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Chapter 4: Problem 67

A common arrangement for heating a large surface area is to move warm air through rectangular ducts below the surface. The ducts are square and located midway between the top and bottom surfaces that are exposed to room air and insulated, respectively. For the condition when the floor and duct temperatures are 30 and \(80^{\circ} \mathrm{C}\), respectively, and the thermal conductivity of concrete is \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), calculate the heat rate from each duct, per unit length of duct. Use a grid spacing with \(\Delta x=2 \Delta y\), where \(\Delta y=0.125 L\) and \(L=150 \mathrm{~mm}\).

Short Answer

Expert verified
The heat rate from each duct per unit length of the duct is \(q' = 1866.67\,\mathrm{W/m}\).

Step by step solution

01

Understanding the given information

We are given: - Floor temperature, \(T_f = 30^{\circ} \mathrm{C}\) - Duct temperature, \(T_d=80^{\circ} \mathrm{C}\) - Thermal conductivity of concrete, \(k = 1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) - Grid spacing, \(\Delta x = 2\Delta y\) - \(\Delta y = 0.125L\) - Length of the duct, \(L = 150\,\mathrm{mm} = 0.15\,\mathrm{m}\) (converted to meters) We'll be using this information to calculate the heat rate from each duct, per unit length of the duct.
02

Calculate temperature difference

First, find the temperature difference (\(\Delta T\)) between the floor and the duct: \(\Delta T = T_d - T_f = 80^{\circ}\mathrm{C} - 30^{\circ}\mathrm{C} = 50^{\circ}\mathrm{C}\)
03

Calculate the grid spacing

Next, find the grid spacing \(\Delta x\) and \(\Delta y\) using the given relations: \(\Delta y = 0.125L = 0.125(0.15\,\mathrm{m}) = 0.01875\,\mathrm{m}\) \(\Delta x = 2 \Delta y = 2(0.01875\,\mathrm{m}) = 0.0375\,\mathrm{m}\)
04

Calculate the conduction heat transfer rate

To find the heat rate from each duct per unit length of the duct (\(q'\)), we can apply the conduction heat transfer equation, which can be written as: \(q' = \dfrac{k \Delta T}{\Delta x}\) Now, substitute the known values: \(q' = \dfrac{1.4\,\mathrm{W/m\cdot K}\cdot 50^{\circ}\mathrm{C}}{0.0375\,\mathrm{m}} = \dfrac{70\,\mathrm{W/m\cdot K}}{0.0375\,\mathrm{m}} = 1866.67\,\mathrm{W/m}\)
05

Final answer

The heat rate from each duct per unit length of the duct is: \(q' = 1866.67\,\mathrm{W/m}\)

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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 material-specific property that measures the ability of a substance to conduct heat. It is an essential factor in understanding how heat transfers through a material. Concretely, it's represented by the symbol \(k\) and is expressed in watts per meter-kelvin (\(W/m \cdot K\)). This means that for each meter of material thickness and each degree Kelvin of temperature difference, a defined amount of heat is conducted. The higher the thermal conductivity, the better the material is at conducting heat.

In the exercise, we're given that concrete has a thermal conductivity of \(1.4 \mathrm{~W/m \cdot K}\). This value reflects how effectively heat is transferred through the concrete from the warm duct to the cooler floor. Generally, for heating applications like the one described, selecting a material with an appropriate thermal conductivity is crucial in maintaining efficient heat transfer. A material with too low thermal conductivity won't effectively transfer heat, while one with very high thermal conductivity might transfer heat too quickly, causing inefficiencies.
Temperature Difference
Temperature difference, denoted as \(\Delta T\), is the driving force behind the conduction of heat. It is the difference in temperature between two points or surfaces, which dictates the flow and rate of heat transfer.

In the given problem, the temperature difference is found between the duct and the floor, which are at \(80^{\circ}C\) and \(30^{\circ}C\) respectively. Calculating \(\Delta T\) is straightforward: subtract the floor temperature from the duct temperature, resulting in a \(\Delta T\) of \(50^{\circ}C\).

This difference is vital because heat transfer proceeds from a higher temperature to a lower temperature. The greater the temperature difference, the more energy is available for transfer, influencing the rate of heat conduction. Understanding this concept helps in engineering applications, as it affects the efficiency and effectiveness of thermal management systems.
Grid Spacing
Grid spacing in thermal conduction problems refers to the distance between calculation points in a model, which helps in breaking down complex heat transfer scenarios into more manageable calculations. In this context, grid spacing is vital for evaluating how heat moves through a medium like concrete.

In the problem, we have grid spacings \(\Delta x\) and \(\Delta y\). The exercise specifies that \(\Delta x = 2 \Delta y\), with \(\Delta y\) being \(0.125L\). Given \(L = 150 \mathrm{~mm}\), this translates to \(\Delta y = 0.01875 \mathrm{~m}\) and \(\Delta x = 0.0375 \mathrm{~m}\).

These spacings are important because they allow us to use numerical methods to approximate the temperature distribution within the concrete. Smaller grid spacing can lead to more accurate results in simulations, but at the cost of more computational resources. Understanding and choosing the right grid spacing helps optimize the balance between accuracy and computational effort.

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Most popular questions from this chapter

Consider an aluminum heat sink \((k=240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), such as that shown schematically in Problem 4.28. The inner and outer widths of the square channel are \(w=20 \mathrm{~mm}\) and \(W=40 \mathrm{~mm}\), respectively, and an outer surface temperature of \(T_{s}=50^{\circ} \mathrm{C}\) is maintained by the array of electronic chips. In this case, it is not the inner surface temperature that is known, but conditions \(\left(T_{\infty}, h\right)\) associated with coolant flow through the channel, and we wish to determine the rate of heat transfer to the coolant per unit length of channel. For this purpose, consider a symmetrical section of the channel and a two-dimensional grid with \(\Delta x=\Delta y=5 \mathrm{~mm}\). (a) For \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the unknown temperatures, \(T_{1}, \ldots, T_{7}\), and the rate of heat transfer per unit length of channel, \(q^{\prime}\). (b) Assess the effect of variations in \(h\) on the unknown temperatures and the heat rate.

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 \(T_{h}=85^{\circ} \mathrm{C}\) ?

A long bar of rectangular cross section is \(60 \mathrm{~mm} \times\) \(90 \mathrm{~mm}\) on a side and has a thermal conductivity of \(1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). One surface is exposed to a convection process with air at \(100^{\circ} \mathrm{C}\) and a convection coeffient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the remaining surfaces are maintained at \(50^{\circ} \mathrm{C}\). (a) Using a grid spacing of \(30 \mathrm{~mm}\) and the Gauss-Seidel iteration method, determine the nodal temperatures and the heat rate per unit length normal to the page into the bar from the air. (b) Determine the effect of grid spacing on the temperature field and heat rate. Specifically, consider a grid spacing of \(15 \mathrm{~mm}\). For this grid, explore the effect of changes in \(h\) on the temperature field and the isotherms.

A hot fluid passes through circular channels of a cast iron platen (A) of thickness \(L_{\mathrm{A}}=30 \mathrm{~mm}\) which is in poor contact with the cover plates (B) of thickness \(L_{\mathrm{B}}=7.5 \mathrm{~mm}\). The channels are of diameter \(D=15 \mathrm{~mm}\) with a centerline spacing of \(L_{o}=60 \mathrm{~mm}\). The thermal conductivities of the materials are \(k_{\mathrm{A}}=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{B}}=75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), while the contact resistance between the two materials is \(R_{t, c}^{r}=2.0 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The hot fluid is at \(T_{i}=150^{\circ} \mathrm{C}\), and the convection coeftient is \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The cover plate is exposed to ambient air at \(T_{\infty}=25^{\circ} \mathrm{C}\) with a convection coeftient of \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The shape factor between one channel and the platen top and bottom surfaces is 4.25.

A plate \((k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is stiffened by a series of longitudinal ribs having a rectangular cross section with length \(L=8 \mathrm{~mm}\) and width \(w=4 \mathrm{~mm}\). The base of the plate is maintained at a uniform temperature \(T_{b}=\) \(45^{\circ} \mathrm{C}\), while the rib surfaces are exposed to air at a temperature of \(T_{\infty}=25^{\circ} \mathrm{C}\) and a convection coeffient of \(h=600 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Using a finite-difference method with \(\Delta x=\Delta y=\) \(2 \mathrm{~mm}\) and a total of \(5 \times 3\) nodal points and regions, estimate the temperature distribution and the heat rate from the base. Compare these results with those obtained by assuming that heat transfer in the rib is one- dimensional, thereby approximating the behavior of a fin. (b) The grid spacing used in the foregoing finitedifference solution is coarse, resulting in poor precision for estimates of temperatures and the heat rate. Investigate the effect of grid refinement by reducing the nodal spacing to \(\Delta x=\Delta y=1 \mathrm{~mm}\) (a \(9 \times 3\) grid) considering symmetry of the center line. (c) Investigate the nature of two-dimensional conduction in the rib and determine a criterion for which the one-dimensional approximation is reasonable. Do so by extending your finite-difference analysis to determine the heat rate from the base as a function of the length of the rib for the range \(1.5 \leq L w \leq 10\), keeping the length \(L\) constant. Compare your results with those determined by approximating the rib as a fin.
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