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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.
