Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2...

31
Last week’s problems excess = 1/2πG × Area under curve 0.0239 × 10 11 in kgs 2 m -3 der curve = -1.8 ×10-6 x 100 m 2 s -2 excess (actually deficit) = -4.3 ×10 5 kg per m anomaly decreases to zero in 10-15 m. This suggests that z is < 5m. area of the car park = excess mass per unit length/density contrast he area = 4.3 ×10 5 /1800 = 239 m 2 g a rectangular shape, the height is 239/100 = 2.39m. Problem 4

Transcript of Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2...

Page 1: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

Last week’s problems

a) Mass excess = 1/2πG × Area under curve1/2πG = 0.0239 × 1011 in kgs2m-3 Area under curve = -1.8 ×10-6 x 100 m2s-2

So Mass excess (actually deficit) = -4.3 ×105 kg per m

b) The anomaly decreases to zero in 10-15 m. This suggests that z is < 5m.

c) The area of the car park = excess mass per unit length/density contrast, hence the area = 4.3 ×105/1800 = 239 m2

Assuming a rectangular shape, the height is 239/100 = 2.39m.

Problem 4

Page 2: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

Problem 5

x1/2 = 2.3 cm = 575 m

For a spherical body zmax = 1.3 x1/2 = 750 m

Page 3: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

Today’s lecture Seismic reflection

OverviewData collection - Sources and receiversTheorySeismic reflection processingTwo new developmentsProblems 6 and 7

Page 4: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

Marine streamer carries 100s of hydrophones. Detect changes in pressureBirds on streamer keep it levelLand receivers detect movement Reflections occur at boundaries where velocity and/or density change

Page 5: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

Horizontal distance ~10 km ~1500 traces

TW

TT

(se

cond

s)

Earth’s surface

~6 kmdepth

Oil explorationcolour = reflection amplitude

0 kmdepth

1 trace

Page 6: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

3D image subsurface

Slice at t = 400 ms

Can also track horizons andgenerate fly-bys

Page 7: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

Seismic stratigraphy

Unravel depositional changes with time

Page 8: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

Normally only interested in p-wave (smallest amplitude/fastest)The higher the frequency the better – reflections appear sharper

Page 9: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.
Page 10: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

Velocity versus density

Page 11: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

Seismic boat

Airgun array10-20 guns

Page 12: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

Tows 6 streamers at once, each 6 km longEach streamer ~480 channels (group of hydrophones)

Page 13: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

5 am in Derbyshire

Dynamite buried ~ 5 m below surface

Packed with filler

Encourage energy to travel down – not up

Ideal source – high frequency

Page 14: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

Truck vibrates up and down to produce a continuous seismic signal

Page 15: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

Horizontal (S-wave)source

Page 16: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

A hammer source

Page 17: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

Geophones

Page 18: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

Ocean-bottom seismometers Broad band seismometers

Page 19: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

i

r

A

AR

i

t

A

AT

1122

1122

vv

vvR

1122

112

vv

vT

For small angle of incidence

Gas beds produce high-amplitude (bright) negative reflections

Page 20: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

QA

At ftexp

0

Rays represent the direction of travel of the seismic wave, and are perpendicular to the 3D wavefront

The energy in the wavefront decreases in with increasing distance from the source (geometrical spreading) and is inversely proportional to 4πr2. The amplitude of a seismic wave is proportional to the square root of energy, hence amplitude is proportional to r-1

Absorption and scattering also decrease the energy in the seismic wave as it passes through the Earth

This decrease in energy with increasing travel time (so-called attenuation) can be written as:

Three effects combined

Page 21: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

White = primaryRed = source ghostBlack = multiple

Page 22: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

Attenuation - decrease in amplitude and frequency content

Multiples and ghosts - wavelet extended

Page 23: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

Seismic reflection processing

Raw data contains noise, refractions, S-wave reflections, surface waves, multiples

The aim of processing is to try to remove all of these and leave in only the primary reflections

Page 24: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

Filtered trace

Filtered trace

Raw data from a broad-band

seismometer

Page 25: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

StaticsIf sources and receivers are not horizontal then reflections from interfaces will not line up correctly in the time section.

After applying statics, horizontal reflectors will appear horizontal

Page 26: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

tx

v

z

vor t

x

vtx x

2

2

2

2

22

2

2 0

24 t

z

v0

2

Reflections are hyperbolic

Stacked trace gets plotted at x position that lies at midpoint of traces being stacked

Page 27: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

NMO t tx

V tx 0

2

2

02

Test velocity (2-5km/s)

Pow

er o

f st

acke

d tr

ace

at e

ach

velo

city

(A

mpl

itude

2 )

Pick velocity that gives highest amplitude (white dots)

Multiples and S-wave reflectionshave lower velocities thanprimaries

Refractions are not hyperbolic

Semblance plot

Page 28: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

Deconvolution

Run a autocorrelation on the seismic tracesto determine the effective wavelet (The wavelet at t = large)

Design an inverse filter (the deconvolution) to convert the wavelet back

Page 29: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

As the distance between two reflectors decreases, their reflections start to overlap, and eventually do not appear as two separate reflections.

The minimum vertical resolution (the closest two reflectors can be and still be identified as two separate reflectors) is assumed to be ~ ¼ seismic wavelength

A reflection does not arrive from one single reflection point – but from circular area. Thus small gaps in a reflector will not be obvious in the seismic data

The minimum horizontal resolution (the smallest observable gap) is assumed to be ~ the width of a Fresnel zone

21

2

1

zw

Page 30: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

Relatively new developments

4D time lapse

Survey 1985and 1999

Track changes in reservoir

Page 31: Last week’s problems a) Mass excess = 1/2πG × Area under curve 1/2πG = 0.0239 × 10 11 in kgs 2 m -3 Area under curve = -1.8 ×10-6 x 100 m 2 s -2 So Mass.

State-of-the-art

Permanent arrayson sea bed

Real-time monitoringof reservoir

e.g. Hydro-fracturing