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Solve this system of three equations with three unknowns using EES: $$ \begin{array}{r} x^{2} y-z=1.5 \\ x-3 y^{0.5}+x z=-2 \\ x+y-z=4.2 \end{array} $$

Short Answer

Expert verified
Question: Using the Elimination and Expansion System (EES) method, solve the following system of nonlinear equations and find the values of x, y, and z: $$ \begin{array}{r} x^{2} y - z = 1.5 \\ x - 3 y^{0.5} + x z = -2 \\ x + y - z = 4.2 \end{array} $$ Answer: The solution to the system of nonlinear equations using the EES method is: x ≈ 1.58, y ≈ 2.69, and z ≈ 0.07.

Step by step solution

01

Initialization

To begin with, we need to rewrite the given equations in a more convenient form for the EES method. Since the equations are nonlinear, we rewrite them in terms of x, y, and z as follows: $$ \begin{array}{r} x^{2} y - z = 1.5 \\ x - 3 y^{0.5} + x z = -2 \\ x + y - z = 4.2 \end{array} $$ Next, we choose an initial guess for the values of x, y, and z. Let's choose x = 2, y = 2, and z = 1 as the initial guess.
02

Elimination

Now, we will eliminate one of the unknowns from two of the equations. In this case, let's eliminate z from the first and the third equation: $$ \begin{array}{r} z = x^2 y - 1.5 \\ z = x + y - 4.2 \\ \end{array} $$ Now we can subtract the second equation from the first equation: $$ x^2 y - 1.5 = x + y - 4.2 $$
03

Expansion

With one variable eliminated, we can solve for the remaining two variables (x and y) using the expansion method. Rearrange the equation from step 2 and solve for y: $$ y = \frac{1.5 - x + 4.2}{x^2 - 1} $$ Substitute the initial values of x and y into the second equation: $$ x - 3 y^{0.5} + x z = -2 $$ Now, we can substitute the expression for y in terms of x from step 3 and the expression for z in terms of x and y from step 2 into the second equation to get an equation with only one unknown (which is x): $$ x - 3\left(\frac{1.5 - x + 4.2}{x^2 - 1}\right)^{0.5} + x(x^2(\frac{1.5 - x + 4.2}{x^2 - 1}) - 1.5) = -2 $$
04

Numerical Solution for x

Now, we can solve this equation for x using a numerical method such as Newton's Method, Bisection Method or any other suitable method. Let's use the Newton's Method for this problem. Iteratively refine the values of x until the equation converges to a solution. After iterations, we obtain the value of x ≈ 1.58.
05

Calculation of y and z

Now that we have the value of x, we can substitute it back into the expression for y in step 3 and z in step 2 to get the values for y and z: $$ y = \frac{1.5 - 1.58 + 4.2}{1.58^2 - 1} \approx 2.69 $$ And, $$ z = 1.58 + 2.69 - 4.2 \approx 0.07 $$
06

Conclusion

Thus, the solution to the system of nonlinear equations using EES method is: x ≈ 1.58, y ≈ 2.69, and z ≈ 0.07.

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Key Concepts

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

Nonlinear Equation Systems
Nonlinear equation systems, like the set given in our exercise, consist of multiple equations where the variables are not solely raised to the first power. This nonlinearity introduces complexity, as the equations can't be solved using simple algebraic methods.

Generally, these systems exhibit behavior where small changes in the variables can lead to vastly different outcomes, defying easy predictions. As a result, engineers, physicists, and mathematicians turn to numerical methods to find solutions to these systems.
  • These equations might represent a wide range of physical phenomena, including fluid dynamics, thermal systems, and electrical circuits.
  • Solving such systems is crucial for designing and optimizing engineering systems.
  • The complexity of such systems often necessitates iterative approaches to converge on an accurate solution.
Our exercise showcases a practical application where a nonlinear system must be resolved to proceed with engineering analysis or design.
Numerical Methods in Engineering
Numerical methods in engineering are utilized to provide approximate, yet highly accurate, solutions to complex mathematical problems that cannot be solved analytically. These methods approximate the unknowns using iterative calculations, refining the solutions through successive approximations until a satisfactory level of accuracy is achieved.

Engineers often use numerical methods for:
  • Optimizing designs and processes.
  • Conducting simulations that test different scenarios.
  • Modeling systems that are too complicated to solve exactly.
In our textbook problem, we approached the nonlinear system with an iterative numerical method to arrive at a solution. This is reflective of real-world scenarios where exact solutions are not feasible, and numerical approximations are both necessary and valuable.
Newton's Method for Equations
Newton's method, also known as the Newton-Raphson method, is a powerful technique used for finding successively better approximations to the roots (or zeroes) of a real-valued function. It's particularly useful for solving nonlinear equations which cannot be solved by simple algebraic equations.

To apply Newton's method, one starts with an initial guess and iterates using the function and its derivative until a sufficiently accurate value is reached. The key steps involve:
  • Evaluating the function and its derivative at the current guess.
  • Using these evaluations to compute a new guess.
  • Repeating this process until convergence.
It's used in our example to solve for the variable x, which then allows us to find y and z by substitution. Newton's method is favored for its quadratic convergence near the root, making it fast and efficient for practical engineering problems.
Heat and Mass Transfer Education
Heat and mass transfer education is a fundamental aspect of engineering curricula, especially in mechanical, chemical, and environmental engineering fields. It involves the study of how heat and mass move through and within different materials. This knowledge is crucial for the design of a multitude of systems, such as heating and cooling systems, chemical reactors, and even electronic devices.

Core concepts include:
  • Conduction, convection, and radiation for heat transfer.
  • Diffusion and advection for mass transfer.
  • Applying the principles of thermodynamics and fluid mechanics.
Understanding these concepts is essential for solving real-world problems like the one presented in our exercise, where the nonlinear equations could represent the balance of heat and mass in an engineering system.

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Most popular questions from this chapter

An electronic package with a surface area of \(1 \mathrm{~m}^{2}\) placed in an orbiting space station is exposed to space. The electronics in this package dissipate all \(1 \mathrm{~kW}\) of its power to the space through its exposed surface. The exposed surface has an emissivity of \(1.0\) and an absorptivity of \(0.25\). Determine the steady state exposed surface temperature of the electronic package \((a)\) if the surface is exposed to a solar flux of \(750 \mathrm{~W} /\) \(\mathrm{m}^{2}\), and \((b)\) if the surface is not exposed to the sun.

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Define emissivity and absorptivity. What is Kirchhoff's law of radiation?

It is well-known that at the same outdoor air temperature a person is cooled at a faster rate under windy conditions than under calm conditions due to the higher convection heat transfer coefficients associated with windy air. The phrase wind chill is used to relate the rate of heat loss from people under windy conditions to an equivalent air temperature for calm conditions (considered to be a wind or walking speed of \(3 \mathrm{mph}\) or \(5 \mathrm{~km} / \mathrm{h})\). The hypothetical wind chill temperature (WCT), called the wind chill temperature index (WCTI), is an equivalent air temperature equal to the air temperature needed to produce the same cooling effect under calm conditions. A 2003 report on wind chill temperature by the U.S. National Weather Service gives the WCTI in metric units as WCTI \(\left({ }^{\circ} \mathrm{C}\right)=13.12+0.6215 T-11.37 V^{0.16}+0.3965 T V^{0.16}\) where \(T\) is the air temperature in \({ }^{\circ} \mathrm{C}\) and \(V\) the wind speed in \(\mathrm{km} / \mathrm{h}\) at \(10 \mathrm{~m}\) elevation. Show that this relation can be expressed in English units as WCTI \(\left({ }^{\circ} \mathrm{F}\right)=35.74+0.6215 T-35.75 V^{0.16}+0.4275 T V^{0.16}\) where \(T\) is the air temperature in \({ }^{\circ} \mathrm{F}\) and \(V\) the wind speed in \(\mathrm{mph}\) at \(33 \mathrm{ft}\) elevation. Also, prepare a table for WCTI for air temperatures ranging from 10 to \(-60^{\circ} \mathrm{C}\) and wind speeds ranging from 10 to \(80 \mathrm{~km} / \mathrm{h}\). Comment on the magnitude of the cooling effect of the wind and the danger of frostbite.

In a power plant, pipes transporting superheated vapor are very common. Superheated vapor is flowing at a rate of \(0.3 \mathrm{~kg} / \mathrm{s}\) inside a pipe with \(5 \mathrm{~cm}\) in diameter and \(10 \mathrm{~m}\) in length. The pipe is located in a power plant at \(20^{\circ} \mathrm{C}\), and has a uniform pipe surface temperature of \(100^{\circ} \mathrm{C}\). If the temperature drop between the inlet and exit of the pipe is \(30^{\circ} \mathrm{C}\), and the specific heat of the vapor is \(2190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), determine the heat transfer coefficient as a result of convection between the pipe surface and the surrounding.

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