# Fluid bulk modulus (Hydraulic fracturing)

Hey all,

Here is a conceptual question:

Lets assume that there is a DFN model (particles that are attached to its neighboring particles via springs) and we inject fluid into one of the pore. Now the injected fluid will cause a change in the pore volume. This change in the pore volume will convert to a change in pore pressure via the fluid bulk modulus. The change in pressure will cause the pore to expand.

Now if the fluid bulk modulus was small (1e5 N/m^2) then the pressure would be small and the pore would not expand much. However, if the fluid bulk modulus was big (1e10 N/m^2) then the pressure would be big which will cause the pore to expand alot. However, because we inject fluid in the pore, we know the exact new volume of the pore.

So how do we get the pore volume to match with the 'theoretical' pore volume without using the fluid bulk modulus as a calibration parameter?

## Question information

- Language:
- English Edit question

- Status:
- Answered

- For:
- Yade Edit question

- Assignee:
- No assignee Edit question

- Last query:
- 2020-09-03

- Last reply:
- 2020-09-04

Robert Caulk (rcaulk) said : | #1 |

I do not understand the question. Tips for improvement:

Does this actually have anything to do with hydraulic fracturing?

Clarify what you mean by "theoretical" pore volume. Are you comparing to an analytical solution? If so, specify which one.

Indicate the problem that you face in Yade.

>So how do we get the pore volume to match with the 'theoretical' pore volume without using the fluid bulk modulus as a calibration parameter?

Why is the fluid bulk modulus a calibration parameter? Why not use the real fluid bulk modulus?

mrhappy (mrhappy) said : | #2 |

Suppose you have a pore that has a volume 0.5 mm^2 (its 2D therefore it's an area). This volume is calculated based on the physical area of the pore. I pump in at a rate Q = 0.05mm^2/s for one time step deltaT = 1 sec. Therefore this pore now has a volume of 0.5 + 0.05*1 = 0.55 mm^2. The change in volume is 0.05mm^2. This change in volume will be converted to a change in pressure via the fluid bulk modulus.

Now is if the fluid bulk modulus is high, then the pressure will be large which will cause a large deformation. This large deformation will cause the pore to have a physical area greater than 0.55mm^2 (the calculated volume after pumping). However, if the fluid bulk modulus is small, then the pressure will be small which will cause a minute deformation. This small deformation will cause the pore to have a physical area smaller than 0.55mm^2.

I can recalculated the bulk fluid modulus to give me the new physical area of the pore to be 0.55mm^2.

Is my understanding correct?

Jan Stránský (honzik) said : | #3 |

Hello,

I have no specific background, but your explanation seems incorrect to me.

> pump in at a rate Q = 0.05mm^2/s

does this mean that you add 0.05 mm^2/s of new liquid inside the pore?

Below I assume that yes.

> this pore now has ... 0.55 mm^2. The change in volume is 0.05mm^2. This change in volume will be converted to a change in pressure via the fluid bulk modulus.

No. The change of volume of the original material causes stress difference, not change of volume of original + added material.

Also, If you have pore and you enlarge it, how can it convert to pore pressure? Normally it would lead to pressure decrease / vanishing.

Or you mean that the new liquid causes pressure? It would be the case for zero (or less then added) volume change (i.e. preserving 0.5 mm^2 also for the new liquid). But if you have original 0.5 volume, you add 0.05 volume and in total you have 0.55 volume, there is no reason for pressure change..

Am I missing something?

In reality, the surrounding structure does not allow the "pressure-free" expansion, making the real volume less than 0.55.

Then yes, the difference between actual volume and "stress-free" volume (dV) causes pressure. And the proportionality parameter between dV and pressure is liquid bulk modulus.

cheers

Jan

Robert Caulk (rcaulk) said : | #4 |

I think you are confounding density changes, flow rates, and physical volume changes. Nearly impossible to discuss without a base level of vocabulary. Try reading through [1] and then reframe the question with respect to yade. I have a feeling that will help everyone.

Cheers,

Robert

mrhappy (mrhappy) said : | #5 |

Let me get back to you on this.

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