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Consider heat transfer in a one-dimensional (radial) cylindrical coordinate system under steady-state conditions with volumetric heat generation. (a) Derive the finite-difference equation for any interior node \(m\). (b) Derive the finite-difference equation for the node \(n\) located at the external boundary subjected to the convection process \(\left(T_{\infty 0}, h\right)\).

Short Answer

Expert verified
The finite-difference equations for (a) an interior node m and (b) an external boundary node n in a one-dimensional radial cylindrical coordinate system with volumetric heat generation under steady-state conditions are as follows: (a) For interior node m: \[ \frac{k_{m + \frac{1}{2}}(T_{m+1} - T_{m}) - k_{m - \frac{1}{2}}(T_{m} - T_{m-1})}{\rho C_p \Delta r^2} = q_g \] (b) For boundary node n: \[ \frac{k_n(T_{n-1}-T_n)}{\Delta r} = h(T_{\infty 0}-T_n) \]

Step by step solution

01

1. Heat transfer equation in cylindrical coordinates

Let's derive the heat transfer equation in one-dimensional cylindrical coordinates. Under steady-state conditions and with volumetric heat generation, the equation is given by: \[ \rho C_p \frac{1}{r} \frac{d}{d r}\left(r k \frac{d T}{d r}\right)= q_g \] Where: - \(\rho\) = density of the material, - \(C_p\) = specific heat capacity, - \(r\) = radial distance, - \(k\) = thermal conductivity, - \(T\) = temperature, - \(q_g\) = volumetric heat generation rate.
02

2. Discretization for interior node m

Now, we will discretize the heat transfer equation using the finite-difference method for an interior node m. Let's consider \(T_m\) as the temperature at node m, \(T_{m-1}\) at node m-1, and \(T_{m+1}\) at node m+1. We can write the equations as follows: For points between the nodes m-1 and m: \[ \rho C_p \frac{1}{r_m} r_{m-\frac{1}{2}} k_{m-\frac{1}{2}} \frac{T_{m}-T_{m-1}}{\Delta r^2}= q_g \] For points between the nodes m and m+1: \[ \rho C_p \frac{1}{r_m} r_{m+\frac{1}{2}} k_{m+\frac{1}{2}} \frac{T_{m+1}-T_{m}}{\Delta r^2}= q_g \]
03

3. Combining equations

Now we combine these equations to obtain the finite-difference equation for an interior node m: \[ \frac{k_{m + \frac{1}{2}}(T_{m+1} - T_{m}) - k_{m - \frac{1}{2}}(T_{m} - T_{m-1})}{\rho C_p \Delta r^2} = q_g \]
04

4. Discretization for boundary node n

For the boundary node n under convection process \((T_{\infty 0}, h)\), we need to consider the heat flux at the external boundary given by Newton's law of cooling: \[ \frac{dq}{dr} = h(T_{\infty 0}-T_n) \] This can be discretized as follows: \[ \frac{k_n(T_{n-1}-T_n)}{\Delta r} = h(T_{\infty 0}-T_n) \]
05

5. Final finite-difference equations

Now we have the finite-difference equations for interior node m and external boundary node n: For interior node m: \[ \frac{k_{m + \frac{1}{2}}(T_{m+1} - T_{m}) - k_{m - \frac{1}{2}}(T_{m} - T_{m-1})}{\rho C_p \Delta r^2} = q_g \] For boundary node n: \[ \frac{k_n(T_{n-1}-T_n)}{\Delta r} = h(T_{\infty 0}-T_n) \] These equations can be used to solve the one-dimensional heat transfer problem in cylindrical coordinates with volumetric heat generation under finite-difference method.

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Key Concepts

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

Heat Transfer
Heat transfer is the process of thermal energy moving from one place to another due to a temperature difference. In this exercise, we focus on the transfer of heat in a cylindrical system. This is a common scenario often found in engineering, especially dealing with pipes or wires. The goal is to analyze how heat flows within a material and how it transfers to surrounding media. In cylindrical coordinates, heat transfer calculations take into account the radial direction. The radial coordinate is crucial since it represents the distance from the center of a cylinder to any point within it. This makes it ideal for problems where the object’s geometry is cylindrical, such as pipes or rods. Using these principles in the finite-difference method involves breaking down the cylinder into discrete nodes and analyzing how heat moves between them. Understanding how heat moves, either by conduction or convection, is vital for predicting temperatures within the material. Heat conduction occurs when there’s direct contact, and the heat moves through the material molecules. Heat convection happens when the heat is transferred through a fluid medium, usually air or water. In this problem, we are also considering convection at the boundary, illustrating how external environments affect heat transfer.
Cylindrical Coordinates
Cylindrical coordinates are a way to express spatial variables in systems that exhibit radial symmetry, such as pipes or cylindrical tanks. These coordinates consist of three components: the radial distance \( r \), the angular coordinate \( \theta \), and the height \( z \). This approach simplifies the complexity of problems where heat transfer occurs in circular or tubular configurations.For heat transfer problems in one-dimensional cylindrical coordinates, the focus is on the radial component \( r \). This implies that temperature changes are analyzed as they occur radially outward from the center of the cylinder rather than along its length or circular direction. By focusing solely on \( r \), the problem becomes greatly simplified.In this exercise, we only deal with radial heat transfer under the assumption of steady-state conditions. This simplification results in an easier-to-manage equation. When applying finite-difference methods, each discrete node represents a specific point along the radial path. The heat transfer between these nodes is calculated to determine the temperature distribution in the cylinder.
Volumetric Heat Generation
Volumetric heat generation refers to heat produced internally within a material or system, due to chemical reactions, electrical currents, or other processes. In the case of a cylindrical system, it is common in situations like nuclear reactors or electronic components, where heat is generated within and needs to be dissipated.For the equations involved, \( q_g \) represents the rate of volumetric heat generation in the system. The presence of \( q_g \) in the heat transfer equation indicates that besides external heat transfer, there is internal heat present as well, complicating the analysis. The finite-difference method addresses this by using interior node equations to incorporate the added heat generation into the calculations. By determining how much heat is generated internally for each discrete part of the system, we can more accurately simulate temperature behaviors. This is crucial for ensuring safety and efficiency in systems with significant internal heat generation, as it helps predict if and where overheating might occur.
Steady-State Conditions
Steady-state conditions refer to a situation where the system’s properties, like temperature, do not change over time. In heat transfer problems, achieving steady-state means that the temperature distribution reached a point where it remains constant unless some external condition changes. In the context of the given exercise, steady-state assumptions simplify the analysis significantly. With no changes over time, the temporal component is ignored, allowing us to focus solely on spatial distribution across the system. Under steady-state, the heat entering any part of the material is equal to the heat leaving it, accounting for any sources like volumetric heat generation. This means our equations only need to satisfy spatial balance rather than dynamic changes. For engineers, steady-state conditions are crucial, as they represent the "equilibrium" or "normal operating" state of many systems, ensuring stability and predictable performance in processes relying on heat transfer.

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Most popular questions from this chapter

Derive the nodal finite-difference equations for the following configurations. (a) Node \((m, n)\) on a diagonal boundary subjected to convection with a fluid at \(T_{\infty}\) and a heat transfer coefficient \(h\). Assume \(\Delta x=\Delta y\). (b) Node \((m, n)\) at the tip of a cutting tool with the upper surface exposed to a constant heat flux \(q_{o}^{\prime \prime}\), and the diagonal surface exposed to a convection cooling process with the fluid at \(T_{\infty}\) and a heat transfer coefficient \(h\). Assume \(\Delta x=\Delta y\).

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 constantan wire of \(1-\mathrm{mm}\) diameter is butt welded to the surface of a large copper block, forming a thermocouple junction. The wire behaves as a fin, permitting heat to flow from the surface, thereby depressing the sensing junction temperature \(T_{j}\) below that of the block \(T_{\sigma}\). Copper block, \(T_{o}\) (a) If the wire is in air at \(25^{\circ} \mathrm{C}\) with a convection coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), estimate the measurement error \(\left(T_{j}-T_{o}\right)\) for the thermocouple when the block is at \(125^{\circ} \mathrm{C}\). (b) For convection coefficients of 5,10 , and \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), plot the measurement error as a function of the thermal conductivity of the block material over the range 15 to \(400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Under what circumstances is it advantageous to use smaller diameter wire?

A hole of diameter \(D=0.25 \mathrm{~m}\) is drilled through the center of a solid block of square cross section with \(w=1 \mathrm{~m}\) on a side. The hole is drilled along the length, \(l=2 \mathrm{~m}\), of the block, which has a thermal conductivity of \(k=150 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The four outer surfaces are exposed to ambient air, with \(T_{\infty, 2}=25^{\circ} \mathrm{C}\) and \(h_{2}=4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while hot oil flowing through the hole is characterized by \(T_{\infty, 1}=300^{\circ} \mathrm{C}\) and \(h_{1}=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the corresponding heat rate and surface temperatures.

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.

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