PENENTUAN AKAR PERSAMAAN TAK LINIER TUNGGAL

• Understanding what roots problems are and where they occur in engineering and science
• Knowing how to determine a root graphically
• Understanding the incremental search method and its shortcomings
• Knowing how to solve a roots problem with the bisection method
• Knowing how to estimate the error of bisection and why it differs from error estimates for other types of root location algorithm
• Understanding false position and how it differs from bisection

  (x value for which f(x) = 0)

  1. analytical method 2. graphical method 3. trial and error

  4. numerical method numerical method numerical method numerical method iterative iterative iterative iterative _►

  Function of f(x): (1) Explicit Explicit Explicit Explicit , (2) Implicit Implicit Implicit Implicit (based on the influence of independent variable on dependent variable)

  GRAPHICAL METHODS GRAPHICAL METHODS GRAPHICAL METHODS GRAPHICAL METHODS

  A simple method for obtaining an estimate of the root of the equation f(x) = 0 is to:

  make a plot of the function , and observe where it crosses the x axis

  Advantages Advantages Advantages Advantages::::

  Roots = zeros _►

  provides a rough approximation of the root → can be employed as starting guesses starting guesses starting guesses starting guesses for numerical methods useful for understanding the properties of the functions useful for anticipating the pitfalls of the numerical methods

  Disadvantage Disadvantage Disadvantage Disadvantage:::: not precise

  (a) (b)

  (c) (d)

  (e) (f)

  (g)

  Method/approach for finding roots: Method/approach for finding roots: Method/approach for finding roots: Method/approach for finding roots:

  INTRODUCTION _► What is a nonlinear equation? What is a nonlinear equation? What is a nonlinear equation? What is a nonlinear equation? _► What are roots? What are roots? What are roots? What are roots?

  Illustration of root location(s) Illustration of root location(s) Illustration of root location(s) Illustration of root location(s)

  1 1 1 ---- Bracketing Methods for Finding the Root of Bracketing Methods for Finding the Root of Bracketing Methods for Finding the Root of Bracketing Methods for Finding the Root of a Single Nonlinear Equation a Single Nonlinear Equation a Single Nonlinear Equation a Single Nonlinear Equation

  PENENTUAN AKAR PERSAMAAN PENENTUAN AKAR PERSAMAAN PENENTUAN AKAR PERSAMAAN PENENTUAN AKAR PERSAMAAN TAK LINIER TUNGGAL TAK LINIER TUNGGAL TAK LINIER TUNGGAL TAK LINIER TUNGGAL Yogyakarta, September 2011 ^{ANALISIS NUMERIK SEMESTER GASAL TAHUN AKADEMIK 2011/2012 PRODI TEKNIK KIMIA – FTI – UPN “VETERAN” YOGYAKARTA}

  Materi Kuliah Materi Kuliah Materi Kuliah Materi Kuliah::::

  PENGANTAR PENGANTAR PENGANTAR PENGANTAR BRACKETING METHODS BRACKETING METHODS BRACKETING METHODS BRACKETING METHODS OPEN METHODS OPEN METHODS OPEN METHODS OPEN METHODS ooooleh leh leh leh:::: Siti Diyar Kholisoh Siti Diyar Kholisoh Siti Diyar Kholisoh Siti Diyar Kholisoh

  MAIN TOPIC & OBJECTIVES ( MAIN TOPIC & OBJECTIVES ( MAIN TOPIC & OBJECTIVES ( MAIN TOPIC & OBJECTIVES (1

  11 1))))

  1

  Specific objectives and topics: Specific objectives and topics: Specific objectives and topics: Specific objectives and topics:

  INTRODUCTION

  MAIN TOPIC & OBJECTIVES ( MAIN TOPIC & OBJECTIVES ( MAIN TOPIC & OBJECTIVES ( MAIN TOPIC & OBJECTIVES (2

  22 2))))

  2

  2 2 2 ---- Open Methods for Finding the Root of Open Methods for Finding the Root of Open Methods for Finding the Root of Open Methods for Finding the Root of a Single Nonlinear Equation a Single Nonlinear Equation a Single Nonlinear Equation a Single Nonlinear Equation

  Objectives: Objectives: Objectives: Objectives:

  INTRODUCTION

  INTRODUCTION

• Recognizing the difference between bracketing and open methods for root location
• Understanding the fixed-point iteration method and how you can evaluate its convergence characteristics
• Knowing how to solve a roots problem with the Newton- Raphson method and appreciating the concept of quadratic convergence
• Knowing how to implement both the secant and the modified secant methods

• – 5 x - 14 = 0 Akar persamaannya: ….?

  2

  4

  2

  Dengan menggunakan rumus abc untuk menentukan akar-akar persamaan kuadrat, diperoleh: a c a b b x

• 3 -2 -1
• 5

  Secara grafik: Secara Analitik: Secara Analitik: Secara Analitik: Secara Analitik:

  Secara analitik: Mudah…!

  2

  8

  7

  5

  6

  4

  3

  2

  1

  15

  10

  5

  Contoh Ilustratif Contoh Ilustratif Contoh Ilustratif Contoh Ilustratif::::

  2

  12 − ± − =

9 Persamaan: f (x) = x

  2

• = − − − + =
• 25
• 20
• 15
• 10
• Nilai tebakan awal

  initial guess):

  ) x ( ' f ) x ( f x x _i ⁱ _{i 1 i}

  − =

  Hasil

  Hasil yang diperoleh dengan Polymath: BRACKETING METHODS BRACKETING METHODS BRACKETING METHODS BRACKETING METHODS and INITIAL GUESSES and INITIAL GUESSES and INITIAL GUESSES and INITIAL GUESSES

  Two major classes Two major classes Two major classes Two major classes of methods for finding the root of a single nonlinear equation (distinguished by the type of

  2. Open methods

  1. Bracketing methods

  Secara numerik:

  Based on two initial guesses that “bracket” the root Always work, but converge slowly (i.e. they typically take more iterations)

  Can involve one or more initial guesses, but there is no need for them to bracket the root Do not always work (i.e. they can diverge), but when they do they usually converge quicker

  Bracketing Methods Bracketing Methods Bracketing Methods Bracketing Methods (Incremental Search Methods) (Incremental Search Methods) (Incremental Search Methods) (Incremental Search Methods)

  1. Metode Bisection

  2. Metode False Position

  Misal, dipilih metode Newton-Raphson :

  2 =

  x Atau, dalam hal ini:

  7

  5 ) 1 (

  2 ) ) 14 ( 1 (

  4 ) 5 (

  2

  1 − = −

  = − − − − = x

  2

  9

  9

  5 ) 1 (

  2 ) ) 14 ( 1 (

  4 ) 5 (

  5

  2

  5

  BISECTION METHOD BISECTION METHOD BISECTION METHOD BISECTION METHOD

  INCREMENTAL SEARCH METHOD = Binary Search Method = Binary Search Method = Binary Search Method = Binary Search Method

  INCREMENTAL SEARCH METHOD

  INCREMENTAL SEARCH METHOD

  INCREMENTAL SEARCH METHOD

  In general, if f(x) is real and continuous in the interval from x to x and f(x ) and f(x ) have opposite signs , that is:

  l u l u

  A “brute force” technique for root solving which is too inefficient for hand computation, but is ideally suited to f(x l ).f(x u ) < 0 machine computation. then there is at least one real root at least one real root at least one real root at least one real root between x and x .

  l u

  An incremental search method in which the interval is the interval is the interval is the interval is

  Incremental search methods Incremental search methods Incremental search methods Incremental search methods capitalize on this observation always divided in half always divided in half . If a function changes sign changes sign over always divided in half always divided in half changes sign changes sign

  by locating an interval where the function changes sign. A an interval, the function value at the midpoint is potential problem with an incremental search is the choice evaluated. The location of the root is then determined as

  of the increment length . If the length is too small , the search

  lying within the subinterval where the sign change can be very time consuming. On the other hand, if the length occurs. The subinterval then becomes the interval for the

  is too great , there is a possibility that closely spaced roots

  next iteration. The process is repeated until the root is might be missed. The problem is compounded by the known to the required precision. possible existence of multiple roots.

  Two initial guesses of x (x l and x u ), START _{and tolerance (tol)}

  Consider: a function f(x) which is known to have Consider: Consider: Consider: one one one one _{x x l u}

• u _{x M =}
₂ f(x) Evaluate: f(x ), f(x ), and f(x ) _{l u M} f(x ) _{u f(x ).f(x ) 0 ? f(x ).f(x ) = 0 ? M u ≤≤≤≤ M u x M → x u M → l N Y Y x x N} f(x ) _M x ^l

  and only one real root and only one real root and only one real root and only one real root in the interval x l < x < x

• x l x u

  x _next x M

  = x _{u iteration} x _M

  2 f(x ) _l midpoint value x x _{N M , present M , previous} − tol ?

  ≤ x ^{M present ,}

• x x _{l u}

  Bisection formula: Bisection formula: Bisection formula: Bisection formula: x = _{M BISECTION METHOD BISECTION METHOD BISECTION METHOD BISECTION METHOD Y}

  2 FLOWCHART FLOWCHART FLOWCHART FLOWCHART _{x = x M END} Two initial guesses of x (x l and x u ),

  START _{and tolerance (tol) f ( x ) x x} FALSE POSITION METHOD FALSE POSITION METHOD FALSE POSITION METHOD FALSE POSITION METHOD _{x x M = u − f ( x ) f ( x ) u ( l − u ) l − u}

  = Linear Interpolation Method _{Evaluate: f(x l ), f(x u ), and f(x M )} = Regula-Falsi Method with the

  It is very similar to bisection method , _{f(x M ).f(x u ) ≤≤≤≤ 0 ? M u Y Y f(x ).f(x ) = 0 ?}

  exception that it uses a different strategy to come _{N N} up with its new root estimate. _{x M → x u M → l x x} False----position formula: False False False position formula: position formula: position formula: _{iteration l u next M = u − x x} ^{f ( x ) x − x} _{f ( x ) − f ( x ) u ( l u )} f ( x ) x x _{u ( l u )} − x x _{M u} = −

  ( ) ( ) x f x f x M _{l − u N} x M present − x M previous _{, ,}

  ≤ tol ? x _{M present ,}

  FALSE POSITION FALSE POSITION FALSE POSITION FALSE POSITION _Y METHOD FLOWCHART METHOD FLOWCHART METHOD FLOWCHART METHOD FLOWCHART _{x = x M END}

  Calculation Results: Calculation Results: Calculation Results: Calculation Results: Example #1: Example #1: Example #1: Example #1:

  Use: (a) bisection method, and (b) false position

• x

  method, to locate the root of: f(x) = e - x Use initial guesses of x = 0 and x = 0,8, and l u iterate until the approximate error falls below 1% _1,2 Graphically: _0,8 ¹ _{) 0,4 f( x -0,4 -0,2 0,2 0,6 0,2 0,4 0,6 0,8} x = 0,5671

• 0,6 _x

  Example # Example #2222:::: Example # Example # Graphical illustration

  Use bisection method bisection method bisection method bisection method and false position method to deter-

  f(m) versus m

  mine the mass of the bungee jumper with a drag coeffi-cient of 0,25 kg/m to have a velocity of 36 m/s after 4 s of free fall.

2 Note: The acceleration of gravity is 9,81 m/s

  Free-fall velocity as a function of time:  

  g m g c _d

   

  v t t

  ( ) = tanh  

  c m _{d  }

  An alternative way to make the equation as a function of

  mass:

   

  g m g c _d

   

  f ( m ) = tanh t − v ( t ) =

   

  c m _{d  }

  Use initial guesses of m = 50 kg and m = 200 kg, and

  l u

  iterate until the approximate error falls below 5% ( ε =

  s

  stopping criterion = 5%)

  Calculation Results (by using MS Excel): Calculation Results (by using MS Excel): Calculation Results (by using MS Excel): Calculation Results (by using MS Excel): Calculation Results (by using Polymath): Calculation Results (by using Polymath): Calculation Results (by using Polymath): Calculation Results (by using Polymath):

• – Metode Dua Kurva

  Starting with an initial guess of x = 0 , this iterative equation can be applied to compute:

• xi xi xi xi

  iiii x x x x iiii eeee

  εεεε a aa a

  , % , % , % , % εεεε tttt

  , % , % , % , % 0,0000 1,0000

• 1 1,0000 0,3679 100,000 76,322 2 0,3679 0,6922 171,828 35,135 3 0,6922 0,5005 46,854 22,050 4 0,5005 0,6062 38,309 11,755 5 0,6062 0,5454 17,447 6,894 6 0,5454 0,5796 11,157 3,835 7 0,5796 0,5601 5,903 2,199 8 0,5601 0,5711 3,481 1,239 9 0,5711 0,5649 1,931 0,705 10 0,5649 0,5684 1,109 0,399
• −

<
• x
• =
• START An initial guess of x (x _i

  ITERATION

  By graphical method, there are two alternatives for determining root of:

  &lt; tol ? x i END next iteration Y

  − = ε ε a

  = x ), tol % 100 . x x x

• _{+ 1 i 1 i i a +}

  ) x ( g x _{i 1 i =}

  ITERATION METHOD METHOD METHOD METHOD flowchart flowchart flowchart flowchart

  Note: The true value of the root = 0,56714329 Two Curves Two Curves Two Curves Two Curves Graphical Method Graphical Method Graphical Method Graphical Method

  (a) Root at the point where it crosses the

  ITERATION

  x axis

  (b) Root at the

  intersection of the component functions ) x e x ( f x

  −

  Two curves graphical method FIXED FIXED FIXED FIXED----POINT POINT POINT POINT

  ITERATION

  − = −

  Solution: Solution: Solution: Solution:

  x 1 i e x

  a formula to predict a new value of x as a function of an old value of x

  Open Methods Open Methods Open Methods Open Methods

  1.Metode Iterasi Satu Titik

  2. Metode Newton-Raphson

  3. Metode Secant

  It is also called:

  One point iteration method , or Successive substitution method SIMPLE FIXED POINT ITERATION METHOD SIMPLE FIXED POINT ITERATION METHOD SIMPLE FIXED POINT ITERATION METHOD SIMPLE FIXED POINT ITERATION METHOD

  Rearranging the function f(x) = 0 so that x is on the

  left-hand side of the equation: This transformation can be accomplished either by:

  Algebraic manipulation, or Simply adding x to both sides of the original equation

  x = g(x) x = g(x) x = g(x) x = g(x)

  The function can be separated directly and then expressed as: ⁱ

  Thus, given an initial guess at root x i , the equation above can be used to compute a new estimate x

  i+1

  as expressed by the iterative formula:

  % 100 . x x x

  1 i i 1 i a

  = ε x i+1

  = g(x i

  ) The approximate error can be determined by:

  Example Example Example Example::::

  Use the simple fixed-point iteration to locate the root of f(x) = e

  N Fixed-point iteration formula Approximate error x _i = x _i+1 CONVERGENCE CONVERGENCE CONVERGENCE CONVERGENCE NEWTON NEWTON NEWTON----RAPHSON METHOD NEWTON RAPHSON METHOD RAPHSON METHOD RAPHSON METHOD

OF SIMPLE OF SIMPLE OF SIMPLE OF SIMPLE

  The most widely used of all root-locating formula FIXED----POINT FIXED FIXED FIXED POINT POINT POINT

  ITERATION

  ITERATION

  ITERATION

  ITERATION

  If the initial guess at the (a) &amp; (b) root is x i , a tangent can be

  convergent

  extended from the point [x ,

  i

  (c) &amp; (d) f(x i )]. The point where this

  divergent tangent crosses the x axis

  usually represents an

  Note that

  improved estimate of the

  convergence root. _׀ occurs when g’(x)׀ &lt; 1 f ( x ) −

  The first derivative at x i is i

  f ' ( x ) = i

  equivalent to the slope:

  x − x

• i i

  1 Starting with an initial guess of x = 0 , this iterative

  which can be rearranged to yield: equation can be applied to compute:

  f ( x ) _i x = x −

• i ^{1 i}

  iiii x x x x εεεε a aa a , % , % , % , % εεεε , % , % , % , %

  f ' ( x ) _i iiii tttt

• 100 Newton-Raphson formula

  Example:::: Example Example Example

  1 0,5 100 11,83886 Use the Newton-Raphson method to estimate the root 2 0,566311003 11,70929 0,146751

• x
• 5

  of f(x) = e – x, employing an initial guess of x = 0 3 0,567143165 0,146729 2,2.10

• 5 -8

  4 0,567143290 2,21.10 7,23.10

  Solution: Solution: Solution: Solution:

  The first derivative of the function can be evaluated

• x

  Comment:

  as: f ’(x) = - e - 1 The approach rapidly converges on the true root. Then, by the Newton-Raphson formula:

  Notice that the true percent relative error at each x

  − i e x iteration decreases much faster than it does in simple

  − i x x

  = −

• i 1 i x − i fixed-point iteration (in previous example)

  e

  1 − −

  FOUR CASES OF POOR CONVERGENCE OF THIS METHOD FOUR CASES OF POOR CONVERGENCE OF THIS METHOD FOUR CASES OF POOR CONVERGENCE OF THIS METHOD FOUR CASES OF POOR CONVERGENCE OF THIS METHOD START An initial guess of x (x = x ), tol _i f ( x ) _i x = x −

• _{+ i 1 i}

  Newton-Raphson formula f ' ( x ) _i x = x _{i i+1} x x

  − _{+ i i 1} = . 100 % ε Approximate error _a next x _{+ i 1} iteration

  N &lt; tol ? ε a

  Y x i

  There is no general convergence criterion for Newton-Raphson method. Its convergence depends on: NEWTON NEWTON----RAPHSON NEWTON NEWTON RAPHSON RAPHSON RAPHSON

  END the nature of the function, and

  flowchart flowchart METHOD METHOD METHOD METHOD flowchart flowchart the accuracy of the initial guess

  Notice that this approach requires two initial

  estimates of x . However, because f(x) is not required to

SECANT METHOD SECANT METHOD SECANT METHOD SECANT METHOD

  change signs between the estimates, it is not classified A potential problem in implementing the Newton- as a bracketing method. Raphson method is: the evaluation of the derivative

  Rather than using two arbitrary values to estimate the There are certain functions whose derivatives may be derivative, an alternative approach involves a fractional difficult or inconvenient to evaluate. For these cases, perturbation of the independent variable to estimate f’(x):

  the derivative can be approximated by a backward

  finite divided difference : ^{i δ i − i} f ' ( x ) _{i ≅} f ( x ) f ( x ) _{i 1 i} −

• f ( x x ) f ( x )

  − x f ' ( x ) ≅ _i

  δ i x − x _{i 1 i}

  − δ perturbation fraction

  where = a small This approximation can be substituted into Newton-

  This approximation can be substituted into Newton- Raphson formula to yield the following iterative Raphson formula to yield the following iterative equation: equation:

  f ( x ) ( x x ) _{i i 1 − i} x f ( x )

  Secant me- δ

  − ^{i i Modified se-} x = x − x x

  = −

• ^{i 1 i}

  thod formula cant method

  − ^{i i i} START

  f ( x ) f ( x ) f ( x x ) f ( x ) _{i 1 i} − ^{i 1 i} − δ

  Hasil Penyelesaian Contoh Soal yang Sama dengan Sebelumnya (Metode Secant): Two initial guesses of x (x &amp; x ), tol _{i-1 i} f ( x ) ( x x ) _{i i − 1 i} − x = x −

• _{+ i 1 i}

  Secant formula f ( x ) f ( x ) _{i − 1 i} − x = x _{i-1 i} x x

  − x = x + i i _{i i+1} ¹

  . 100 % Approximate error ε = _{a +} x _{i 1} next iteration

  N ε &lt; tol ? _a

  Y x _i

  SECANT METHOD SECANT METHOD SECANT METHOD SECANT METHOD

  END flowchart flowchart flowchart flowchart

  Konvergen! START

  Example Example Example Example:::: An initial guess of x (x = x ), δ , tol _i

  Use the modified secant method the modified secant method to determine the mass

  the modified secant method the modified secant method

  of the bungee jumper with a drag coefficient of 0,25

  x f ( x ) δ i i kg/m to have a velocity of 36 m/s after 4 s of free fall. x = x − _{+ i 1 i}

  Modified secant formula

• 2

  f ( x x ) f ( x ) _{i δ i i} −

  Note: The acceleration of gravity is 9,81 m/s . Use an

• 6

  initial guess of 50 kg and a value of 10 for the

  x = x _{i i+1} x − x _{+ i i 1 perturbation factor.}

  . 100 % ε = Approximate error _{a +} x _{i 1} Solution: Solution: Solution: Solution: next iteration

  N &lt; tol ? ε a

  First iteration: Y x = 50 f(x ) = -4,57938708 x i x x = 50,00005 f(x x ) = -4,5793381118 + +

  MODIFIED MODIFIED

  δ δ

  MODIFIED MODIFIED SECANT METHOD SECANT METHOD SECANT METHOD SECANT METHOD END

  flowchart flowchart flowchart flowchart

  The calculation can be continued to yield: − ⁶

  10 ( 50 )( − 4 , 57938708 ) ׀ ׀ ׀ ׀

  εεεε t εεεε a ₁ i x x x f (x ) f (x + x x )

• + (%) (%)

  = 50 − = 88 , 39931 i i δδδδ i i i δδδδ i

  − 4 , 579381118 − ( − 4 , 57938708 )

  50 50,00005 -4,57938708 -4,579381118 64,97 -

  ׀ ׀ ׀ ε = 38,07%: ׀ ε = 43,44% t a

  1 88,39931 88,39940 -1,692207707 -1,692203516 38,07 43,44 2 124,08970 124,08982 -0,432369881 -0,43236662 13,06 28,76

  Second iteration:

  3 140,54172 140,54186 -0,045550483 -0,045547526 1,54 11,71

  x = 88,39931 f(x ) = -1,69220771

  1

  1

  4 142,70719 142,70733 -0,000622927 -0,000620007 0,02 1,52

  x δ + x = 88,39940 f(x δ + x ) = -1,692203516

  1

  1

  1

  1

• _{-7 -6}
₆ 5 142,73763 142,73777 -1,19176.10 2,80062.10 0,00 0,02 −
• ^{-14 -6}

  6 142,73763 142,73778 9,9476.10 2,9198.10 0,00 0,00 10 ( 88 , 39931 )( 1 , 69220771 )

  − x

  88 , 39931 124 , 08970 ₂ = − = 1 , 692203516 ( 1 , 69220771 )

  − − − Comment:

  δ The choice of a proper value for is not automatic.

  ׀ ׀ ׀ = 13,06%: ׀ = 28,76% ε ε t a

  If δ is too small : …………………

  δ If is too big : ………………..

  Problem # Problem Problem Problem #1111:::: # # (a) f x x x x

  ( ) sin ,

  5 = − =

  ( ) PROBLEMS

  2

  3

  f x x x x (b) ( )

  11

  22

  17 2 ,

  5 = + − − −

  Carilah nilai x pada interval x = 1 dan x = 3 Problem #2222:::: Problem # Problem # Problem #3333:::: Problem # Problem # Problem # Problem #

  Using x =1 as the starting point, find a root Using x = 4 as the starting point, find a root of the following equation: of the following equation to three significant

  figures: 2 x f x x e x e

  ( ) = − 5 − 5 = f ( x ) x e

• x x

  1 = − = using: using: a. Newton’s method

  a. successive substitution

  b. the secant method (use x = 4,1 as your

  b. Newton’s method second point) c. the secant method (use x = 1,01 as

  a. the regula falsi method your second point)

  Problem Problem Problem # Problem #5555:::: # # Water is flowing in a trapezoidal channel at a rate of Q = 20

  Problem #4444:::: Problem # Problem # Problem #

  3 m /s. The critical depth y for such a channel must satisfy the ₂ Q equation:

  = 1 − B ₃ Consider the following nonlinear equation: g A _c

  2

  2

  2 x where g = 9,81 m/s , A = the cross-sectional area (m ), and B =

  c f x x e

  ( ) = − = the width of the channel at the surface (m). For this case, the width and the cross-sectional area can be related to depth y

  B =

  3 y

• by:

  Show at least three cycles of search using ² y and

  A = y _c a starting point of x = 1 for:

• 3

  2

a. Newton’s method Solve for the critical depth using: (a) the graphical method,

  (b) bisection, and (c) false position . For (b) and (c), use initial initial initial initial

  b. regula falsi method

  guesses of guesses of guesses of guesses of y y y y = = = = 0000,,,,5 5 5 and and and and y y y y = = 2222,,,,5555 = = , and iterate until the

  5

  llll u uu u approximate error falls below 1% or the number of iterations exceeds 10. Discuss your results.

  Problem # Problem # Problem #7777:::: Problem # Problem # Problem # Problem #6666:::: Problem #

  R T a The Redlich-Kwong p = −

  In a chemical engineering process, water vapor (H O) is

  equation of state is given by: v b

  − v v b T ( )

• 2

  heated to sufficiently high temperatures that a significant portion of the water dissociates, or splits apart, to form oxygen (O ) and hydrogen (H ):

  2

  2 H O H + O

  2

  2

  2 If it assumed that this is the only reaction involved, the mole

  2 2 ,

  5

  fraction x of H O that dissociates can be represented by:

  2 R T T _c c and a , 427 b , 0866 R

  = = x

  2 P _t p p c

  K = c

  2 x − where p = 4600 kPa and T = 191 K. As a chemical engineer, c c you are asked to determine the amount of methane fuel that

• 1 x

  where K is the reaction’s equilibrium constant and P is

  t 3 o can be held in a 3-m tank at a temperature of -40 C with a

  the total pressure of the mixture. If P = 3 atm and K = 0,05,

  t pressure of 65000 kPa. Use a root locating method of your determine the value of x that satisfies the equation above . choice to calculate v and then determine the mass of methane contained in the tank.

  Problem # Problem # Problem # Problem #8888:::: Then the partial pressures are given as: Determine the equilibrium conversion for:

  N 2 x −

  2 CO + O

  2 CO CO

  2

  2 p = =

  CO if stoichiometric amounts of CO and air are reacted at

  N 6 , 76 , 5 x −

  T 2000 K and 1 atmosphere pressure. At 2000 K the

  6 -1 equilibrium constant for this reaction is 62,4 x 10 atm .

  As a basis, consider 2 gmoles of CO. Then there would N O

  1 , 5 x ₂ − be 1 gmole of O and 3,76 gmole of N . Performing a

  2

  2 p

  = = O ₂ mole balance on each species and defining x as the

  N 6 , 76 − , 5 x T amount of CO that reacts yields: N = 2 – x

  CO N x

  CO N = 1 – 0,5 x ² O2 p

  = = CO ₂ N = x

  CO2 N 6 , 76 , 5 x

  − T

  N = 3,76 N2

  2 p P _{CO T 2} The equilibrium relationship is given by: K = _{2 Problem #9999::::} Problem # Problem # Problem # p p _{CO O 2} Van der Waals equation of state is given as:

    a

  where P is the total pressure and remembering that the ^T

  V b R T − =

• P

   ^{2 } ( ) standard state fugacities of CO , CO, and O are unity. _{2 2  }

  V Substituting yields: ₂ where: P ≡ pressure (10 atm), T ≡ temperature (250 K) x ( 6 , 76 , 5 x )

  − ⁶ R ≡ gas constant (0,082 liter.atm/gmole.K), V ≡ specific ₂ = 62 , 4 .

  10 ( 1 − , 5 x ) ( 2 − x )

  volume (liter/gmole) Rearranging into a normalized form: ₂ Determine the specific volume for ammonia using:

  a. successive substitution

  x ( 6 , 76 , 5 x ) −

  1 _{6 2} − =

  b. Newton’s method

  62 , 4 . 10 ( 1 , 5 x ) ( 2 x ) − −

  c. Secant method

  a. Solve for the equilibrium composition using Newton’s The Van der Waals constants for ammonia are: _{6 3 2 3} method with a starting point of x = 1,0 gmole. a = 4,19 x 10 atm (cm /gmole) and b = 37,2 cm /gmole.

  b. Solve this problem using the regula falsi method. (Beware of the units!)

  Problem # Problem Problem Problem #10 # #

  10

10 Problem # Problem #11 Problem # Problem #

  11 11 11:::: 10::::

  The saturation concentration of dissolved oxygen in fresh

  Pendirian suatu pabrik kimia memerlukan water can be calculated with the equation: fixed

  (FC) = Rp 700 milyar dan capital working capital ^{5 7 10 11}

  1 , 575701 x

  10 6 , 642308 x

  10 1 , 243800 x

  10 8 , 621949 x

  10

• ln o sf = − 139 , 34411 − −
_{2 3 4} (WC) = Rp 300 milyar.

  T a T a T a T a Nilai (C) = Rp 250 milyar. annual cash flow

  where o = the saturation concentration of dissolved oxygen _{sf -1}

  Umur pabrik diperkirakan selama 10 tahun dengan

  in fresh water at 1 atm (mg L ); and T = absolute _a temperature (K). Remember that T = T + 273.15, where T =

  salvage value (SV) = Rp 70 milyar. _o ^a

  temperature (

  C). According to this equation, saturation

  Jika i menyatakan nilai suku bunga investasi ini yang

  decreases with increasing temperature. For typical natural waters in temperate climates, the equation can be used to

  diekivalensikan dengan jika disimpan di bank,

  determine that oxygen concentration ranges from 14.621

  tentukan nilai i! _{o o}

  mg/L at 0 C to 6.949 mg/L at 35

  C. Given a value of oxygen

  Jika digunakan , maka nilai i present value analysis concentration, this formula and the bisection method can be _o

  used to solve for temperature in

  C. If the initial guesses are

  dapat dihitung dari persamaan berikut ini: _o

  set as 0 and 35

  C, how many bisection iterations would be

  C C C C WC SV _o ...

• required to determine temperature to an absolute error of

FC WC

  _{2 3 10}

• = + + + + +
₁₀ 0.005 C? 1 i (1 i ) (1 i ) (1 i ) (1 i<

  Selamat Belajar!