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How do numerical solution methods differ from analytical ones? What are the advantages and disadvantages of numerical and analytical methods?

Short Answer

Expert verified
In summary, analytical methods are used to solve problems exactly using algebraic manipulations and well-established mathematical formulas, while numerical methods are techniques used to approximate the solution to a mathematical problem using iterative calculations and estimations. Analytical methods offer exact solutions, better understanding of the problem, and less computational resources, but are limited in applicability, can be complex, and rely on assumptions. On the other hand, numerical methods provide wider applicability, flexibility, and simplicity in some cases, but produce approximate solutions, require more computational resources, and may have convergence and stability issues.

Step by step solution

01

Definition of Analytical Methods

Analytical methods are techniques used to solve problems exactly using algebraic manipulations and well-established mathematical formulas. These methods produce accurate and closed-form solutions, provided that the problem being solved can be expressed using known mathematical relationships and structures.
02

Definition of Numerical Methods

Numerical methods are techniques used to approximate the solution to a mathematical problem using iterative calculations and estimations. These methods are applied when it is impossible or very difficult to find an exact closed-form solution using analytical techniques, often because the problem is too complex or involves variables that are not easy to isolate or eliminate.
03

Advantages of Analytical Methods

Some advantages of using analytical methods include: 1. Exact solutions: analytical methods provide exact solutions when possible, which is highly desirable in situations where high precision is required. 2. Better understanding of the problem: using analytical methods often requires a deep understanding of the underlying mathematical structures, which can lead to a better comprehension of the problem itself. 3. Less computational resources: analytical solutions do not rely on extensive computations, but rather on algebraic manipulations, making them generally less demanding on computational resources like processing power and memory.
04

Disadvantages of Analytical Methods

Some disadvantages of using analytical methods include: 1. Limited applicability: analytical methods can only be used for problems that can be expressed through known mathematical relationships, making them inapplicable to problems involving complex or unknown relationships. 2. Complexity: some problems may have analytical solutions that are very difficult to obtain or even too complex to be practical, requiring the use of numerical methods as an alternative. 3. Assumptions: analytical methods often rely on assumptions and idealizations, which may not hold true in real-world situations.
05

Advantages of Numerical Methods

Some advantages of using numerical methods include: 1. Applicability: numerical methods can be used for a wider range of problems, including those that cannot be solved analytically. 2. Flexibility: numerical methods can be tailored to specific problems and adapted to handle different degrees of complexity and precision, depending on the requirements of the problem. 3. Simplicity: some problems may have more straightforward numerical solutions than analytical ones, making them easier to implement and understand.
06

Disadvantages of Numerical Methods

Some disadvantages of using numerical methods include: 1. Approximate solutions: numerical methods only provide approximate solutions, which may not be acceptable in situations where high precision is required. 2. Computational resources: numerical methods generally rely on extensive computations, often iterative in nature, which can be demanding on computational resources like processing power and memory. 3. Convergence and stability: numerical methods may not always converge to the correct solution or may be unstable, depending on the problem and the specific method used.

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Most popular questions from this chapter

What is the cause of the discretization error? How does the global discretization error differ from the local discretization error?

Consider a large plane wall of thickness \(L=0.4 \mathrm{~m}\), thermal conductivity \(k=2.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=20 \mathrm{~m}^{2}\). The left side of the wall is maintained at a constant temperature of \(95^{\circ} \mathrm{C}\), while the right side loses heat by convection to the surrounding air at $T_{\infty}=15^{\circ} \mathrm{C}\( with a heat transfer coefficient of \)h=18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Assuming steady one-dimensional heat transfer and taking the nodal spacing to be \(10 \mathrm{~cm}\), (a) obtain the finite difference formulation for all nodes, (b) determine the nodal temperatures by solving those equations, and (c) evaluate the rate of heat transfer through the wall.

A nonmetal plate is connected to a stainless steel plate by long ASTM A437 B4B stainless steel bolts \(9.5 \mathrm{~mm}\) in diameter. The portion of the bolts exposed to convection heat transfer with a cryogenic fluid is \(5 \mathrm{~cm}\) long. The fluid temperature for convection is at \(-50^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of $100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. The thermal conductivity of the bolts is known to be \)23.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. Both the nonmetal and stainless steel plates maintain a uniform temperature of \(0^{\circ} \mathrm{C}\). According to the ASME Code for Process Piping (ASME B31.3-2014, Table A-2M), the minimum temperature suitable for ASTM A437 B4B stainless steel bolts is \(-30^{\circ} \mathrm{C}\). 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. Compare the numerical results with the analytical solution. Plot the temperature distribution along the bolt. Would any part of the ASTM A437 B4B bolts be lower than the minimum suitable temperature of \(-30^{\circ} \mathrm{C}\) ?

The roof of a house consists of a \(15-\mathrm{cm}\)-thick concrete slab \(\left(k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\) and \(\left.\alpha=0.69 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) that is \(18 \mathrm{~m}\) wide and \(32 \mathrm{~m}\) long. One evening at $6 \mathrm{p} . \mathrm{m}$., the slab is observed to be at a uniform temperature of \(18^{\circ} \mathrm{C}\). The average ambient air and the night sky temperatures for the entire night are predicted to be \(6^{\circ} \mathrm{C}\) and \(260 \mathrm{~K}\), respectively. The convection heat transfer coefficients at the inner and outer surfaces of the roof can be taken to be $h_{i}=5 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\( and \)h_{o}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, respectively. The house and the interior surfaces of the walls and the floor are maintained at a constant temperature of \(20^{\circ} \mathrm{C}\) during the night, and the emissivity of both surfaces of the concrete roof is \(0.9\). Considering both radiation and convection heat transfers and using the explicit finite difference method with a time step of \(\Delta t=5 \mathrm{~min}\) and a mesh size of $\Delta x=3 \mathrm{~cm}$, determine the temperatures of the inner and outer surfaces of the roof at 6 a.m. Also, determine the average rate of heat transfer through the roof during that night.

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

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