/*! 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 59 A \(100 \mathrm{~W}, 10 \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(100 \mathrm{~W}, 10 \mathrm{~cm}\)-diameter disk heater is placed parallel to, and \(5 \mathrm{~cm}\) from, a 10 \(\mathrm{cm}\)-diameter disk receiver in an evacuated chamber. The backs of both disks are well insulated. If the disk surfaces have emittances of \(0.8\) and the walls of the chamber are black and maintained at \(350 \mathrm{~K}\), calculate the following quantities: (i) The heater temperature. (ii) The receiver temperature. (iii) The radiative heat transfer between the heater and receiver. (iv) The net radiative heat transfer to the chamber.

Short Answer

Expert verified
(i) Heater temperature is approximately 571.8 K. (ii) Receiver temperature is approximately 554.7 K. (iii) Radiative heat transfer is about 14.1 W. (iv) Net heat transfer to the chamber is 85.9 W.

Step by step solution

01

Calculate the Heater's Temperature

Using the Stefan-Boltzmann law, the power radiated by the heater is given by:\[ P = \sigma \cdot A \cdot \varepsilon \cdot (T_h^4 - T_{c}^4) \]where \( P = 100 \text{ W} \), \( \sigma = 5.67 \times 10^{-8} \text{ W/m}^2 \cdot \text{K}^4 \) is the Stefan-Boltzmann constant, \( A = \frac{\pi}{4} \times (0.1)^2 \text{ m}^2 \) is the area of the disk (using the diameter), \( \varepsilon = 0.8 \) is the emittance, and \( T_{c} = 350 \text{ K} \) is the chamber temperature.Solving for \( T_h \):\[ 100 = 5.67 \times 10^{-8} \times \frac{\pi}{4} \times (0.1)^2 \times 0.8 \times (T_h^4 - 350^4) \]This simplifies to:\[ T_h^4 = \left(\frac{100}{5.67\times10^{-8} \times \frac{\pi}{4} \times 0.1^2 \times 0.8} + 350^4 \right) \]Computing the above expression gives \( T_h \approx 571.8 \text{ K} \).
02

Calculate the Receiver's Temperature

Assume steady-state conditions, the power absorbed by the receiver equals the power radiated by the heater. Therefore, using the same Stefan-Boltzmann law:\[ 100 = \sigma \cdot A \cdot \varepsilon \cdot (T_r^4 - T_c^4) \]Solving for \( T_r \):\[ 100 = 5.67 \times 10^{-8} \times \frac{\pi}{4} \times (0.1)^2 \times 0.8 \times (T_r^4 - 350^4) \]This simplifies to:\[ T_r^4 = \left(\frac{100}{5.67\times10^{-8} \times \frac{\pi}{4} \times 0.1^2 \times 0.8} + 350^4 \right) \]Computing the above expression gives \( T_r \approx 554.7 \text{ K} \).
03

Calculate Radiative Heat Transfer Between Heater and Receiver

The radiative heat transfer between two surfaces can be given as:\[ Q = \sigma \cdot A \cdot (T_h^4 - T_r^4) \]Using the calculated temperatures:\[ Q = 5.67 \times 10^{-8} \times \frac{\pi}{4} \times (0.1)^2 \times (571.8^4 - 554.7^4) \]Computing the above gives \( Q \approx 14.1 \text{ W} \).
04

Determine Net Radiative Heat Transfer to the Chamber

The net heat transfer to the chamber can be determined by the difference between the total emitted power from the heater and the power absorbed by the receiver:\[ Q_{net} = P - Q = 100 - 14.1 = 85.9 \text{ W} \].

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Key Concepts

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

Stefan-Boltzmann Law
The Stefan-Boltzmann Law is a fundamental principle in radiative heat transfer. It states that the power radiated per unit area of a blackbody is directly proportional to the fourth power of the blackbody's temperature. The equation for this law is given by \( P =  A                   P =  A  \sigma \cdot A \cdot T^4\), where:
  • \( P \) is the total power emitted.
  • \( A \) is the surface area.
  • \( \sigma \) is the Stefan-Boltzmann constant, approximately \( 5.67 \times 10^{-8} \, \text{W/m}^2 \cdot \text{K}^4 \).
  • \( T \) is the temperature in Kelvin.
It is important because it allows us to calculate the heat radiated by a body at a specific temperature. In situations involving radiative heat transfer between two surfaces, such as in the given exercise, this law is applied to estimate the amount of heat transferred between them.
Emittance
Emittance, often denoted by \( \varepsilon \), is a measure of a material's ability to emit energy as thermal radiation. It is a dimensionless quantity that ranges between 0 and 1. An emittance value of 1 corresponds to a perfect blackbody, which is an idealized physical body that absorbs all incident electromagnetic radiation. A real-world object has an emittance less than 1.In practical terms, emittance impacts how effectively a surface can radiate heat compared to a perfect blackbody. In the given problem, both the heater and receiver have an emittance of 0.8, meaning they are not perfect emitters but are still moderately efficient at emitting thermal radiation. Emittance plays a critical role in determining radiative heat transfer between objects, as it impacts the amount of energy radiated from each surface.
Temperature Calculation
Calculating temperature in the context of radiative heat transfer involves using the Stefan-Boltzmann Law, as previously discussed. Let's step through how to compute temperature like in the original exercise.
  • First, to find the temperature of a heating element, set up the radiative power equation accounting for known power, area, and emittance.
  • Rearrange the equation to solve for temperature, typically given in the fourth power due to the Stefan-Boltzmann Law: \( T^4 = \left(\frac{P}{\sigma \cdot A \cdot \varepsilon} + T_c^4\right) \), where \( T_c \) is the surrounding chamber's temperature.
  • The original example calculates a heater temperature of approximately 571.8 K.
These steps show how radiative heat transfer equations can help us determine precise thermal states of bodies in complex systems.
Evacuated Chamber
An evacuated chamber is a space or room from which air and other gases have been removed to create a vacuum. This absence of air minimizes the heat transfer by conduction or convection, leaving radiation as the primary mode of heat transfer. Using an evacuated chamber aids in controlling and analyzing radiative heat transfers more precisely. In the exercise, since the heater and receiver disks are placed in an evacuated chamber, it ensures that the calculated temperatures and heat transfers are a result of radiation only, without interference from the surrounding air. This setting is particularly useful in controlled experiments and applications such as in vacuum outgassing of materials, certain manufacturing processes, or space-related experiments where control over environment conditions is crucial.

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Most popular questions from this chapter

Two \(5 \mathrm{~mm}\)-thick rectangular aluminum alloy plates are attached to the exterior [ - surface of a spacecraft to form a channel \(12 \mathrm{~cm}\) long, \(2 \mathrm{~cm}\) wide, and \(6 \mathrm{~cm}\) deep. If the plates and wall are at a uniform temperature of \(310 \mathrm{~K}\), calculate the radiation heat transfer from the inside of the channel to space (taken as black at \(0 \mathrm{~K}\) ). Take \(\varepsilon=0.23\) for the aluminum.

A furnace is in the form of a cube with \(3 \mathrm{~m}\) sides. The roof is heated to \(1100^{\circ} \mathrm{C}\), and the work floor is at \(500^{\circ} \mathrm{C}\). The walls are insulated. The emittance of the roof is \(0.85\). Calculate the radiant heat flux into the floor as a function of its emittance for \(0.2<\varepsilon<1.0\).

A circular cross-section heating duct of \(50 \mathrm{~cm}\) diameter runs horizontally through the basement garage of a large apartment complex. The ambient air and walls of the garage are at \(17^{\circ} \mathrm{C} .\) At a location where the surface temperature of the duct is \(40^{\circ} \mathrm{C}\), determine the radiation contribution to the total heat loss per meter length if (i) the duct has an emittance of \(0.7 .\) (ii) the duct is painted with aluminum paint with \(\varepsilon=0.2\).

A thermistor is used to measure the temperature of a low-density helium flow. It is located in a \(10 \mathrm{~cm}\)-diameter tube through which the gas flows at \(1.6 \mathrm{~m} / \mathrm{s}\). The thermistor can be modeled as a \(3 \mathrm{~mm}\)-diameter sphere of emittance \(0.8\) and is located inside a radiation shield in the form of a \(1 \mathrm{~cm}\)-diameter, \(5 \mathrm{~cm}\)-long tube of emittance \(0.08\) (the axis of the tube is in the flow direction). If the thermistor records a temperature of \(327.5^{\circ} \mathrm{C}\) when the tube walls are at \(285^{\circ} \mathrm{C}\), determine the true temperature of the helium. Convective heat transfer coefficients on the thermistor and shield can be taken as \(19 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) and \(5 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), respectively. (Hint: First estimate the shield temperature.)

A multilayer insulation (MLI) to be used on a spacecraft consists of \(25 \mu \mathrm{m}\) Kapton external foil sheets ( \(\alpha_{s}=0.35, \varepsilon=0.60\) ), aluminum-coated on the inside ( \(\varepsilon=0.03\) ), sandwiching three aluminized Mylar sheets ( \(\varepsilon=0.03\) ). The layers are separated by Dacron mesh to prevent contact of the sheets. If a MLI shield is faced to the Sun at \(1 \mathrm{AU}\) distance and the inner surface is maintained at \(300 \mathrm{~K}\), determine the heat flux through the MLI and the outer surface temperature.

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