Finding a rock porosity change after an acid mix treatment

Click For Summary
SUMMARY

The discussion focuses on calculating the porosity change in a sandstone reservoir after a treatment involving a 3 wt% HF and 12 wt% HCl acid mix, specifically in the context of CaCO3 dissolution and CaF2 precipitation. The participant outlines their approach to determine the volumetric dissolving power of the acids and the resulting changes in porosity due to the reactions. Key calculations involve the molecular weights of HCl, HF, and CaCO3, and the participant seeks clarification on the densities of the acids to finalize their volumetric power calculations. The discussion emphasizes the importance of understanding the stoichiometry of the reactions involved.

PREREQUISITES
  • Understanding of acid-base reactions, particularly involving HF and HCl.
  • Knowledge of stoichiometry and molar relationships in chemical reactions.
  • Familiarity with volumetric and gravimetric dissolving power calculations.
  • Basic principles of porosity in geological formations.
NEXT STEPS
  • Research the densities of HCl and HF solutions at various concentrations.
  • Study the effects of acid treatments on sandstone reservoirs, focusing on CaCO3 and CaF2 interactions.
  • Explore the use of property charts for estimating physical properties of acid solutions.
  • Learn about advanced techniques for predicting porosity changes in geological formations post-treatment.
USEFUL FOR

Geologists, petroleum engineers, and chemical engineers involved in reservoir management and acidizing treatments will benefit from this discussion, particularly those focusing on the chemical interactions affecting porosity in sandstone formations.

reed2100
Messages
49
Reaction score
1

Homework Statement


A 3 wt% HF / 12 wt% HCL treatment is injected without an HCL preflush into a sandstone reservoir containing 10% CaCO3. If half of the HF is consumed in reaction with CaCO3 to form CaF2, what will be the next porosity change, considering both CaCO3 dissolution and CaF2 precipitation? Assume that all CaCO3 is dissolved in the region contacted by the acid. The density of CaF2 is 2.5 g/cm^3.

Homework Equations


gravimetric dissolving power for 100% concentrated acid with some mineral = (mineral coefficient * Molec weight of mineral) / (acid coefficient * Molec weight of acid) = answer in lbm of mineral / lbm of acid

gravimetric dissolving power of a given concentration = 100% gravimetric dissolving power * % concentration

volumetric dissolving power X = gravimetric dissolving power at whatever concentration * (density of acid solution / density of mineral)

The Attempt at a Solution


So my mental approach is to figure out how much CaF2 is formed, and how much CaCO3 is lost in this reaction on a volume per acid injected basis.

First thing I did was find the dissolving power of the HF and HCL components. I'm taking the molecular weights of HCL, HF, and CaCO3 to be 37, 20, and 100.1. These numbers were given by the textbook.

12%HCL =.12 *100% HCL=.12* (100.1/(2*37))=.1644 lbmCaCO3 / lbmHCL
3%HF = .03*100%HF= .03*(100.1/(2*20)) = .075 lbm CaCO3 / lbm HF

Then to get the volumetric dissolving power to get the volume of mineral I can dissolve per given volume of acid. I would again do this by components. However I'm unsure about this part because I don't know the density of the HCL or HF, or the specific gravity. I'm also not given any temperature conditions. I guess I could estimate from property charts or something. But if I knew those values, I imagine it would go something like this?

Volumetric power of 12% HCL = gravimetric power of 12% HCL * (HCL density / (169 lbm/ft^3) ) = .1644 * ( HCL spec grav * 62.4 lb/ft^3 / (169 lbm/ft^3) )

Volumetric power of 3% HF = .075 * ( HF spec grav * 62.4 lb/ft^3 / (169 lbm/ft^3) )

So I was thinking that once I had each of those volumetric powers I would add them, saying that on the basis of a ft^3 of 12%HCL / 3%HF, I'd dissolve (sum of their dissolving powers) ft^3 of CaCO3.

But to account for the precipitation, for every 2 moles of HF reacting with a mole of CaCO3, I'd get one mole of CaF2. So on a molar basis, if half the original HF yields CaF2, and the ratio of yield to reactant here is .5, then I get a 4th of my HF back as CaF2.

To think about the porosity change now...if 10% of the reservoir is CaCO3 and it's all dissolved, the porosity increases by 10%. But some of that is replaced with the newly precipitated CaF2, the question is how much. Half the HF is consumed, and on a molar basis I get back half as many moles of CaF2. Then I could say... moles CaF2 * (mass per mole) / density = volume of CaF2 produced from this complete dissolution of CaCO3.

Then on a basis of 1 volume of HF/HCL mix I'd lose some calculated amount of CaCO3 and gain the above amount of CaF2. The porosity change would basically be a sum of these 2 amounts?

...At this point I'm tired and have little idea what I'm doing, feel like I'm running in circles. Any and all advice is greatly appreciated, thank you.
 
Physics news on Phys.org

Similar threads

Checking Answers of Chemistry Assignment
  • · Replies 1 ·
Replies
1
Views
8K