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A heated 6-mm-thick Pyroceram plate $\left(\rho=2600 \mathrm{~kg} / \mathrm{m}^{3}\right.\(, \)c_{p}=808 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=3.98 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, and \)\left.\alpha=1.89 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)$ is being cooled in a room with air temperature of \(25^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(13.3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heated Pyroceram plate had an initial temperature of \(500^{\circ} \mathrm{C}\), and it is allowed to cool for \(286 \mathrm{~s}\). If the mass of the Pyroceram plate is \(10 \mathrm{~kg}\), determine the heat transfer from the Pyroceram plate during the cooling process using the analytical one-term approximation method.

Short Answer

Expert verified
The heat transfer from the Pyroceram plate during the cooling process is 3,826,200 J.

Step by step solution

01

Calculate the Biot Number (Bi)

The Biot number (Bi) is a dimensionless number that characterized the ratio of internal conductive resistance to external convective resistance. It's calculated as: $$Bi=\frac{hL_{c}}{k}$$ Where: \(h\) = convection heat transfer coefficient \((13.3 \,\mathrm{W/m^2K})\) \(L_{c}\) = characteristic length (thickness of the plate, \(0.006 \,\mathrm{m}\)) \(k\) = thermal conductivity of Pyroceram \((3.98\,\mathrm{W/mK})\) Now, calculate the Biot number: $$Bi=\frac{13.3(0.006)}{3.98} = 0.02$$
02

Calculate the Fourier Number (Fo) times time (t)

The Fourier number (Fo) is a dimensionless number characterized by the ratio of the heat conduction rate to the heat storage rate. Multiply the Fourier number by the cooling time to obtain the product: $$Fo\cdot t=\frac{\alpha t}{L_{c}^{2}}$$ Where: \(\alpha\) = thermal diffusivity of Pyroceram \((1.89\times 10^{-6}\,\mathrm{m^2/s})\) \(t\) = cooling time \((286\,\mathrm{s})\) \(L_{c}\) = characteristic length \((0.006\,\mathrm{m})\) Now, calculate the product Fo and t: $$Fo\cdot t=\frac{1.89\times 10^{-6}(286)}{(0.006)^{2}}=2.694$$
03

Obtain the one-term approximation dimensionless temperature (\(\Theta\)) Using Bi and Fo

The one-term approximation dimensionless temperature is obtained from the chart or table provided in textbooks where the X-axis represents the Biot number (Bi) and the Y-axis represents the product of Fourier number (Fo) and time (t): $$\Theta=1.0$$
04

Calculate the actual temperature change and total heat transfer

Calculate the actual temperature change of the plate using the initial temperature (\(T_{i}\)), room temperature (\(T_{\infty}\)), and dimensionless temperature (\(\Theta\)): $$\Delta T=(T_{i}-T_{\infty})\Theta$$ Next, use the mass (\(m\)), specific heat (\(c_p\)), and actual temperature change to find the total heat transfer (\(Q\)): $$Q=mc_{p}\Delta T$$ Plug in the values: $$\Delta T=(500-25)1.0=475\,\mathrm{K}$$ $$Q=(10)(808)(475)=3,826,200\,\mathrm{J}$$ The heat transfer from the Pyroceram plate during the cooling process is 3,826,200 J using the one-term approximation method.

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Most popular questions from this chapter

What is a refrigerated shipping dock? How does it reduce the refrigeration load of the cold storage rooms?

Consider the engine block of a car made of cast iron $\left(k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\( and \)\left.\alpha=1.7 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)$. The engine can be considered to be a rectangular block whose sides are \(80 \mathrm{~cm}\), \(40 \mathrm{~cm}\), and \(40 \mathrm{~cm}\). The engine is at a temperature of \(150^{\circ} \mathrm{C}\) when it is turned off. The engine is then exposed to atmospheric air at \(17^{\circ} \mathrm{C}\) with a heat transfer coefficient of $6 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\(. Determine \)(a)$ the center temperature of the top surface whose sides are \(80 \mathrm{~cm}\) and \(40 \mathrm{~cm}\) and \((b)\) the corner temperature after 45 min of cooling. Solve this problem using the analytical one-term approximation method.

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A \(10-\mathrm{cm}\)-inner-diameter, 30-cm-long can filled with water initially at \(25^{\circ} \mathrm{C}\) is put into a household refrigerator at $3^{\circ} \mathrm{C}\(. The heat transfer coefficient on the surface of the can is \)14 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Assuming that the temperature of the water remains uniform during the cooling process, the time it takes for the water temperature to drop to \(5^{\circ} \mathrm{C}\) is (a) \(0.55 \mathrm{~h}\) (b) \(1.17 \mathrm{~h}\) (c) \(2.09 \mathrm{~h}\) (d) \(3.60 \mathrm{~h}\) (e) \(4.97 \mathrm{~h}\)

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