Jim Lambers ENERGY 8 Spring Quarter 7-8 Homework Assignment 3 Solution Plot the exponential integral solution p D (r D,t D ) = ) ( Ei r D, 4t D and the late time approximation to this solution, p D (r D,t D ) = ( ln t ) D rd + 897, on the same axes for a range of t D values and r D = It is recommended that you use values of t D that vary over several orders of magnitude For what values of t D does the late time approximation closely approximate the exponential integral solution? What real time values does this correspond to? In field units, t D is defined as where t is in hours Choose t D = 64kt φµc t rw, k = md, φ = 3, µ = cp, c t = 6 psi, r w = 3 ft The quickest, easiest way to do this is in a tool such as Matlab, Mathematica or Maple Consider a linear one-dimensional reservoir The inner boundary condition is constant pressure (p w ) with skin (S) The outer boundary
condition is infinite acting The initial-boundary value problem to be solved is p x = φµc t p k t, with initial condition p(x,) = p i, inner boundary condition p(,t) S p x = p w, x= and condition at infinity lim p(x,t) = p i x (a) First, write the problem in dimensionless form Use the same transformations for x and t as in the Lecture notes For p, use the same transformation as in the case of a constant pressure boundary condition Non-dimensionalize the skin using the transformation S D = S/L, where L is the length scale of the reservoir (b) Next, solve the dimensionless differential equation in Laplace space (c) Finally, invert the transform Use either a table, or the Maple function invlaplace If you use a table, indicate which table If you use Maple, submit your code and output Solution p w = p(x = ) S p x x= + Choose a dimensionless pressure and length: Also nondimensionalise the skin: p D = p i p p i p w x D = x L S D = S L
Simplifying this gives p w = p D (p i p w ) + p i + S D L p D p i p w L p w p i p i p w = p D + S D p D p D S D p D = The governing differential equation is p D t D = p D t D The infinite acting condition is p(,t) = p i p D (,t D ) = ˆp D (,t D ) = The inner boundary condition is p D S D p D xd = = ˆp D S D ˆp D xd = = s The initial condition is p(x,) = p i p D (x D,) = When the differential equation is transformed it becomes ˆp D x D = sˆp D The general solution is ˆp D = c e sx D + c e sx D To ensure this solution remains bounded as x D, the constant c is set to zero Now use the inner condition to find c : c + c ssd = s c = s( + ss D ) 3
ˆp D = e sxd s 3/ S D + s The transform can be inverted by using the Maple function invlaplace: x D p D = e t D S D e 3 Solve the Cauchy problem S D erfc( td S D + x D t D ) ( ) xd + erfc t D T x = α T, < x <, t >, t where α is a constant, with initial condition T(x,) = e πx, < x < Use both the Fourier transform and Laplace transform together to obtain an algebraic equation that can easily be solved, and then invert both transforms to obtain T(x, t) Be careful with your notation to keep track of your transforms 4 Consider the following sampled signal g(t) on the interval [,): t < 5 5 5 t < 5 g(t) = 75 5 t < 75 5 75 t < This signal is shown in Figure (a) This signal can be expressed as a sum of Haar smoothing functions at a particular scale M, g(t) = φ M,n (t)s M,n provided that this scale is sufficiently fine What is the largest value of M for which this is possible? In other words, what is the largest M such that g V M? Recall that larger values of M correspond to coarser scales Solution Because the signal is piecewise constant on intervals of width /4 =, it is a linear combination of scaling functions of the form ( ) t n φ,n (t) = φ 4
5 75 g(t) 5 5 5 5 75 t Figure : Signal for Problem 4 that are also piecewise constant on intervals of width, so M = (b) Perform the Haar decomposition of g to compute the detail coefficients {d M+,n }, {d M+,n }, and {s M+,n }, where M is the scale determined in part (a) Solution Because the signal is supported only on [,), and [ (n+/) m ] (n+) m d m,n = m/ g(t)dt g(t)dt, n m (n+/) m it follows that for m =, we only need to compute coefficients 5
for n = and n = We have d, = [ /4 g(t) d, = [ 3/4 / Using the formula / /4 g(t) dt ] = 4 ( 5) = 8, ] g(t) g(t) dt = 3/4 4 (75 5) = 8 we also obtain (n+) m s m,n = m/ g(t)dt, n m s, = / g(t)dt = 4 ( + 5) = 3 8, s, = g(t)dt = / 4 (75 + 5) = 4, which yields a smoothed signal belonging to V, { 3/4 x < / g (t) = s, φ, (t) + s, φ, () = / / x < We can work with this simpler signal to complete the decomposition We have d, = / s, = g (t)dt / g (t)dt = (3/4 /) = 8, g (t)dt = (3/4 + /) = 5 8 (c) Compress the signal by setting to zero the detail coefficients which satisfy d m,n 5 Reconstruct the compressed signal using the relation s m,n φ m,n (t) = s m+,n φ m+,n (t)+ d m+,n ψ m+,n (t) 6
Solution All of the detail coefficients have magnitude less than 5, so all of them are dropped, leaving only the approximation coefficient s, = 5/8 It follows that the reconstructed signal is g(t) = g (t) = s, φ, (t) = { 5/8 x < otherwise The relation given above shows that because no detail is added to this signal, the compressed signal g(t) at the finest scale of is equal to the signal at scale, which is g (t) (d) If g is a compression of a signal g, then the compression ratio is given by g d L m, m,n g =, L m, d m,n where { d m,n } is the set of detail coefficients for the compressed signal Compute and interpret the compression ratio for the compression performed in part (c) Hint: dm+3, = d M+3, = sm+, Solution We have d, = d, = 5 /8 It follows that the compression ratio S is S = m, d m,n m, d m,n (5 /8) = (5 /8) + (/8) + ( /8) + ( /8) = 999 The interpretation of this ratio is that a little more than 9% of the data in the signal is lost in the compression Typical compression algorithms should preserve close to % of the data, so the threshold from part (c) should be lowered 7