Question

Air is flowing between two infinitely large parallel plates. The upper plate is at \(500 \mathrm{~K}\) and has an emissivity of \(0.7\), while the lower plate is a black surface with temperature at \(330 \mathrm{~K}\). If the air temperature is \(290 \mathrm{~K}\), determine the convection heat transfer coefficient associated with the air.

Short answer

Based on the given information, the convection heat transfer coefficient between the two parallel plates and the air flowing between them is approximately: \(h = \frac{5.67 \times 10^{-8} \cdot 0.7 \cdot 1 \cdot (500^4 - 330^4)}{(500 - 290)}\) Calculating the above expression gives: \(h \approx 49.12 \mathrm{~W/m^2K}\)

Step-by-step solution

Calculate the radiative heat transfer

To find the radiative heat transfer, we need to use the Stefan-Boltzmann law. The formula for radiative heat transfer is given by: \(Q_{rad} = \sigma \cdot \varepsilon_{1} \cdot \varepsilon_{2} \cdot (T_{1}^{4} - T_{2}^{4})\) where: - \(\sigma\) is the Stefan-Boltzmann constant, \(\sigma = 5.67 \times 10^{-8} \mathrm{~W/m^2K^4}\) - \(\varepsilon_{1}\) and \(\varepsilon_{2}\) are the emissivities of the upper and lower plates respectively - \(T_{1}\) and \(T_{2}\) are the temperatures of the upper and lower plates respectively. Plugging in the given values: \(Q_{rad} = 5.67 \times 10^{-8} \cdot 0.7 \cdot 1 \cdot (500^4 - 330^4)\) Now, calculate the radiative heat transfer CURLOPT_MATH_OUTPUT_OTHER_THAN_MATH_OUTPUT.

Overall heat transfer

The overall heat transfer between the plates equals the sum of radiative and convective heat transfer. Using this, we can set up the following equation: \(Q_{total} = Q_{rad} + Q_{conv}\) Where \(Q_{conv}\) is the convective heat transfer: \(Q_{conv} = h \cdot A \cdot (T_{1} - T_{air})\) where: - \(h\) is the convection heat transfer coefficient (which we want to find) - \(A\) is the surface area of the plates (which is assumed to be the same for both plates since they are parallel and infinitely large) - \(T_{air}\) is the air temperature. Considering the flows are in steady state, then \(Q_{rad} = Q_{total} = Q_{conv}\) So, \(h \cdot A \cdot (T_{1} - T_{air}) = \sigma \cdot \varepsilon_{1} \cdot \varepsilon_{2} \cdot (T_{1}^{4} - T_{2}^{4})\)

Calculate the convection heat transfer coefficient

We can now solve the equation for the convection heat transfer coefficient (\(h\)): \(h = \frac{\sigma \cdot \varepsilon_{1} \cdot \varepsilon_{2} \cdot (T_{1}^{4} - T_{2}^{4})}{A \cdot (T_{1} - T_{air})}\) Plugging in the given values: \(h = \frac{5.67 \times 10^{-8} \cdot 0.7 \cdot 1 \cdot (500^4 - 330^4)}{(500 - 290)}\) Now, divide the radiative heat transfer semiclass=filedrawer_escape_for_latex_environment_TeX-Live_open_proc>}CUTPUT_MATH_OUTPUT_OTHER_THAN_MATH_OUTPUT{semiclass=filedrawer_escape_for_latex_environment_TeX-Live_close_proc>}' by the temperature difference to find the convection heat transfer coefficient, \(h\): \(h = \mathrm{heat \transfer \coefficient \@ provided \values} \mathrm{~W/m^2K}\)