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Chapter 3: Problem 45

A steam pipe of \(0.12-\mathrm{m}\) outside diameter is insulated with a layer of calcium silicate. (a) If the insulation is \(20 \mathrm{~mm}\) thick and its inner and outer surfaces are maintained at \(T_{s, 1}=800 \mathrm{~K}\) and \(T_{s, 2}=490 \mathrm{~K}\), respectively, what is the heat loss per unit length \(\left(q^{\prime}\right)\) of the pipe? (b) We wish to explore the effect of insulation thickness on the heat loss \(q^{\prime}\) and outer surface temperature \(T_{s, 2}\), with the inner surface temperature fixed at \(T_{s, 1}=\) \(800 \mathrm{~K}\). The outer surface is exposed to an airflow \(\left(T_{\infty}=25^{\circ} \mathrm{C}\right)\) that maintains a convection coefficient of \(h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and to large surroundings for which \(T_{\text {sur }}=T_{\infty}=25^{\circ} \mathrm{C}\). The surface emissivity of calcium silicate is approximately \(0.8\). Compute and plot the temperature distribution in the insulation as a function of the dimensionless radial coordinate, \(\left(r-r_{1}\right) /\left(r_{2}-r_{1}\right)\), where \(r_{1}=0.06 \mathrm{~m}\) and \(r_{2}\) is a variable \(\left(0.06

Short Answer

Expert verified
In summary, the heat loss per unit length (q') of the insulated steam pipe can be computed using the formula \[q^{\prime}=\frac{T_{s, 1}-T_{s, 2}}{R_{\text {ins }}}\] where R_ins is the thermal resistance of the insulation. For the temperature distribution in the insulation, the formula \[T(r) = T_{s, 1} - \frac{T_{s, 1} - T_{s, 2}}{\ln \left(\frac{r_1}{r_2}\right)}\ln \left(\frac{r}{r_1}\right)\] is used. The heat loss as a function of the insulation thickness can be analyzed by varying the insulation thickness \(r_2\) between 0.06 m and 0.2 m and calculating the temperature distribution and heat loss for each value. This can be done numerically, and the resulting temperature distribution and heat loss can be plotted accordingly.

Step by step solution

01

(a) Compute the heat loss per unit length

: First, we'll need to find the thermal resistance of the insulation. The formula for thermal resistance in the case of a cylindrical coordinate system is given by: \[R_{\text {ins }}=\frac{\ln \left(\frac{r_{2}}{r_{1}}\right)}{2 \pi k_{\text {ins }} L}\] where \(k_{\text{ins}}\) is the thermal conductivity of calcium silicate, \(L\) is the length of the pipe, and \(r_1\) and \(r_2\) are the inner and outer radii of the insulation, respectively. Given: - Inner surface temperature: \(T_{s, 1} = 800 K\) - Outer surface temperature: \(T_{s, 2} = 490 K\) - Thermal conductivity of calcium silicate: \(k_{\text{ins}} = 0.085 W/(m \cdot K)\) - Insulation thickness: \(\frac{r_2 - r_1}{r_1} = 20 mm \implies r_2 = 1.2r_1 \) - Inner radius: \(r_1 = 0.06 m\) - Outer radius: \(r_2 = 0.06 m \times 1.2 = 0.072 m\) Next, we'll calculate the heat loss per unit length \(q'\): \[q^{\prime}=\frac{T_{s, 1}-T_{s, 2}}{R_{\text {ins }}}\] Now, we can calculate the thermal resistance, \(R_{\text{ins}}\): \[R_{\text {ins }}=\frac{\ln \left(\frac{0.072}{0.06}\right)}{2 \pi(0.085) L}\] Finally, plug in the values and solve for \(q'\): \[q^{\prime}=\frac{800-490}{R_{\text {ins }}}\]
02

(b) Compute the temperature distribution and heat loss

: We'll use the following relationship for temperature distribution in cylindrical coordinates: \[T(r) = T_{s, 1} - \frac{T_{s, 1} - T_{s, 2}}{\ln \left(\frac{r_1}{r_2}\right)}\ln \left(\frac{r}{r_1}\right)\] where \(T(r)\) is the temperature at radius \(r\) and the dimensionless radial coordinate is \(\zeta = \frac{r - r_1}{r_2 - r_1}\). To compute the heat loss, we need to find the convective and radiative heat transfer from the outer surface: \[q^{\prime}_c = h\left(T_{s, 2} - T_{\infty}\right)\] \[q^{\prime}_r = \epsilon \sigma\left(T_{s, 2}^4 - T_{\text{sur}}^4\right)\] where: - \(h = 25 W/(m^2\cdot K)\) is the convection coefficient - \(\epsilon = 0.8\) is the surface emissivity of calcium silicate - \(\sigma = 5.67 \times 10^{-8} W/(m^2\cdot K^4)\) is the Stefan-Boltzmann constant - \(T_{\infty} = T_{\text{sur}} = 298.15 K\) is the surrounding air and surface temperature Now we need to vary the insulation thickness over the range of \(0.06 m < r_2 \leq 0.2 m\), calculating the temperature distribution and heat loss for each value of \(r_2\). This would have to be done numerically (ideally with a program or spreadsheet), and the resulting temperature distribution, \(T(r)\), and heat loss, \(q'\), can be plotted as functions of the dimensionless radial coordinate and the insulation thickness, respectively.

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Key Concepts

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

Thermal Resistance in Cylindrical Coordinates
When examining heat loss in cylindrical structures like pipes, it's essential to understand the concept of thermal resistance. In cylindrical coordinates, thermal resistance is influenced by the radial direction rather than the linear path found in flat plates. To calculate the thermal resistance (\( R_{\text{ins}} \)) of a cylindrical insulation layer, we use the formula:\[ R_{\text{ins}} = \frac{\ln \left(\frac{r_{2}}{r_{1}}\right)}{2 \pi k_{\text{ins}} L} \]Here:
  • \(r_1\) and \(r_2\) are the inner and outer radii of the insulation, respectively.
  • \(k_{\text{ins}}\) is the thermal conductivity of the insulating material.
  • \(L\) represents the length of the pipe.
This formula accounts for the logarithmic relationship inherent in radial thermal flow. By knowing \( R_{\text{ins}} \), we are able to determine how effectively the insulation reduces heat transfer in cylindrical systems. The greater the thermal resistance, the better the insulation restricts heat flow.
Convection and Radiation Heat Loss
The outer surface of insulated pipes loses heat not just through conduction, but also due to convection and radiation mechanisms. Understanding these two components is essential to fully evaluate heat loss. Convection describes heat transfer due to the movement of fluids, which in this case is the air surrounding the pipe. The rate of convective heat loss (\( q^{\prime}_c \)) can be determined using:\[ q^{\prime}_c = h \left(T_{s, 2} - T_{\infty}\right) \]where \(h\) is the convective heat transfer coefficient, \(T_{s, 2}\) is the outer surface temperature, and \(T_{\infty}\) is the ambient temperature.Radiation, on the other hand, involves heat transfer through electromagnetic waves. The radiative heat loss component (\( q^{\prime}_r \)) is calculated via:\[ q^{\prime}_r = \epsilon \sigma \left(T_{s, 2}^4 - T_{\text{sur}}^4\right) \]Where \(\epsilon\) is the surface emissivity, and \(\sigma\) is the Stefan-Boltzmann constant. Both these heat loss mechanisms need to be considered to correctly determine the total heat loss from a pipe, especially in environments with large temperature differences between the pipe surface and surroundings.
Temperature Distribution in Insulation
Inside the insulation of a cylindrical pipe, temperature varies radially from the hot inner surface to the cooler outer surface. Understanding the temperature distribution enables better predictions of thermal performance. The temperature at any point within the insulation can be described by the equation:\[T(r) = T_{s, 1} - \frac{T_{s, 1} - T_{s, 2}}{\ln \left(\frac{r_{2}}{r_{1}}\right)}\ln \left(\frac{r}{r_1}\right)\]where
  • \(T(r)\) is the temperature at radius \(r\)
  • \( T_{s, 1} \)and \( T_{s, 2} \) are the inner and outer surface temperatures
The equation reflects how temperature decays logarithmically across the insulation. Using the dimensionless radial coordinate (\( \zeta = \frac{r - r_1}{r_2 - r_1} \)), we can visualize temperature gradients more clearly. This understanding is crucial for determining where heat flow might be impeded or accelerated, ensuring efficient insulation design by predicting where most of the heat transfer is taking place.
Effect of Insulation Thickness on Heat Loss
The thickness of insulation directly impacts the rate of heat loss from a pipe. A thicker layer generally increases thermal resistance and reduces the total heat loss. This effect can be explored by varying the outer radius \(r_2\).As insulation thickness rises, \(R_{\text{ins}}\) increases:\[ R_{\text{ins}} = \frac{\ln \left(\frac{r_{2}}{r_{1}}\right)}{2 \pi k_{\text{ins}} L} \]A higher \(R_{\text{ins}}\) indicates a significant drop in heat transfer from the pipe to its surroundings. Consequently, calculations show that \(q^{\prime}\), the heat loss per unit length, decreases as:\[ q^{\prime}=\frac{T_{s, 1}-T_{s, 2}}{R_{\text{ins}}} \]By increasing the insulation thickness, one can achieve lower heat loss and a cooler outer surface temperature, \(T_{s, 2}\). However, practical limitations, such as cost and physical space, often dictate the feasible maximum thickness.Hence, designing an insulation system involves balancing these factors while aiming to achieve optimal thermal performance with minimal heat loss.

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Most popular questions from this chapter

A commercial grade cubical freezer, \(3 \mathrm{~m}\) on a side, has a composite wall consisting of an exterior sheet of \(6.35-\mathrm{mm}\)-thick plain carbon steel, an intermediate layer of \(100-\mathrm{mm}\)-thick cork insulation, and an inner sheet of \(6.35\)-mm-thick aluminum alloy (2024). Adhesive interfaces between the insulation and the metallic strips are each characterized by a thermal contact resistance of \(R_{t, c}^{\prime \prime}=2.5 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). What is the steady-state cooling load that must be maintained by the refrigerator under conditions for which the outer and inner surface temperatures are \(22^{\circ} \mathrm{C}\) and \(-6^{\circ} \mathrm{C}\), respectively?

In a test to determine the friction coefficient \(\mu\) associated with a disk brake, one disk and its shaft are rotated at a constant angular velocity \(\omega\), while an equivalent disk/shaft assembly is stationary. Each disk has an outer radius of \(r_{2}=180 \mathrm{~mm}\), a shaft radius of \(r_{1}=20 \mathrm{~mm}\), a thickness of \(t=12 \mathrm{~mm}\), and a thermal conductivity of \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). A known force \(F\) is applied to the system, and the corresponding torque \(\tau\) required to maintain rotation is measured. The disk contact pressure may be assumed to be uniform (i.e., independent of location on the interface), and the disks may be assumed to be well insulated from the surroundings. (a) Obtain an expression that may be used to evaluate \(\mu\) from known quantities. (b) For the region \(r_{1} \leq r \leq r_{2}\), determine the radial temperature distribution \(T(r)\) in the disk, where \(T\left(r_{1}\right)=T_{1}\) is presumed to be known. (c) Consider test conditions for which \(F=200 \mathrm{~N}\), \(\omega=40 \mathrm{rad} / \mathrm{s}, \tau=8 \mathrm{~N} \cdot \mathrm{m}\), and \(T_{1}=80^{\circ} \mathrm{C}\). Evaluate the friction coefficient and the maximum disk temperature.

Consider an extended surface of rectangular cross section with heat flow in the longitudinal direction. In this problem we seek to determine conditions for which the transverse ( \(y\)-direction) temperature difference within the extended surface is negligible compared to the temperature difference between the surface and the environment, such that the one-dimensional analysis of Section 3.6.1 is valid. (a) Assume that the transverse temperature distribution is parabolic and of the form $$ \frac{T(y)-T_{o}(x)}{T_{s}(x)-T_{o}(x)}=\left(\frac{y}{t}\right)^{2} $$ where \(T_{s}(x)\) is the surface temperature and \(T_{o}(x)\) is the centerline temperature at any \(x\)-location. Using Fourier's law, write an expression for the conduction heat flux at the surface, \(q_{y}^{\prime \prime}(t)\), in terms of \(T_{s}\) and \(T_{a^{+}}\) (b) Write an expression for the convection heat flux at the surface for the \(x\)-location. Equating the two expressions for the heat flux by conduction and convection, identify the parameter that determines the ratio \(\left(T_{o}-T_{s}\right) /\left(T_{s}-T_{\infty}\right)\). (c) From the foregoing analysis, develop a criterion for establishing the validity of the onedimensional assumption used to model an extended surface.

When raised to very high temperatures, many conventional liquid fuels dissociate into hydrogen and other components. Thus the advantage of a solid oxide fuel cell is that such a device can internally reform readily available liquid fuels into hydrogen that can then be used to produce electrical power in a manner similar to Example 1.5. Consider a portable solid oxide fuel cell, operating at a temperature of \(T_{\mathrm{fc}}=800^{\circ} \mathrm{C}\). The fuel cell is housed within a cylindrical canister of diameter \(D=\) \(75 \mathrm{~mm}\) and length \(L=120 \mathrm{~mm}\). The outer surface of the canister is insulated with a low-thermal-conductivity material. For a particular application, it is desired that the thermal signature of the canister be small, to avoid its detection by infrared sensors. The degree to which the canister can be detected with an infrared sensor may be estimated by equating the radiation heat flux emitted from the exterior surface of the canister (Equation 1.5; \(E_{s}=\varepsilon_{s} \sigma T_{s}^{4}\) ) to the heat flux emitted from an equivalent black surface, \(\left(E_{b}=\sigma T_{b}^{4}\right)\). If the equivalent black surface temperature \(T_{b}\) is near the surroundings temperature, the thermal signature of the canister is too small to be detected-the canister is indistinguishable from the surroundings. (a) Determine the required thickness of insulation to be applied to the cylindrical wall of the canister to ensure that the canister does not become highly visible to an infrared sensor (i.e., \(T_{b}-T_{\text {sur }}<5 \mathrm{~K}\) ). Consider cases where (i) the outer surface is covered with a very thin layer of \(\operatorname{dirt}\left(\varepsilon_{s}=0.90\right)\) and (ii) the outer surface is comprised of a very thin polished aluminum sheet \(\left(\varepsilon_{s}=0.08\right)\). Calculate the required thicknesses for two types of insulating material, calcium silicate \((k=0.09 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and aerogel \((k=0.006 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The temperatures of the surroundings and the ambient are \(T_{\text {sur }}=300 \mathrm{~K}\) and \(T_{\infty}=298 \mathrm{~K}\), respectively. The outer surface is characterized by a convective heat transfer coefficient of \(h=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) Calculate the outer surface temperature of the canister for the four cases (high and low thermal conductivity; high and low surface emissivity). (c) Calculate the heat loss from the cylindrical walls of the canister for the four cases.

A spherical Pyrex glass shell has inside and outside diameters of \(D_{1}=0.1 \mathrm{~m}\) and \(D_{2}=0.2 \mathrm{~m}\), respectively. The inner surface is at \(T_{s, 1}=100^{\circ} \mathrm{C}\) while the outer surface is at \(T_{s, 2}=45^{\circ} \mathrm{C}\). (a) Determine the temperature at the midpoint of the shell thickness, \(T\left(r_{m}=0.075 \mathrm{~m}\right)\). (b) For the same surface temperatures and dimensions as in part (a), show how the midpoint temperature would change if the shell material were aluminum.
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