PGE 310: Formulation and Solution of Geosystems Engineering Problems

Dr. Matthew T. BalhoffSpring 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)

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

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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.

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

2( ) 4 3 0f x x x

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.

2( ) 4 3 0f x x x

22 4 4 4(1)(3)4 1,32 2(1)

b b acxa

Example 1. Roots of Equations

• Ideal gas law doesn’t always apply: iPV RT

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)

iPV RT

2 22i i i

RT aPV b V bV b

2 20.457 2.3 6

0.077824.7

c

c

c

c

R Ta EPRT

bP

Methane

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)

iPV RT

2 22i i i

RT aPV b V bV b

2 20.457 2.3 6

0.077824.7

c

c

c

c

R Ta EPRT

bP

Methane

2

39325 2.3 6( ) 50 024.7 49.4 611i i i

Ef VV V V

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?

• How can we get very close to the “exact” solution by performing very few calculations?

83.14 473786.5

50iRTVP

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?

83.14 473786.5

50iRTVP

2

2

2

2

39325 2.3 6(786) 50 1.8724.7 49.4 611

39325 2.3 6(750) 50 0.38924.7 49.4 611

39325 2.3 6(768) 50 0.751824.7 49.4 611

39325 2.3 6(759) 50 0.18824.7 49.4 611

(754.5

i i i

i i i

i i i

i i i

EfV V V

EfV V V

EfV V V

EfV V V

f

2

39325 2.3 6) 50 0.098824.7 49.4 611i i i

EV V V

Root ~ 755

Could have plotted points

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

2 4 3y x x

• 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

342 xxy• Q: What is the derivative (dydx) at x = 1?

42 xdxdy 24)1(21 x

dxdy

• But how do we find the derivative of a really complicated function – or one that isn’t 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 2

04 3x x dx

Example 3: Integration

• Integral: The area under the curve

• But then we learned some fancy tricks in Calculus

• Find the Integral:

3432

3134

1

0

231

0

2 xxxdxxx

• These “tricks” don’t always work in the real world and we need APPROXIMATE methods

w1 = 1/4

H1 = y(0)Area1 = H1*w1

Add areas of triangles to approximate area under the curve

Area2 = H2*w2

w1 = 1/4

H1 = y(0)Area1 = H1*w1

Add areas of triangles to approximate area under the curve

Area2 = H2*w2

Some error

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 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