4-115 A semi-infinite aluminum cylinder \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\left.\alpha=9.71 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\) of diameter \(D=15 \mathrm{~cm}\) is initially at a uniform temperature of \(T_{i}=115^{\circ} \mathrm{C}\). The cylinder is now placed in water at \(10^{\circ} \mathrm{C}\), where heat transfer takes place by convection with a heat transfer coefficient of \(h=140 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperature at the center of the cylinder \(5 \mathrm{~cm}\) from the end surface 8 min after the start of cooling. 4-116 A 20-cm-long cylindrical aluminum block \((\rho=\) \(2702 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=0.896 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=236 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=\) \(\left.9.75 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right), 15 \mathrm{~cm}\) in diameter, is initially at a uniform temperature of \(20^{\circ} \mathrm{C}\). The block is to be heated in a furnace at \(1200^{\circ} \mathrm{C}\) until its center temperature rises to \(300^{\circ} \mathrm{C}\). If the heat transfer coefficient on all surfaces of the block is \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine how long the block should be kept in the furnace. Also, determine the amount of heat transfer from the aluminum block if it is allowed to cool in the room until its temperature drops to \(20^{\circ} \mathrm{C}\) throughout.

Step by step solution

01

List the given parameters

- Initial temperature of the block, \(T_i = 20^{\circ}\mathrm{C}\) - Length of the block, \(L = 20 \,\mathrm{cm} = 0.2 \,\mathrm{m}\) - Diameter of the block, \(D = 15 \,\mathrm{cm} = 0.15 \,\mathrm{m}\) - Density of aluminum, \(\rho = 2702 \,\mathrm{kg/m^3}\) - Specific heat capacity, \(c_p = 0.896 \,\mathrm{kJ/kg\cdot K} = 896 \,\mathrm{J/kg\cdot K}\) - Thermal conductivity, \(k = 236\,\mathrm{W/m\cdot K}\) - Thermal diffusivity, \(\alpha = 9.75 \times 10^{-5}\,\mathrm{m^2/s}\) - Furnace temperature, \(T_f = 1200^{\circ}\mathrm{C}\) - Desired center temperature, \(T_c = 300^{\circ}\mathrm{C}\) - Heat transfer coefficient, \(h = 80\,\mathrm{W/m^2\cdot K}\)

02

Find the Bi and Fourier Numbers

We need to find the Biot Number \((\mathrm{Bi})\) and Fourier number \((\mathrm{Fo})\) in order to use the heat transfer equation effectively. The Biot Number is given by: $$ \mathrm{Bi}=\frac{h \cdot L_c}{k}=\frac{h(D/2)}{k} $$ The Fourier number is given by: $$ \mathrm{Fo}=\frac{\alpha \cdot t}{L_c^2}=\frac{\alpha \cdot t}{(D/2)^2} $$ We need to find the value of time \(t\) that satisfies both the Biot and Fourier numbers such that the desired center temperature \(T_c = 300^{\circ}\mathrm{C}\) is achieved.

03

Apply the lumped heat capacity method

Since we don't know the required time \(t\), we can use the lumped heat capacity method, which states that: $$ \frac{T(t)-T_i}{T_f-T_i} = e^{-\mathrm{Bi\cdot Fo}} $$ We can rearrange the formula to solve for time \(t\): $$ t = \frac{(D/2)^2}{\alpha}ln\left(\frac{T_f-T_i}{T_c-T_i}\right) $$

04

Calculate the time

Substitute the given values into the equation: $$ t = \frac{(0.15/2)^2}{9.75 \times 10^{-5}}ln\left(\frac{1200-20}{300-20}\right) $$ Calculate the result: $$ t = 808.3 \,\mathrm{s} $$

05

Calculate the total heat transfer

To find the total heat transfer from the block, we use the formula: $$ Q = mc_p\Delta T = \rho Vc_p(T_i - T_c) $$ Where: $$ V = \pi(D/2)^2L $$ Calculate the volume and heat transfer: $$ V = \pi(0.15/2)^2(0.2) = 0.00353 \,\mathrm{m^3} $$ $$ Q = 2702\cdot 0.00353\cdot 896\cdot(20-300) = -8.98\times10^{6} \,\mathrm{J} $$ #Results# The block should be kept in the furnace for \(808.3\,\mathrm{s}\) in order for the center temperature to rise to \(300^{\circ}\mathrm{C}\). The total amount of heat transfer from the block if it is allowed to cool down to \(20^{\circ}\mathrm{C}\) would be \(-8.98\times 10^{6}\,\mathrm{J}\).

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