Question

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.

Short answer

Question: Calculate the efficiency of a solar absorber with given parameters, such as solar radiation rate, absorber area, absorption percentage, emissivity, surface temperature, surrounding temperature, ambient temperature, and convection heat transfer coefficient. Answer: The efficiency of the solar absorber is approximately 73.45%.

Step-by-step solution

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: $$ Q_\text{absorbed} = (\text{Radiation Rate}) \times (\text{Absorber Area}) \times (\text{Absorption Percentage}) $$ $$ Q_\text{absorbed} = 800 \frac{\text{W}}{\text{m}^2} \times 5 \text{m}^2 \times 0.93 $$ $$ Q_\text{absorbed} = 3720 \text{W} $$

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. $$ Q_\text{radiation} = 0.9 \times 5.67 \times 10^{-8} \text{W m}^{-2} \text{K}^{-4} \times 5 \text{m}^2 (313.15^4 - 268.15^4) $$ $$ Q_\text{radiation} \approx 81.95 \text{W} $$

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. $$ Q_\text{convection} = 7 \frac{\text{W}}{\text{m}^2 \cdot \text{K}} \times 5 \text{m}^2 (40 - 20) $$ $$ Q_\text{convection} = 700 \text{W} $$

Calculate the usable heat collected by the solar absorber

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

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%.

Key concepts

Radiation Heat Transfer

Radiation heat transfer involves the transfer of heat energy by electromagnetic waves. Every object emits radiation energy, and the amount depends on its temperature. In the context of solar absorbers, radiation exchange occurs between the absorber plate and surrounding bodies.

For example, the solar absorber in this exercise radiates heat energy to its colder surroundings. The amount of radiation heat loss can be critical for calculating the efficiency of the system. The Stefan-Boltzmann law is used to determine the radiation heat loss, considering factors like the emissivity of the material and the difference in temperature with the surroundings.

Convective Heat Transfer

Convective heat transfer is another important mechanism through which heat is lost or gained. It involves the movement of heat through a fluid (such as air) in contact with a solid surface. The heat transfer from the solar absorber to the surrounding air is an example of this.

The convective heat loss is calculated using the convective heat transfer coefficient and the temperature difference between the surface and the ambient air. This loss needs to be minimized or accounted for to improve the system's efficiency.

Stefan-Boltzmann Law

The Stefan-Boltzmann law is crucial in understanding radiation heat transfer. It states that the total power radiated per unit area by a black body is proportional to the fourth power of its absolute temperature.

The equation is: \[ Q_{\text{radiation}} = \varepsilon \sigma A (T_{\text{surface}}^4 - T_{\text{surrounding}}^4) \] where:

• \(\varepsilon\) is the emissivity of the material
• \(\sigma\) is the Stefan-Boltzmann constant \(5.67 \times 10^{-8} \text{W m}^{-2} \text{K}^{-4}\)
• \(A\) is the area
• \(T\) values are temperatures in Kelvin

This law allows us to quantify the radiation energy exchange for designing efficient thermal systems.

Solar Radiation Absorption

Solar radiation absorption is a measure of how much solar energy is captured by a surface, like a solar absorber. It is essential to maximize absorption for effective energy utilization. In our exercise, 93% of the incident solar radiation is absorbed by the absorber plate.

The calculation for absorbed solar radiation involves multiplying the incident energy rate by the absorber area and the absorption percentage. This gives the total energy but not all translates to usable energy due to heat losses.

Heat Transfer Coefficient

The heat transfer coefficient is a measure of the heat transfer rate per unit area and temperature difference. It plays a vital role in both radiation and convective heat transfer calculations.

In the solar absorber example, the convection heat transfer coefficient is used to determine the convective heat loss. It is characterized by factors like air velocity, surface area, and temperature.

Knowing the heat transfer coefficients allows engineers to calculate and understand the efficiency of thermal systems accurately.