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.

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, \(\Delta T = T_{env} - T_{box} = 700°C - 150°C = 550\,K\). 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°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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