/*! 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 41 Many products are processed in a... [FREE SOLUTION] | 91Ó°ÊÓ

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Many products are processed in a manner that requires a specified product temperature as a function of time. Consider a product in the shape of long, 10-mmdiameter cylinders that are conveyed slowly through a processing oven as shown in the schematic. The product exhibits near-black behavior and is attached to the conveying apparatus at the product's ends. The surroundings are at \(300 \mathrm{~K}\), while the radiation panel heaters are each isothermal at \(500 \mathrm{~K}\) and have surfaces that exhibit near-black behavior. An engineer proposes a novel oven design with a tilting top surface so as to be able to quickly change the thermal response of the product. (a) Determine the radiation per unit length incident upon the product at \(x=0.5 \mathrm{~m}\) and \(x=1 \mathrm{~m}\) for \(a=0\). (b) Determine the radiation per unit length incident upon the product at \(x=0.5 \mathrm{~m}\) and \(x=1 \mathrm{~m}\) for \(\alpha=\pi / 15\). Hint: The view factor from the cylinder to the left-hand side surroundings can be found by summing the view factors from the cylinder to the two surfaces shown as red dashed lines in the schematic.

Short Answer

Expert verified
To determine the radiation per unit length incident upon the product, we first find the view factors for the given geometry and tilt angles (\(\alpha = 0\) and \(\alpha = \frac{\pi}{15}\)). We then use the view factors to calculate the radiation per unit length using the equation: \[q''=A_{cylinder}F_{cylinder→ surroundings}(T_{panel}^4 - T_{surroundings}^4)\] where \(T_{panel} = 500\mathrm{~K}\) and \(T_{surroundings} = 300\mathrm{~K}\). For the cases when \(x=0.5\mathrm{~m}\) and \(x=1\mathrm{~m}\), we replace \(A_1\) and \(A_2\) to find \(q''_{x=0.5}\) and \(q''_{x=1}\) for \(\alpha = 0\) and \(\alpha = \frac{\pi}{15}\), respectively. The resulting values will show the change in radiation per unit length for each position due to the varying tilt angles.

Step by step solution

01

Determine view factors when \(\alpha = 0\)

The first step is to find the view factors for the geometry of the problem, which can be determined using the View Factor Summation Rule. When \(\alpha = 0\), the view factor from the cylinder to the left-hand side \(A_1\) and right-hand side \(A_2\) surroundings can be found by summing the view factors from the cylinder to the two surfaces outlined by the red dashed lines in the schematic. \[A_{1}F_{cylinder→ left- side} = A_{2}F_{cylinder→ right-side}\] Since the tilting top surface has an angle \(\alpha = 0\), the view factor from the cylinder to the left and right surroundings would be equal. Let this view factor be \(F_{cylinder}\), and we can write the equations as follows: \[A_1F_{cylinder} = A_2F_{cylinder}\]
02

Determine view factors when \(\alpha = \frac{\pi}{15}\)

Similar to the previous step when \(\alpha=\mask{\frac{pi}{15}}{0}\), we need to determine the new view factors for the cylinder to the tilted surroundings. The hint states that the view factor from the cylinder to the left-hand side surroundings can be found by summing the view factors from the cylinder to the two surfaces outlined by the red dashed lines. \[A_1'F_{cylinder→ tilted-left-side} = A_2F_{cylinder→ right-side}\]
03

Calculate the radiation per unit length when \(x = 0.5\) and \(1\mathrm{~m}\)

To calculate the radiation per unit length incident upon the product, we need to use these view factors. The equation for radiation per unit length is given by: \[q''=A_{cylinder}F_{cylinder→ surroundings}(T_{panel}^4 - T_{surroundings}^4)\] Note that, \(T_{panel} = 500\mathrm{~K}\) and \(T_{surroundings} = 300\mathrm{~K}\). For the case when \(x=0.5\mathrm{~m}\) and \(x=1\mathrm{~m}\), replace \(A_1\) and \(A_2\) to find \(q''_{x=0.5}\) and \(q''_{x=1}\) for \(\alpha = 0\) and \(\alpha = \frac{\pi}{15}\), respectively. Calculate these values and note the difference in radiation per unit length.

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Key Concepts

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

View Factor
Understanding the view factor is essential in radiation heat transfer problems. The view factor, also known as the configuration factor or shape factor, represents the fraction of radiation leaving a surface that reaches another designated surface. Mathematically, it's the proportion of radiative energy that leaves one surface and impinges directly on another.

For instance, in a processing oven scenario, as the long cylinder receives radiation from the heaters, the detailed characteristics of the oven, including its orientation and proximity to the cylinder, must be accounted for through view factors. Determining view factors accurately is crucial for predicting the intensity of radiation that an object, such as the product cylinder, would receive from the panel heaters.

When calculating view factors, geometric relationships and symmetries play pivotal roles. In exercises like these, it's important to apply the view factor summation rule, which states that the sum of view factors from a surface to all surrounding surfaces, including itself, must equal one. Furthermore, adjustments such as the angle \( \alpha \) require the recalculation of view factors to reflect the altered heat exchange characteristics.
Thermal Radiation
Thermal radiation is the transfer of heat energy by electromagnetic waves, typically infrared rays, emitted by the surfaces of objects due to their temperature. It's one of the three fundamental modes of heat transfer, the others being conduction and convection. Unlike these other modes, thermal radiation doesn't require a medium to propagate and can occur in a vacuum.

In the context of industrial processes, such as the heating of products in an oven, thermal radiation plays a pivotal role. The energy radiated by the panel heaters and absorbed by the product is significant for achieving the desired temperature distribution within the material. Materials exhibit various levels of emissivity, a measure of an object's effectiveness in emitting thermal radiation. In the provided exercise, the product and panel heaters have 'near-black' behavior, indicating that they are almost perfect emitters and absorbers of thermal radiation, simplifying the heat transfer calculations.
Heat Transfer Calculations
Heat transfer calculations are at the heart of thermal engineering and are necessary for designing systems like ovens or radiators. They allow engineers to predict and control the temperature changes and distributions in materials and environments. The steps involve applying equations based on thermodynamics and heat transfer principles to calculate radiation, conduction, or convection behaviors.

In problems dealing with radiation heat transfer, the Stefan-Boltzmann Law is often used, which states that the power radiated per unit area of a black body is proportional to the fourth power of the temperature of the body. Calculations must account for real-world factors like surface emissivity and view factors, which affect the total heat exchanged between surfaces. In the exercise, equation \(q''=A_{cylinder}F_{cylinder\rightarrow surroundings}(T_{panel}^4 - T_{surroundings}^4)\) exemplifies this method, where \(q''\) is the radiative heat flux, \(F_{cylinder\rightarrow surroundings}\) is the view factor, \(T_{panel}\) is the temperature of the panel heaters, and \(T_{surroundings}\) is the temperature of the surroundings.
Temperature Distribution
Temperature distribution in a material or system reflects how temperature varies across different points or regions. It's a critical aspect that affects material properties, efficiency, and product quality in many industrial processes. Achieving a uniform or specific temperature profile can be essential for processes such as tempering, annealing, or even baking.

In the exercise, the temperature distribution in the cylinders as they pass through the oven depends on factors like the intensity and distribution of thermal radiation reaching the product. By calculating radiation at different points (e.g., \(x=0.5 \mathrm{m}\) and \(x=1 \mathrm{m}\)), engineers can predict the resulting temperature profile of the product. The proposed oven design with the tilting top surface allows for rapid changes to this radiation pattern and therefore to the temperature distribution, highlighting the intertwined relationship between view factors and temperature outcomes in radiation heat transfer scenarios.

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Most popular questions from this chapter

Consider two diffuse surfaces \(A_{1}\) and \(A_{2}\) on the inside of a spherical enclosure of radius \(R\). Using the following methods, derive an expression for the view factor \(F_{12}\) in terms of \(A_{2}\) and \(R\). (a) Find \(F_{12}\) by beginning with the expression \(F_{i j}=\) \(q_{i \rightarrow j} / A_{i} J_{i} .\) (b) Find \(F_{12}\) using the view factor integral, Equation 13.1.

Two plane coaxial disks are separated by a distance \(L=0.20 \mathrm{~m}\). The lower disk \(\left(A_{1}\right)\) is solid with a diameter \(D_{\mathrm{o}}=0.80 \mathrm{~m}\) and a temperature \(T_{1}=300 \mathrm{~K}\). The upper disk \(\left(A_{2}\right)\), at temperature \(T_{2}=1000 \mathrm{~K}\), has the same outer diameter but is ring-shaped with an inner diameter \(D_{i}=0.40 \mathrm{~m}\). Assuming the disks to be blackbodies, calculate the net radiative heat exchange between them.

A long, cylindrical heating element of \(20-\mathrm{mm}\) diameter operating at \(700 \mathrm{~K}\) in vacuum is located \(40 \mathrm{~mm}\) from an insulated wall of low thermal conductivity. (a) Assuming both the element and the wall are black, estimate the maximum temperature reached by the wall when the surroundings are at \(300 \mathrm{~K}\). (b) Calculate and plot the steady-state wall temperature distribution over the range \(-100 \mathrm{~mm} \leq\) \(x \leq 100 \mathrm{~mm}\).

A wall-mounted natural gas heater uses combustion on a porous catalytic pad to maintain a ceramic plate of emissivity \(\varepsilon_{c}=0.95\) at a uniform temperature of \(T_{c}=1000 \mathrm{~K}\). The ceramic plate is separated from a glass plate by an air gap of thickness \(L=50 \mathrm{~mm}\). The surface of the glass is diffuse, and its spectral transmissivity and absorptivity may be approximated as \(\tau_{\lambda}=0\) and \(\alpha_{\lambda}=1\) for \(0 \leq \lambda \leq 0.4 \mu \mathrm{m}, \tau_{\lambda}=1\) and \(\alpha_{\lambda}=0\) for \(0.4<\lambda \leq 1.6 \mu \mathrm{m}\), and \(\tau_{\lambda}=0\) and \(\alpha_{\lambda}=0.9\) for \(\lambda>1.6 \mu \mathrm{m}\). The exterior sarface of the glass is exposed to quiescent ambient air and large surroundings for which \(T_{m}=T_{a x}=300 \mathrm{~K}\). The height and width of the heater are \(H=W=2 \mathrm{~m}\). (a) What is the total transmissjvity of the glass to irradiation from the ceramic plate? Can the glass be approximated as opaque and gray? (b) For the prescribed conditions, evaluate the glass temperature, \(T_{R}\), and the rate of heat transfer from the heater, \(q_{k}\). (c) A fan may be used to control the convection coefficient \(h_{n}\) at the exterior surface of the glass. Compute and plot \(T_{g}\) and \(q_{\mathrm{b}}\) as a function of \(h_{e}\) for \(10 \leq h_{0} \leq 100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

An opaque, diffuse, gray ( \(200 \mathrm{~mm} \times 200 \mathrm{~mm})\) plate with an emissivity of \(0.8\) is placed over the opening of a furnace and is known to be at \(400 \mathrm{~K}\) at a certain instant. The bottom of the fumace, having the same dimensions as the plate, is black and operates at \(1000 \mathrm{~K}\). The sidewalls of the fumace are well insulated. The top of the plate is exposed to ambient air with a convection coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and to large surroundings. The air and surroundings are each at \(300 \mathrm{~K}\). (a) Evaluate the net radiative heat transfer to the bottom surface of the plate. (b) If the plate has mass and specific heat of \(2 \mathrm{~kg}\) and \(900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively, what will be the change in temperature of the plate with time, \(d T_{p} / d r\) ? Assume convection to the bottom surface of the plate to be negligible. (c) Extending the analysis of part (b), generate a plot of the change in temperature of the plate with time, \(d T_{\rho} / d l\), as a function of the plate temperature for \(350 \leq T_{p} \leq 900 \mathrm{~K}\) and all other conditions remaining the same. What is the steady-state tem-

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