Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A furnace is shaped like a long equilateral-triangular duct where the width of each side is \(2 \mathrm{~m}\). Heat is supplied from the base surface, whose emissivity is \(\varepsilon_{1}=0.8\), at a rate of \(800 \mathrm{~W} / \mathrm{m}^{2}\) while the side surfaces, whose emissivities are \(0.5\), are maintained at \(500 \mathrm{~K}\). Neglecting the end effects, determine the temperature of the base surface. Can you treat this geometry as a two-surface enclosure?

Short Answer

Expert verified
Answer: The base surface temperature is approximately 612.7 K, and this geometry cannot be treated as a two-surface enclosure.

Step by step solution

01

Apply the Stefan-Boltzmann Law

The Stefan-Boltzmann Law relates the heat transfer (power emitted) by a surface to its temperature and emissivity: \(q = \varepsilon \sigma(T^{4} - T_{s}^4)\) where: - \(q\) is the heat transfer per unit area [W/m²] - \(\varepsilon\) is the emissivity of the surface - \(\sigma\) is the Stefan-Boltzmann constant, \(5.67 \times 10^{-8} \mathrm{W/m^{2}K^4}\) - \(T\) is the temperature of the surface [K] - \(T_s\) is the surrounding surface temperature [K] The base surface has an emissivity of \(\varepsilon_{1}=0.8\) and receives a heat transfer of \(800 \mathrm{~W}/\mathrm{m}^{2}\). We will call the base surface temperature \(T_{1}\). The side surfaces have an emissivity of \(\varepsilon_{2}=0.5\) and are maintained at \(500 \mathrm{~K}\) (i.e., \(T_{s} = 500 \mathrm{~K}\)).
02

Energy balance

We can set up an energy balance equation using the Stefan-Boltzmann Law for the base surface and side surfaces: \(q_{1} = \varepsilon_{1} \sigma (T_{1}^4 - T_{s}^4)\) Substitute the given values: \(800 = 0.8 \times 5.67 \times 10^{-8} \times (T_{1}^4 - 500^4)\)
03

Solve for the base surface temperature, \(T_1\)

Solve the energy balance equation for \(T_{1}\): \(800 = 4.54 \times 10^{-8} \times (T_{1}^4 - 500^4)\) \(800 = 4.54 \times 10^{-8} \times (T_{1}^4 - 62500000000)\) \(T_{1}^4 = \frac{800}{4.54 \times 10^{-8}} + 62500000000\) \(T_{1}^4 = 17624382989\) \(T_{1} = \sqrt[4]{17624382989}\) \(T_{1} \approx 612.7 \mathrm{~K}\)
04

Determine if we can treat this as a two-surface enclosure

Calculate the angle factor (view factor) between the base surface and the side surfaces, \(F_{12}\): \(F_{12} = \frac{1}{2} \times (1 - \cos(60^{\circ}))\) \(F_{12} = 0.25\) Since \(F_{12} = 0.25\) is not close to zero, we cannot treat this geometry as a two-surface enclosure. It means that radiant heat exchange between surfaces should be taken into account. To summarize, the base surface temperature is approximately \(612.7 \mathrm{~K}\), and this geometry cannot be treated as a two-surface enclosure.

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 Vaia!

Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics dealing with heat, work, and energy. At the heart of thermodynamics lie the four fundamental laws which govern the behavior of thermal energy in systems. The first law, also known as the law of energy conservation, states that energy cannot be created or destroyed, only transformed from one form to another. The second law introduces the concept of entropy, indicating that systems naturally progress towards disorder. The third law addresses the impossibility of reaching absolute zero temperature. Lastly, there's the zeroth law which involves thermal equilibrium and introduces the concept of temperature.

These principles provide the foundation for understanding heat transfer mechanisms, including radiation, conduction, and convection, which are essential for solving problems related to the temperature of objects, such as the exercise given where the temperature of a furnace surface is determined.
Heat Transfer
Heat transfer is the movement of heat from a region of higher temperature to one of lower temperature and can occur via three primary mechanisms: conduction, convection, and radiation. Conduction is the transfer of heat through a solid or between solids in direct contact, driven by a temperature gradient. Convection is the transfer of heat through a fluid, which could be liquid or gas, and is influenced by the fluid's motion. Radiation, however, does not require a medium and transfers energy through electromagnetic waves.

In the described exercise, we focus on radiative heat transfer, particularly how the Stefan-Boltzmann Law describes the rate at which heat is emitted by the surfaces of the furnace. The application of thermodynamics principles, specifically concerning heat transfer, enables us to calculate the base surface temperature within the given conditions.
Radiative Heat Exchange
Radiative heat exchange refers to the transfer of heat energy through electromagnetic radiation. This process is fundamental to the behavior of the furnace in our problem. Objects emit radiation according to their temperature and emissivity—a measure of how effectively they radiate energy. The Stefan-Boltzmann Law quantitatively describes this relationship, showing that the power radiated per unit area by a surface is proportional to the fourth power of the surface’s temperature and the surface’s emissivity.

In the furnace problem, we see a practical application of radiative heat exchange. The heated base surface radiates heat energy outwards, and this transfer of heat is modeled by the equation provided by the Stefan-Boltzmann Law. The law takes into account the temperatures of both the emitting and surrounding surfaces, thereby allowing us to solve for the unknown temperature given the other parameters.
Emissivity
Emissivity is a dimensionless quantity that measures the efficiency with which a surface emits thermal radiation compared to a perfect black body. A black body is an idealized physical entity that absorbs all incident radiation, and it has an emissivity of 1. Real-world objects have emissivities ranging from 0 to 1; the closer the value is to 1, the more effectively the object emits radiation.

In the context of our exercise, different surfaces of the furnace have distinct emissivities (\( \boldsymbol{\frac{1}{0.8}} \) for the base and \( \boldsymbol{0.5} \) for the sides). This difference in emissivity is crucial when we apply the Stefan-Boltzmann Law to determine the rate of radiative heat exchange and, consequently, the temperature. By acknowledging that each surface has its own emissivity, we can more accurately predict the flow of heat between surfaces in the furnace.
Temperature Calculation
Temperature calculation, especially in the field of thermodynamics, involves determining an object's temperature based on various factors like heat transfer rate, material properties, and surrounding conditions. In the example provided, the calculation involves an iterative process. An initial equation derived from the Stefan-Boltzmann Law relates heat transfer rate, emissivity, and temperatures of the interacting surfaces.

The step-by-step solution demonstrates how to isolate the unknown temperature (the base surface temperature of the furnace) from the equation then solve for it using algebra and available data (emissivity and the rate of heat transfer). Temperature calculation often requires understanding and manipulation of these types of formulas to predict how a system will behave under given thermal conditions.

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

\(3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and both natural convection and radiation heat transfer occur at the outer shield surface. To keep the ammonia inside the tube in its liquid state, determine the surrounding temperature that would maintain temperatures of the outer tube and the shield at the specified values. 13-68 PtD A hot liquid is being transported inside a long tube with a diameter of \(25 \mathrm{~mm}\). The hot liquid causes the tube surface temperature to be \(150^{\circ} \mathrm{C}\). To prevent thermal burn hazards, the tube is enclosed with a concentric outer cylindrical cover of \(5 \mathrm{~cm}\) in diameter allowing a vacuumed gap in between the two surfaces. The concentric outer cover has an emissivity of \(0.6\) and the outer surface is exposed to natural convection with a heat transfer coefficient of \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and radiation heat transfer with the surrounding at a temperature of \(20^{\circ} \mathrm{C}\). Determine the necessary emissivity of the inside tube so that the outer cover temperature is below \(45^{\circ} \mathrm{C}\) to prevent thermal burns.

Consider a \(3-\mathrm{m} \times 3-\mathrm{m} \times 3-\mathrm{m}\) cubical furnace. The base surface is black and has a temperature of \(400 \mathrm{~K}\). The radiosities for the top and side surfaces are calculated to be \(7500 \mathrm{~W} / \mathrm{m}^{2}\) and \(3200 \mathrm{~W} / \mathrm{m}^{2}\), respectively. If the temperature of the side surfaces is \(485 \mathrm{~K}\), the emissivity of the side surfaces is (a) \(0.37\) (b) \(0.55\) (c) \(0.63\) (d) \(0.80\) (e) \(0.89\)

How does radiation transfer through a participating medium differ from that through a nonparticipating medium?

The base surface of a cubical furnace with a side length of \(3 \mathrm{~m}\) has an emissivity of \(0.80\) and is maintained at \(500 \mathrm{~K}\). If the top and side surfaces also have an emissivity of \(0.80\) and are maintained at \(900 \mathrm{~K}\), the net rate of radiation heat transfer from the top and side surfaces to the bottom surface is (a) \(194 \mathrm{~kW}\) (b) \(233 \mathrm{~kW}\) (c) \(288 \mathrm{~kW}\) (d) \(312 \mathrm{~kW}\) (e) \(242 \mathrm{~kW}\)

Consider two infinitely long concentric cylinders with diameters 20 and \(25 \mathrm{~cm}\). The inner surface is maintained at \(700 \mathrm{~K}\) and has an emissivity of \(0.40\), while the outer surface is black. If the rate of radiation heat transfer from the inner surface to the outer surface is \(2400 \mathrm{~W}\) per unit area of the inner surface, the temperature of the outer surface is (a) \(605 \mathrm{~K}\) (b) \(538 \mathrm{~K}\) (c) \(517 \mathrm{~K}\) (d) \(451 \mathrm{~K}\) (e) \(415 \mathrm{~K}\)

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.

Sign-up for free