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

Expert verified
Answer: The surface temperature of the street sign is approximately 32.03°C.

Step by step solution

01

Understand the given data and formulate the energy balance equation

We are given the following data: - \(A = 0.2 \mathrm{~m} \times 0.2 \mathrm{~m}\) (area of the street sign) - \(\alpha = 0.6\) (absorptivity) - \(\epsilon = 0.7\) (emissivity) - \(G = 200 \mathrm{~W} / \mathrm{m}^{2}\) (solar radiation rate) - \(T_{\infty} = 25^{\circ} \mathrm{C}\) (surrounding air temperature) - \(T_{f} = 30^{\circ} \mathrm{C}\) (film temperature) Now, let's formulate the energy balance equation: The energy absorbed by the street sign is equal to the energy radiated and convected to the surroundings. Energy absorbed by the street sign can be calculated as: \(\alpha GA\). Energy radiated to the surroundings can be calculated as : \(q_{r} = \epsilon \sigma A (T_{s}^4 - T_{\infty}^4)\), where \(\sigma\) is the Stefan-Boltzmann constant and \(T_{s}\) is the unknown surface temperature we need to find. Energy convected to the surroundings can be calculated as : \(q_{c} = hA(T_{s} - T_{\infty})\), where \(h\) is the heat transfer coefficient. So our energy balance equation is: \(\alpha GA=\epsilon \sigma A (T_{s}^4 - T_{\infty}^4) + hA(T_{s} - T_{\infty})\).
02

Find the natural convection heat transfer coefficient (h)

We need to find the heat transfer coefficient (h) to proceed. We will use the dimensions of the street sign and the film temperature to find h using the following correlation for natural convection from a vertical plate: \(h = 1.32 (T_{f} - T_{\infty}) / L^{1/3}\), where \(L\) is the height of the street sign. In this case, \(L = 0.2 \mathrm{~m}\), \(T_{f} = 30^{\circ} \mathrm{C}\), and \(T_{\infty} = 25^{\circ} \mathrm{C}\). \(h = 1.32 (30 - 25) / 0.2^{1/3}\). Calculating h: \(h \approx 7.634 \mathrm{~W} / (\mathrm{m}^{2} \cdot \mathrm{K})\).
03

Substitute values in the energy balance equation and solve for street sign surface temperature

Now we have all the necessary values to solve the energy balance equation: \(\alpha GA=\epsilon \sigma A (T_{s}^4 - T_{\infty}^4) + hA(T_{s} - T_{\infty})\). Plug in the given values and the value of h we just calculated: \(0.6 \cdot 200 \cdot (0.2 \cdot 0.2) = 0.7 \cdot \sigma \cdot (0.2 \cdot 0.2) (T_{s}^4 - (25 + 273.15)^4) + 7.634 \cdot (0.2 \cdot 0.2)(T_{s} - (25 + 273.15))\). Now, we have an equation in terms of the one unknown, \(T_{s}\). Solving for \(T_{s}\), we get: \(T_{s} \approx 305.18 \mathrm{~K}\). Finally, we obtain the surface temperature of the street sign: \(T_{s} = 305.18 \mathrm{~K} - 273.15 \mathrm{~K} \approx 32.03^{\circ} \mathrm{C}\). So, the surface temperature of the street sign is approximately \(32.03^{\circ} \mathrm{C}\).

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