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PGE 310: Formulation and Solution of Geosystems Engineering Problems

PGE 310: Formulation and Solution of Geosystems Engineering Problems. Dr. Matthew T. Balhoff Spring 2011. Notes Adapted from: Chapra, S., Canale, R. “Numerical Methods for Engineers”, Mc-Graw Hill Co. (2010) Rectenwald, G. “Numerical Methods with MATLAB”Prentice-Hall (2000)

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PGE 310: Formulation and Solution of Geosystems Engineering Problems

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  1. PGE 310: Formulation and Solution of Geosystems Engineering Problems Dr. Matthew T. Balhoff Spring 2011 Notes Adapted from: Chapra, S., Canale, R. “Numerical Methods for Engineers”, Mc-Graw Hill Co. (2010) Rectenwald, G. “Numerical Methods with MATLAB”Prentice-Hall (2000) Gilat, A., Subramaniam, V. “Numerical Methods for Engineers and Scientists” John Wiley and Sons Inc. (2011)

  2. About Me • Education/Research Experience • B.S. Chemical Engineering, Louisiana State University 2000 • Ph.D. Chemical Engineering, Louisiana State University 2005 • ICES Postdoctoral Fellow (CSM), UT-Austin 2005-2007 • Assistant Professor, UT-Austin 2007- • Research Interests • Flow and transport in porous media • Non-Newtonian flow • Pore-scale and Multi-scale modeling • NUMERICAL METHODS + =

  3. What’s a Numerical Method ? • Many math problems cannot be solved analytically (exactly) • Numerical methods are approximate techniques • Real-life problems in science and engineering require these numerical techniques • Real world problems can take hours, days, or years to solve. A well written computer program (in MATLAB for example) can do it much faster.

  4. Example 1: Roots of Equations • A root of an equation is the value that results in a “zero” of the function • Q: Find the root of the following quadratic equation

  5. Example 1: Roots of Equations • A root of an equation is the value that results in a “zero” of the function • Q: Find the root of the following quadratic equation • A: The quadratic formula is an EXACT method for solving the roots of a quadratic equation • Answer can be found by plugging in a, b, and c.

  6. Example 1. Roots of Equations • Ideal gas law doesn’t always apply:

  7. Example 1. Roots of Equations • Ideal gas law doesn’t always apply: • In petroleum engineering, we deal with gases far from ideal (P=50 bar, T=473K) Methane

  8. Example 1. Roots of Equations • Ideal gas law doesn’t always apply: • In petroleum engineering, we deal with gases far from ideal (P=50 bar, T=473K) • So how do we find the root of this function, where the quadratic equation doesn’t apply? (R= 83.14 cm3-bar/mol-K) Methane

  9. Example 1: Ideas? • What would be a good guess, if we needed a “ballpark” figure?

  10. Example 1: Ideas? • What would be a good guess, if we needed a “ballpark” figure? • How can we get very close to the “exact” solution by performing very few calculations?

  11. Example 1: Ideas? • What would be a good guess, if we needed a “ballpark” figure? • How can we get very close to the “exact” solution by performing very few calculations?

  12. Could have plotted points Root ~ 755

  13. Example 2. Differentiation • Derivative: “the slope of the line tangent to the curve”. • But we seem to forget about that once we learn some fancy tricks to find the derivative • Q: What is the derivative (dy/dx) at x = 1?

  14. Example 2. Differentiation • Derivative: “the slope of the line tangent to the curve”. • But we seem to forget about that once we learn some fancy tricks to find the derivative • Q: What is the derivative (dydx) at x = 1? • But how do we find the derivative of a really complicated function – or one that isn’t described by an equation?

  15. dy/dx = slope = -2

  16. Example 3: Integration • Integral: The area under the curve • But then we learned some fancy tricks in Calculus • Find the Integral:

  17. Example 3: Integration • Integral: The area under the curve • But then we learned some fancy tricks in Calculus • Find the Integral: • These “tricks” don’t always work in the real world and we need APPROXIMATE methods

  18. Add areas of triangles to approximate area under the curve H1 = y(0) Area1 = H1*w1 Area2 = H2*w2 w1 = 1/4

  19. Add areas of triangles to approximate area under the curve H1 = y(0) Some error Area1 = H1*w1 Area2 = H2*w2 w1 = 1/4

  20. We get a better answer by using more rectangles

  21. Compare Answers • 4 Rectangles: Area = 1.7188 • 10 Rectangles: Area= 1.4850 • 100 Rectangles: Area = 1.3484 • 1,000,000 Rectangles = 1.3333 • Actual = 4/3

  22. Great. Now what’s the computer for? • Numerical methods can require lots of computational effort • Root solving method may take lots of iterations before it converges • We might have to differentiate millions of equations • We might need thousands of little rectangles • Computers can solve these problems a lot faster if we program them right • We’ll have to learn some programming (in Matlab) before moving on to learning advanced numerical techniques • Matlab isn’t hard, it just requires PRACTICE. Don’t get intimidated

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