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Chapter 12: Problem 61

A blackbody plate \(\left(A_{1}=5 \mathrm{~cm}^{2}\right)\) is subjected to a uniform heat flux of \(1000 \mathrm{~W} / \mathrm{m}^{2}\) on the bottom, while the top surface is exposed to an ambient surrounding at \(5^{\circ} \mathrm{C}\). The heat transfer coefficient due to natural convection on the plate surface is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). A radiometer is placed \(1 \mathrm{~m}\) above the plate normal to the direction of viewing from the plate. Determine the irradiation that the radiometer would detect.

Short Answer

Expert verified
Based on the given information and calculations, the radiometer placed 1 meter above the blackbody plate would detect an irradiation of approximately \(0.02193~W/m^2\).

Step by step solution

01

Calculate the total heat transfer to the plate (Q)

First, we need to convert the given heat flux \(q''=1000~W/m^2\) to find the total heat transfer to the plate (in Watts). To do this, we multiply the heat flux by the area of the plate \(A_1\) in meters squared. The area of the plate \(A_1 = 5~cm^2\) must be converted to meters squared. $$A_1 = 5 \times 10^{-4}~m^2$$ Now we can find the total heat transfer (Q) by multiplying the heat flux and area: $$Q = q'' \cdot A_1 = 1000~W/m^2 \cdot 5 \times 10^{-4}~m^2 = 0.5~W$$
02

Calculate the plate temperature

Next, we can determine the plate's temperature. We know that the heat transfer due to natural convection (Q_conv) is described by the following formula: $$Q_{conv} = h \cdot A_1 \cdot (T_p - T_{\infty})$$ Where \(h = 5~W/m^2 K\) is the heat transfer coefficient, \(T_p\) is the plate's temperature in Kelvin, and \(T_{\infty} = 5^\circ C = 278.15~K\) is the ambient temperature. From step 1, we know Q=0.5W. So we can rearrange the equation for Q_conv and solve for \(T_p\): $$T_p = \frac{Q}{h \cdot A_1} + T_{\infty} = \frac{0.5~W}{5~W/m^2K \cdot 5 \times 10^{-4}~m^2} + 278.15~K = 428.15~K$$
03

Calculate heat transfer by radiation

Now that we have the plate's temperature, we can calculate the heat transfer by radiation (Q_rad), using the Stefan-Boltzmann Law: $$Q_{rad} = \sigma \cdot A_1 \cdot (T_p^4 - T_{\infty}^4)$$ Where \(\sigma = 5.67 \times 10^{-8}~W/m^2K^4\) is the Stefan-Boltzmann constant. Plugging in the values, we get: $$Q_{rad} = 5.67 \times 10^{-8}~W/m^2K^4 \cdot 5 \times 10^{-4}~m^2 \cdot (428.15^4 - 278.15^4~K^4) = 0.27569~W$$
04

Calculate the irradiation detected by the radiometer

Finally, we can calculate the irradiation (E) detected by the radiometer at a distance of 1 meter using the following equation: $$E = \frac{Q_{rad}}{A_2}$$ Where \(A_2\) is the area of the sphere with a radius of 1 meter: $$A_2 = 4\pi r^2 = 4\pi (1~m)^2 = 12.566~m^2 $$ Now, we can find the irradiation detected by the radiometer: $$E = \frac{0.27569~W}{12.566~m^2} = 0.02193~W/m^2$$ Thus, the radiometer would detect an irradiation of approximately \(0.02193~W/m^2\).

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