Introducción a la Geofísica

Size: px

Start display at page:

Download "Introducción a la Geofísica"

Verity Abigayle Dixon
5 years ago
Views:

1 Introducción a la Geofísica TAREA 7 1) FoG. A plane seismic wave, travelling vertically downwards in a rock of density 2200 kg m -3 with seismic velocity 2,000 m s -1, is incident on the horizontal top surface of a rock layer of density 2400 kg m 3 and seismic velocity 3300 m s -1. (a) What are the amplitude ratios of the transmitted and reflected waves? The amplitude ratios for vertical incidence are described by the reflection coefficient R and the transmission coefficient I: R = Z 1 ; T = 2Z 1 where Z1 = ρ1v1 and Z2 = ρ 2V2 are seismic impedances, and ρ and V are the density and seismic velocity, respectively, in each medium In the upper layer, Z 1 = ρ 1 V 1 = ( 2200) ( 2000) = kg m -2 s -1 and in the lower layer = ρ 2 V 2 = ( 2400) ( 3300) = kg m -2 s -1 thus R = Z = = 0.29 T = 2Z 1 = 2(4.4) = 0.71 (b) What fraction of the energy of the incident wave is transmitted into the lower medium? The fraction ER of the incident energy reflected at the interface is ER = R 2 ; the fraction ET of the incident energy transmitted through the interface is equal to (1 ER). ET = (1 ER)= (1 R2 ) = % of the incident energy is transmitted into the lower medium.

2 2) FoG A plane seismic wave travels vertically downwards at a velocity of 4800 m s 1 through a salt layer with density 2100 kg m 3. The wave is incident upon the top surface of a sandstone layer with density 2400 kg m 3. The phase of the reflected wave is changed by 180 and the reflected amplitude is 2% of the incident amplitude. What is the seismic velocity of the sandstone? The 180 phase change at the surface of the sandstone layer implies that the seismic impedance Z2 of the sandstone is less than the seismic impedance Z1.of the salt layer. The reflection coefficient of 2% is thus negative. R = Z 1 = 0.02 ( )( 4800) ( )( 4800) = Solving gives = ρ 2 V 2 = ( 2100) ( 4800) = km m -2 s -1 and by substituting the density ρ2 = 2400 kg m 3 the seismic velocity of the sandstone is found to be V2 = 4000 m s 1. 3) FoG a) Calculate the minimum arrival times for seismic reflections from each of the reflecting interfaces in the following section. Consider the base of the lowermost bed to be also a reflector.

3 The formation thicknesses and velocities give the following two-way travel-times in each formation, and cumulative two-way travel-times for each reflecting interface: b) What is the average velocity of the section for a reflection from the base of the dolomite? V = s t = 2 ( 3000 ) = 3160 m s c) Using the listed densities calculate the reflection coefficient for each interface (except the base of the dolomite). Which interface gives the strongest reflection and which the weakest? At which interfaces does a change in phase occur? What does this mean? The reflection coefficient for each interface between successive rock types is computed with the formula for R in exercises 9 and 10:

4 The strongest reflection is at the interface between the alluvium and shale; the weakest is between the shale and the sandstone. The negative reflection coefficient at the interface between the limestone and salt indicates a phase change of π radians. 4) FoG The following table gives two-way travel times of seismic waves reflected from different reflecting interfaces in a horizontally layered medium. (a) Draw a plot of (travel-time) 2 against (distance) 2.

5 The best-fit straight lines (1), (2) and (3) to the data in the diagram have the following equations, with x = (distance [km]) 2 and y = (travel-time [s]) 2 : (1) y = x (2) y = x (3) y = x (b) Determine the vertical two-way travel-time ( echo-time ) and average velocity to each reflecting interface. The equation of each line in the (t2 x2) diagram has the form t 2 = t x 2 V, where t = 2d 2 0 is the vertical travel-time ("echo-time") and V is the V average (or stacking) velocity to a reflector. Applied to the equations of the three straight lines: (1) average velocity V = =1.80 km s -1 vertical two-way travel-time t 0 = = s. (2) average velocity V = = 2.36 km s -1 vertical two-way travel-time t 0 = = s. (3) average velocity V = = 3.64 km s -1 vertical two-way travel-time t 0 = = s. (c) Compute the depth of each reflector and the thickness of each layer. From the vertical two-way travel-time t0 and the average velocity V the depth d to each reflector is equal to (Vt0)/2 (1) depth d1 = (0.5)(0.111)(1800) = 100 m thickness of layer (1) is 100 m. (2) depth d2 = (0.5)(0.297)(2360) = 350 m thickness of layer (2) is ( ) = 250 m. (3) depth d3 = (0.5)(0.576)(3640) = 1050 m thickness of layer (3) is ( ) = 700 m. (d) Compute the true velocity (interval velocity) of each layer. The interval velocities of the layers require the layer thicknesses d as well as the interval travel-time in each layer, which is found from the differences between the successive "echo-times". Thus, V i 2d i ( t 0 ) 2 ( t 0 ) 1 (1) the interval velocity of the top layer is the same as the first average velocity:

6 V1 = 1800 m s -1 = 1.8 km s -1 2( 250) (2) the interval velocity of layer (2) is V i = = 2700ms-1 = 2.7km s -1 Vi 2( 700) (3) the interval velocity of layer (3) is V i = = 5000ms-1 = 5.0km s -1 (e) Verify your results by computing the total vertical travel-time for a wave reflected from the deepest interface. The total travel-time for a reflection from the deepest interface is the sum of the two-way travel-times in each layer. t = = s, 5000 which is equal to the vertical "echo time" from the deepest reflector (solution (3) in part (b) above).

2, from which K = # " 2 % 4 3 $ ) ( )

2, from which K = # 2 % 4 3 $ ) ( ) Introducción a la Geofísica 2010-01 TAREA 6 1) FoG. Calculate the bulk modulus (K), the shear modulus (µ) and Poisson s ratio (ν) for the lower crust, upper mantle and lower mantle, respectively, using