Hydro-mechanical Calculation

I try to understand how 3DEC performs hydro-mechanical calculations. From manual of 3DEC 5.20 I understood a little, but I will be grateful if someone can simply explain the computation loop from start point to the end point. For example, if a constant injection into a flow-know is prescribed, how 3DEC deals with this problem to obtain new values of hydraulic aperture, fluid pressure, flow rate and so on along the flow plane in the next time-steps?

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It is probably better to ask the question step by step. If a constant flow rate of Q is prescribed to a flow-knot, an extra fluid pressure, in consequence, will be generated which will change the flow-knot volume as well depending on the bulk modulus of the fluid as \deltaP=\K_f*\deltaV/\V_0.
\deltaV\ at this problem will be Q (assuming that the time-step is 1 second). Then new P will be old P plus \K_f*Q/\V_0.
If it is so, then what is the third term in right hand side of equation (4.18) at page 4-12 of “Fluid in 3DEC” version 5.20 manual (or equation 18 on webpage: Joint Fluid Flow — 3DEC 7.0 documentation)? and why the denominator of the second term of RHS is V instead of V_0?
Any help will be highly appreciated.

The third term is change in pressure caused by a change in volume (a decrease in volume increases the pressure, and vice versa). This term is calculated first, then the volume is updated, and the second term is calculated using the updated volume.

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Thanks @jhazzard

If the third term is change in pressure caused by change in volume, then what is the second term? The reason behind my question is that the second term itself is change in pressure caused by a change in volume because Q*deltat is in fact change in volume.
And why the denominator or third term is average volume instead of initial volume (based on the definition of bulk modulus)?

The second term is the change in pressure due to the change in fluid volume. The third term is change in pressure due to change in mechanical volume.


Understood! Many thanks @jhazzard.

One last question, what is the reason behind use of average flow-knot volume in denominator of the third term instead of initial flow-knot volume?

Hi @DariushJ !

My guess is that this is due to the rightmost term refering to a relative change in pore space that requires a reference volume.
There is no physical reason for having V0 or V as the reference and a complete solving would require non-linear equations, hence drastically dampening computation times.
To keep things simple, you then have to select a reference volume that remains consistant with the physics at stake. You could choose V0 or V as the reference but since we are dealing with linearized equations, the mean volume between the two timesteps seems to be a reasonable and less impacting choice.

Just my thoughts!


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Thanks @Theophile for sharing your thoughts.