Question

Solar radiation is incident on the front surface of a thin plate with direct and diffuse components of 300 and $250\mathrm{W}/{\mathrm{m}}^2$, respectively. The direct radiation makes a ${30}^{\circ }$ angle with the normal of the surface. The plate surfaces have a solar absorptivity of $0.63$ and an emissivity of $0.93$. The air temperature is $5^{\circ }\mathrm{C}$ and the convection heat transfer coefficient is $20\mathrm{W}/{\mathrm{m}}^2\cdot \mathrm{K}$. The effective sky temperature for the front surface is $-{33}^{\circ }\mathrm{C}$ while the surrounding surfaces are at $5^{\circ }\mathrm{C}$ for the back surface. Determine the equilibrium temperature of the plate.

Short answer

Question: Determine the equilibrium temperature of the plate given the solar radiation, absorptivity, emissivity, air temperature, convection heat transfer coefficient, and effective sky temperature values provided in the solution.

Step-by-step solution

Calculate the absorbed solar radiation

To calculate the absorbed solar radiation, we have to consider both the direct and diffuse components. Since the direct radiation makes a ${30}^{\circ }$ angle with the normal of the surface, we will use the cosine of this angle to obtain the component normal to the surface: Absorbed direct solar radiation = solar absorptivity × direct radiation × cos(${30}^{\circ }$) Absorbed direct solar radiation = 0.63 × 300 × cos(${30}^{\circ }$) For the diffuse radiation, we assume that it is coming uniformly from all directions, and hence, we should multiply it by $0.5$: Absorbed diffuse solar radiation = solar absorptivity × diffuse radiation × 0.5 Absorbed diffuse solar radiation = 0.63 × 250 × 0.5 Add the absorbed direct and diffuse solar radiation to get the total absorbed solar radiation: Total absorbed solar radiation = Absorbed direct solar radiation + Absorbed diffuse solar radiation

Calculate the heat loss by convection

The heat loss by convection is given by the following formula: Heat loss by convection = convection heat transfer coefficient × surface area × (plate temperature - air temperature) Since we are not given the surface area of the plate, we will leave it as a variable (A) and find the expression for the heat loss by convection: Heat loss by convection = 20 × A × (plate temperature - 5)

Calculate the heat loss by radiation

The heat loss by radiation is given by the following formula: Heat loss by radiation = emissivity × Stefan-Boltzmann constant × surface area × (plate temperature^4 - effective sky temperature^4) Heat loss by radiation = 0.93 × 5.67 × 10^(-8) × A × (plate temperature^4 - (-33)^4)

Energy balance

At equilibrium, the total absorbed solar radiation will be equal to the total heat loss (convection + radiation). Using the expressions obtained in the previous steps, we can write the energy balance equation as follows: Total absorbed solar radiation = Heat loss by convection + Heat loss by radiation Solve this equation for the plate temperature to get the equilibrium temperature. Note that the surface area (A) will cancel out since it appears in both terms of the energy balance equation.

Key concepts

Solar Radiation Absorption

When an object is exposed to sunlight, it absorbs some of the energy as heat. This process is known as solar radiation absorption. It's essential to distinguish between direct and diffuse solar radiation. Direct solar radiation is sunlight that reaches the object's surface without being scattered, whereas diffuse solar radiation is sunlight that has been scattered by molecules and particles in the atmosphere, arriving at the surface from various angles.

Solar absorptivity is a measure of how well a material absorbs solar energy. It varies between 0 and 1, with 1 indicating that the surface absorbs all incoming solar energy and 0 meaning none is absorbed. In our problem, the plate's solar absorptivity is given as 0.63, suggesting it absorbs 63% of the solar radiation that strikes it. To calculate the total absorbed solar radiation, we combine the absorbed direct and diffuse components, taking into account the orientation of the surface and the proportion of diffuse radiation absorbed.

Convection Heat Transfer

Convection heat transfer involves the movement of heat between a solid surface and a fluid, such as air or water, that is moving across it. The rate at which heat is lost or gained through convection depends on several factors including the convection heat transfer coefficient, surface area, and the temperature difference between the surface and the fluid.

In our textbook example, the convection heat transfer coefficient is given as 20 W/m²·K, which indicates how effective the air is at removing heat from the plate per square meter and per degree Kelvin temperature difference between the plate and the air. To calculate the heat loss from the plate due to convection, we would multiply this coefficient by the area of the plate and the temperature difference between the plate's surface and the ambient air.

Radiative Heat Loss

Radiative heat loss is another critical form of heat transfer where energy is emitted from the surface of an object in the form of electromagnetic radiation. Objects at a temperature above absolute zero all emit thermal radiation, and hotter objects emit more radiation than cooler ones.

The rate of energy emission is a function of the object’s emissivity, the Stefan-Boltzmann constant, the object’s surface area, and the temperature difference between the object and its surroundings. Emissivity, similar to solar absorptivity, is a measure on a scale from 0 to 1 of a material's ability to emit thermal radiation. A perfect black body, which emits the maximum possible radiation, has an emissivity of 1. In our example, the emissivity of the plate is 0.93, which means the plate is a very good emitter of thermal radiation.

Stefan-Boltzmann Constant

The Stefan-Boltzmann constant is a physical constant that is essential in the realm of thermodynamics and heat transfer. It appears in the Stefan-Boltzmann law, which states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time (also known as black-body radiant emittance) is directly proportional to the fourth power of the black body's thermodynamic temperature.

The constant is denoted by the symbol $\sigma$ and has a value of approximately 5.67 × 10${}^{-8}$ watts per square meter per Kelvin to the fourth power (W/m²·K${}^4$). In the context of our exercise, the Stefan-Boltzmann constant is used to calculate the rate of radiative heat loss from the plate's surface.

Energy Balance in Heat Transfer

Energy balance is a fundamental principle in heat transfer that states that the energy into a system must equal the energy out of the system plus the change in the energy within the system over time. For a system in equilibrium, it means that there is no net change in internal energy over time, and hence, the energy in equals the energy out.

In our exercise, we use this principle to determine the equilibrium temperature of the plate. At equilibrium, the total heat (energy) that the plate absorbs from the solar radiation must be equal to the sum of the heat (energy) it loses through convection and radiation. This equality allows us to set up an equation where the absorbed solar radiation is set equal to the sum of convective and radiative heat losses, and by solving this equation, we find the temperature at which the plate neither gains nor loses heat in net terms. This temperature is referred to as the equilibrium temperature.