Tuesday, December 1, 2009

Why you should not use heat flow as input in your basin model

It is still common practice to use heat flow as input to basin models. It is really a bad practice, especially when the heat flow is supplied at the base of the sediment column.  Modelers usually fit a heat flow to match temperature data and then use the same heat flow in the kitchen or even over geological time. The problem is heat flow is a function of deposition rate (so called transient effects) , which changes laterally (the kitchen area usually has higher sedimentation rates), as well as in time. Yes I am talking about basement heat flow. We recommend using 1330 °C at base of lithosphere as the boundary condition. This will automatically determine the heat flow in the kitchen and its variation in time. Here are a couple of examples:


This figure shows heat flow (at base of sediment column) vs time for the Gulf of Mexico deep water areas. The rapid drop in heat flow in Miocene is caused by rapid deposition. By assuming 1330 °C at base of lithosphere, basin models will automatically determine heat flow based on sedimentation rate as well as the conductivity of the rocks being deposited. Faster deposition rates with a lower conductivity rock will depress heat flow more. Heat flow will slowly equilibrate to steady state if there has been no deposition for about 40 million years.


The second example here is from North Africa, which has undergone significant uplift and erosion during the Tertiary. Heat flow calculated based on 1330 °C at base lithosphere shows how heat flow increases during periods of erosion.
You may check out this earlier post to see how this approach can help determine heat flow in the kitchen area without wells. In most situations, vitrinite reflectance data (and other thermal indicators) are not sensitive enough in determining the paleo-heat flow as deeper burial at present day overprints any impact of cooler temperatures in the past.

10 comments:

  1. Are you saying that for the entire range of typical sedimentation rates, if you
    spontaneously stopped sedimentation, it will always take 40 million years before
    a steady state is achieved ? How large perturbations of Q vertically (let say from 3-5km) in the column are you calculating for let say a sedimentation rate of 1km per 50million years of mud ?

    I am curious about the details of those calculations. How is your lithosphere behaving in this model, relative e.g., stretching / compression and sediment load ? (As I understand you, you assume that the top of the asthenosphere is buffered by crystallization/melting (latent heat) and if that interface does not move too fast that may be OK.) What kind of heat capacities are you assuming for the sediments and basement part of the lithosphere ? I have a hard time seeing how you relate
    heat flow into the sediment column completely by deposition rate and sediment thermal properties in this manner (what are the "couplings" and balances here ?). Is your basement lithosphere all the time of a fixed thickness ?

    At first glance it looks slightly like a brute-force approach ?

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  2. The perturbation in theory will take infinity to equilibrate regardless of magnitude, like any diffusion model. After 40 my, the disequilibrium is reduced to less than 5% of the initial difference.

    2 km of mud deposited in 1 my will reduce the heat flow from 60 to 35 mW/m2 which is about what we observe in the GoM.

    The boundary condition is a fixed 1300 °C temperature at base lithosphere at about 120 km. The temperature, and instantaneous heat flow, in the sediments and rest of the lithosphere are solved using a finite difference model. In the GoM case, about 20,000 ft of sediment is deposited in about 10 million years. However, the sediment mainly replaced the salt that was there, and the thickness of the lithosphere remained about the same.

    In a passive margin basin, often the lithosphere thickness increases due to cooling while sediment is added.

    Cheers,

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  3. Unfortunately, I am still confused and worried about your results and statements. There are still many unclear and strange issues. But before moving on to those, we need to clarify your first point because without that we will not get anywhere. I do not see where you are going with "The perturbation in theory will take infinity to equilibrate regardless of magnitude, like any diffusion model.".

    How does the theory in any way prohibit a transient heat diffusing system to reach a steady state ("equilibrium in an open system") after a finite amount of time ? (and with many properties and settings; virtually no time at all ?) Sure there are theoretical buggers with the heat equation (Infinite transmission speed: see e.g., Luikov (1966) in Intl. Jour. Heat & Mass Transfer.) but nothing relevant to applied use at our time-scale, or which is not eliminated by our numerical methods.

    Please clarify.

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  4. The equilibration process slows down as the disequilibrium diminishes. I will add a separate post with a picture to show.

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  5. What you are saying here is not in dispute: a non-steady gradient will decay slower with time, but
    that has absolutely nothing to do with your statement that I quoted: "The perturbation in theory will take infinity to equilibrate regardless of magnitude, like any diffusion model.".

    Theoretically, a diffusion problem is a parabolic problem; with fixed boundaries it will reach a steady state at a finite amount of time. An everyday example would be heating a brick on a stove. After a finite amount of time the thermal gradient will stabilize, and no longer change. What you said would imply it would never develop a steady gradient, i.e., the gradient would never stop changing ! The statement of yours which I quoted is fundamentally incorrect. I just wanted that to be clear before we try to dig deeper.

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  6. Sorry for not replying. I admit that statement was a bit outlandish without any support information. So let me try to explain by this example. Say that we have a U shaped glass tube, with a plug of permeability K, and length L at the bottom. If we fill the tube with different water levels on two two sides. We can relate the rate the water level equilibrates with the Darcy's law:
    dh/dt = -K* h/L
    where h is the difference in water level. Solving this gives the difference in water level after time t.
    h = H*exp[-Kt/L]
    where H is the initial water level difference. Yes, practically after a finite period of time, the water level is essentially the same if the permeability is high enough. But if we ask the question, how long does it take to equalize the water level (h=0)? The answer is infinity. I think this also applies to the question of how long will it take to dissipate overpressure?

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  7. Whoops, there should be a 2 in there. With a U tube, it should be twice as fast: h = H*exp[-2Kt/L]

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    ReplyDelete