By Richard G. Rice, Duong D. Do
This moment version of the go-to reference combines the classical research and glossy purposes of utilized arithmetic for chemical engineers. The e-book introduces conventional ideas for fixing traditional differential equations (ODEs), including new fabric on approximate resolution equipment equivalent to perturbation options and hassle-free numerical strategies. it is also analytical easy methods to care for vital sessions of finite-difference equations. The final part discusses numerical answer options and partial differential equations (PDEs). The reader will then be outfitted to use arithmetic within the formula of difficulties in chemical engineering. just like the first variation, there are lots of examples supplied as homework and labored examples
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Additional info for Applied Mathematics And Modeling For Chemical Engineers, Second Edition
Finite difference) processes, is Riccati’s equation k2 A À! B À! C hence, the original equation is now linear in v 35 ð2:60Þ QðuÞ ¼ 0; RðuÞ ¼ k1 k1 CA0 exp À u k2 k2 to see that dCB þ C 2B ¼ RðuÞ du ð2:63Þ 36 SOLUTION TECHNIQUES FOR MODELS YIELDING ORDINARY DIFFERENTIAL EQUATIONS If we make the Riccati transformation CB ¼ 1 du u du we finally obtain d 2 uðuÞ k1 k1 À C exp À u uðuÞ ¼ 0 A0 k2 k2 du2 ð2:64Þ CASE (c) We have thus transformed a nonlinear first-order equation to a solvable, linear second-order equation.
13 MODEL HIERARCHY AND ITS IMPORTANCE IN ANALYSIS Solutions of Eqs. 25) T I ¼ T 1 þ A cosh½nðx þ L1 Þ ð1:147Þ and T II ¼ T 0 þ B cosh½mðL2 À xÞ ð1:148Þ where m is defined in Eq. 137 and a new ratio is rﬃﬃﬃﬃﬃﬃﬃ 2hL n¼ ð1:149Þ Rk The constants of integration, A and B, can be found by substituting Eqs. 146a, b) to finally get B¼ ðT 1 À T 0 Þ ½cosh ðmL2 Þ þ ðm=nÞ ðsinh ðmL2 Þ=sinh ðnL1 ÞÞ cosh ðnL1 Þ ð1:150Þ A¼À ðT 1 À T 0 Þ coshðnL ½ 1 Þ þ ðn=mÞ ðsinhðnL1 Þ=sinh ðmL2 ÞÞ coshðmL2 Þ ð1:151Þ The rate of heat transfer can be obtained by using either of the two ways mentioned earlier, that is, using flux at x ¼ 0, or by integrating around the lateral surface.
This is done by augmenting the matrix A with an identity matrix I. After the elimination process in converting the matrix A to an identity matrix, the right-hand side identity matrix will become the inverse AÀ1. To show this, we use the following example: 2 1 1 6 6 2 À1 4 1 À2 1 1 0 1 0 1 20 0 0 2 À1 6 61 4 1 1 À2 1 0 1 1 1 0 20 0 7 07 5 ð1:110Þ 1 0 1 À2 1 6 6 1 61 4 1 À2 0 11 20 1 2 1 2 3 0 0 1 6 6 60 4 0 1 À2 1 0 0 1 2 3 3 1 12 1 2 3 7 7 07 5 1 1 À3 1 À2 0 2 1 À3 3 7 1 À6 17 À67 1 À2 1 2 5 ð1:114Þ Obtaining the matrix inverse using the Gauss–Jordan method provides a compact way of solving linear equations.