/*! 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 98 Boiler tubes exposed to the prod... [FREE SOLUTION] | 91Ó°ÊÓ

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Boiler tubes exposed to the products of coal combustion in a power plant are subject to fouling by the ash (mineral) content of the combustion gas. The ash forms a solid deposit on the tube outer surface, which reduces heat transfer to a pressurized wated'steam mixture flowing through the tubes. Consider a thin-walled boiler tube \(\left(D_{t}=0.05 \mathrm{~m}\right)\) whose surface is maintained at \(T_{t}=600 \mathrm{~K}\) by the boiling process. Combustion gases flowing over the tube at \(T_{-}=1800 \mathrm{~K}\) provide a convection coefficient of \(\bar{h}=100 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\), while radiation from the gas and boiler walls to the tube may be approximated as that originating from large surroundings at \(T_{\text {arr }}=1500 \mathrm{~K}\). (a) If the tube surface is diffuse and gray, with \(\varepsilon_{t}=0.8\), and there is no ash deposit layer, what is the rate of heat transfer per unit length, \(q^{\prime}\), to the boiler tube? (b) If a deposit layer of diameter \(D_{d}=0.06 \mathrm{~m}\) and thermal conductivity \(k=1 \mathrm{~W} / \mathrm{m}\), \(\mathrm{K}\) forms on the tube, what is the deposit surface temperature, \(T_{d} ?\) The deposït is diffuse and gray, with \(\varepsilon_{a}=0.9\), and \(T_{m} T_{\ldots}, T_{\text {}}\), and \(T_{\text {sar }}\) remain unchanged. What is the net rate of heat transfer per wnit length, \(y^{\prime}\) ', to the boxler tube? (c) Explore the effect of variations in \(D_{d}\) and \(\bar{h}\) on \(q^{r}\), as well as on relative contributions of convection and radiation to the net heat transfer fate. Represent your results graphically.

Short Answer

Expert verified
In short: 1. The heat transfer rate per unit length without ash deposit layer, \(q'\), can be computed using convection and radiation equations: \(q'=q_{conv}+q_{rad}\). 2. The deposit surface temperature, \(T_d\), can be calculated by finding the temperature difference across the deposit layer and adding it to the tube surface temperature, \(T_t\). 3. The net heat transfer rate per unit length with ash deposit, \(q'_d\), can be computed using modified convection and radiation equations: \(q'_{d}=q_{conv,d}+q_{rad,d}\). 4. By varying values of deposit diameter, \(D_d\), and convection coefficient, \(\bar{h}\), you can analyze their effect on heat transfer rate per unit length and the relative contributions of convection and radiation by plotting the results.

Step by step solution

01

Calculate heat transfer per unit length without ash deposit layer

Use the convection equation and radiation equation to find the heat transfer rate per unit length \(q'\). For convection: \(q_{conv} = \bar{h}A\left(T_{g} - T_{t}\right)\) For radiation: \(q_{rad} = \varepsilon_t\sigma A\left(T_{g}^4 - T_{t}^4\right)\), where \(\sigma\) is the Stefan-Boltzmann constant \(\left(5.67\times 10^{-8}\mathrm{~W/m^2K^4}\right)\). The total heat transfer per unit length is: \(q'=q_{conv}+q_{rad}\)
02

Calculate deposit surface temperature with ash deposit

First, compute the temperature difference across the deposit layer: \(\Delta T=\dfrac{q'}{kA}\), where \(k\) is the thermal conductivity of the deposit layer and \(A\) is the surface area. Next, compute the deposit surface temperature: \(T_d = T_t + \Delta T\)
03

Calculate net heat transfer per unit length with ash deposit layer

Calculate the heat transfer rate per unit length due to convection with deposit layer: \(q_{conv,d} = \bar{h}A\left(T_{g} - T_{d}\right)\). Calculate the heat transfer rate per unit length due to radiation with deposit layer: \(q_{rad,d} = \varepsilon_d\sigma A\left(T_{g}^4 - T_d^4\right)\), where \(\varepsilon_d\) is the emissivity of the deposit layer. The net heat transfer per unit length with ash deposit is: \(q'_{d}=q_{conv,d}+q_{rad,d}\)
04

Analyze the effect of variations

For variations in deposit diameter \(D_d\) and convection coefficient \(\bar{h}\), you can calculate the heat transfer rate per unit length with the ash deposit layer and analyze the relative contributions of convection and radiation by varying the values of \(D_d\) and \(\bar{h}\) and plotting the results. This will give a graphical representation of the effect of these variations on the overall heat transfer rate and the relative contributions of convection and radiation to the net heat transfer rate.

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Key Concepts

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

Boiler Tube Fouling
Fouling in boiler tubes is a common challenge in power plants, where ash and other particles from the combustion of fuels like coal build up on the surfaces of the tubes. This deposit reduces the efficiency of heat transfer by acting as a layer of insulation. The ash layer increases the thermal resistance, leading to higher operational costs and requiring more frequent maintenance due to the reduced exchange of heat. To illustrate, consider a boiler tube with a specific outer diameter and surface temperature, where the combustion gases flow over the tube, offering a certain convective heat transfer coefficient.

When there are no deposits, the rate of heat transfer per unit length, denoted as \(q'\), can be calculated using the sum of convective and radiative heat transfer. However, when a deposit layer with a specified diameter and thermal conductivity forms, it alters the surface temperature and reduces the net rate of heat transfer, \(q'_d\). This is a critical consideration for power plant operators, as fouling can significantly impact the overall efficiency and cost of plant operation.
Convective Heat Transfer
Convective heat transfer is one of the primary mechanisms of heat transfer in power plants, especially around boiler tubes where a fluid, such as combustion gas, moves across the surface. It involves the movement of heat from the hotter gas to the cooler boiler tube surface. The rate of convective heat transfer per unit area can be described by Newton's law of cooling, which states that heat transfer is proportional to the temperature difference between the gas and the tube surface and the convective heat transfer coefficient \( \bar{h} \). This coefficient depends on the properties of the fluid and the flow conditions.

In the absence of fouling, the formula \(q_{conv} = \bar{h}A(T_g - T_t)\) is used, where \(A\) is the surface area. However, when fouling occurs, the fouling layer affects the convective heat transfer by altering the effective contact area and by changing the thermal resistance of the system. As the layer grows thicker, it can significantly diminish the rate at which heat is transferred, thus necessitating calculations that reflect the presence of the fouling layer.
Radiative Heat Transfer
Radiative heat transfer is another key method by which heat is transferred in power plants. Unlike convection, radiation does not require a medium to transfer heat and can occur even through a vacuum. In the context of boiler tubes, radiant heat is emitted from the hot gases and the surrounding boiler walls to the tube's surface. The rate of radiative heat transfer can be modeled using the Stefan-Boltzmann law, which relates the emitted heat to the fourth power of the absolute temperature.

When there is no fouling, the radiative heat transfer rate per unit length is given by \(q_{rad} = \varepsilon_t\sigma A(T_{g}^4 - T_{t}^4)\), where \(\varepsilon_t\) represents the emissivity of the boiler tube, and \(\sigma\) is the Stefan-Boltzmann constant. Emissivity is a measure of how effectively a surface emits or absorbs radiant energy compared to a perfect black body. The presence of ash deposit changes not only the surface emissivity but also adds to the radial path of thermal resistance, changing the effective heat transfer rate due to radiation, \(q'_{rad,d}\).
Thermal Conductivity
Thermal conductivity is a material property that quantifies the ability to conduct heat. In power plants, the thermal conductivity of boiler tubes and any fouling material is pivotal in determining the rate at which heat is conveyed from the hot gas outside the tubes to the fluid, such as water/steam mixture, inside the tubes.

The impact of fouling on thermal conductivity becomes apparent when we consider the formation of an ash deposit layer. This layer with thermal conductivity \(k\) alters the overall heat transfer process by introducing additional thermal resistance. For a given heat transfer rate per unit length, the temperature drop across the fouling layer is computed using \(\Delta T=\frac{q'}{kA}\). As \(k\) decreases with fouling materials that have lower thermal conductivities than the boiler tube material, the temperature drop increases, reducing the deposit surface temperature \(T_d\), which directly relates to the net thermal efficiency of the system. It's important to regularly monitor and manage this fouling for maintaining the efficiency of heat transfer in the power plant.

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Most popular questions from this chapter

A row of regularly spaced, cylindrical heating elements is used to maintain an insulated furnace wall at \(500 \mathrm{~K}\). The opposite wall is at a uniform temperature of \(300 \mathrm{~K}\). The insulated wall experiences convection with air at \(450 \mathrm{~K}\) and a convection coefficient of \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming the walls and elements are black, estimate the required operating temperature for the elements.

A special surface coating on a square panel that is \(5 \mathrm{~m} \times 5 \mathrm{~m}\) on a side is cured by placing the panel directly under a radiant heat source having the same dimensions. The heat source is diffuse and gray and operates with a power input of \(75 \mathrm{~kW}\). The top surface of the heater, as well as the bottom surface of the panel. may be assumed to be well insulated, and the ammangement exists in a large room with air and wall temperatures of \(25^{\circ} \mathrm{C}\). The surface coating is diffuse and gray, with an emissivity of \(0.30\) and an upper temperature limit of \(400 \mathrm{~K}\). Neglecting convection effects, what is the minimum spacing that may be maintained between the heater and the panel to ensure that the panel temperature will not exceed \(400 \mathrm{~K}\) ? Allowing for convection effects at the costed surface of the panel. what is the minimum spacing?

Options for thermally shielding the top ceiling of a large furnace include the use of an insulating material of thickness \(L\) and thermal conductivity \(k\), case (a), or an air space of equävalent thickness formed by installing a steel sheet above the ceiling, case \((b)\). (a) Develop mathematical models that could be used to assess which of the two approaches is better. In both cases the interior surface is maintained at the same temperature, \(T_{s,}\) and the ambient air and surroundings are at equivalent temperatures \(\left(T_{w}=T_{\text {ue }}\right)\). (b) If \(k=0.090 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L=25 \mathrm{~mm}, h_{a}=25\) W/ \(\mathrm{m}^{2} \cdot \mathrm{K}\), the surfaces are diffuse and gray with \(\varepsilon_{i}=\varepsilon_{o}=0.50, T_{s u}=900 \mathrm{~K}\), and \(T_{w}=T_{\mathrm{ar}}=\) \(300 \mathrm{~K}\), what is the outer surface temperature \(T_{\text {a }}\) and the heat loss per unit surface area associated with each option? (c) For each case, assess the effect of surface radiative properties on the outer surface temperature and the heat loss per unit area for values of

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