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Lithospheric Plates The lithosphere can be defined thermally by an isotherm at the base of the lithosphere which should be around 1350 o C. Mantle rocks below this isotherm are cool and behave rigidly Rocks above this isotherm are hotter and may deform How are plates created ?

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Heat Flow Through the Oceanic Lithosphere Measuring heat flow on the seafloor Thermal Conductivity profiles in sediments

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Heat Flow Through the Oceanic Lithosphere Heat flow measurements on the seafloor are found to decrease steadily with increasing distance from the spreading center. Deviations from a theoretical curve for heat flow indicate complications such as hydrothermal circulation.

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Heat Flow (Q) Log Heat Flow (Q) Age (Ma) Log Age (Ma) Heat Flow Above the Oceanic Lithosphere Heat flow is highest for what age lithosphere ? How does heat flow vary with seafloor age ? Is it linear ? What kind of mathematical expression does this resemble ? On a log-log plot, the data falls on a slope of -1/2. How can you express this in an equation ?

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Heat Flow Above the Oceanic Lithosphere Log Heat Flow (Q) Log Age (Ma) log Q = log A -1/2 Q ~ 1/sqrt(t) Heat flow is inversely proportional to the square root of age (time). Lithosphere is defined by heat flow Why does heat flow decrease with age ? log Q = -1/2 log A (y = mx) (1) 10 (log Q) = 10 (log A -1/2 ) (2 ) Q = A -1/2 (3 )

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Depth of the Lithosphere (Topography or Bathymetry) Sonar techniques taken from ships passing over the ocean surface measure the water depth to the seafloor with excellent accuracy (at least within a few meters).

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Depth of the Lithosphere (Topography or Bathymetry) Seafloor bathymetry across a spreading ridge is shown here. Do you notice any differences between the Atlantic (A) and Pacific (E) spreading centers ? What causes these differences ?

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Depth of the Lithosphere Depth (m) Age (Ma) The depth to the seafloor (from sea level) is not constant... The seafloor is shallow at the rise axis and gets deeper away from it. Why is this ?

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Depth of the Lithosphere Depth (m) Sqrt(Age (Ma)) Plotting seafloor depth versus the sqrt(Age) shows a slope of -1/3 – regardless of spreading rate ! How can we write the equation that describes this ? d = -1/3 sqrt(A) (1) d is proportional to sqrt(A) -or sqrt(time) d ~ sqrt(t) The lithosphere is described by it's depth in the ocean What does this mean ?

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Depth of the Lithosphere Depth (m) Sqrt(Age (Ma)) The lithosphere can be described by conductive cooling Other factors have a smaller effect (upwelling volcanism at spreading axis etc...) How does thermal conductivity vary over time ? The time-dependent heat conduction equation – (notes on in class and on board)

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Class notes on board...

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Time-Dependent Heat Conduction dT/dt = d 2 T/dx 2 (Known as the “Heat flow Equation”) Where = k/ C p is thermal diffusivity (m 2 /s). describes the diffusion of temperature or heat across a body of material

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Time-Dependent Heat Conduction dT/dt = d 2 T/dx 2 Charcteristic diffusion time (t) can be described using where t = d 2 / This gives the time for heat to diffuse across a distance, d.

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Time-Dependent Heat Conduction dT/dt = d 2 T/dx 2 Charcteristic diffusion distance (d) can be described using where This gives the distance temperature will propogate through the material in a given time period. d = sqrt( t

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Time-Dependent Heat Conduction dT/dt = d 2 T/dx 2 Charcteristic diffusion distance (d) can be described using where This gives the distance temperature will propogate through the material in a given time period. d = sqrt( t

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Activity Zhao et al., 1997 P wave tomography image of the Tonga trench subduction zone High velocity subducting slab is clearly visible (blue) extending down to at least 660 km depth.

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Activity Seismic tomography image of the Pacific plate subducting beneath Japan. Scientists argue about whether subducting plates penetrate through the 660 km discontinuity into the lower mantle. Fukao et al., 2001

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Activity Some authors say some slabs just rest at the 660 and may “thermally assimilate” over time. Calculate how long it would take such a slab to thermally assimilate. Use the thickness of the slab you observe in the images above Assume thermal conductivity of peridotite, k = 3.0 Wm -1 K -1, density = 3250 kg m -3, and heat capacity, Cp = 0.8 kJ/kg K Fukao et al., 2001

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