Solar radiation is incident on a \(5 \mathrm{~m}^{2}\) solar absorber plate surface at a rate of \(800 \mathrm{~W} / \mathrm{m}^{2}\). Ninety-three percent of the solar radiation is absorbed by the absorber plate, while the remaining 7 percent is reflected away. The solar absorber plate has a surface temperature of \(40^{\circ} \mathrm{C}\) with an emissivity of \(0.9\) that experiences radiation exchange with the surrounding temperature of \(-5^{\circ} \mathrm{C}\). In addition, convective heat transfer occurs between the absorber plate surface and the ambient air of \(20^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the efficiency of the solar absorber, which is defined as the ratio of the usable heat collected by the absorber to the incident solar radiation on the absorber.

Step by step solution

01

Calculate the absorbed solar radiation

To find the absorbed solar radiation, we have to multiply the incident solar radiation rate by the area of the solar absorber plate and then by the absorption percentage: $$ Q_\text{absorbed} = (\text{Radiation Rate}) \times (\text{Absorber Area}) \times (\text{Absorption Percentage}) $$ $$ Q_\text{absorbed} = 800 \frac{\text{W}}{\text{m}^2} \times 5 \text{m}^2 \times 0.93 $$ $$ Q_\text{absorbed} = 3720 \text{W} $$

02

Calculate the radiation heat loss

To find the radiation heat loss, we will use the Stefan-Boltzmann law. First, convert the temperatures from Celsius to Kelvin: $$ T_\text{surface} = 40 + 273.15 = 313.15 \text{K} $$ $$ T_\text{surrounding} = -5 + 273.15 = 268.15 \text{K} $$ Now, apply the Stefan-Boltzmann law: $$ Q_\text{radiation} = \varepsilon \sigma A (T_\text{surface}^4 - T_\text{surrounding}^4) $$ Where \(\varepsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4}\)), and \(A\) is the area of the solar absorber plate. $$ Q_\text{radiation} = 0.9 \times 5.67 \times 10^{-8} \text{W m}^{-2} \text{K}^{-4} \times 5 \text{m}^2 (313.15^4 - 268.15^4) $$ $$ Q_\text{radiation} \approx 81.95 \text{W} $$

03

Calculate the convective heat loss

We can find the convective heat loss by using the following equation: $$ Q_\text{convection} = h A (T_\text{surface} - T_\text{ambient}) $$ Where \(h\) is the convection heat transfer coefficient and \(T_\text{ambient}\) is the ambient temperature. $$ Q_\text{convection} = 7 \frac{\text{W}}{\text{m}^2 \cdot \text{K}} \times 5 \text{m}^2 (40 - 20) $$ $$ Q_\text{convection} = 700 \text{W} $$

04

Calculate the usable heat collected by the solar absorber

Now, we can calculate the usable heat collected by the solar absorber: $$ Q_\text{usable} = Q_\text{absorbed} - Q_\text{radiation} - Q_\text{convection} $$ $$ Q_\text{usable} = 3720 \text{W} - 81.95 \text{W} - 700 \text{W} $$ $$ Q_\text{usable} \approx 2938.05 \text{W} $$

05

Determine the efficiency of the solar absorber

Finally, we can find the efficiency by dividing the usable heat collected by the incident solar radiation on the absorber plate: $$ \eta = \frac{Q_\text{usable}}{Q_\text{incident}} $$ $$ \eta = \frac{2938.05 \text{W}}{4000 \text{W}} $$ $$ \eta \approx 0.7345 \times 100\% $$ $$ \eta \approx 73.45\% $$ The efficiency of the solar absorber is approximately 73.45%.

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