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Chapter 4: Problem 36

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

Short Answer

Expert verified
For any interior node, use a central difference scheme to derive the equation. For the external boundary, include convective heat loss in the balance.

Step by step solution

01

Understand the System and Setup the General Equation

In a cylindrical coordinate system, under steady-state conditions with volumetric heat generation but no time-dependent change, the differential equation for heat transfer is given by:\[ \frac{1}{r}\frac{d}{dr}\left(r\frac{dT}{dr}\right) + \frac{q}{k} = 0\]where \(r\) is the radial coordinate, \(q\) is the volumetric heat generation per unit volume, and \(k\) is the thermal conductivity. This is the basis for deriving finite-difference equations.
02

Discretize the Radial Domain

Consider a grid with nodes spaced radially outward, indexed by \(m\) where \(r_m\) is the radial position. Discretization involves replacing the differential terms with finite-difference approximations. For an interior node, the central difference approximation for the second derivative is used.
03

Derive the Finite-Difference Equation for an Interior Node

For an interior node \(m\), the central difference approximation of the radial heat conduction equation can be expressed as:\[ \frac{T_{m+1} - T_m}{\Delta r_m} + \frac{T_{m-1} - T_m}{\Delta r_{m-1}} + \frac{q \Delta r_m}{k} = 0\]This equation represents the balance of conduction and generation at node \(m\). Rearrange this to solve for \(T_m\) using the specific discretization scheme depending on the radial grid spacing \(\Delta r\).
04

Setup Boundary Conditions for the External Node

For the node at the external boundary, where convective heat transfer occurs (node \(n\)), the heat balance includes convection. The boundary condition is given by the heat loss to the environment:\[ -k\left(\frac{T_{n-1} - T_n}{\Delta r}\right) = h(T_n - T_w)\]where \(h\) is the convective heat transfer coefficient and \(T_w\) is the ambient temperature.
05

Derive the Finite-Difference Equation for the External Boundary Node

Rearrange the boundary condition equation for the external node in terms of finite differences:\[ -k\frac{T_{n-1} - T_n}{\Delta r} = h(T_n - T_w)\]Simplify and solve for \(T_n\) to express the equation using only grid terms, incorporating the convection term into the discretized equation. This can be cast into the same form as interior nodes but adjusted for boundary flux.

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Key Concepts

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

Understanding Heat Transfer
Heat transfer refers to the movement of thermal energy from one area to another. It occurs due to the temperature difference between the regions. There are three primary modes of heat transfer: conduction, convection, and radiation.
In the context of the exercise, we are focusing on conduction and convection. Conduction is the heat transfer through solid materials. It happens due to direct contact of molecules. The heat flows from the hotter region to the cooler region. In mathematical terms, conduction in solids can be represented by Differential Equations (in this exercise, through a radial system).
Convection, on the other hand, involves the transfer of heat between a solid surface and an adjacent fluid (like air or water). This process occurs at the external boundary, where the heat is ultimately lost to the surroundings due to airflow, represented here by the convection boundary conditions.
Cylindrical Coordinates in Heat Transfer
Cylindrical coordinates are a way of representing the position in a 3D space, particularly useful for objects that have circular or cylindrical symmetry. This system defines a point by a radial distance, an angle, and a linear distance along a cylinder's axis.
In heat transfer problems, especially involving pipes or cylinders, using cylindrical coordinates is more natural and simplifies the mathematics. The radial distance in cylindrical coordinates is denoted as \( r \), and it indicates how far a point is from the center axis of the cylinder.
Solving heat transfer problems in this coordinate system involves understanding how temperature varies along the radial direction. By using finite-difference methods, which discretize the differential equations, we can approximate how heat flows through the material in sections such as an internal node and an external boundary node.
Convection Boundary Conditions
Convection boundary conditions describe the heat exchange between a solid surface and a fluid in motion. This is critical in engineering applications where surface temperatures and heat loss rates need to be controlled.
The convection process at the boundary node, as shown in the exercise, involves parameters like the conductive heat transfer coefficient \( h \) and the surrounding ambient temperature \( T_w \). It is given by Newton's law of cooling, which states that the rate of heat loss of a body is proportional to the difference in temperatures between the body and its surroundings.
In finite-difference computations, convection boundary conditions become essential as they affect the heat behavior at the system's limits. They ensure that the mathematical models reflect real-world heat transfer accurately, balancing the conduction in the material with the convection at its surface.

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Most popular questions from this chapter

Consider one-dimensional conduction in a plane composite wall. The exposed surfaces of materials \(\mathrm{A}\) and \(\mathrm{B}\) are maintained at \(T_{1}=600 \mathrm{~K}\) and \(T_{2}=\) \(300 \mathrm{~K}\) respectively. Material \(\mathrm{A}\). of thickness \(L_{c}=\) \(20 \mathrm{~mm}\), has a temperature- dependent thermal conductivity of \(k_{e}=k_{v}\left[1+\alpha\left(T-T_{e}\right)\right]\), where \(k_{e}=4.4\) W/m \(\cdot \mathrm{K}, a=0.008 \mathrm{~K}^{-1}, T_{e}=300 \mathrm{~K}\) and \(T\) is in kelvins. Material B is of thickness \(L_{b}=5 \mathrm{~mm}\) and has a thermal conductivity of \(k_{b}=1 \mathrm{~W} / \mathrm{m} \cdot \mathbf{K}\). (a) Calculate the heat flex through the composite wall by assuming material A to have a uniform thermal conductivity evaluated at the average temperature of the section. (b) Using a spoce increment of \(1 \mathrm{~mm}\), obtain the finitedifference equations for the internal nodes and calculate the heat flux considering the temperaturedependent thermal conductivity for Material \(A\). If the IHT software is employed, call-up functions from ToolsFinire-Difference Equations may be used to obtain the nodal equations. Compare your resilt with that obtained in part (a). (c) As an altemative to the finite-difference method of pant (b), use the finite-element method of FEIIT to calculate the heat flux, and compare the result with that from part ( \(\mathrm{a}\) ). Hinr: In the Specify/Moterial Properties box, properties may be entered as a function of temperature \((T)\), the spoce coordinates \((x, y)\), or time \((t)\). See the Heip section for more details.

A long conducting rod of rectangular cross section \((20 \mathrm{~mm} \times 30 \mathrm{~mm})\) and thermal conductivity \(k=\) \(20 \mathrm{~W} / \mathrm{n}-\mathrm{K}\) experiences uniform heat generation at a rate \(\dot{q}=5 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\), while its surfaces are maintained at \(300 \mathrm{~K}\). (a) Using a finite-difference method with a grid spocing of \(5 \mathrm{~mm}\), determine the temperature distribetion in the rod. (b) With the boundary conditions unchanged, what heat generation rate will cause the midpoint temperature to reach \(600 \mathrm{~K}\) ?

Small diameter electrical heating clements dissipating 50 Whan (length norunal to the sketch) are used to heat a ceramic plate of thermal conductivity 2 Whm \(+\mathrm{K}\). The upper surface of the plate is cxposed to ambient air at \(30^{\circ} \mathrm{C}\). with a convection coefficient of \(100 \mathrm{~W} / \mathrm{m}^{3} \cdot \mathrm{K}\), while the lower surface is well insulated. (a) Using the Gauss-Seidel method with a grid spacing of \(\mathrm{At}=6 \mathrm{~mm}\) and \(\Delta \mathrm{y}=2 \mathrm{~mm}\), 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) Calculute the heat loss by convection from the plate to the ftuid. Compare this value with the clement dissipation rate, (d) What advantage, if any, is there in not making \(\Delta r=\Delta y\) for thik situation? (c) With \(\Delta x=\Delta y=2 \mathrm{~mm}\), calculate the icmperatare field within the plate and the rate of heal transfer from the plate. Under no circumstances may the temperature at any location in the plate excecd \(400^{\circ} \mathrm{C}\). Would this limit be excecded if the air flow were terminated and heat transfer to the air was by natural convection with \(h=10 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) ?

A heat sink for cooling computer chips is fabricated from copper \(\left(k_{3}=400 \mathrm{~W} / \mathrm{m}-\mathrm{K}\right)\), with machined microchannels passing a cooling fluid for which \(T_{v}=25^{\circ} \mathrm{C}\) and \(h=30.000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The lower side of the sink experiences no heat removal, and a preliminary heat sink design calls for dimensicns of \(a=b=w_{s}=w_{f}=200 \mu \mathrm{m}\). A symmetrical element of the heat path from the chip to the fluid is shown in the inset. (a) Using the symmetrical element with a square nodal network of \(\Delta x=\Delta y=100 \mu \mathrm{m}\), determine the corresponding temperature field and the heat rate \(q\) ' to the coolant per unit channel length \((W / m)\) for a maximum allowable chip termerature \(T_{\text {cmat }}=75^{\circ} \mathrm{C}\). Estimate the corresponding thermal resistance between the chip surface and the fluid, \(R_{\text {U- }-}(\mathrm{m} \cdot \mathrm{K} / \mathrm{W})\). What is the maximum allowable heat dissipation for a chip that measures \(10 \mathrm{~mm} \times 10 \mathrm{~mm}\) cn a side? (b) The grid spacing wed in the foregoing finite-difference solution is coarse, resulting in poor precision for the temperature distribution and heat removal rate. Investigate the effect of grid spacing by considering spatial increments of 50 and \(25 \mu \mathrm{m}\). (c) Consistent with the requirement that \(a+b=400 \mu \mathrm{m}\), can the heat sink dimensions be altered in a manner that reduces the overall thermal resistance?

A cubical glass melting furnace has exterior dimensions of width \(W=5 \mathrm{~m}\) on a side and is constructed from refractory brick of thickness \(L=0.35 \mathrm{~m}\) and thermal conductivity \(k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The sides and top of the furnace are exposed to ambient air at \(25^{\circ} \mathrm{C}\), with free convection chanseterized by an average coefficient of \(h=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The bottom of the furnace rests on a framed platform for which much of the surface is exposed to the ambient air, and a convection coefficient of \(h=5 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) may be assumed as a first approximaticn. Under operating conditions for which combustion gases maintain the inner surfaces of the fumace at \(1100^{\circ} \mathrm{C}\). what is the heat loss from the furmace?
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