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I came across a question in my notes about why the actual thickness of permafrost is much less than the theoretical maximum possible permafrost thickness value.

The actual permafrost measure on sites were much less thick than the maximum possible thickness.

So the actual permafrost thickness had values ranging from 0-15m. But the maximum possible permafrost thickness had values froom 75, to 375m. We determined the maximum possible thickness by using the formula $z=(T_s)(k/q)$. With $T_s$ which was -3 and we took the absolute value of this so 3. Then K was the thermal conductivity $W/m$ and Q was the heat flux which was $.04w/m^2$. So why was the actual thickness much less than the theoretical max possible? Clearly the max possibly max is going to be higher than the actual thickness because this gives a end range for the thickness of the permafrost but besides that I am not quite sure

Edit: as an example, the following values are for Newfoundland and Labrador. The maximum possible thickness was calculated by that equation I gave up above.

Material, Max Possible Thickness, Actual Permafrost Thickness:
granite, 75m, 0m
sandstone, 150m, 1m,
glacial till, 113m,1m
esker gravel, 188m,0m
marine silt, 375m,15m
river sand, 225m, 0m
fen peat, 300m, 10m

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  • $\begingroup$ It would be great to have more information. With the parameters you provide I am not sure how to obtain 75 to 375 m for the depth. Also - I have never seen a permafrost body with a mean annual surface temperature of -3 °C reaching 375 m depth. Let's set the numbers straight before going further. What area is this? $\endgroup$ – Etienne Godin Oct 22 '16 at 0:41
  • $\begingroup$ Permafrost can be thinner under lakes if they are deep enough not to freeze to the bottom. $\endgroup$ – Keith McClary Nov 17 '16 at 4:29
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I could not recover my account so here is some additional information for you. This area is Newfoundland and Labrador. The maximum possible thickness was calculated by that equation I gave up above

Material, Max Possible Thickness, Actual Permafrost Thickness:
granite, 75m, 0m
sandstone, 150m, 1m,
glacial till, 113m,1m
esker gravel, 188m,0m
marine silt, 375m,15m
river sand, 225m, 0m
fen pea,t 300m, 10m

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  • $\begingroup$ @ Etienne Godin $\endgroup$ – user6834 Oct 22 '16 at 12:35
  • $\begingroup$ Hello, and welcome to Stack Exchange. I've edited this text into your original question. $\endgroup$ – Daniel Griscom Oct 22 '16 at 14:48
  • $\begingroup$ @ Etienne Godin we were actually given the values of k too sorry, they are granite 1, sandstone 2, glacial till 1.5, esker gravel 2.5, marine silt 5, river sand 3 and fen peat 4 $\endgroup$ – user6834 Oct 22 '16 at 15:25
  • $\begingroup$ but yes this formula is leaving out a lot of other factors that's what I was thinking but just wanted to check $\endgroup$ – user6834 Oct 22 '16 at 15:34
  • $\begingroup$ OK but if you have those k values, the equation cannot balance with these huge maximum depths you provided. For instance if we take granite, k=1 give z = 3 x 1 / 0.4 = 7.5 m, which seem reasonable. Consistent with sandstone at 15 m. Maybe you made an error with the decimals when initially calculating ? $\endgroup$ – Etienne Godin Oct 22 '16 at 17:06
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I resolved the following based on your input:

$z = (T_s)(k/q)$ for the granite case,

where

$75 = (3)(k/0.04)$ where $k = 1$

so based on this site here, typical granite thermal conductivity is in the range of 1.7 to 4 $W/m K$.

We could say your site is special with $k=1$ maybe local conditions..

Let's test the Esker gravel, where the aforementioned site propose $k=0.7$ for a typical value for gravel. Let's see your model

$188 = (3)(k/0.04)$ where $k = 2.5$

to balance the equation here you need a value of $k$ that is larger, can't be local factors explaining that.

So how to approach this ?

  1. the soil temperature $T_s$ cannot be fixed. It should vary a lot depending on your environment, and also even in a single environment you could have variations when considering other factors such as the moisture, the slope, the albedo, the snow conditions, the vegetation (eg: n-factors)
  2. the heat flux at 0.04 is usually a regional estimation. While not impossible, I found this value low. In the Appalachians (Chic-Chocs) values where estimated to be around $2 W/m^2$
  3. You need better estimations for $k$ for this this to work

Overall, your equation to resolve the depth is quite simple and do not take into account a lot of factors, and some important parameters which are generalized /estimated broadly weaken your results.

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