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Chapter 1 Mathematical Modeling and Engineering Problem Solving Lecture Notes Dr. Rakhmad Arief Siregar Universiti Malaysia Perlis Applied Numerical Method for Engineers 1 Background Over years and years observation and experiment, engineers and scientists have noticed that certain aspects of their empirical studies occur repeatedly. To understand this behavior, fundamental knowledge is required to develop a mathematical model. 2 Engineering Problemsolving process 3 Research data and Mathematical model 4 A Simple Mathematical Model A mathematical model can be broadly defined as a formulation or equation that expresses the essential features of a physical system or process in mathematical term In general, it can be represented as a functional relationship of the form Dependent variable = f( independent variable, parameters, Forcing functions ) 5 A falling Parachutist Problem The net force: where: F FD FU FD = mg FU = -cv For free falling body near the earth’s surface, the Newton’s second law can be used: F a m or dv F dt m 6 A falling Parachutist Problem Then the relation of acceleration of a falling object to forces is: dv mg cv dt m dv c g v dt m By using advanced techniques in calculus solves: (v=0 at t=0) gm ( c / m ) t v(t ) (1 e ) c 7 Ex. 1.1 Analytical solution A parachutist of mass 68.1 kg jumps out of stationary hot air balloon. Use Eq. (1.10) to compute velocity prior to opening the chute. The drag coefficient is equal to 12.5 kg/s gm v(t ) (1 e ( c / m )t ) c (9.8)(68.1) v(t ) (1 e ((12.5) /(68.1)t ) 12.5 v(t ) 53.39(1 e0.18355t ) 8 Ex. 1.1 Analytical solution After computing, the result is: v(t ) 53.39(1 e0.18355t ) 9 Ex. 1.1 Analytical solution 10 Numerical Methods In these methods, the mathematical problem is reformulated so it can be solved by arithmetic operation. 11 Numerical Methods The change of velocity can be approximated by: dv v v(ti 1 ) v(ti ) dt t ti 1 ti This is called as a finite divided difference approximation 12 Numerical Methods Then substitute the net force and it will yield: v(ti 1 ) v(ti ) c g v(ti ) ti 1 ti m Be rearranged to yield c v(ti 1 ) v(ti ) g v(ti ) (ti 1 ti ) m 13 Ex. 1.2 Numerical Solution Perform the same computation as in Example 1.1 but use Eq. (1.12) to compute the velocity. c v(ti 1 ) v(ti ) g v(ti ) (ti 1 ti ) m 12.5 (0) (1) 9.80 t=1 v 0 9.8 68.1 12.5 v 9 . 80 9 . 8 ( 9 . 80 ) (1) 17.80 t=2 68.1 14 Ex. 1.2 Numerical Solution 15 Conservation Laws and Engineering Aside from Newton’s second law, there are other major organizing principles in engineering The basics one: Change increases decreases Flow in Flow out 16 17 For exercise Compute Questions 1.1, 1.2, 1.3, and 1.4 Question 1.11 will be discussed in the lab if time permitted 18