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I'm looking to calculate the total stress at a depth of 10 km in a rock column, where porosity varies with depth.

I believe I can figure this out, however I might have copied down an equation wrong. In the equation for total stress I need to know the density of the sediment, so I have to calculate that first. The equation in question is as follows:

Density of Sediment = Density of water * porosity + density of grain * (1 - porosity) * .052 m3/s

The .0052 m3/s is whats throwing me off, as I end up trying to add the units kg/m3 to kg/s, which can't happen. What I'm thinking is that he might have said "Cubic meters per section" or I'm just way off. Any help would be appreciated

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  • $\begingroup$ I'll keep it to your one relevant question. "Section" as in the 10 km rock column has differing sections of porosity, becoming less porous as depth increases. If you'd pay attention, you'd have noticed that I didn't ask what porosity is or how to calculate it. I asked about the appropriate units for the last term in the equation. So the equation would look something like: Kg/m^3 * dimensionless quantity + Kg/m^3 * (1 - dimensionless quantity) * m^3/s. This leaves me stuck at Kg/m^3 + Kg/s. I'm wondering what the last term in the equation is supposed to represent and if my units are incorrect. $\endgroup$ – user327 Feb 28 '17 at 23:41
  • $\begingroup$ Calculating stress at the bottom of 10Km tall column of rock requires knowing what this column weighs. For that, you'd need acceleration due due to gravity, in units of m/sec*2. Maybe you heard "section" when he said "seconds"? $\endgroup$ – Knob Scratcher Mar 1 '17 at 0:48
  • $\begingroup$ That's what I'm questioning. I thought I heard "seconds" when he possibly said "sections". He's not a native speaker of English and I've never had a problem understanding him, but it's not beyond the realm of possibilities that he said something I misconstrued. But I'm agreeing with your line of thinking. The total stress I'm looking to calculate would go as follows: total stress = density of sediment * gravity * column of sediment. The only variable I haven't found out is the density of the sediment. I need to find this out with only water density, grain density, and porosity given. $\endgroup$ – user327 Mar 1 '17 at 4:35
  • $\begingroup$ I have no notes or recollection as to what the last term, 052 m3/s, represents. I apologize if I'm being vague, but I would like to complete this class work by my own means. The professor does not post any class content on the university course management website and the book has no information on this either. So I have no idea whether or not that equation is correct. I realize it's not an easy task to decipher all this, regardless, I thank you for your effort. $\endgroup$ – user327 Mar 1 '17 at 4:45
  • $\begingroup$ Calculating effective stress at the bottom of a column whose density varies is relatively complicated. You could assume two different end members: one at 0 percent porosity and another at 100 percent porosity (solid rock vs. just water), and interpolate a value based on typical porosities between 3 and 8 percent. $\endgroup$ – Knob Scratcher Mar 1 '17 at 5:11

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