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How is an insulated boundary handled in the finite difference formulation of a problem? How does a symmetry line differ from an insulated boundary in the finite difference formulation?

Short Answer

Expert verified
Answer: In the finite difference formulation, the treatment of heat flux and temperature distribution differs for insulated boundary conditions and symmetry lines. Insulated boundary conditions involve approximating spatial derivatives and setting them equal to zero, corresponding to zero heat flux at the boundary. On the other hand, symmetry lines imply that the temperature distribution is symmetric with respect to that line, which allows using the symmetric temperature distribution to reduce problem size or fix the boundary condition for the temperature at the symmetry line.

Step by step solution

01

Insulated Boundary Condition

An insulated boundary condition, also called a Neumann boundary condition, is when the heat transfer through a boundary is zero, which means there is no heat flux at the boundary. In a 1D problem, the heat flux is the derivative of the temperature with respect to the space coordinate. In the finite difference formulation, we approximate this derivative by calculating the difference in the temperature between neighboring nodes. So, to apply an insulated boundary condition, we will need to express the derivative at the boundary using a finite difference approximation. Consider the following steps for a 1D problem: 1. Write down the finite difference equation using forward, backward, or central differences. 2. For the insulated boundary, express the first derivative as a finite difference approximation. 3. Set the heat flux at the boundary equal to zero. 4. Solve the resulting system of finite difference equations for the unknown temperatures.
02

Symmetry Line

A symmetry line is a boundary where the temperature distribution is symmetric. It means that the temperature distribution will be the same on both sides of the symmetry line, but with opposite spatial gradients due to the symmetry. In the finite difference formulation, we do not impose symmetry explicitly as in an insulated boundary. Instead, we can take advantage of the symmetric temperature distribution to reduce the problem's size and decrease the computational effort. Consider the following steps for a 1D problem with a symmetry line: 1. Write down the finite difference equation using forward, backward, or central differences. 2. Identify the symmetry line and then find its corresponding node index or location in the finite difference mesh. 3. Since, due to symmetry, the temperature distribution is identical on both sides of the symmetry line, you can either remove the symmetry line node and only work with half of the mesh or use the symmetry line temperature value in the finite difference equations as a fixed boundary condition. 4. Solve the resulting system of finite difference equations for the unknown temperatures.
03

Comparison of Insulated Boundary and Symmetry Line

The primary difference in applying insulated boundary and symmetry line in finite difference formulation is how heat flux and temperature distribution are treated. Insulated boundary refers to zero heat flux at the boundary, whereas symmetry line implies that the temperature distribution is symmetric with respect to that line. Insulated boundary conditions involve approximating spatial derivatives and setting them equal to zero, whereas symmetry lines involve employing the symmetric temperature distribution to reduce problem size or fix the boundary condition for the temperature at the symmetry line.

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Most popular questions from this chapter

Consider a 2-m-long and \(0.7-\mathrm{m}\)-wide stainless steel plate whose thickness is \(0.1 \mathrm{~m}\). The left surface of the plate is exposed to a uniform heat flux of \(2000 \mathrm{~W} / \mathrm{m}^{2}\), while the right surface of the plate is exposed to a convective environment at $0^{\circ} \mathrm{C}\( with \)h=400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. The thermal conductivity of the stainless steel plate can be assumed to vary linearly with temperature range as \(k(T)=k_{o}(1+\beta T)\) where $k_{o}=48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( and \)\beta=9.21 \times 10^{-4 \circ} \mathrm{C}^{-1}$. The stainless steel plate experiences a uniform volumetric heat generation at a rate of $8 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}$. Assuming steady-state one-dimensional heat transfer, determine the temperature distribution along the plate thickness.

Why do the results obtained using a numerical method differ from the exact results obtained analytically? What are the causes of this difference?

Express the general stability criterion for the explicit method of solution of transient heat conduction problems.

What causes the round-off error? What kinds of calculations are most susceptible to round-off error?

A stainless steel plate is connected to a copper plate by long ASTM B98 copper-silicon bolts \(9.5 \mathrm{~mm}\) in diameter. The portion of the bolts exposed to convection heat transfer with hot gas is \(5 \mathrm{~cm}\) long. The gas temperature for convection is at \(500^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of $50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. The thermal conductivity of the bolt is known to be \)36 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. The stainless steel plate has a uniform temperature of \(100^{\circ} \mathrm{C}\), and the copper plate has a uniform temperature of \(80^{\circ} \mathrm{C}\). According to the ASME Code for Process Piping (ASME B31.3-2014, Table A-2M), the maximum use temperature for an ASTM B98 copper-silicon bolt is \(149^{\circ} \mathrm{C}\). Using the finite difference method with a uniform nodal spacing of \(\Delta x=5 \mathrm{~mm}\) along the bolt, determine the temperature at each node. Plot the temperature distribution along the bolt. Compare the numerical results with the analytical solution. Would any part of the ASTM B 98 bolts be above the maximum use temperature of \(149^{\circ} \mathrm{C}\) ?

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