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CHE_250_&acirc;€“..

# CHE_250_&acirc;€“.. - CHE 250 Exam 1 Review...

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CHE 250 – Exam 1 Review Solution of a Single Non-linear Equation - cannot be rearranged to solve explicitly for the variable being calculated -Example: v t = 4 g r p - r ( ) D p 3 C D r where v t is the settling velocity of a spherical particle and C D is a function of Reynolds number Re = D p v t r / m . Since v t is a non-linear function of C D and C D is ultimately a function of v t , v t is a non-linear function of itself and cannot be rearranged to solve for v t alone. Successive Substitution (fixed-point iteration) - (1) rearrange equation f(x) into the form x = g(x) - in the example case, the equation is already presented in this manner - (2) select an initial value x 1 and calculated g(x 1 ) - set x 2 = g(x 1 ) and then calculated g(x 2 ) and set x 3 = g(x 2 ) - continue until the calculated value of g(x) is approximately equal to x - advantage: can be started with a single point - disadvantage: does not always converge Wegstein Method - set up equation in same form as fixed-point iteration - new approximations of x are arrived at by equation: x n +1 = x n - 1 g x n ( ) - x n g x n - 1 ( ) x n - 1 - g x n - 1 ( ) - x n + g x n ( ) Linear Interpolation (false-position method) - graph equation to get approximation for roots - pick a point on either side of the root; x 1 and x 2 and calculate f(x 1 ) and f(x 2 ) - calculate x 3 using x 3 = x 1 - f x 1 ( ) x 2 - x 1 ( ) f x 2 ( ) - f x 1 ( ) - if x 3 is positive, replace the positive of x 1 or x 2 and repeat calculations using x n = x + - f x + ( ) x + - x - ( ) f x + ( ) - f x - ( ) Newton-Raphson Method - most widely used, based on Taylor-Series expansion of f(x) around an initial

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estimate - Taylor-Series expansion of f(x) is f x ( ) = f x 1 ( ) + f x 1 ( ) x - x 1 ( ) + f x 1 ( ) x - x 1 ( ) 2 2! + ...... - Set f(x) equal to 0 and solve for x to get new initial guess - If taylor expansion is truncated to first two terms, the new approximation is equal to x n +1 = x n - f x n ( ) f x n ( ) Solving System of Non-Linear Equations - f 1 (x 1 ,x 2 ,….x n ) = 0, f 2 (x 1 ,x 2 ,….x n ) = 0, f n (x 1 , x 2 ,….x n ) = 0 - solution to the set of equations is values of x 1 through x n that makes all equations equal to zero Fixed-Point Iteration - rearrange to x 1 = g(x 1 ), x 2 = g(x 2 ) through x n = g(x n ) - pick initial guesses for x values and substitute to calculate g(x) values - g(x) values are set equal to new x values and iteration is continued Newton-Raphson Method - single equation form: f x n +1 ( ) = f x n ( ) + x n +1 + x n ( ) f x n ( ) -
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CHE_250_&acirc;€“.. - CHE 250 Exam 1 Review...

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