Wellbore Damage, Skin Factor, and Acidizing: 24-Pet-A3 Fundamental Reservoir Engineering, May 2014, Q6

Question 6: Wellbore Damage, Skin Factor, and Acidizing (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Undamaged permeability $k$	150 mD
Damaged-zone radius $r_s$	3 ft
Damaged-zone permeability $k_s$	$150/3=50$ mD
External radius $r_e$ / pressure $p_e$	3000 ft / 2000 psia
Net pay $h$	10 ft
Well radius $r_w$	0.3 ft
$B_o$ / $\mu_o$	1.25 bbl/STB / 12 cp
Bottom-hole flowing pressure $p_{wf}$	1500 psia (both cases)

Find. (a) $q_o$ before the acid job (damaged, with skin); (b) $q_o$ after the acid job (damage fully removed, $k_s=k$, $S=0$).

Approach. Compute the skin factor from Hawkins' formula for a damaged annular zone, then apply the steady-state radial-flow equation with skin (before) and without skin (after, since restoring $k_s=k$ makes $S=0$).

Skin factor (Hawkins' formula). $$S=\left(\frac{k}{k_s}-1\right)\ln\frac{r_s}{r_w}=\left(\frac{150}{50}-1\right)\ln\frac{3}{0.3}=2\times\ln10$$ so $\boxed{S=4.605}$.
Part (a) — rate before the acid job. $\ln(r_e/r_w)=\ln(3000/0.3)=\ln10{,}000=9.210$. Using the steady-state radial-flow formula with skin, $k=0.150$ Darcy: $$q=\frac{7.08\,k\,h\,(p_e-p_{wf})}{\mu_o B_o\left[\ln(r_e/r_w)+S\right]}=\frac{7.08\times0.150\times10\times500}{12\times1.25\times(9.210+4.605)}$$ $$q=\frac{5310}{15\times13.816}=\frac{5310}{207.2}$$ so $\boxed{q_{\text{before}}=25.6\text{ STB/d}}$.
Part (b) — rate after the acid job. Full removal of damage means $k_s=k$, so $S=0$ (Hawkins' formula gives zero when $k/k_s=1$): $$q=\frac{7.08\times0.150\times10\times500}{12\times1.25\times9.210}=\frac{5310}{138.2}$$ so $\boxed{q_{\text{after}}=38.4\text{ STB/d}}$ — a productivity gain of about 50% from the acid treatment.

Quantity	Value
Skin factor (damaged well)	4.605
(a) Oil rate before acid job	25.6 STB/d
(b) Oil rate after acid job	38.4 STB/d
Productivity gain from acidizing	+50%

Topic: Hawkins' skin-factor formula and steady-state radial flow with skin.

Key relations: $S=(k/k_s-1)\ln(r_s/r_w)$ (Hawkins); $q=\dfrac{7.08kh(p_e-p_{wf})}{\mu_o B_o\left[\ln(r_e/r_w)+S\right]}$.

Why this works: Hawkins' formula treats near-wellbore damage as an extra, concentrated pressure loss equivalent to lengthening the effective flow path by $S$ log-cycles; adding $S$ directly into the denominator of the steady-state flow equation is mathematically equivalent to solving the damaged and undamaged zones as two resistances in series.

Common pitfall: Forgetting that a NEGATIVE skin (a stimulated, undamaged well) reduces the denominator and increases rate, while positive skin (as here, a damaged well) increases it and reduces rate — sign errors here silently reverse the acid job's benefit.

Source: Craft & Hawkins, Applied Petroleum Reservoir Engineering, 3rd ed., Ch. 3 (skin and productivity index); Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 6.

Question 6 of 7: Wellbore Damage, Skin Factor, and Acidizing

Question 6: Wellbore Damage, Skin Factor, and Acidizing (20 marks)