Question

A \(0.2-\mathrm{m} \times 0.2-\mathrm{m}\) street sign surface has an absorptivity of \(0.6\) and an emissivity of \(0.7\). Solar radiation is incident on the street sign at a rate of \(200 \mathrm{~W} / \mathrm{m}^{2}\), and the surrounding quiescent air is at \(25^{\circ} \mathrm{C}\). Determine the surface temperature of the street sign. Assume the film temperature is $30^{\circ} \mathrm{C}$.

Short answer

The surface temperature of the street sign is approximately \(36.15^{\circ} \mathrm{C}\).

Step-by-step solution

Convert the temperature values to Kelvin

Given temperatures are in Celsius, and in heat transfer problems, it's preferable to work in Kelvin. $$T_{surrounding} = 25^{\circ} \mathrm{C} + 273.15 = 298.15 \mathrm{K}$$ $$T_{film} = 30^{\circ} \mathrm{C} + 273.15 = 303.15 \mathrm{K}$$

Calculate heat absorbed by the surface due to solar radiation

Given the area of the street sign \(A_s\), absorptivity \(\alpha\), and solar radiation rate \(q_{solar}\), the heat absorbed by the surface due to solar radiation can be calculated as: $$q_{absorbed} = A_s \times \alpha \times q_{solar} = 0.2 \times 0.2 \times 0.6 \times 200 = 4.8\ \mathrm{W}$$

Calculate convection heat transfer coefficient

We can use the film temperature for natural convection in quiescent air to find the convection heat transfer coefficient \(h\). A common empirical correlation for the average Nusselt number over a vertical flat plate is given by: $$Nu = 0.15 (Gr \times Pr)^{1/3}$$ Where: Gr is the Grashof number, given by $$Gr = \dfrac{g \beta (T_s - T_{\infty}) L^3}{\nu^2}$$ g is the acceleration due to gravity $$g = 9.81 \mathrm{m/s^2}$$ \(\beta\) is the volume expansion coefficient $$\beta \approx \dfrac{1}{T_f}$$ \(T_s\) is the surface temperature (unknown) \(T_{\infty}\) is the surrounding temperature (298.15 K) \(L\) is the characteristic length, in our case, $$L = 0.2 \mathrm{m}$$ \(\nu\) is the kinematic viscosity of air $$\nu \approx 15.69 \times 10^{-6} \mathrm{m^2/s}$$ at \(T_f = 303.15\ \mathrm{K}\), taken from air properties table Pr is the Prandtl number $$Pr \approx 0.7$$ for air Since we need to solve for the unknown surface temperature \(T_s\), we will use an iterative method. Let's start with an initial surface temperature, say $$T_s = 30^{\circ} \mathrm{C} + 273.15 = 303.15\ \mathrm{K}$$(same as the film temperature) and calculate the Grashof number: $$Gr = \dfrac{9.81 \times \dfrac{1}{303.15} \times (303.15 - 298.15) \times 0.2^3}{(15.69 \times 10^{-6})^2} \approx 3.58 \times 10^5$$ Now, calculate the Nusselt number: $$Nu = 0.15 (3.58 \times 10^5 \times 0.7)^{1/3} \approx 12.31$$ Finally, find the convection heat transfer coefficient: $$h = \dfrac{Nu \times k}{L} \approx \dfrac{12.31 \times 0.02624}{0.2} = 1.60 \mathrm{W/m^2K}$$

Set up the heat balance equation

The heat absorbed by the street sign due to solar radiation must be equal to the heat transfer away from the street sign by both radiation and convection: $$q_{absorbed} = q_{radiation} + q_{convection}$$ With known values: $$q_{radiation} = A_s \times \varepsilon \times \sigma (T_s^4 - T_{\infty}^4)$$ $$q_{convection} = A_s \times h (T_s - T_{\infty})$$ Combine the two above equations: $$q_{absorbed} = A_s \times \varepsilon \times \sigma (T_s^4 - T_{\infty}^4) + A_s \times h (T_s - T_{\infty})$$ With given values, $$4.8\ \mathrm{W} = 0.2 \times 0.2 \times 0.7 \times 5.669 \times 10^{-8} (T_s^4 - 298.15^4) + 0.2 \times 0.2 \times 1.60 (T_s - 298.15)$$ Solving for \(T_s\) using an iterative method (or numerical solver) yields: $$T_s \approx 309.3\ \mathrm{K}$$

Convert the surface temperature to Celsius

Convert the calculated surface temperature back to Celsius: $$T_s \approx 309.3\ \mathrm{K} - 273.15 = 36.15^{\circ} \mathrm{C}$$ The surface temperature of the street sign is approximately \(36.15^{\circ} \mathrm{C}\).