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Chapter 4: Problem 14

Two parallel pipelines spaced \(0.5 \mathrm{~m}\) apart are buried in soil having a thermal conductivity of \(0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The pipes have outer diameters of 100 and \(75 \mathrm{~mm}\) with surface temperatures of \(175^{\circ} \mathrm{C}\) and \(5^{\circ} \mathrm{C}\), respectively. Estimate the heat transfer rate per unit length between the two pipelines.

Short Answer

Expert verified
In order to estimate the heat transfer rate per unit length between the two parallel pipelines, we first convert all given values to SI units, calculate the temperature difference and mean radius values, and then determine the equivalent thermal resistance. Using the formula \(Q = \frac{\Delta T}{R_{eq}}\), we find that the heat transfer rate per unit length between the two pipelines is approximately \(\frac{170}{R_{eq}}\), after calculating the value of \(R_{eq}\).

Step by step solution

01

Identify the given information

We are given: - Distance between the pipelines: 0.5 m - Thermal conductivity of the soil: \(k = 0.5 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}}\) - Outer diameters of the pipelines: - Pipeline 1: 100 mm - Pipeline 2: 75 mm - Surface temperatures of the pipelines: - Pipeline 1: \(T_1 = 175^{\circ} \mathrm{C}\) - Pipeline 2: \(T_2 = 5^{\circ} \mathrm{C}\)
02

Convert all values to SI units

First, we need to convert all values into SI units. - Outer diameters of the pipelines: - Pipeline 1: \(D_1 = \frac{100}{1000} \mathrm{m} = 0.1 \mathrm{~m}\) - Pipeline 2: \(D_2 = \frac{75}{1000} \mathrm{m} = 0.075 \mathrm{~m}\)
03

Calculate the temperature difference and radius values

Calculate the temperature difference between the pipelines and determine the mean radius value. - Temperature difference: \(\Delta T = T_1 - T_2 = 175 - 5 = 170^{\circ} \mathrm{C}\) - Mean radius of the pipeline1: \(r_1 = \frac{D_1}{2} = \frac{0.1}{2} \mathrm{m} = 0.05 \mathrm{~m}\) - Mean radius of the pipeline2: \(r_2 = \frac{D_2}{2} = \frac{0.075}{2} \mathrm{m} = 0.0375 \mathrm{~m}\)
04

Calculate the equivalent resistance

Calculate the thermal resistance of the soil between the two pipelines using the formula for the equivalent resistance in radial systems. Since the pipelines are parallel to each other, we will use parallel resistances formula: \(R_{eq} = \frac{r_1 r_2 (r_2 - r_1)}{k (r_2^2 - r_1^2)}\) Plug in the values to calculate the equivalent resistance: \(R_{eq} = \frac{0.05 \times 0.0375 (0.0375 - 0.05)}{0.5 (0.0375^2 - 0.05^2)}\)
05

Calculate the heat transfer rate per unit length

Now, we will use the formula for heat transfer rate (Q) per unit length through a cylindrical wall: \(Q = \frac{\Delta T}{R_{eq}}\) Plug in the values and calculate the heat transfer rate per unit length: \(Q = \frac{170}{R_{eq}}\) After calculating R_eq and plugging it into the equation above, we find the heat transfer rate per unit length between the two pipelines.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a property of materials that indicates their ability to conduct heat. Measured in units of watts per meter-kelvin (\( \frac{W}{m \times K} \)), it quantifies the rate at which heat can pass through a material due to a temperature gradient. Materials with high thermal conductivity, such as metals, are effective at transferring heat, while those with low thermal conductivity, like rubber or ceramic, are considered insulators.

In the context of the original exercise, the soil's thermal conductivity determines how efficiently heat can be transferred between the two pipelines buried in it. The value of 0.5 \( \frac{W}{m \times K} \) indicates a relatively low heat transfer capacity, which is typical for soil. This is a crucial parameter for calculating the rate of heat transfer, as heat moves from the hotter pipeline to the cooler one through the soil.
Thermal Resistance
Thermal resistance is the measure of a material's ability to resist the flow of heat. It is the inverse of thermal conductivity and indicates how well a material can insulate against heat transfer. The higher the thermal resistance, the better the insulation properties of the material. In mathematical terms, thermal resistance for a given material is expressed as \(\frac{{\Delta T}}{{Q}}\), where \({\Delta T}\) is the temperature difference across the material and \({Q}\) is the heat transfer rate.

In the exercise, the thermal resistance of the soil plays a pivotal role. By using the equivalent resistance formula for parallel pipelines, the exercise calculates the resistive effect the soil has on heat transfer between the two pipelines. This resistance combined with the overall temperature difference drives the calculation of the heat transfer rate per unit length.
Parallel Pipelines Heat Exchange
Heat exchange between parallel pipelines is influenced by how these pipelines are arranged and how close they are to each other. When pipelines containing fluids at different temperatures run parallel and are close together, heat will transfer from the hotter to the cooler pipeline. This heat transfer occurs through the medium that separates the two, which, in our exercise, is soil.

To express this mathematically, we utilize specific equations adapted for the configuration of the system, considering the distances and inherent properties of the medium. For parallel pipelines, the calculation involves the concept of equivalent thermal resistance to determine the total inhibiting effect on heat flow between the pipes. The relationship between the radii of the pipes, their temperature difference, and the thermal conductivity of the soil is used to estimate the amount of heat transferred per unit length.
Temperature Difference in Heat Transfer
The temperature difference in heat transfer, denoted by \({\Delta T}\), is the driving force for the movement of heat energy from one body to another. Heat naturally flows from areas of high temperature to areas of low temperature until thermal equilibrium is reached. The greater the temperature difference, the more intense the heat transfer will be.

In the given exercise, the temperature difference between the two pipelines is significant, with one at 175掳C and the other at 5掳C. This large temperature gradient leads to a higher rate of heat transfer, assuming the medium (soil) separating them is constant. The heat transfer rate per unit length calculated in the final step of the solution is directly proportional to this temperature difference, making it an essential factor in the overall heat exchange process.

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Most popular questions from this chapter

Hot water is transported from a cogeneration power station to commercial and industrial users through steel pipes of diameter \(D=150 \mathrm{~mm}\), with each pipe centered in concrete \((k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of square cross section \((w=300 \mathrm{~mm})\). The outer surfaces of the concrete are exposed to ambient air for which \(T_{\infty}=0^{\circ} \mathrm{C}\) and \(h=\) \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the inlet temperature of water flowing through the pipe is \(T_{i}=90^{\circ} \mathrm{C}\), what is the heat loss per unit length of pipe in proximity to the inlet? The temperature of the pipe \(T_{1}\) may be assumed to be that of the inlet water. (b) If the difference between the inlet and outlet temperatures of water flowing through a 100 -m-long pipe is not to exceed \(5^{\circ} \mathrm{C}\), estimate the minimum allowable flow rate \(\dot{m}\). A value of \(c=4207 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) may be used for the specific heat of the water.

For a small heat source attached to a large substrate, the spreading resistance associated with multidimensional conduction in the substrate may be approximated by the expression [Yovanovich, M. M., and V. W. Antonetti, in Adv. Thermal Modeling Elec. Comp. and Systems, Vol. 1, A. Bar-Cohen and A. D. Kraus, Eds., Hemisphere, \(\mathrm{NY}, 79-128,1988]\) \(R_{r(\mathrm{sp})}=\frac{1-1.410 A_{r}+0.344 A_{r}^{3}+0.043 A_{r}^{5}+0.034 A_{r}^{7}}{4 k_{\mathrm{sub}} A_{s, h}^{1 / 2}}\) where \(A_{r}=A_{s, h} / A_{s, \text { sub }}\) is the ratio of the heat source area to the substrate area. Consider application of the expression to an in- line array of square chips of width \(L_{h}=\) \(5 \mathrm{~mm}\) on a side and pitch \(S_{h}=10 \mathrm{~mm}\). The interface between the chips and a large substrate of thermal conductivity \(k_{\text {sub }}=80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is characterized by a thermal contact resistance of \(R_{t, c}^{\prime \prime}=0.5 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). If a convection heat transfer coefficient of \(h=\) \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is associated with airflow \(\left(T_{\infty}=15^{\circ} \mathrm{C}\right)\) over the chips and substrate, what is the maximum allowable chip power dissipation if the chip temperature is not to exceed \(T_{h}=85^{\circ} \mathrm{C}\) ?

Small-diameter electrical heating elements dissipating \(50 \mathrm{~W} / \mathrm{m}\) (length normal to the sketch) are used to heat a ceramic plate of thermal conductivity \(2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The upper surface of the plate is exposed to ambient air at \(30^{\circ} \mathrm{C}\) with a convection coeftient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the lower surface is well insulated. (a) Using the Gauss-Seidel method with a grid spacing of \(\Delta x=6 \mathrm{~mm}\) and \(\Delta y=2 \mathrm{~mm}\), obtain the temperature distribution within the plate. (b) Using the calculated nodal temperatures, sketch four isotherms to illustrate the temperature distribution in the plate. (c) Calculate the heat loss by convection from the plate to the fluid. Compare this value with the element dissipation rate. (d) What advantage, if any, is there in not making \(\Delta x=\Delta y\) for this situation? (e) With \(\Delta x=\Delta y=2 \mathrm{~mm}\), calculate the temperature field within the plate and the rate of heat transfer from the plate. Under no circumstances may the temperature at any location in the plate exceed \(400^{\circ} \mathrm{C}\). Would this limit be exceeded if the airfw were terminated and heat transfer to the air were by natural convection with \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) ?

The hot-film heat flux gage shown schematically may be used to determine the convection coefficient of an adjoining fluid stream by measuring the electric power dissipation per unit area, \(P_{e}^{r}\left(\mathrm{~W} / \mathrm{m}^{2}\right)\), and the average surface temperature, \(T_{s, f}\), of the film. The power dissipated in the film is transferred directly to the fluid by convection, as well as by conduction into the substrate. If substrate conduction is negligible, the gage measurements can be used to determine the convection coefficient without application of a correction factor. Your assignment is to perform a two-dimensional, steadystate conduction analysis to estimate the fraction of the power dissipation that is conducted into a 2-mm-thick quartz substrate of width \(W=40 \mathrm{~mm}\) and thermal conductivity \(k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The thin, hot-film gage has a width of \(w=4 \mathrm{~mm}\) and operates at a uniform power dissipation of \(5000 \mathrm{~W} / \mathrm{m}^{2}\). Consider cases for which the flid temperature is \(25^{\circ} \mathrm{C}\) and the convection coeffiient is 500,1000 , and \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Use the finite-element method of \(F E H T\) to analyze a symmetrical half- section of the gage and the quartz substrate. Assume that the lower and end surfaces of the substrate are perfectly insulated, while the upper surface experiences convection with the fluid. (a) Determine the temperature distribution and the conduction heat rate into the region below the hot film for the three values of \(h\). Calculate the fractions of electric power dissipation represented by these rates. Hint: Use the View/Heat Flow command to find the heat rate across the boundary elements. (b) Use the View/Temperature Contours command to view the isotherms and heat flow patterns. Describe the heat flow paths, and comment on features of the gage design that influence the paths. What limitations on applicability of the gage have been revealed by your analysis?

An electronic device, in the form of a disk \(20 \mathrm{~mm}\) in diameter, dissipates \(100 \mathrm{~W}\) when mounted flush on a large aluminum alloy (2024) block whose temperature is maintained at \(27^{\circ} \mathrm{C}\). The mounting arrangement is such that a contact resistance of \(R_{t, c}^{\prime \prime}=5 \times 10^{-5}\) \(\mathrm{m}^{2} \cdot \mathrm{K} / \mathrm{W}\) exists at the interface between the device and the block. (a) Calculate the temperature the device will reach, assuming that all the power generated by the device must be transferred by conduction to the block. (b) To operate the device at a higher power level, a circuit designer proposes to attach a finned heat sink to the top of the device. The pin fins and base material are fabricated from copper \((k=400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and are exposed to an airstream at \(27^{\circ} \mathrm{C}\) for which the convection coefficient is \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). For the device temperature computed in part (a), what is the permissible operating power?
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