Question

Solar radiation is incident on the glass cover of a solar collector at a rate of \(700 \mathrm{~W} / \mathrm{m}^{2}\). The glass transmits 88 percent of the incident radiation and has an emissivity of \(0.90\). The entire hot water needs of a family in summer can be met by two collectors \(1.2 \mathrm{~m}\) high and \(1 \mathrm{~m}\) wide. The two collectors are attached to each other on one side so that they appear like a single collector \(1.2 \mathrm{~m} \times 2 \mathrm{~m}\) in size. The temperature of the glass cover is measured to be \(35^{\circ} \mathrm{C}\) on a day when the surrounding air temperature is \(25^{\circ} \mathrm{C}\) and the wind is blowing at \(30 \mathrm{~km} / \mathrm{h}\). The effective sky temperature for radiation exchange between the glass cover and the open sky is \(-40^{\circ} \mathrm{C}\). Water enters the tubes attached to the absorber plate at a rate of \(1 \mathrm{~kg} / \mathrm{min}\). Assuming the back surface of the absorber plate to be heavily insulated and the only heat loss to occur through the glass cover, determine \((a)\) the total rate of heat loss from the collector, \((b)\) the collector efficiency, which is the ratio of the amount of heat transferred to the water to the solar energy incident on the collector, and \((c)\) the temperature rise of water as it flows through the collector.

Short answer

The total rate of heat loss from the solar collector is 1355.37 W, the collector efficiency is 7.28%, and the temperature rise of the water as it flows through the collector is 1.76 °C.

Step-by-step solution

Calculate the incident solar energy absorbed by the collector

To find the absorbed solar energy, we first need to calculate the area and the transmitted radiation. We are given the size of the collectors and the percentage of the radiation transmitted through the glass. Let's calculate the incident solar energy (I) and the area of the collector (A_collector). I = 700 W/m^2 A_collector = 1.2 m × 2 m = 2.4 m^2 Now, we will calculate the transmitted radiation (Q_transmitted) using the percentage of incident radiation transmitted through the glass (88 %). Q_transmitted = 0.88 × I × A_collector = 0.88 × 700 W/m^2 × 2.4 m^2 = 1478.4 W

Calculate the heat loss due to convection and radiation

We are given the emissivity (ε) and the temperatures of the glass cover (T_glass) and the environment (T_air). We also have the wind speed (V), which will affect the convective heat transfer coefficient (h_c). For simplicity, we will use the following empirical correlation for an external air flow: h_c = 10 + 4 × V where V is in m/s. First, we need to convert the wind speed from km/h to m/s. V = 30 km/h × (1000 m/km) / (3600 s/h) = 8.33 m/s Now, let's calculate h_c. h_c = 10 + 4 × 8.33 = 43.32 W/m^2C Next, we'll calculate the heat loss due to convection (Q_conv) using h_c and the temperature difference between the glass cover and the air. Q_conv = h_c × A_collector × (T_glass - T_air) = 43.32 W/m^2C × 2.4 m^2 × (35°C - 25°C) = 1039.68 W Now, let's calculate the heat loss due to radiation (Q_rad). For this, we will use the Stefan-Boltzmann constant (σ) and the effective sky temperature (T_sky). σ = 5.67 × 10^(-8) W/m^2K^4 T_sky = -40°C = 233.15 K Using the emissivity and the temperatures of the glass T_glass and sky T_sky, we can calculate Q_rad. Q_rad = ε × σ × A_collector × (T_glass^4 - T_sky^4) = 0.9 × 5.67 × 10^(-8) W/m^2K^4 × 2.4 m^2 × (308.15 K)^4 - (233.15 K)^4) = 315.69 W

Calculate the total heat loss from the collector

To find the total heat loss, we will add the heat losses due to convection and radiation. Q_total = Q_conv + Q_rad = 1039.68 W + 315.69 W = 1355.37 W

Determine the collector efficiency

The collector efficiency is the ratio of the amount of heat transferred to the water to the incident solar energy on the collector. We find the heat transfer to the water (Q_water) by subtracting the total heat loss from the transmitted radiation. Q_water = Q_transmitted - Q_total = 1478.4 W - 1355.37 W = 123.03 W Now, let's calculate the collector efficiency (η). η = (Q_water / (I × A_collector)) × 100 = (123.03 W / (700 W/m^2 × 2.4 m^2)) × 100 = 7.28 %

Calculate the temperature rise of water as it flows through the collector

To find the temperature rise, we will use the mass flow rate of water (m_dot) and the specific heat capacity of water (C_p). m_dot = 1 kg/min × (1 min / 60 s) = 1/60 kg/s C_p = 4.186 × 10^3 J/kgK Now, let's calculate the temperature rise (ΔT). ΔT = Q_water / (m_dot × C_p) = 123.03 W / ((1/60) kg/s × 4.186 × 10^3 J/kgK) = 1.76 K The total rate of heat loss from the collector is 1355.37 W, the collector efficiency is 7.28 %, and the temperature rise of water as it flows through the collector is 1.76 °C

Key concepts

Understanding Heat Loss in Solar Collectors

When solar collectors harness the sun's energy, they're not just gathering heat but also contending with losing some of it. Heat loss is mainly attributed to two factors: radiative heat transfer and convective heat transfer.

In the context of our solar collector problem, calculating the heat loss due to these factors helps determine the collector's overall efficiency. Insulation plays a crucial role in minimizing heat loss. However, even with the best insulation, some energy will inevitably escape, especially from the collector's surface. Understanding and mitigating these losses can significantly increase the effectiveness of solar collectors used for heating water or air.

Radiative Heat Transfer in Solar Collectors

Radiative heat transfer is an electromagnetic process and doesn't require a medium like air or water. Objects emit energy in the form of infrared radiation, which is dependent on their temperature and emissivity—a measure of a material's ability to emit absorbed energy.

Applying this to our solar collector scenario, the glass cover, with an emissivity of 0.90, loses energy to the surroundings and the cold sky. The temperature difference between the glass, at 35°C, and the effective sky temperature, at -40°C, creates a thermal gradient driving radiative heat loss. By applying the Stefan-Boltzmann law, we quantify this loss, which is essential for calculating overall efficiency. In real-world applications, materials with lower emissivity are favored for surfaces facing the environment to reduce such losses.

Convective Heat Transfer in Solar Collectors

Convective heat transfer occurs when a fluid, such as air or water, carries heat from one place to another. This transfer is chiefly influenced by the fluid's movement, which can be natural due to buoyancy forces or forced by external factors like wind. The properties of the fluid, like viscosity and thermal conductivity, also affect the rate of convective heat transfer.

As in our example, wind speed has a direct effect on convective heat loss. A high wind speed of 30 km/h increases the convective heat transfer coefficient. This coefficient, combined with the temperature difference between the collector's glass surface and ambient air, gives us the heat lost due to convection. When designing or optimizing solar collectors, reducing convective loss can involve strategies such as wind shields or using selective surfaces that have different emissive properties in the infrared range compared to the visible range.