Question

Consider a flat-plate solar collector placed horizontally on the flat roof of a house. The collector is \(5 \mathrm{ft}\) wide and \(15 \mathrm{ft}\) long, and the average temperature of the exposed surface of the collector is \(100^{\circ} \mathrm{F}\). The emissivity of the exposed surface of the collector is \(0.9\). Determine the rate of heat loss from the collector by convection and radiation during a calm day when the ambient air temperature is \(70^{\circ} \mathrm{F}\) and the effective sky temperature for radiation exchange is \(50^{\circ} \mathrm{F}\). Take the convection heat transfer coefficient on the exposed surface to be \(2.5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\).

Short answer

Answer: The total rate of heat loss from the solar collector during a calm day is 354 Btu/h.

Step-by-step solution

Compute the area of the solar collector

To calculate the heat loss, we first need to determine the area of the solar collector. The collector is given to be 5 ft wide and 15 ft long, so the area can be computed as: Area = width × length Area = 5 ft × 15 ft Area = 75 ft²

Calculate the heat loss due to convection

To calculate the heat loss due to convection, we use the convection heat transfer equation: Q_conv = h×A×(T_surface - T_ambient) Where: h = convection heat transfer coefficient = 2.5 Btu/h·ft²·°F A = area of the solar collector = 75 ft² T_surface = average temperature of the exposed surface = 100°F T_ambient = ambient air temperature = 70°F Plugging the values, we can compute the heat loss due to convection: Q_conv = 2.5 Btu/h·ft²·°F × 75 ft² × (100°F - 70°F) Q_conv = 187.5 Btu/h

Calculate the heat loss due to radiation

To calculate the heat loss due to radiation, we use the Stefan-Boltzmann law for the rate of radiation heat transfer: Q_rad = ε×σ×A×(T_surface^4 - T_sky^4) Where: ε = emissivity = 0.9 (dimensionless) σ = Stefan-Boltzmann constant = 5.67×10^{-8} W/m²·K⁴ = 1.714×10^{-9} Btu/h·ft²·°R⁴ (conversion 1 W = 3.412 Btu/h) A = area of the solar collector = 75 ft² T_surface and T_sky are the surface and sky temperatures, respectively, in °R: T_surface = 100°F + 460 = 560°R T_sky = 50°F + 460 = 510°R Plugging the values, we can compute the heat loss due to radiation: Q_rad = 0.9 × 1.714×10^{-9} Btu/h·ft²·°R⁴ × 75 ft² × (560^4°R - 510^4°R) Q_rad ≈ 166.5 Btu/h

Calculate the total heat loss

Finally, we can add the heat loss due to convection and radiation to find the total heat loss: Q_total = Q_conv + Q_rad Q_total = 187.5 Btu/h + 166.5 Btu/h Q_total = 354 Btu/h The total rate of heat loss from the solar collector during a calm day is 354 Btu/h.

Key concepts

Solar Collector

A solar collector is a device used to capture solar energy and convert it into usable heat. This heat can then be used for various purposes such as heating water, space heating, or even generating electricity. Solar collectors are typically installed on rooftops or open areas where they can receive maximum sunlight exposure.
The solar collector in this exercise is a flat-plate type, meaning it has a flat surface to absorb solar radiation. Its dimensions are 5 feet wide and 15 feet long, resulting in a total area of 75 square feet. This design is efficient for collecting heat over a large area. The absorbed heat is then either transferred to a heat transfer fluid or directly used by the application.
Flat-plate solar collectors are commonly used in residential and commercial solar thermal systems due to their simplicity and effectiveness. They work best in environments with consistent sunlight exposure and are relatively affordable, making them a popular choice for solar energy harnessing.

Convection

Convection is the process of heat transfer based on the movement of fluid or air. In the context of a solar collector, convection occurs when warm air near the surface of the collector is replaced by cooler air from the environment.
This exercise calculated heat loss through convection using the formula: \[ Q_{\text{conv}} = h \times A \times (T_{\text{surface}} - T_{\text{ambient}}) \]where:

• \( h \) is the convection heat transfer coefficient.
• \( A \) is the area of the collector.
• \( T_{\text{surface}} \) and \( T_{\text{ambient}} \) are the temperatures of the collector's surface and the ambient air, respectively.

The coefficient \( h \) provided is 2.5 Btu/h·ft²·°F, indicating how effectively the heat is transferred from the surface to the air. The calculated convection heat loss is 187.5 Btu/h, representing the energy lost due to air movement around the collector.

Radiation

Radiation is another form of heat transfer that occurs through electromagnetic waves without the need for a medium, meaning it can even happen in a vacuum. For the solar collector, radiation involves the emission of heat from the collector's surface as infrared radiation.
The heat loss due to radiation in this exercise is calculated using the Stefan-Boltzmann law:\[ Q_{\text{rad}} = \varepsilon \times \sigma \times A \times (T_{\text{surface}}^4 - T_{\text{sky}}^4) \]where:

• \( \varepsilon \) is the surface emissivity, a measure of how efficiently a surface emits thermal radiation.
• \( \sigma \) is the Stefan-Boltzmann constant.
• \( A \) is the area of the solar collector.
• \( T_{\text{surface}} \) and \( T_{\text{sky}} \) are the surface and sky temperatures in Rankine.

This process resulted in a heat loss of approximately 166.5 Btu/h. Radiation mechanisms are significant when there are large temperature differences and clear sky conditions, influencing the overall heat balance of solar collectors.

Thermal Emissivity

Thermal emissivity is a property of surfaces that measures their efficiency in emitting thermal radiation. It is a dimensionless value between 0 and 1, representing the ratio of radiation emitted by a surface compared to that emitted by a perfect black body at the same temperature.
A higher emissivity means a surface is more effective at emitting radiative energy. For example, in this exercise, the solar collector's surface has an emissivity of 0.9, indicating it is highly efficient at radiating heat. This property significantly influences the calculation of radiative heat loss using the Stefan-Boltzmann law.
Understanding emissivity is crucial in solar thermal applications because it affects the efficiency and heat loss of solar collectors. Surfaces with higher emissivity may need to be optimized or balanced with other materials or coatings to better achieve desired thermal outcomes. In practice, selecting or designing materials with an appropriate emissivity is essential for improving the performance of solar thermal systems.