Question

A \(0.5-\mathrm{m}\)-long thin vertical plate is subjected to uniform heat flux on one side, while the other side is exposed to cool air at $5^{\circ} \mathrm{C}\(. The plate surface has an emissivity of \)0.73$, and its midpoint temperature is \(55^{\circ} \mathrm{C}\). Determine the heat flux on the plate surface.

Short answer

Answer: The heat flux on the surface of the thin vertical plate is approximately \(1372.43\,\mathrm{W/m^2}\).

Step-by-step solution

List the given data

The exercise provided us with the following information: Length of the plate, L = \(0.5\,\mathrm{m}\) Midpoint temperature of the plate, T_m = \(55^{\circ}\mathrm{C}\) Temperature of cool air, T_a = \(5^{\circ}\mathrm{C}\) Emissivity of the plate surface, ε = 0.73 First, we need to convert the temperatures to Kelvin: T_m(K) = T_m(C) + 273.15 = \(55 + 273.15 = 328.15\,\mathrm{K}\) T_a(K) = T_a(C) + 273.15 = \(5 + 273.15 = 278.15\,\mathrm{K}\)

Calculate the radiated heat flux

The radiated heat flux is given by the Stefan-Boltzmann law for a black body, which states that \(q_{rad} = εσ(T_m^4 - T_a^4)\), where σ is the Stefan-Boltzmann constant (\(5.67\times10^{-8}\,\mathrm{W/m^2K^4}\)). Now we can calculate the radiated heat flux: \(q_{rad} = 0.73\times5.67\times10^{-8}(328.15^4 - 278.15^4) = 1257.16\,\mathrm{W/m^2}\)

Calculate the convective heat flux

We will use Newton's Law of cooling to find the convective heat transfer coefficient (h): \(q_{conv} = h(T_m - T_a)\). However, we need to find h first. To do this, we will assume a correlation for the Nusselt number - a dimensionless number, which can be expressed as \(Nu = \frac{hL}{k}\), where L is the characteristic length and k is the thermal conductivity. For vertical surfaces, a rough estimate can be obtained using the correlation: \(Nu = 0.68 + 0.67Ra^{1/4}Pr^{1/3}\), where Ra is the Rayleigh number and Pr is the Prandtl number. The Rayleigh number is given by \(Ra = Gr \times Pr\), where Gr is the Grashof number. The Prandtl number is a material property and for air at room temperature is approximately 0.7. The Grashof number is given by: \(Gr = \frac{g\beta(T_m - T_a)L^3}{\nu^2}\), where g is the acceleration due to gravity (\(9.81\,\mathrm{m/s^2}\)), β is the coefficient of thermal expansion of air (approximately \(3.56\times10^{-3}\,\mathrm{K^{-1}}\)) and ν is the kinematic viscosity of air (approximately \(15.7\times10^{-6}\,\mathrm{m^2/s}\)). Now, we can calculate Gr: \(Gr = \frac{9.81\times3.56\times10^{-3}(328.15-278.15)(0.5)^3}{(15.7\times10^{-6})^2} = 2.595\times10^8\) Next, we can find Ra: \(Ra = Gr\times Pr = 2.595\times10^8\times0.7 = 1.817\times10^8\). Now we can calculate Nu: \(Nu = 0.68 + 0.67(1.817\times10^8)^{1/4}(0.7)^{1/3} = 44.11\) Finally, we can determine the heat transfer coefficient h: \(h = \frac{Nu\times k}{L} = \frac{44.11\times 0.026}{0.5} = 2.31\,\mathrm{W/m^2K}\) Now, using Newton's Law of Cooling, we can find the convective heat flux, \(q_{conv}\): \(q_{conv} = h(T_m - T_a) = 2.31(328.15 - 278.15) = 115.27\,\mathrm{W/m^2}\)

Determine the total heat flux

The total heat flux on the plate surface is the sum of the radiated and convective heat fluxes: \(q_{total} = q_{rad} + q_{conv} = 1257.16 + 115.27 = 1372.43\,\mathrm{W/m^2}\) The heat flux on the plate surface is approximately \(1372.43\,\mathrm{W/m^2}\).