Question

A large ASTM B152 copper plate is placed in parallel near a large ceramic plate. The ceramic plate is at a temperature of \(520^{\circ} \mathrm{C}\). The copper and ceramic plates have emissivity values of \(0.15\) and \(0.92\), respectively. For the ASTM B152 copper plate, the ASME Code for Process Piping specifies the maximum use temperature at \(260^{\circ} \mathrm{C}\) (ASME B31.3-2014, Table A-1M). If the net radiation heat flux between the two parallel plates is \(2000 \mathrm{~W} / \mathrm{m}^{2}\), determine whether the ASTM B 152 copper plate would comply with the ASME code.

Short answer

Answer: No, the ASTM B152 copper plate does not comply with the ASME code under these conditions, as the calculated temperature of \(410.99^{\circ} \mathrm{C}\) is greater than the maximum use temperature specified by ASME code (\(260^{\circ} \mathrm{C}\)).

Step-by-step solution

Write down the formula for net radiation heat transfer between two parallel plates

We will use the following formula to calculate the net radiation heat flux between the two parallel plates: \[q = \sigma (e_A T_A^4 - e_B T_B^4)\] Where: \(q\) is the net radiation heat flux (\(\mathrm{W/m^2}\)) \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W/(m^2 K^4)}\)) \(e_A\) and \(e_B\) are the emissivities of the two plates (dimensionless) \(T_A\) and \(T_B\) are the absolute temperatures of the two plates in Kelvin We will use this formula to find the temperature of the copper plate, \(T_A\).

Convert temperatures to Kelvin and insert the given data into the formula

First, we need to convert the temperature of the ceramic plate to Kelvin: \(520^{\circ} \mathrm{C} + 273.15 = 793.15 \mathrm{K}\) Now, insert the given data into the formula: \(2000 \mathrm{~W/m^2} = 5.67 \times 10^{-8} \mathrm{W/(m^2 K^4)} (0.15 \cdot T_A^4 - 0.92 \cdot (793.15 \mathrm{K})^4)\)

Solve the equation for the temperature of the copper plate, \(T_A\)

Rearrange the formula to solve for \(T_A\): \(T_A^4 = \frac{2000 \mathrm{~W/m^2} + 0.92 \cdot (793.15 \mathrm{K})^4 \times 5.67 \times 10^{-8} \mathrm{W/(m^2 K^4)}}{0.15 \times 5.67 \times 10^{-8} \mathrm{W/(m^2 K^4)}}\) Calculate the value of \(T_A^4\): \(T_A^4 = \frac{2000 + 0.92 \cdot (793.15)^4 \times 5.67 \times 10^{-8}}{0.15 \times 5.67 \times 10^{-8}} = 2.229 \times 10^9\) Now, find the temperature of the copper plate by taking the 4th root of the previous result: \(T_A = (2.229 \times 10^9)^{\frac{1}{4}} = 684.14 \mathrm{K}\)

Convert the temperature to Celsius and compare to ASME code's maximum use temperature

To determine if the copper plate complies with the ASME code, convert the temperature to Celsius and compare it to the maximum use temperature specified in the ASME code: \(684.14 \mathrm{K} - 273.15 = 410.99^{\circ} \mathrm{C}\) Since \(410.99 ^{\circ} \mathrm{C}\) is greater than the maximum use temperature specified by ASME code, which is \(260^{\circ} \mathrm{C}\), the ASTM B152 copper plate would not comply with the ASME code under these conditions.