/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 111 Solar flux of \(900 \mathrm{~W} ... [FREE SOLUTION] | 91Ó°ÊÓ

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Solar flux of \(900 \mathrm{~W} / \mathrm{m}^{2}\) is incident on the top side of a plate whose surface has a solar absorptivity of \(0.9\) and an emissivity of \(0.1\). The air and surroundings are at \(17^{\circ} \mathrm{C}\) and the convection heat transfer coefficient between the plate and air is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming that the bottom side of the plate is insulated, determine the steady-state temperature of the plate.

Short Answer

Expert verified
The steady-state temperature of the plate can be found by setting up an energy balance equation and solving for the temperature T in Kelvin using numerical methods. The equation is \(αQ_S = ε σ (T^4 - T_\text{air}^4) + h (T - T_\text{air})\). After obtaining the temperature in Kelvin, convert it to Celsius by subtracting 273.15 K.

Step by step solution

01

Define given values and constants

- Solar flux (Q_S), \(Q_S = 900 \frac{\text{W}}{\text{m}^2}\) - Solar absorptivity (α), \(α = 0.9\) - Emissivity (ε), \(ε = 0.1\) - Convection heat transfer coefficient (h), \(h = 20 \frac{\text{W}}{\text{m}^2 \cdot \text{K}}\) - Air temperature (T_air), \(T_\text{air} = 17^{\circ} \mathrm{C}\) We also need the Stefan-Boltzmann constant (σ), which is approximately \(5.67 \times 10^{-8} \frac{\text{W}}{\text{m}^2 \cdot \text{K}^4}\).
02

Interpret the energy balance

We need to set up an energy balance equation for the plate in steady-state. The absorbed solar energy, \(Q_\text{abs}\), must equal the sum of emitted radiation, \(Q_\text{rad}\), and convective heat transfer, \(Q_\text{conv}\). So: \(Q_\text{abs} = Q_\text{rad} + Q_\text{conv}\)
03

Express heat transfer terms

- Absorbed solar energy: \(Q_\text{abs} = αQ_S\) - Emitted radiation: \(Q_\text{rad} = ε \times σ \times A \times (T^4 - T_\text{air}^4)\), where A is the surface area and T is the plate's temperature in K - Convective heat transfer: \(Q_\text{conv} = h \times A \times (T - T_\text{air})\)
04

Set up the energy balance equation

Now we can set up the energy balance equation. Since the bottom side of the plate is insulated, we only need to consider the top side. Hence, we can cancel the surface area (A) from the equation: \(αQ_S = ε σ (T^4 - T_\text{air}^4) + h (T - T_\text{air})\)
05

Solve for the plate's temperature

We need to solve the energy balance equation for T, the plate's temperature in steady-state. However, this equation is nonlinear due to the \(T^4\) term. We need to solve it numerically or iteratively. You may use numerical methods such as the Newton-Raphson method or any suitable programming language to find the temperature T in Kelvin. Once you obtain the temperature in Kelvin, convert it to Celsius by subtracting 273.15 K. The steady-state temperature of the plate should be within a reasonable range, considering the given conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Solar Flux
Solar flux refers to the power per unit area received from the sun in the form of electromagnetic radiation. In the context of this problem, solar flux is given as 900 W/m², which is a measure of the intensity of solar energy striking the plate. A crucial factor in this scenario is the solar absorptivity, a dimensionless coefficient typically denoted by the Greek letter alpha (\(\alpha\)). It represents the fraction of incident solar energy that a surface absorbs; a higher absorptivity means the surface will absorb more solar energy. In our case, the absorptivity is 0.9, meaning the plate absorbs 90% of the incoming solar flux.

To illustrate how this plays into the overall energy balance of the plate, we can calculate the absorbed solar energy (\(Q_\text{abs}\)) as the product of solar flux (\(Q_S\)) and solar absorptivity (\(\alpha\)). Thus, \(Q_\text{abs} = \alpha Q_S = 0.9 \times 900\frac{\text{W}}{\text{m}^2} = 810\frac{\text{W}}{\text{m}^2}\).
Deciphering the Energy Balance Equation
At the heart of this problem lies the energy balance equation, which is crucial for determining the steady-state temperature of an object. Energy balance is a concept from thermodynamics, stating that energy cannot be created or destroyed, only transferred or transformed. For the plate to reach a steady-state temperature, the incoming energy absorption must equal the outgoing energy loss, meaning the absorbed solar energy must equal the total of emitted radiation and convective heat loss.

This balance is articulated through the equation \(Q_\text{abs} = Q_\text{rad} + Q_\text{conv}\). In this equation, \(Q_\text{rad}\) stands for the radiant energy emitted by the plate, which depends on its emissivity (\(\epsilon\)) and temperature (T), following the Stefan-Boltzmann law. On the other hand, \(Q_\text{conv}\) represents the heat transferred to the surrounding air through convection, dependent on the convection heat transfer coefficient (h) and the temperature difference between the plate and the air. The true challenge involves solving this equation for T when it contains both linear (\(T - T_\text{air}\)) and non-linear (\(T^4 - T_\text{air}^4\)) terms.
Unraveling Convection Heat Transfer
Convection heat transfer is one of the three modes of heat transfer, along with conduction and radiation. It involves the transfer of heat between a solid surface and a fluid (in this case, air) moving over the surface. The rate of convective heat transfer can be quantified by the convection heat transfer coefficient, denoted by h, with the units of W/m²K. This coefficient tells us how effective the convective process is at moving heat away from or toward the surface.

In our scenario, the coefficient is provided as 20 W/m²K, meaning that for every degree Celsius temperature difference between the plate and the ambient air, 20 joules of heat energy are transferred per square meter per second. By multiplying this coefficient by the temperature difference and the area, we get the convective heat loss (\(Q_\text{conv}\)).

A higher h value typically corresponds to a more effective convective cooling or heating process, whereas a lower value indicates that the convection is less efficient at transferring heat. Also, the surrounding temperature (\(T_\text{air}\)), which is 17°C in this problem, is a crucial parameter, since the temperature difference drives the convective heat transfer process.

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Most popular questions from this chapter

A roof-cooling system, which operates by maintaining a thin film of water on the roof surface, may be used to reduce air-conditioning costs or to maintain a cooler environment in nonconditioned buildings. To determine the effectiveness of such a system, consider a sheet metal roof for which the solar absorptivity \(\alpha_{5}\) is \(0.50\) and the hemispherical emissivity \(\varepsilon\) is \(0.3\). Representative conditions correspond to a surface convection coefficient \(h\) of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), a solar irradiation \(G_{S}\) of \(700 \mathrm{~W} / \mathrm{m}^{2}\), a sky temperature of \(-10^{\circ} \mathrm{C}\), an atmospheric temperature of \(30^{\circ} \mathrm{C}\), and a relative humidity of \(65 \%\). The roof may be assumed to be well insulated from below. Determine the roof surface temperature without the water film. Assuming the film and roof surface temperatures to be equal, determine the surface temperature with the film. The solar absorptivity and the hemispherical emissivity of the film-surface combination are \(\alpha_{S}=\) \(0.8\) and \(s=0.9\), respectively.

A thermograph is a device responding to the radiant power from the scene, which reaches its radiation detector within the spectral region 9-12 \(\mu \mathrm{m}\). The thermograph provides an image of the scene, such as the side of a furnace, from which the surface temperature can be determined. (a) For a black surface at \(60^{\circ} \mathrm{C}\), determine the emissive power for the spectral region \(9-12 \mu \mathrm{m}\). (b) Calculate the radiant power (W) received by the thermograph in the same range \((9-12 \mu \mathrm{m})\) when viewing, in a normal direction, a small black wall area, \(200 \mathrm{~mm}^{2}\), at \(T_{s}=60^{\circ} \mathrm{C}\). The solid angle \(\omega\) subtended by the aperture of the thermograph when viewed from the target is \(0.001 \mathrm{sr}\). (c) Determine the radiant power \((\mathrm{W})\) received by the thermograph for the same wall area \(\left(200 \mathrm{~mm}^{2}\right)\) and solid angle \((0.001 \mathrm{sr})\) when the wall is a gray, opaque, diffuse material at \(T_{x}=60^{\circ} \mathrm{C}\) with emissivity \(0.7\) and the surroundings are black at \(T_{\text {sur }}=23^{\circ} \mathrm{C}\).

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