Rock Exhumation Exercise

Stephen Hurst
University of Illinois
Department of Geology
1301 W Green St.
Urbana, Il 61801
shurst@illinois.edu

An exercise to build a one-dimensional STELLA model of heat flow during rock uplift and exhumation. The model exemplifies the Laplacian world view which simplifies this model immensely. difficulties in building the model are typically related to boundary conditions, time steps, and correct use of units.

Goals are to see how a change of view can simplify modeling, how to analyse graphs and see how feedbacks create a non-linear system.

Rock Exhumation Exercise - Background
Pressure - temperature paths in the Earth and what they mean.

Many rocks currently at the surface of the earth have been buried many kilometers, sometimes up to 40 or so km. For example the rocks atop Mt. Everest are deep-sea limestones that have been uplifted therefore at least 12 km (8 km currently above sea level, plus 4 km to the bottom of the ocean, plus whatever depth they were once buried to). A nearby fault brings metamorphic rocks to the surface that have come from even deeper down. Geologists have been trying to unravel how, and how fast rocks are uplifted and exposed (called exhumation) to get a better understanding of how the earth works. Recent analytical advances have allowed crude paleotemperature and paleo-pressure measurements to be made on some kinds of rocks. Together, these measurements can be plotted on what are called P-T-t (Pressure-Temperature-time) graphs to give a history of the rocks.
 Fig. 1 - a Pressure-Temp graph showing a clockwise (typical) path for an uplifted rock.

If we make the assumption that the time to take a step along the path is constant, then we interpret this as first a quick decrease in pressure, then a fast decrease in temperature. Many people also assume that the geothermal gradient is constant and linear at about 30° per km (average for continental crust). If so, then the fast decrease in temperature then indicates a very fast exhumation rate, often much faster than seems reasonable. For example the Nanga Parbat region of the western Himalaya has been said to be exhuming at rates of 20+ mm/yr (20 km of exhumation/million years). Although erosion rates (which is one of the main causes of exhumation) are some of the highest in the world, the highest rate is still unlikely.

Why rocks are uplifted and exhumed is not the topic of this exercise, but basically crustal rocks are lighter than deeper mantle rocks and therefore “float”. In areas of uplift, generally crustal rocks are thicker than normal because of thrust faulting and igneous activity.

So the object of this exercise is to model the geothermal gradient and the P-T-t path of a typical rock in the crust in an area of relatively fast exhumation. This may be difficult to model in what we think of as a common sense Stella way (Eulerian method). 

So, some more background first:

This shows the standard graphs that depict the data and inferred exhumation path.

If we allow a non-linear geotherm then the exhumation rate can be quite different.

The reason for having a non-linear geotherm in an uplifting terrain is “advection”. In addition to rocks conducting heat they also carry it with them. So although heat is lost through the surface of the earth, rocks are very good insulators and carry most of it with them wherever they go. So any model of uplift and exhumation must take into account both conduction of heat and advection.

Therefore advection has the effect of shifting the geotherm (flattening it in the right two graphs above. If you don’t take it into account then you will tend to overestimate the rate of exhumation.

The appropriate equations for an analytical or numerical solution of the problem are those above. For our purposes we will ignore heat production (the last term in the equation above) in the rocks themselves. Also we will make this a 1-D model and ignore topography caused by the uplift (although more advanced modeling shows that in areas of fast uplift the topography does make a difference).

The Exercise

The objective of this exercise is to follow a piece of rock from deep in the crust (say start at 20 km) and model the temperature and pressure history of the rock given that it rises at various rates. Plot the geotherm for that area where the rock is exposed at the time just before the rock is exposed at the surface.

You will have to keep track of the depth of the piece of rock and its temperature (really Heat but we are going to convert to T on the fly). Worry about units - uplift rates are in mm but depths are in km or meters.

Start the rock (its bottom) at about 20 km depth and bring it to the surface. Assume the temperature at 20 km depth is 1000 ° C (1273 °K) and the temperature at the surface is 0° C. Run the model for several uplift rates (in the range of 1 to 100 mm/yr) and plot Temperature vs. Depth as above. Note any significant differences with rate and in particular, at what point the uplift rate might cause the inferred exhumation rate to be higher than is actually true.

Run the model for several different thicknesses of rock (why do we need a rock thickness in this problem?), say from 500 meters to 5 kilometers. Why does this make a difference? Does it negate the use of the model in the real world?

For rocks: K = thermal diffusivity = 10- 6 m2s-1 (note units)

Hints: This is a one-dimensional problem. Keep track of heat entering the rock from below and leaving it from the top in one section. Keep track of the depth of the rock in another. 

Example: Heat out top = thermal diffusivity /rockThickness*(T - T_surface)/(rockDepth - 0)

Note that this whole part on the right of the equal sign is the diffusion part of the equation above. We are going to ignore heat production and the STELLA model will take care of the advection. So the equation above is all that is needed for one flow.

End points are going to be problems in this model. Note that there are possibilities of dividing by zero (or even negative numbers) unless you are careful. You are going to need a lot of DT steps – STELLA limits you to 32,000 steps. This is related to the end-point problem.

Make a Temp vs. Depth plot – this is different from the usual Time series plot but is an option in STELLA<