/*! 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 19 A circular cross-section heating... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(\varepsilon=0.2\).

Short Answer

Expert verified
For emittance 0.7, heat loss is about 61.84 W/m; for emittance 0.2, it is about 17.67 W/m.

Step by step solution

01

Understanding Radiation Heat Loss

Radiation heat loss per unit length of a duct is calculated using the Stefan-Boltzmann law. The formula is \[ q = \varepsilon \sigma (T_s^4 - T_a^4) \pi D \]where \( q \) is the radiation heat loss per unit length, \( \varepsilon \) is the emittance of the duct, \( \sigma = 5.67 \times 10^{-8} \mathrm{~W/m^2K^4} \) is the Stefan-Boltzmann constant, \( T_s \) is the surface temperature of the duct, \( T_a \) is the ambient temperature, and \( D \) is the diameter of the duct.
02

Convert Temperatures to Kelvin

Convert the given temperatures to Kelvin, since the Stefan-Boltzmann law requires temperatures in absolute units:Initial temperatures are given as:\[ T_s = 40^{\circ} \mathrm{C} \] \[ T_a = 17^{\circ} \mathrm{C} \] We use the formula: \[ T(K) = T(\degree C) + 273.15 \]Converting:\[ T_s = 40 + 273.15 = 313.15 \text{ K} \]\[ T_a = 17 + 273.15 = 290.15 \text{ K} \]
03

Calculate the Radiation Heat Loss for Emittance 0.7

Substitute the values into the radiation heat loss formula for an emittance of 0.7, and a duct diameter of 0.5 meters:\[ \varepsilon = 0.7 \]\[ D = \frac{50}{100} = 0.5 \text{ m} \]Calculate:\[ q_1 = 0.7 \times 5.67 \times 10^{-8} \left((313.15)^4 - (290.15)^4\right) \times \pi \times 0.5 \text{ W/m} \]
04

Calculate the Radiation Heat Loss for Emittance 0.2

Substitute the values into the same radiation formula for an emittance of 0.2:\[ \varepsilon = 0.2 \]Calculate:\[ q_2 = 0.2 \times 5.67 \times 10^{-8} \left((313.15)^4 - (290.15)^4\right) \times \pi \times 0.5 \text{ W/m} \]
05

Obtain Final Numerical Results

Perform the calculations:For emittance 0.7,\[ q_1 \approx 61.84 \text{ W/m} \]For emittance 0.2,\[ q_2 \approx 17.67 \text{ W/m} \]

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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 fundamental in understanding radiation heat transfer. It relates the power radiated by a black body to its temperature. The law states that the total energy radiated per unit surface area is proportional to the fourth power of the black body's absolute temperature. This is expressed in the formula: \[ E = ext{const} imes T^4 \]where \( E \) is the emissive power, \( T \) is the absolute temperature, and the constant is known as the Stefan-Boltzmann constant, \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \). When applied to a real object, the law incorporates the emittance \( \varepsilon \), modifying the formula to:\[ q = \varepsilon \sigma T^4 \].This shows us how material properties and surface conditions impact radiation heat loss. Every material has a specific emittance, which affects how much energy it emits compared to a perfect black body. Understanding this law helps calculate heat lost by radiation effectively.
Emittance
Emittance is a measure of a material's ability to emit thermal radiation. It ranges from 0 to 1, with 1 representing a perfect emitter, or black body, that emits maximum radiation. It is a critical factor in calculating radiation heat loss using the Stefan-Boltzmann Law. In practical scenarios: - **Higher emittance:** More radiation is emitted, leading to increased heat loss. For instance, in our example, the duct with emittance 0.7 loses more heat than when painted with aluminum, which has lower emittance. - **Lower emittance:** Lesser radiation is emitted, minimizing heat loss. Materials like polished metals usually have a lower emittance, making them suitable for reducing unwanted heat loss. The duct in the exercise shows how changing the emittance by applying different coatings affects the heat lost by radiation. By choosing materials with appropriate emittance values, you can influence thermal efficiency in heating systems.
Heat Transfer Calculation
Calculating heat transfer, particularly radiation heat loss, involves several key steps:1. **Formula Setup:** Use the Stefan-Boltzmann law tailored for real-world objects: \[ q = \varepsilon \sigma (T_s^4 - T_a^4) \pi D \] where: - \( q \) is the heat loss per unit length. - \( \varepsilon \) is the emittance. - \( \sigma \) is the Stefan-Boltzmann constant. - \( T_s \) and \( T_a \) are the surface and ambient temperatures respectively, in Kelvin. - \( D \) is the diameter of the duct.2. **Temperature Conversion:** Convert all temperatures from Celsius to Kelvin using the formula: \[ T(\text{K}) = T(\degree \text{C}) + 273.15 \]3. **Substitution and Calculation:** Plug the values into the formula and solve. This provides the radiation heat loss per meter.These calculations help in optimizing the design and operation of heating systems by accurately determining energy losses. The exercise illustrates how changing emittance influences heat loss, demonstrating the practical application of these calculations in engineering and building science.

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Most popular questions from this chapter

A \(1 \mathrm{~cm}\)-thick redwood \((k=0.1 \mathrm{~W} / \mathrm{m} \mathrm{K})\) patio cover is surfaced with a layer of black tar paper ( \(\varepsilon=0.90 . \alpha_{s}=0.95\) ), and the underside is stained with redwood sealer-stain \((\varepsilon=0.88)\). Typical summer noon conditions include a solar irradiation of \(950 \mathrm{~W} / \mathrm{m}^{2}\), a clear sky, and ambient air at \(27^{\circ} \mathrm{C}\) and \(50 \% \mathrm{RH}\). Wind blowing through the patio gives an estimated convective heat transfer coefficient on both sides of the cover of \(8 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). An important factor determining the comfort of people sitting on the patio is the radiosity of the underside of the cover, since it is this radiosity that determines the radiant heating experienced by the people. Estimate the radiosity for the following situations: (i) For the cover as designed. (ii) If the underside is painted with aluminum paint ( \(\varepsilon=0.22\) ). (iii) If, instead, the top of the cover is painted with a white paint \(\left(\alpha_{s}=0.30, \varepsilon=0.85\right)\) and kept clean. (iv) The combination of (ii) and (iii). $$ T_{e}=27^{2} \mathrm{C} .50 \mathrm{~cm} \mathrm{RH} $$

A \(10 \mathrm{~cm}-\mathrm{square}\) horizontal plate coated with a black paint \((\varepsilon=0.94)\) is exposed to a clear night sky when the ambient air is still at \(290 \mathrm{~K}, 1\) atm, and \(80 \% \mathrm{RH}\). If the backside of the plate is perfectly insulated, will dew form on the plate? Use the Brunt formula for sky emittance.

Calculate the blackbody emissive power \(E_{b}\) for surfaces at \(300 \mathrm{~K}, 1000 \mathrm{~K}, 2000 \mathrm{~K}\), and \(5000 \mathrm{~K}\). Also estimate the peak wavelength \(\lambda_{\max }\) for each surface using: (i) \(\lambda_{\max } T=2898 \mu \mathrm{m} \mathrm{K}\) (from Eq. \(6.6\) and Fig. \(6.3 a\) ) (ii) \(\lambda_{\max } T=5100 \mu \mathrm{m} \mathrm{K}\) (from Fig. 6.3b)

A long furnace used for an enameling process has a \(3 \mathrm{~m} \times 3 \mathrm{~m}\) cross section with the roof maintained at \(1900 \mathrm{~K}\), and the side walls are well insulated. What is the radiant heat transfer to the floor when it is at \(600 \mathrm{~K}\) ? All surfaces may be taken to be gray and diffuse, with emittances of \(0.5\). (i) Assume the side walls are isothermal. (ii) Divide the side walls into two surfaces and allow these surfaces to be isothermal at different temperatures.

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 \(0.2<\varepsilon<1.0\).

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