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 MATLABPrentice-Hall (2000) Gilat, A., Subramaniam, V. Numerical Methods for Engineers and Scientists John Wiley and Sons Inc. (2011) 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 +

= Whats 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. 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 f ( x) x 2 4 x 3 0

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 f ( x) x 2 4 x 3 0 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.

b b 2 4ac 4 42 4(1)(3) x 1,3 2a 2(1) Example 1. Roots of Equations Ideal gas law doesnt always apply: PVi RT Example 1. Roots of Equations PVi RT

Ideal gas law doesnt always apply: In petroleum engineering, we deal with gases far from ideal (P=50 bar, T=473K) RT a P 2 Vi b Vi 2bVi b 2 Methane 0.457 R 2Tc2 a 2.3E 6 Pc

b 0.0778 RTc 24.7 Pc Example 1. Roots of Equations PVi RT Ideal gas law doesnt always apply: In petroleum engineering, we deal with gases far from ideal (P=50 bar, T=473K) RT

a P 2 Vi b Vi 2bVi b 2 Methane 0.457 R 2Tc2 a 2.3E 6 Pc b 0.0778 RTc 24.7 Pc

So how do we find the root of this function, where the quadratic equation doesnt apply? (R= 83.14 cm3-bar/mol-K) f (V ) 50 39325 2.3E 6 2 0 Vi 24.7 Vi 49.4Vi 611 Example 1: Ideas? What would be a good guess, if we needed a ballpark figure? Example 1: Ideas?

What would be a good guess, if we needed a ballpark figure? Vi RT 83.14 473 786.5 P 50 How can we get very close to the exact solution by performing very few calculations? Example 1: Ideas? What would be a good guess, if we needed a ballpark figure? Vi

RT 83.14 473 786.5 P 50 How can we get very close to the exact solution by performing very few calculations? f (786) 50 39325 2.3E 6 2 1.87 Vi 24.7 Vi 49.4Vi 611

f (750) 50 39325 2.3E 6 2 0.389 Vi 24.7 Vi 49.4Vi 611 f (768) 50 39325 2.3E 6 2 0.7518 Vi 24.7 Vi 49.4Vi 611 f (759) 50 39325

2.3E 6 2 0.188 Vi 24.7 Vi 49.4Vi 611 f (754.5) 50 39325 2.3E 6 2 0.0988 Vi 24.7 Vi 49.4Vi 611 Could have plotted points Root ~ 755 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 y x 2 4 x 3 Q: What is the derivative (dy/dx) at x = 1? 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 y x 2 4 x 3 Q: What is the derivative (dydx) at x = 1? dy 2 x 4 dx dy x 1 2(1) 4 2 dx

But how do we find the derivative of a really complicated function or one that isnt described by an equation? dy/dx = slope = -2 Example 3: Integration Integral: The area under the curve But then we learned some fancy tricks in Calculus Find the Integral:

1 x 0 2 4 x 3 dx Example 3: Integration Integral: The area under the curve But then we learned some fancy tricks in Calculus

Find the Integral: 1 1 3 4 2 0 x 4 x 3 dx 3 x 2 x 3x 0 3 1 2

These tricks dont always work in the real world and we need APPROXIMATE methods Add areas of triangles to approximate area under the curve H1 = y(0) Area1 = H1*w1 w1 = 1/4 Area2 = H2*w2 Add areas of triangles to approximate area under the curve

H1 = y(0) Area1 = H1*w1 w1 = 1/4 Some error Area2 = H2*w2 We get a better answer by using more rectangles 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 Great. Now whats 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 Well have to learn some programming (in Matlab) before moving on to learning advanced numerical techniques Matlab isnt hard, it just requires PRACTICE. Dont get intimidated