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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 \(0.20\)-m-diameter, thin-walled steel pipe is used to transport saturated steam at a pressure of 20 bars in a room for which the air temperature is \(25^{\circ} \mathrm{C}\) and the convection heat transfer coefficient at the outer surface of the pipe is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What is the heat loss per unit length from the bare pipe (no insulation)? Estimate the heat loss per unit length if a 50 -mm-thick layer of insulation (magnesia, \(85 \%\) is added. The steel and magnesia may each be assumed to have an emissivity of \(0.8\), and the steam-side convection resistance may be neglected. (b) The costs associated with generating the steam and installing the insulation are known to be \(\$ 4 / 10^{9} \mathrm{~J}\) and \(\$ 100 / \mathrm{m}\) of pipe length, respectively. If the steam line is to operate \(7500 \mathrm{~h} / \mathrm{yr}\), how many years are needed to pay back the initial investment in insulation?

The energy transferred from the anterior chamber of the eye through the cornea varies considerably depending on whether a contact lens is worn. Treat the eye as a spherical system and assume the system to be at steady state. The convection coefficient \(h_{o}\) is unchanged with and without the contact lens in place. The cornea and the lens cover one-third of the spherical surface area. Values of the parameters representing this situation are as follows: \(\begin{array}{ll}r_{1}=10.2 \mathrm{~mm} & r_{2}=12.7 \mathrm{~mm} \\\ r_{3}=16.5 \mathrm{~mm} & T_{\infty, o}=21^{\circ} \mathrm{C} \\ T_{\infty \infty, i}=37^{\circ} \mathrm{C} & k_{2}=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ k_{1}=0.35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} & h_{o}=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \\ h_{i}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} & \end{array}\) (a) Construct the thermal circuits, labeling all potentials and flows for the systems excluding the contact lens and including the contact lens. Write resistance elements in terms of appropriate parameters. (b) Determine the heat loss from the anterior chamber with and without the contact lens in place. (c) Discuss the implication of your results.

Consider cylindrical and spherical shells with inner and outer surfaces at \(r_{1}\) and \(r_{2}\) maintained at uniform temperatures \(T_{s, 1}\) and \(T_{s, 2}\), respectively. If there is uniform heat generation within the shells, obtain expressions for the steady-state, one-dimensional radial distributions of the temperature, heat flux, and heat rate. Contrast your results with those summarized in Appendix C.

A spherical vessel used as a reactor for producing pharmaceuticals has a 10 -mm-thick stainless steel wall \((k=17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and an inner diameter of \(1 \mathrm{~m}\). The exterior surface of the vessel is exposed to ambient air \(\left(T_{\infty}=25^{\circ} \mathrm{C}\right)\) for which a convection coefficient of \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) may be assumed. (a) During steady-state operation, an inner surface temperature of \(50^{\circ} \mathrm{C}\) is maintained by energy generated within the reactor. What is the heat loss from the vessel? (b) If a 20 -mm-thick layer of fiberglass insulation \((k=0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is applied to the exterior of the vessel and the rate of thermal energy generation is unchanged, what is the inner surface temperature of the vessel?

A brass rod \(100 \mathrm{~mm}\) long and \(5 \mathrm{~mm}\) in diameter extends horizontally from a casting at \(200^{\circ} \mathrm{C}\). The rod is in an air environment with \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=30\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the temperature of the rod 25,50 , and \(100 \mathrm{~mm}\) from the casting?
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