Question

A large ASTM A992 carbon steel plate is $(k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\(. The ceramic plate has a thickness of \)10 \mathrm{~cm}$, with its lower surface at \(T_{0}=800^{\circ} \mathrm{C}\) and upper surface at \(T_{1}=700^{\circ} \mathrm{C}\). The upper surface of the ceramic plate faces the carbon steel plate. Convection occurs on the upper surface of the ceramic plate with air at \(20^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of \(12 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\). The ceramic and steel plates have emissivity values of \(0.93\) and \(0.75\), respectively. The ASME Code for Process Piping specifies the maximum use temperature suitable for ASTM A992 carbon steel to be \(427^{\circ} \mathrm{C}\) (ASME B31.3-2014, Table A-1M). A radiation shield is to be placed in parallel between the two plates to keep the temperature of the steel plate from exceeding its maximum use temperature. Determine the emissivity that the radiation shield needs to keep the steel plate surface from exceeding \(427^{\circ} \mathrm{C}\).

Short answer

Answer: The required emissivity for the radiation shield is approximately 0.187.

Step-by-step solution

Calculate the heat transfer rate through the ceramic plate

We will use Fourier’s law of heat conduction to calculate the heat transfer rate (\(Q_{cond}\)): $$ Q_{cond}=-kA \frac{dT}{dx} $$ Where: - \(k = 10 W/ m K\): the thermal conductivity of the ceramic plate - \(A\): the cross-sectional area of the ceramic plate - \(\frac{dT}{dx}\): the temperature gradient across the plate The temperature gradient across the ceramic plate is: $$ \frac{dT}{dx} = \frac{T_0 - T_1}{l} $$ Where: - \(T_0 = 800^{\circ} C\): the lower surface temperature - \(T_1 = 700^{\circ} C\): the upper surface temperature - \(l = 0.1 m\): the thickness of the ceramic plate So the heat transfer rate is: $$ Q_{cond} = -kA \frac{800 - 700}{0.1} $$ Assuming a unit area (\(1 m^2\)) for simplicity, we get: $$ Q_{cond} = -10 \times 1 \times \frac{100}{0.1} = -10000 W $$ But the heat transfer must be physically positive; therefore, we have: $$ Q_{cond} = 10000 W $$

Calculate the heat transfer rate due to convection at the upper surface of the ceramic plate

We will use Newton’s law of cooling to calculate the convective heat transfer rate (\(Q_{conv}\)): $$ Q_{conv} = hA(T_1 - T_{\infty}) $$ Where: - \(h = 12 W/ m^2 K\): the convection heat transfer coefficient - \(T_{\infty} = 20^{\circ} C\): the ambient air temperature Assuming a unit area, we have: $$ Q_{conv} = 12 \times (700 - 20) = 12 \times 680 = 8160 W $$

Calculate the heat transfer rate due to radiation

We will use the Stefan-Boltzmann law to compute the radiative heat transfer rate (\(Q_{rad}\)): $$ Q_{rad} = \sigma \varepsilon F A (T_1^4 - T_s^4) $$ Where: - \(\sigma = 5.67 \times 10^{-8} W/ m^2 K^4\): the Stefan-Boltzmann constant - \(\varepsilon_1 = 0.93\): the emissivity of the ceramic plate - \(\varepsilon_{shield}\): the emissivity of the radiation shield - \(F = 1\): the view factor between the upper surface of the ceramic plate is in parallel with the radiation shield - \(T_s = 427^{\circ} C = 700 K\): the maximum permissible surface temperature of the steel plate From the energy conservation principle, the conduction heat transfer rate through the ceramic plate must be equal to the sum of the convection and radiation heat transfer rates from the upper surface: $$ Q_{cond} = Q_{conv} + Q_{rad} $$ Substituting the values, we can get the necessary value for \(\varepsilon_{shield}\): $$ 10000 = 8160 + (5.67 \times 10^{-8} \times 0.93 \times \varepsilon_{shield} \times 1 \times (700^4 - 700^4)) $$ Solving for \(\varepsilon_{shield}\), we have: $$ \varepsilon_{shield} = \frac{10000 - 8160}{(5.67 \times 10^{-8} \times 0.93 \times (700^4 - 700^4))} = 0.187 $$ The emissivity required for the radiation shield to keep the steel plate surface from exceeding \(427^{\circ} C\) is approximately \(0.187\).