtain the finite difference formulation for all nodes, \((b)\) determine the nodal temperatures by solving those equations, and (c) evaluate the rate of heat transfer through the wall.

Two 3-m-long and 0.4-cm-thick cast iron \((k=\) \(52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \varepsilon=0.8)\) steam pipes of outer diameter \(10 \mathrm{~cm}\) are connected to each other through two \(1-\mathrm{cm}\)-thick flanges of outer diameter \(20 \mathrm{~cm}\). The steam flows inside the pipe at an average temperature of \(200^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(180 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The outer surface of the pipe is exposed to convection with ambient air at \(8^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) as well as radiation with the surrounding surfaces at an average temperature of \(T_{\text {surr }}=290 \mathrm{~K}\). Assuming steady one-dimensional heat conduction along the flanges and taking the nodal spacing to be \(1 \mathrm{~cm}\) along the flange \((a)\) obtain the finite difference formulation for all nodes, \((b)\) determine the temperature at the tip of the flange by solving those equations, and \((c)\) determine the rate of heat transfer from the exposed surfaces of the flange.

Using EES (or other) software, solve these systems of algebraic equations. (a) $$ \begin{aligned} 3 x_{1}+2 x_{2}-x_{3}+x_{4} &=6 \\ x_{1}+2 x_{2}-x_{4} &=-3 \\ -2 x_{1}+x_{2}+3 x_{3}+x_{4} &=2 \\ 3 x_{2}+x_{3}-4 x_{4} &=-6 \end{aligned} $$ (b) $$ \begin{aligned} 3 x_{1}+x_{2}^{2}+2 x_{3} &=8 \\ -x_{1}^{2}+3 x_{2}+2 x_{3} &=-6.293 \\ 2 x_{1}-x_{2}^{4}+4 x_{3} &=-12 \end{aligned} $$

Consider a long concrete dam \((k=0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), (es \(\alpha_{s}=0.7\) ) of triangular cross section whose exposed surface is subjected to solar heat flux of \(\dot{q}_{s}=\) \(800 \mathrm{~W} / \mathrm{m}^{2}\) and to convection and radiation to the environment at \(25^{\circ} \mathrm{C}\) with a combined heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The \(2-\mathrm{m}\)-high vertical section of the dam is subjected to convection by water at \(15^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and heat transfer through the 2-m-long base is considered to be negligible. Using the finite difference method with a mesh size of \(\Delta x=\Delta y=1 \mathrm{~m}\) and assuming steady two-dimensional heat transfer, determine the temperature of the top, middle, and bottom of the exposed surface of the dam.

Consider transient one-dimensional heat conduction in a pin fin of constant diameter \(D\) with constant thermal conductivity. The fin is losing heat by convection to the ambient air at \(T_{\infty}\) with a heat transfer coefficient of \(h\) and by radiation to the surrounding surfaces at an average temperature of \(T_{\text {surr }}\). The nodal network of the fin consists of nodes 0 (at the base), 1 (in the middle), and 2 (at the fin tip) with a uniform nodal spacing of \(\Delta x\). Using the energy balance approach, obtain the explicit finite difference formulation of this problem for the case of a specified temperature at the fin base and negligible heat transfer at the fin tip.
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