91Ó°ÊÓ

A row of regularly spaced, cylindrical heating elements (1) is used to cure a surface coating that is applied to a large panel (2) positioned below the elements. A second large panel (3), whose top surface is well insulated, is positioned above the elements. The elements are black and maintained at \(T_{1}=600 \mathrm{~K}\). while the panel has an emissivity of \(\varepsilon_{2}=0.5\) and is maintained at \(T_{2}=400 \mathrm{~K}\). The cavity is filled with a nonparticipating gas and convection heat transfer occurs at surfaces 1 and 2 , with \(\bar{h}_{1}=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(\bar{h}_{2}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (Convection at the insulated panel may be neglected.) (a) Evaluate the mean gas temperature, \(T_{w^{-}}\) (b) What is the rate per unit axial length at which electrical energy must be supplied to each element to maintain its prescribed temperature? (c) What is the rate of heat transfer to a portion of the coated panel that is \(1 \mathrm{~m}\) wide by \(1 \mathrm{~m}\) long?

Step by step solution

01

(Step 1: Determine mean gas temperature)

(To determine the mean gas temperature, we can find the average temperature between the heating elements (1) and the large panel (2) using: \(T_{w^{-}} = \frac{T_1 + T_2}{2}\))

02

(Step 2: Calculate mean gas temperature)

(Now, plug in the given values to find the mean gas temperature: \(T_{w^{-}} = \frac{600 + 400}{2} = 500 \mathrm{K}\)) The mean gas temperature is 500 K.

03

(Step 3: Calculate the rate of radiant heat transfer)

(We can use the Stefan-Boltzmann Law to find the rate of radiant heat transfer between surfaces 1 and 2: \(q_{rad} = \varepsilon_1 \sigma (T_1^4 - T_2^4)\) Since the heating elements are black, they have an emissivity of 1 (\(\varepsilon_1 = 1\)). Thus, we have: \(q_{rad} = \sigma (T_1^4 - T_2^4)\))

04

(Step 4: Plug in values for rate of radiant heat transfer)

(Using the given temperatures and the Stefan-Boltzmann constant, \(\sigma = 5.67 \times 10^{-8} \, \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4}\), we find the rate of radiant heat transfer: \(q_{rad} = 5.67 \times 10^{-8} (600^4 - 400^4) = 110784 \, \mathrm{W} / \mathrm{m}^{2}\))

05

(Step 5: Calculate the rate of convection heat transfer)

(We can use the convection heat transfer equation to find the rate of convection heat transfer for surfaces 1 and 2: \(q_{conv_1} = \bar{h}_1 (T_1 - T_{w^{-}})\) \(q_{conv_2} = \bar{h}_2 (T_2 - T_{w^{-}})\))

06

(Step 6: Plug in values for rate of convection heat transfer)

(Now, we can use the given convection coefficients and mean gas temperature to find the rate of convection heat transfer for both surfaces: \(q_{conv_1} = 10 \times (600 - 500) = 1000 \, \mathrm{W} / \mathrm{m}^{2}\) \(q_{conv_2} = 2 \times (400 - 500) = -200 \, \mathrm{W} / \mathrm{m}^{2}\))

07

(Step 7: Calculate the rate of electrical energy supply)

(The rate at which electrical energy must be supplied to each element is equal to the total heat transfer by both radiation and convection at surface 1: \(q_{elec} = q_{rad} + q_{conv_1}\))

08

(Step 8: Plug in values for rate of electrical energy supply)

(Using the previously calculated values, we find the rate of electrical energy supply: \(q_{elec} = 110784 + 1000 = 111784 \, \mathrm{W} / \mathrm{m}^{2}\)) The rate at which electrical energy must be supplied to each element is 111,784 W/m².

09

(Step 9: Calculate the heat transfer to the coated panel)

(The rate of heat transfer to a portion of the coated panel that is 1m x 1m can be calculated as: \(q_{panel} = q_{conv_2} \cdot 1\:m \cdot 1\:m\))

10

(Step 10: Plug in values for heat transfer to the coated panel)

(Using the previously calculated convection heat transfer value for surface 2, we find the heat transfer to the coated panel: \(q_{panel} = (-200) \cdot 1\:m \cdot 1\:m = -200 \, \mathrm{W}\)) The heat transfer to a portion of the coated panel is -200 W. Note that the negative sign indicates that heat is being removed from the panel.

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.

Radiation

Radiation is one of the key modes of heat transfer and plays a crucial role in systems exposed to high temperatures. It involves the transfer of energy through electromagnetic waves. This means no medium is required for radiation to occur, unlike conduction and convection, which depend on a material medium like the air or walls.

In the context of the heating elements and panels in our exercise, radiation is responsible for transferring heat across the gap between the elements and the panel below. The energy radiates in all directions and is absorbed by the panel based on its surface properties. The effectiveness of this transfer depends greatly on the temperature difference between the surfaces and the nature of their interaction, such as their emissivity, which we'll discuss later.

• Heat is transferred via electromagnetic waves.
• Does not require a physical medium.
• Depends on surface temperature differences.

Convection

Convection is another mode of heat transfer which occurs in fluids where heat is carried by the movement of currents within the fluid itself. This concept can be divided into two types: natural convection and forced convection. Natural convection happens due to temperature-induced density differences, while forced convection involves an external factor like a fan or pump to move the fluid.

In the exercise, convection occurs alongside radiation. Both the heating elements and the coating panel experience a convection heat transfer with the surrounding gas, although the insulated panel is noted to have negligible convection. The convection rate is impacted by the convection heat transfer coefficient, denoted as \( \bar{h} \), and the temperature differences between the surfaces and the gas.

• Transfer of heat through fluid movement.
• Includes both natural and forced convection.
• Influenced by convection coefficients and temperature differences.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is fundamental in quantifying the rate of radiant heat transfer. It defines how much energy is radiated by a blackbody in terms of its temperature. The law is expressed through the Stefan-Boltzmann equation, which calculates the radiant power per unit area from a surface: \[ q_{rad} = \varepsilon \sigma T^4 \] where \( q_{rad} \) is the heat transfer per unit area, \( \varepsilon \) is the emissivity of the surface, \( \sigma \) is the Stefan-Boltzmann constant \( (5.67 \times 10^{-8} \mathrm{W/m^2 \cdot K^4}) \), and \( T \) is the absolute temperature of the surface.

In this application, the heating element being black has an emissivity of 1, which means it behaves like a perfect blackbody. Thus, it relinquishes heat effectively through radiation. The Stefan-Boltzmann Law here helps us calculate how much energy the element radiates based on its higher temperature.

• Describes radiant heat transfer for blackbodies.
• Depends on the fourth power of temperature.
• The emissivity value modifies the equation for real objects.

Emissivity

Emissivity is a measure of a surface's ability to emit thermal radiation and is a key player in the Stefan-Boltzmann Law. It's essentially the efficiency with which a surface radiates energy compared to an ideal blackbody, which is assigned an emissivity value of 1. A higher emissivity means an object radiates energy more effectively.

In the exercise, the black heating elements have an emissivity of 1, making them very efficient radiators of heat. Meanwhile, the coated panel has an emissivity of 0.5, which means it emits only half of the energy a perfect blackbody would at the same temperature. Emissivity influences the rate of heat transfer by radiation, directly affecting temperature regulation in thermal management systems like the one in this problem.

• Measures radiative efficiency relative to a blackbody.
• Varies between 0 and 1, with 1 being ideal.
• Impacts rate and effectiveness of heat transfer through radiation.