/*! 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

91Ó°ÊÓ

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An opaque surface, \(2 \mathrm{~m} \times 2 \mathrm{~m}\), is maintained at \(400 \mathrm{~K}\) and is simultaneously exposed to solar irradiation with \(G_{S}=1200 \mathrm{~W} / \mathrm{m}^{2}\). The surface is diffuse and its spectral absorptivity is \(\alpha_{\lambda}=0,0.8,0\), and \(0.9\) for \(0 \leq \lambda \leq\) \(0.5 \mu \mathrm{m}, 0.5 \mu \mathrm{m}<\lambda \leq 1 \mu \mathrm{m}, 1 \mu \mathrm{m}<\lambda \leq 2 \mu \mathrm{m}\), and \(\lambda>2 \mu \mathrm{m}\), respectively. Determine the absorbed irradiation, emissive power, radiosity, and net radiation heat transfer from the surface.

The energy flux associated with solar radiation incident on the outer surface of the earth's atmosphere has been accurately measured and is known to be \(1368 \mathrm{~W} / \mathrm{m}^{2}\). The diameters of the sun and earth are \(1.39 \times 10^{9}\) and \(1.27 \times 10^{7} \mathrm{~m}\), respectively, and the distance between the sun and the earth is \(1.5 \times 10^{11} \mathrm{~m}\). (a) What is the emissive power of the sun? (b) Approximating the sun's surface as black, what is its temperature? (c) At what wavelength is the spectral emissive power of the sun a maximum? (d) Assuming the earth's surface to be black and the sun to be the only source of energy for the earth, estimate the earth's surface temperature.

The extremely high temperatures needed to trigger nuclear fusion are proposed to be generated by laserirradiating a spherical pellet of deuterium and tritium fuel of diameter \(D_{p}=1.8 \mathrm{~mm}\). (a) Determine the maximum fuel temperature that can be achieved by irradiating the pellet with 200 lasers, each producing a power of \(P=500 \mathrm{~W}\). The pellet has an absorptivity \(\alpha=0.3\) and emissivity \(\varepsilon=0.8\). (b) The pellet is placed inside a cylindrical enclosure. Two laser entrance holes are located at either end of the enclosure and have a diameter of \(D_{\mathrm{LEH}}=2 \mathrm{~mm}\). Determine the maximum temperature that can be generated within the enclosure.

Isothermal furnaces with small apertures approximating a blackbody are frequently used to calibrate heat flux gages, radiation thermometers, and other radiometric devices. In such applications, it is necessary to control power to the furnace such that the variation of temperature and the spectral intensity of the aperture are within desired limits. (a) By considering the Planck spectral distribution, Equation \(12.30\), show that the ratio of the fractional change in the spectral intensity to the fractional change in the temperature of the furnace has the form $$ \frac{d I_{\lambda} / I_{\lambda}}{d T / T}=\frac{C_{2}}{\lambda T} \frac{1}{1-\exp \left(-C_{2} / \lambda T\right)} $$ (b) Using this relation, determine the allowable variation in temperature of the furnace operating at \(2000 \mathrm{~K}\) to ensure that the spectral intensity at \(0.65 \mu \mathrm{m}\) will not vary by more than \(0.5 \%\). What is the allowable variation at \(10 \mu \mathrm{m}\) ?

Square plates freshly sprayed with an epoxy paint must be cured at \(140^{\circ} \mathrm{C}\) for an extended period of time. The plates are located in a large enclosure and heated by a bank of infrared lamps. The top surface of each plate has an emissivity of \(\varepsilon=0.8\) and experiences convection with a ventilation airstream that is at \(T_{\infty}=27^{\circ} \mathrm{C}\) and provides a convection coefficient of \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The irradiation from the enclosure walls is estimated to be \(G_{\text {wall }}=450 \mathrm{~W} / \mathrm{m}^{2}\), for which the plate absorptivity is \(\alpha_{\text {wall }}=0.7\). (a) Determine the irradiation that must be provided by the lamps, \(G_{\text {lamp. }}\). The absorptivity of the plate surface for this irradiation is \(\alpha_{\text {Lamp }}=0.6\). (b) For convection coefficients of \(h=15,20\), and \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), plot the lamp irradiation, \(G_{\text {lamp, as a }}\) function of the plate temperature, \(T_{s}\), for \(100 \leq\) \(T_{x} \leq 300^{\circ} \mathrm{C}\). (c) For convection coefficients in the range from 10 to \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and a lamp irradiation of \(G_{\text {lmp }}=\) \(3000 \mathrm{~W} / \mathrm{m}^{2}\), plot the airstream temperature \(T_{x}\) required to maintain the plate at \(T_{x}=140^{\circ} \mathrm{C}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.