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Modelling of subsidence in an Underground Coal Gasification

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I am currently working on the effects of underground coal gasification on the nearby strata. I have been trying to simulate the same using COMSOL. I made a model consisting of a host rock (1000m x 1200m) and a rectangular cavity (7.5m x 100m) within it, signifying the cavity left after burning of coal.

1. First I made the model having no cavity and allowed it to settle on its own by gravity. The top boundary of the model was kept free and roller were placed on the remaining three boundaries. The total displacement noticed was 15.7 mm downwards.
2. Next I simulated the cavity and noticed the strata behaviour. It gave expected results showing a subsidence on the surface.
3. Now, to simulate the heating caused by burning of coal, I places four temperature source of 1273.15 K at the boundaries of the cavity.

The problem started here as the whole model started behaving like a single body showing thermal expansion. The expansion was visible on the top boundary of the model, as the other three boundaries were simulated with rollers. The total displacement of the top surface is about 750 mm upwards.

I think, the problem lies in the type of heating, the coupling of physics and/ or in the boundary conditions. I have tried every other combination, but nothing seems to work. There is an upheaval on the top surface no matter what.

It would really helpful if someone helps me out on this.
.................................
Nikhil Ninad Sirdesai
Research Scholar
Department of Earth Sciences
IITB-Monash Research Academy
IIT Bombay,Powai
Mumbai - 400076.
Mobile:91-9594972974

3 Replies Last Post 2015年4月26日 GMT-4 09:22

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Posted: 9 years ago 2015年4月25日 GMT-4 23:08
You specify the temperature in the cavity but do you have any other temperature boundary conditions? You need to specify an outer boundary at ambient temperature.

The thermal resistance of a shell at radius r from a point heat source is proportional to 1/r2, so integrating this yields 1/r, so the temperature on the limit of large r falls off 1/r in steady state far enough from the cavity that it appears to be a point source, at least until you hit the surface. So if you put your thermal boundary too close you may overestimate the thermal cooling. An adjustment can be to add a lumped element thermal resistor to an outer boundary and connect that to ambient if you can estimate that thermal resistance.
You specify the temperature in the cavity but do you have any other temperature boundary conditions? You need to specify an outer boundary at ambient temperature. The thermal resistance of a shell at radius r from a point heat source is proportional to 1/r2, so integrating this yields 1/r, so the temperature on the limit of large r falls off 1/r in steady state far enough from the cavity that it appears to be a point source, at least until you hit the surface. So if you put your thermal boundary too close you may overestimate the thermal cooling. An adjustment can be to add a lumped element thermal resistor to an outer boundary and connect that to ambient if you can estimate that thermal resistance.

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Posted: 9 years ago 2015年4月26日 GMT-4 01:49

You specify the temperature in the cavity but do you have any other temperature boundary conditions? You need to specify an outer boundary at ambient temperature.

The thermal resistance of a shell at radius r from a point heat source is proportional to 1/r2, so integrating this yields 1/r, so the temperature on the limit of large r falls off 1/r in steady state far enough from the cavity that it appears to be a point source, at least until you hit the surface. So if you put your thermal boundary too close you may overestimate the thermal cooling. An adjustment can be to add a lumped element thermal resistor to an outer boundary and connect that to ambient if you can estimate that thermal resistance.


@Daniel

Thanks for your input.
1. I applied room temperature condition to the external boundaries.
2. The temperature distribution looks fine. As the top boundary is still at room temperature.
3. The reason I am confused is, if the top boundary is at room temperature why is there a upheaval.

I am attaching the pictures and the report created along with this. Please help
.................................
Nikhil Ninad Sirdesai
Research Scholar
Department of Earth Sciences
IITB-Monash Research Academy
IIT Bombay,Powai
Mumbai - 400076.
Mobile:91-9594972974
[QUOTE] You specify the temperature in the cavity but do you have any other temperature boundary conditions? You need to specify an outer boundary at ambient temperature. The thermal resistance of a shell at radius r from a point heat source is proportional to 1/r2, so integrating this yields 1/r, so the temperature on the limit of large r falls off 1/r in steady state far enough from the cavity that it appears to be a point source, at least until you hit the surface. So if you put your thermal boundary too close you may overestimate the thermal cooling. An adjustment can be to add a lumped element thermal resistor to an outer boundary and connect that to ambient if you can estimate that thermal resistance. [/QUOTE] @Daniel Thanks for your input. 1. I applied room temperature condition to the external boundaries. 2. The temperature distribution looks fine. As the top boundary is still at room temperature. 3. The reason I am confused is, if the top boundary is at room temperature why is there a upheaval. I am attaching the pictures and the report created along with this. Please help ................................. Nikhil Ninad Sirdesai Research Scholar Department of Earth Sciences IITB-Monash Research Academy IIT Bombay,Powai Mumbai - 400076. Mobile:91-9594972974


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Posted: 9 years ago 2015年4月26日 GMT-4 09:22
I'm not sure about upheaval, but I recommen making the domain wider (maybe 4x) and using symmetric boundary conditions on the left and right. You've got a lot of weirdness going on there. Also I think you'll get insight looking at the trace of the strain tensor (hydrostatic strain).
I'm not sure about upheaval, but I recommen making the domain wider (maybe 4x) and using symmetric boundary conditions on the left and right. You've got a lot of weirdness going on there. Also I think you'll get insight looking at the trace of the strain tensor (hydrostatic strain).

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