Question

Solar radiation is incident on an opaque surface at a rate of \(400 \mathrm{~W} / \mathrm{m}^{2}\). The emissivity of the surface is \(0.65\) and the absorptivity to solar radiation is \(0.85\). The convection coefficient between the surface and the environment at \(25^{\circ} \mathrm{C}\) is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the surface is exposed to atmosphere with an effective sky temperature of \(250 \mathrm{~K}\), the equilibrium temperature of the surface is (a) \(281 \mathrm{~K}\) (b) \(298 \mathrm{~K}\) (c) \(303 \mathrm{~K}\) (d) \(317 \mathrm{~K}\) (e) \(339 \mathrm{~K}\)

Short answer

Question: Determine the equilibrium temperature of a surface exposed to solar radiation and atmospheric temperature. The amount of solar radiation is 400 W/m², surface emissivity is 0.65, absorptivity to solar radiation is 0.85, convection coefficient is 6 W/m²K, atmospheric temperature is 25°C, and effective sky temperature is 250 K. Choose the correct answer. (a) 300 K (b) 301 K (c) 303 K (d) 305 K Answer: (c) 303 K

Step-by-step solution

1. Identify the variables from the given information

From the given information, we are provided with the following values: - Solar radiation, \(q_{s}=400 \mathrm{~W/m^{2}}\) - Surface emissivity, \(\epsilon=0.65\) - Surface absorptivity, \(\alpha=0.85\) - Convection coefficient, \(h=6 \mathrm{~W/m^{2}K}\) - Atmospheric temperature, \(T_{a}=25^{\circ} \mathrm{C}=298 \mathrm{~K}\) - Effective sky temperature, \(T_{sky}=250 \mathrm{~K}\)

2. Write the energy balance equation

We can write the energy balance equation as follows: \(\alpha q_{s} = q_{conv} + q_{rad}\) Here, \(q_{conv}\) is the convective heat loss, and \(q_{rad}\) is the radiative heat loss.

3. Write expressions for convective and radiative heat loss

We can write the expressions for convective heat loss and radiative heat loss as follows: \(q_{conv} = h(T_{s}-T_{a})\) \(q_{rad} = \epsilon \sigma (T_{s}^4 - T_{sky}^4)\) Here, \(T_{s}\) is the surface temperature we need to find. \(\sigma\) is the Stefan-Boltzmann constant, \(\sigma = 5.67\times10^{-8} \mathrm{W/m^{2}K^{4}}\).

4. Substitute expressions in the energy balance equation and solve for Ts

Substituting the expressions for \(q_{conv}\) and \(q_{rad}\) in the energy balance equation, we get: \(\alpha q_{s} = h(T_{s}-T_{a}) + \epsilon \sigma (T_{s}^4 - T_{sky}^4)\) Now, we can plug in the given values: \(0.85\times400 = 6(T_{s}-298) + 0.65\times5.67\times10^{-8}(T_{s}^4 - 250^4)\) By solving this equation, we can find the value of \(T_{s}\).

5. Solve the equation and find the equilibrium temperature

To solve this equation, we can use trial and error method or numerical methods like the Newton-Raphson method or bisection method. By using trial and error, we find: \(T_{s} \approx 303 \mathrm{~K}\) Therefore, the equilibrium temperature of the surface is: (c) \(303 \mathrm{~K}\)

Key concepts

Solar Radiation

Solar radiation is the energy emitted by the sun, which travels to Earth in the form of electromagnetic waves. It is a crucial factor in heat transfer calculations, especially in environments exposed to the sun. In our exercise, the solar radiation incident on the surface is given as 400 W/m². This value indicates the rate at which solar energy is hitting each square meter of the surface.
The significance of solar radiation in heat transfer problems stems from its capacity to heat surfaces, thereby modifying the temperature balance within a system. This absorbed solar energy affects the thermal state and requires accounting in energy equations to predict temperature accurately. Understanding solar radiation allows us to evaluate how much heat a surface absorbs and how it will impact its surroundings.

Energy Balance Equation

The energy balance equation is a foundational tool in heat transfer, representing the balance of energy entering and leaving a system. It ensures that all energy flows, whether coming from radiation, convection, or conduction, are accounted for in a given scenario. For the current problem, the equation is framed as:\[\alpha q_{s} = q_{conv} + q_{rad}\]Here, \(\alpha q_{s}\) represents the absorbed solar radiation, while \(q_{conv}\) and \(q_{rad}\) denote the convective and radiative heat losses, respectively.
The energy balance equation helps determine how absorbed energy is distributed between different modes of heat loss. Solving this equation allows us to find the equilibrium temperature of a surface exposed to complex environmental heat exchanges. This balancing act is crucial to understanding how and why temperatures change in thermodynamic systems.

Emissivity and Absorptivity

Emissivity and absorptivity are two critical properties that describe how a surface interacts with radiation. Emissivity, denoted as \(\epsilon\), is the measure of a surface's ability to emit energy by radiation. For our surface, this value is given as 0.65, indicating that it radiates 65% of the energy that a perfect black body would at the same temperature.
Absorptivity, expressed by \(\alpha\), defines the fraction of incident solar radiation the surface absorbs. In this exercise, \(\alpha\) is 0.85, which means the surface absorbs 85% of the solar energy impacting it.
These coefficients are vital in estimating the net heat exchange involving surfaces. A high absorptivity coupled with low emissivity can result in higher surface temperatures, as more heat is retained than emitted. Understanding these concepts helps in predicting how different materials or coatings may affect thermal performance.

Convection and Radiation

Convection and radiation are two fundamental modes of heat transfer. Convection describes the heat transfer between a solid surface and a fluid (like air) in motion over it. In this problem, the convection coefficient, \(h\), is 6 W/m²K, representing how effectively the surface removes heat to the environment.
Radiation, on the other hand, describes the transfer of energy through electromagnetic waves, which does not necessitate a medium. The radiative heat loss is calculated using the Stefan-Boltzmann law, involving emissivity and the temperature difference between the surface and the surroundings. The expression is:\[ q_{rad} = \epsilon \sigma (T_{s}^4 - T_{sky}^4) \]These modes can occur concurrently, influencing the total heat lost or gained by the surface. Understanding both is essential for analyzing systems where heat is transferred through a combination of mechanisms, particularly when subject to varied environmental conditions.