The Transient Electromagnetic (TEM) Method, sometimes referred to as Time domain EM (TDEM)

Transcription

1 The Transient Electromagnetic (TEM) Method, sometimes referred to as Time domain EM (TDEM)

2 From diffusion to wave E z, 100 ohm-m, K=30 Induction (EM, MT) Wave (GPR)

3 The basics Primary field Secondary field I Current flow moment = current x area x winds of wire

4 Typical ground based configurations Wind mils

5 Induction in the Earth Secondary magnetic field Induced EMF Current

6 TEM response geologic structure

7 Noise Natural background noise Noise from the power distribution grid

8 Noise Natural background noise Noise from the power distribution grid

9 Sensitivity relative contribution from various depths

10 Fields in the ground full space, plane wave Diffusion depth Z max =(t/) 1/

11 Penetration depth Diffusion depth z d t 1. 6 t Diffusion time t d z d 0.68 zd

12 Half space curves t 5/ voltage decay Resistive Earth o Small response at early times o Large response at late times Conductive Earth o Large response at early times o Small response at late times layer Earth 50 m 100 ohm-m 1 ohm-m

13 High resistivity equivalence Thickness resistivity product kept constant Modeller 1 = 3 m t 1 = 16 m = m t = 0,5 0, m 3 = 8 m Models Rhoa [ohmm] Resistivity Geoelektriske modelkurver Højmodstandsækvivalens L/ [m] 5% STD 5% STD TEM TEM mode 1E-004 Time

14 Low resistivity equivalence Thickness divided by resistivity kept constant (conductivity thickness product) Models 1 = 18 m t 1 = 16 m = 0, m t = m 3 = 51 m Modeller Rhoa [ohmm] Geoelektriskemodelkurver Resistivity 5% STD Lavmodstandsækvivalens , L/ [m] ,5 TEM modelkurv 5% 1E-004 1E Time [s]

15 Resistivity equivalence Resistivity changed in layer Models Modeller 1 = 3 m t 1 = 16 m = m t = m 3 = 8 m Rhoa [ohmm] Geoelektriske Resistivity modelkurver 5% STD Modstandsækvivalens TEM modelk 5% STD L/ [m] 1E-004 Time [s]

16 Layer supression decending Resistivity decreasing with depth thickness resistivity product constant Modeller Models Geoelektriskemodelkurver Resistivity TEM modelkur 1 = 51 m t 1 = 3 m = m t = m 3 = 8 m 1 = 51 m t 1 = 3 m = 8 m Rhoa [ohmm] % STD Lagundertrykkelse, dobbelt nedadstigende 5% STD L/ [m] 1E Time [s]

17 Layer supression ascending Resistivity decreasing with depth thickness divided by resistivity constant Modeller Models 1 = 3 m t 1 = 16 m = m t = m 3 = 51 m 1 = 3 m t 1 = 16 m = 51 m Rhoa [ohmm] Geoelektriske Resistivity modelkurver L/ [m] 5% STD Lagundertrykkelse, dobbelt opadstigende TEM modelkurve TEM 5% 1E-004 1E- Time [s]

18 TX Coupling in TEM Data Distortion response due to transmitterinduced currents in manmade conductors Cannot be removed by stacking Two types of coupling o capacitive LCR circuit o galvanic LR circuit

19 Capacitive Coupling V Capacitive return path L C C R Where? o long isolated wires, telephones cables, buried cables LCR oscillating, decaying circuit db/ dt (V/ m ) TEM coupled undisturbe d Time (s)

20 Capacitive coupling field example db/ dt (volt/ m ) db/ dt (volt/ m ) db/ dt (volt/ m ) time (s) time (s) time (s) profile db/ dt (volt/ m ) road profile tics

21 Galvanic Coupling L Galvanic return path R V LR decaying circuit Where? o high voltage wires, metal road guards and fences db/dt (V/m) TEM coupled undisturbed Time (s)

22 Galvanic coupling field example db/ dt (volt/ m ) db/ dt (volt/ m ) db/ dt (volt/ m ) time (s) time (s) time (s) profile db/ dt(volt/ m ) profile tics

23 Galvanic coupling inverse modelling 10-6 Voltage data normalized db/ dt (v/ m ) S1 S time (s) depth (m) Inverse models S S resistivity (ohm-m)

24 Coupling Distance Dependency Infinite wire 1/a (Tx to wire) * 1/a (wire to Rx) = 1/a Finite length wire (grounded dipole) 1/a (Tx to wire) * 1/a (wire to Rx) = 1/a 4 Small loop 1/a 3 (Tx to wire) * 1/a 3 (wire to Rx) = 1/a 6

25 Basic equations Vertical magnetic dipole Schelkunoff vector potential Field is obtained by differentiating d J u e r e m z z F h z u h z u ) ( 4 ˆ ), ( ) ( TE z F z k z z H ˆ 1

26 The forward response schematic d J u e r e m z z F h z u h z u ) ( 4 ˆ ), ( ) ( TE 0 0 0

27 Basic equations For a circular loop on a half space u zh u (zh) e r e J ( a)d Ia 0 0 H z 0 TE 1 u H z Ia 0 J1 u ( a) d 0

28 Basic equations By inverse Laplace transform in the time domain a) erf( a 3 1 e a 3 a I b a 1 0 z a 1 3 z e a 3 a - a) 3erf( a I t b

29 Half space curves Half space response from previous Late times z a t dt b 0 Ia a 1 3 z e a 3 a - a) 3erf( a I t b z t 0 a I t b

30 An airborne hydrogeophysical study of San Christobal, Galapagos slands Auken, Esben, University of Aarhus, Denmark; N. d Ozouville, S. Violette, Sophie, & G. de Marsily, Université Pierre et Marie Curi rance; B. Deffontaines, Universtié de Marne-la-Vallée, France; A. Vizello & K.I Sørensen, University of Aarhus, Denmark

31 Example : Galápagos Archipelago 13 islands, 61 islets Total land area: 7880 km Population: 7,000 inhabitants Tourism: 10,000 per year San Cristobel Fishing: > 1000 fishermen

32 DEM San Christobal Faults mapped from DEM 714 m 18 x 30 km 0 m

33 Flight Lines Flight lines 18 x 30 km ~800 line km collected over 4 days

34 Average Resistivity 0 0 mbs 18 x 30 km

35 Average Resistivity mbs South-Eastern tradewinds

36 Average Resistivity mbs

37 Average Resistivity mbs 18 x 30 km

38 Average Resistivity mbs 18 x 30 km

39 Cross Section 18 x 30 km

40 Cross Section Fault and basis of springs Unweathered basalts S Newer volcanic structure Saltwater N

41 Cross Section 18 x 30 km

42 Cross Section S Unweathered basalts Fault N Newer volcanic structure Saltwater

43 3D Animation