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Can the global (accumulated) discretization error be less than the local error during a step? Explain.

Short Answer

Expert verified
Explain your answer. Answer: Yes, it is possible for the global discretization error to be smaller than the local error during a step in a numerical method. This can occur when local errors compensate for each other to some extent due to their distribution in different directions. The specific behavior of the numerical method and the errors it introduces during its operation will determine if this occurs.

Step by step solution

01

Define global and local discretization errors

The global discretization error is the accumulated error of a method when solving a differential equation numerically from the initial point up to a certain point. It is caused by the way the numerical method approximates the exact solution. On the other hand, the local discretization error is the amount by which the method deviates from the exact solution within just one small step of the numerical method.
02

Analyze local and global discretization errors' interplay

In general, the global error increases as the number of steps increases. However, every step introduces a new local error, which is compounded into the accumulated error. These local discretization errors can accumulate in the same direction or they can be distributed in different directions. This means that there might be cases where the errors compensate each other partially, which makes the total accumulated global discretization error smaller than the local error in one of the steps.
03

Provide an explanation

During each time step, a local discretization error may be introduced due to various factors such as truncation errors in the method, inaccuracies in the input data, or round-off errors. If the local errors are such that they tend to cancel each other (for example, one positive error is followed by a similar negative error), the global discretization error will be less than the sum of the local errors. There may also be cases where the global error at a specific time is less than the local error during one step because some of the previous local errors have compensated each other. In conclusion, it is possible for the global discretization error to be less than the local error during a step, due to the fact that local errors can compensate for each other to some extent. However, this depends on the specific behavior of the numerical method and the errors it introduces during its operation.

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Most popular questions from this chapter

A nonmetal plate \((k=0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached on the upper surface of an ASME SB-96 copper-silicon plate $(k=36 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$. The nonmetal plate and the ASME SB-96 plate have thicknesses of \(20 \mathrm{~mm}\) and \(30 \mathrm{~mm}\), respectively. The bottom surface of the ASME SB-96 plate (surface 1) is subjected to a uniform heat flux of \(150 \mathrm{~W} / \mathrm{m}^{2}\). The top nonmetal plate surface (surface 2) is exposed to convection at an air temperature of \(15^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Copper- silicon alloys are not always suitable for applications where they are exposed to certain median and high temperatures. Therefore, the ASME Boiler and Pressure Vessel Code (ASME BPVC.IV-2015, HF-300) limits equipment constructed with ASME SB-96 plate to be operated at a temperature not exceeding \(93^{\circ} \mathrm{C}\). Using the finite difference method with a uniform nodal spacing of \(\Delta x=5 \mathrm{~mm}\) along the the temperature distribution as a function of \(x\). Would the use of the ASME SB- 96 plate under these conditions be in compliance with the ASME Boiler and Pressure Vessel Code? What is the lowest value of the convection heat transfer coefficient for the air so that the ASME SB-96 plate is below $93^{\circ} \mathrm{C}$ ?

Consider transient heat conduction in a plane wall whose left surface (node 0 ) is maintained at \(80^{\circ} \mathrm{C}\) while the right surface (node 6) is subjected to a solar heat flux of \(600 \mathrm{~W} / \mathrm{m}^{2}\). The wall is initially at a uniform temperature of \(50^{\circ} \mathrm{C}\). Express the explicit finite difference formulation of the boundary nodes 0 and 6 for the case of no heat generation. Also, obtain the finite difference formulation for the total amount of heat transfer at the left boundary during the first three time steps.

Quench hardening is a mechanical process in which the ferrous metals or alloys are first heated and then quickly cooled down to improve their physical properties and avoid phase transformation. Consider a $40-\mathrm{cm} \times 20-\mathrm{cm}\( block of copper alloy \)\left(k=120 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=3.91 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)$ being heated uniformly until it reaches a temperature of \(800^{\circ} \mathrm{C}\). It is then suddenly immersed into the water bath maintained at \(15^{\circ} \mathrm{C}\) with $h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ for quenching process. However, the upper surface of the metal is not submerged in the water and is exposed to air at $15^{\circ} \mathrm{C}\( with a convective heat transfer coefficient of \)10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Using an explicit finite difference formulation, calculate the temperature distribution in the copper alloy block after 10 min have elapsed using \(\Delta t=10 \mathrm{~s}\) and a uniform mesh size of \(\Delta x=\Delta y=10 \mathrm{~cm}\).

Consider a \(2-\mathrm{cm} \times 4-\mathrm{cm}\) ceramic strip $(k=3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, \)\rho=1600 \mathrm{~kg} / \mathrm{m}^{3}\(, and \)\left.c_{p}=800 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)$ embedded in very high conductivity material as shown in Fig. P5-131. The two sides of the ceramic strip are maintained at a constant temperature of \(300^{\circ} \mathrm{C}\). The bottom surface of the strip is insulated, while the top surface is exposed to a convective environment with \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and ambient temperature of \(50^{\circ} \mathrm{C}\). Initially at \(t=0\), the ceramic strip is at a uniform temperature of \(300^{\circ} \mathrm{C}\). Using the implicit finite difference formulation and a time step of \(2 \mathrm{~s}\), determine the nodal temperatures after \(12 \mathrm{~s}\) for a uniform mesh size of $1 \mathrm{~cm}$.

Consider steady one-dimensional heat conduction in a composite plane wall consisting of two layers \(A\) and \(B\) in perfect contact at the interface. The wall involves no heat generation. The nodal network of the medium consists of nodes 0,1 (at the interface), and 2 with a uniform nodal spacing of $\Delta x$. Using the energy balance approach, obtain the finite difference formulation of this problem for the case of insulation at the left boundary (node 0 ) and radiation at the right boundary (node 2 ) with an emissivity of \(\varepsilon\) and surrounding temperature of \(T_{\text {sarr }}\).

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