/*! 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 107 Solar irradiation of \(1100 \mat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 107

Solar irradiation of \(1100 \mathrm{~W} / \mathrm{m}^{2}\) is incident on a large, flat, horizontal metal roof on a day when the wind blowing over the roof causes a convection heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The outside air temperature is \(27^{\circ} \mathrm{C}\), the metal surface absorptivity for incident solar radiation is \(0.60\), the metal surface emissivity is \(0.20\), and the roof is well insulated from below. (a) Estimate the roof temperature under steady-state conditions. (b) Explore the effect of changes in the absorptivity, emissivity, and convection coefficient on the steady-state temperature. 12.108 Neglecting the effects of radiation absorption, emission, and scattering within their atmospheres, calculate the average temperature of Earth, Venus, and Mars assuming diffuse, gray behavior. The average distance from the sun of each of the three planets, \(L_{s p}\), along with their measured average temperatures, \(\bar{T}_{p}\), are shown in the table below. Based upon a comparison of the calculated and measured average temperatures, which planet is most affected by radiation transfer in its atmosphere? \begin{tabular}{lcc} \hline Planet & \(L_{x-p}(\mathbf{m})\) & \(\bar{T}_{p}(\mathbf{K})\) \\ \hline Venus & \(1.08 \times 10^{11}\) & 735 \\ Earth & \(1.50 \times 10^{11}\) & 287 \\ Mars & \(2.30 \times 10^{11}\) & 227 \\ \hline \end{tabular}

Short Answer

Expert verified
(a) The steady-state roof temperature, \(T_s\), is calculated by balancing the energy gained from solar radiation and the energy lost through convection and radiation. We use an iterative method or a solver to find the numerical value of \(T_s\). (b) To explore the effect of changes in absorptivity, emissivity, and convection coefficient on the steady-state temperature, we analyze their individual impacts on the heat gain from solar radiation and the heat losses due to convection and radiation. We qualitatively discuss how changing each of these factors will affect the steady-state temperature, thus understanding the sensitivity of the roof's steady-state temperature to these parameters.

Step by step solution

01

Part (a): Estimate the roof temperature

To estimate the steady-state roof temperature, we must first balance the energy gains and losses for the surface of the metal roof. The energy balance equation is given by: \(Q_{in} = Q_{out}\) Where \(Q_{in}\) is the energy gained from solar radiation and \(Q_{out}\) is the energy lost through convection and radiation. We will use the given values for solar irradiation, convective heat transfer coefficient, air temperature, surface absorptivity, and surface emissivity. 1) Calculate the absorbed solar radiation: \[Q_{solar} = \rho I_{sun}\] Where \(\rho\) is the absorptivity and \(I_{sun}\) is the solar irradiation. 2) Calculate the heat loss due to convection: \[Q_{conv} = h_{conv} A (T_s - T_{air})\] Where \(h_{conv}\) is the convection heat transfer coefficient, \(A\) is the roof surface area, \(T_s\) is the surface temperature, and \(T_{air}\) is the outside air temperature. 3) Calculate the heat loss due to radiation: \[Q_{rad} = \epsilon \sigma A T_s^4 - \sigma A T_{air}^4\] Where \(\epsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant, and \(T_s\) and \(T_{air}\) are the surface and air temperatures, respectively. 4) Equate the energy balance: \[Q_{solar} = Q_{conv} + Q_{rad}\] 5) Rearrange the equation to solve for \(T_s\): \[T_s^4 = \frac{\rho I_{sun} - h_{conv} A (T_s - T_{air})}{\epsilon \sigma} + T_{air}^4\] 6) Find the numeric value of \(T_s\) using an iterative method or a solver. Once we have estimated the steady-state roof temperature, we can then proceed to analyze the effect of changes in absorptivity, emissivity, and the convection coefficient.
02

Part (b): Explore the effect of changes

In order to explore the effect of changes in absorptivity, emissivity, and the convection coefficient, we need to analyze how each of these factors affects the steady-state temperature. We can study each change separately and look at how it affects the energy balance equation. 1) Effect of absorptivity changes: Examine the effect of an increase or decrease in absorptivity on the heat gain from solar radiation and on the steady-state temperature. 2) Effect of emissivity changes: Analyze the effect of an increase or decrease in emissivity on the heat loss due to radiation and the effect on the steady-state temperature. 3) Effect of the convection coefficient changes: Consider how changes in the convection coefficient will affect the heat loss due to convection and the steady-state temperature. For each of these three factors, we can qualitatively discuss how changing them will affect the steady-state temperature. Comparing these hypothetical scenarios will help us understand the sensitivity of the roof's steady-state temperature to these parameters.

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 91影视!

Key Concepts

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

Solar Irradiation
Understanding solar irradiation is crucial since it represents the power per unit area received from the Sun. In our context, it directly influences the temperature of surfaces like roofs. Simply put, solar irradiation is a measure of how much solar energy is hitting a given area, and it is often measured in watts per square meter ((W/m^2))
.
The exercise provided talks about a specific amount of solar irradiation, 1100 W/m^2, hitting a metal roof. This value is key when estimating the steady-state temperature because it represents the initial energy input. The roof absorbs this energy based on its absorptivity, which is a factor of how 'dark' or 'light' the surface is in terms of solar energy. A surface with high absorptivity will absorb more energy and thus potentially reach higher temperatures. In our case, the metal roof has an absorptivity of 0.60, meaning it absorbs 60% of the incident solar irradiation.
To fully grasp the potential impact of solar irradiation on the roof temperature, it's essential to understand that not the entire amount of incident solar energy will heat the roof. Some percentage is reflected away, depending on the absorptivity. This concept emphasizes the relationship between solar irradiation and the potential for heat gain by a surface. Furthermore, if the solar irradiation were to increase, the energy absorbed by the roof would also increase, leading to a higher steady-state temperature, all other factors being constant.
Heat Transfer Coefficient
The heat transfer coefficient is another vital concept in understanding how heat is exchanged between surfaces and their surroundings. It measures how effectively heat is transferred by convection, which is the process of heat transfer through a fluid (such as air) that flows over a surface.
In our exercise, the convection heat transfer coefficient is given as 25 W/m^2路K. This means that for every square meter of roof surface, 25 watts of heat is transferred for each degree Kelvin difference in temperature between the roof and the surrounding air. It serves as a scaling factor: a higher coefficient signifies that heat is lost more rapidly to the surroundings, while a lower coefficient means heat is retained more effectively.
When the outside air temperature is cooler than the roof, the heat transfer coefficient helps us determine how much heat the roof is losing to the environment. It's part of the equation that balances the energy coming in from solar irradiation and the energy going out by convection and radiation. As the roof gets hotter, the heat loss to the surrounding air will increase, assuming the air temperature remains constant. Evaluating the effects of changes in the convection coefficient on the steady-state temperature is essential. An increase would lead to higher heat loss and a cooler stable temperature, while a decrease would do the opposite.
Energy Balance Equation
The energy balance equation is at the heart of our exercise. It is a statement of the conservation of energy principle, which implies that energy can neither be created nor destroyed, only transferred. Essentially, it's the mathematical representation of the battle between the energy inputs and outputs of a system.
In this case, the energy balance equation equates the energy gained through absorbed solar irradiation with the energy lost through convection and radiation. This equation is fundamental to predicting the steady-state temperature of the roof.
When we set the incoming heat flux equal to the outgoing heat flux, we can solve for the unknown roof temperature. The steady-state is reached when these energy flows are in equilibrium, meaning the incoming and outgoing heat are balanced, and the temperature doesn't change.
The exercise uses an iterative approach to solve for the steady-state roof temperature since the heat loss due to radiation is dependent on the fourth power of the temperature. By adjusting different parameters such as absorptivity, emissivity, and the convection heat transfer coefficient, we can see their effect on the steady-state temperature. Small changes in these values can lead to significant changes in the temperature, demonstrating the sensitivity of the system to these parameters. This highlights the importance of accurately determining and understanding these variables when designing or assessing the thermal performance of structures.

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

Plant leaves possess small channels that connect the interior moist region of the leaf to the environment. The channels, called stomata, pose the primary resistance to moisture transport through the entire plant, and the diameter of an individual stoma is sensitive to the level of \(\mathrm{CO}_{2}\) in the atmosphere. Consider a leaf of corn (maize) whose top surface is exposed to solar irradiation of \(G_{S}=600 \mathrm{~W} / \mathrm{m}^{2}\) and an effective sky temperature of \(T_{\text {sky }}=0^{\circ} \mathrm{C}\). The bottom side of the leaf is irradiated from the ground which is at a temperature of \(T_{g}=20^{\circ} \mathrm{C}\). Both the top and bottom of the leaf are subjected to convective conditions characterized by \(h=35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, T_{\infty}=25^{\circ} \mathrm{C}\) and also experience evaporation through the stomata. Assuming the evaporative flux of water vapor is \(50 \times 10^{-6} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\) under rural atmospheric \(\mathrm{CO}_{2}\) concentrations and is reduced to \(5 \times 10^{-6} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\) when ambient \(\mathrm{CO}_{2}\) concentrations are doubled near an urban area, calculate the leaf temperature in the rural and urban locations. The heat of vaporization of water is \(h_{f g}=2400 \mathrm{~kJ} / \mathrm{kg}\) and assume \(\alpha=\varepsilon=0.97\) for radiation exchange with the sky and the ground, and \(\alpha_{S}=0.76\) for solar irradiation.

Growers use giant fans to prevent grapes from freezing when the effective sky temperature is low. The grape, which may be viewed as a thin skin of negligible thermal resistance enclosing a volume of sugar water, is exposed to ambient air and is irradiated from the sky above and ground below. Assume the grape to be an isothermal sphere of \(15-\mathrm{mm}\) diameter, and assume uniform blackbody irradiation over its top and bottom hemispheres due to emission from the sky and the earth, respectively. (a) Derive an expression for the rate of change of the grape temperature. Express your result in terms of a convection coefficient and appropriate temperatures and radiative quantities. (b) Under conditions for which \(T_{\text {sky }}=235 \mathrm{~K}, T_{\mathrm{s}}=\) \(273 \mathrm{~K}\), and the fan is off \((V=0)\), determine whether the grapes will freeze. To a good approximation, the skin emissivity is 1 and the grape thermophysical properties are those of sugarless water. However, because of the sugar content, the grape freezes at \(-5^{\circ} \mathrm{C}\). (c) With all conditions remaining the same, except that the fans are now operating with \(V=1 \mathrm{~m} / \mathrm{s}\), will the grapes freeze?

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.

A diffuse, opaque surface at \(700 \mathrm{~K}\) has spectral emissivities of \(\varepsilon_{\lambda}=0\) for \(0 \leq \lambda \leq 3 \mu \mathrm{m}, \varepsilon_{\lambda}=0.5\) for \(3 \mu \mathrm{m}<\lambda \leq 10 \mu \mathrm{m}\), and \(\varepsilon_{\lambda}=0.9\) for \(10 \mu \mathrm{m}<\lambda<\infty\). A radiant flux of \(1000 \mathrm{~W} / \mathrm{m}^{2}\), which is uniformly distributed between 1 and \(6 \mu \mathrm{m}\), is incident on the surface at an angle of \(30^{\circ}\) relative to the surface normal. Calculate the total radiant power from a \(10^{-4} \mathrm{~m}^{2}\) area of the surface that reaches a radiation detector positioned along the normal to the area. The aperture of the detector is \(10^{-5} \mathrm{~m}^{2}\), and its distance from the surface is \(1 \mathrm{~m}\).

A two-color pyrometer is a device that is used to measure the temperature of a diffuse surface, \(T_{s}\). The device measures the spectral, directional intensity emitted by the surface at two distinct wavelengths separated by \(\Delta \lambda\). Calculate and plot the ratio of the intensities \(I_{\lambda+\Delta, e}\left(\lambda+\Delta \lambda, \theta, \phi, T_{s}\right)\) and \(I_{\lambda, e}\left(\lambda, \theta, \phi, T_{s}\right)\) as a function of the surface temperature over the range \(500 \mathrm{~K} \leq T_{s} \leq 1000 \mathrm{~K}\) for \(\lambda=5 \mu \mathrm{m}\) and \(\Delta \lambda=0.1,0.5\), and \(1 \mu \mathrm{m}\). Comment on the sensitivity to temperature and on whether the ratio depends on the emissivity of the surface. Discuss the tradeoffs associated with specification of the various values of \(\Delta \lambda\). Hint: The change in the emissivity over small wavelength intervals is modest for most solids, as evident in Figure 12.17.
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Translational Dynamics

Read Explanation

Modern Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Linear Momentum

Read Explanation

Rotational Dynamics

Read Explanation
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.