Question 2 of 5: Standing's Flow-Efficiency Method — Hydraulic-Fracturing Evaluation

Check: $P_b$ (4600 psig) is above $\bar P_R$ (4500 psig), so the reservoir is already saturated (two-phase) at every pressure in this problem — Standing's flow-efficiency-corrected Vogel equation applies over the full range, with no above-$P_b$ straight-line segment.

Approach. Convert the known skin to a flow efficiency $FE=X/(X+S')$, $X=\ln(0.472\,r_e/r_w)$; use it with the single pre-frac test point in Standing's equation $q_o/(q_o)_{max,FE=1}=FE(1-R)\left[1.8-0.8\,FE(1-R)\right]$, $R=P_{wf}/\bar P_R$, to solve for the ideal ($FE=1$) reference AOF, then apply the same $FE$ and reference AOF at the requested pressure.

1. Flow efficiency from skin. $X=\ln(0.472\,r_e/r_w)=\ln(0.472\times3000/0.5)=\ln(2832)=7.949$. $FE_{before}=\dfrac{X}{X+S'}=\dfrac{7.949}{7.949+8}$. $\boxed{FE_{before}=0.498}$ — a moderately damaged completion.
2. Ideal ($FE=1$) reference AOF from the pre-frac test. At the test point, $R=4000/4500=0.889$, $1-R=0.111$, so $FE(1-R)=0.498\times0.111=0.0554$ and $\text{ratio}=0.0554\left(1.8-0.8\times0.0554\right)=0.0972$. $(q_o)_{max,FE=1}=q_{o,test}/\text{ratio}=500/0.0972$. $\boxed{(q_o)_{max,FE=1}=5143\ \text{STBO/day}}$.
3. Rate at $P_{wf}=2000$ psig, same $FE$. $R=2000/4500=0.444$, $1-R=0.556$, $FE(1-R)=0.498\times0.556=0.277$, $\text{ratio}=0.277(1.8-0.8\times0.277)=0.437$. $q_o=5143\times0.437$. $\boxed{q_o=2248\ \text{STBO/day}}$.

Quantity	Value
Pre-frac flow efficiency, $FE_{before}$	0.498
Ideal ($FE=1$) reference AOF	5143 STBO/day
(a) $q_o$ at $P_{wf}=2000$ psig, pre-frac	2248 STBO/day

(b) Post-frac performance and success evaluation

Given. A single post-frac test with a new water cut; the reservoir's own delivery potential (the ideal $FE=1$ AOF from part (a)) is a rock/fluid property and is unaffected by the near-wellbore stimulation.

Post-frac water cut, $f_w$	0.25
Post-frac test: $P_{wf}$, $q_L$	3662 psig, 2000 STBL/day

Find. The anticipated oil rate at $P_{wf}=2000$ psi after the frac job, and whether the job succeeded.

Approach. Convert the post-frac test's liquid rate to an oil rate via $f_w$, hold the reservoir's ideal ($FE=1$) reference AOF fixed at the 5143 STBO/day found in part (a), and solve Standing's equation backward for the post-frac $FE$ at the single post-frac test point; then apply that $FE$ at $P_{wf}=2000$ psig.

1. Post-frac test oil rate. $q_{o,test}=q_L(1-f_w)=2000\times0.75$. $q_{o,test}=1500$ STBO/day, at $R=3662/4500=0.814$.
2. Solve for $FE_{after}$. $\dfrac{q_{o,test}}{(q_o)_{max,FE=1}}=\dfrac{1500}{5143}=0.2917=FE(1-R)\left[1.8-0.8\,FE(1-R)\right]$ with $1-R=0.186$. Solving this quadratic in $FE(1-R)$ (bisection) gives $\boxed{FE_{after}=0.944}$ — skin has fallen from $S'=8$ to an equivalent $S'\approx0.47$.
3. Compare and judge the job. $\text{FOI}=FE_{after}/FE_{before}=0.944/0.498$. $\boxed{\text{FOI}=1.89}$ — the fracturing job raised effective flow efficiency by 89%, so it is judged a successful stimulation.
4. Anticipated oil rate at $P_{wf}=2000$ psig, post-frac. $R=2000/4500=0.444$, $1-R=0.556$, $FE(1-R)=0.944\times0.556=0.525$, $\text{ratio}=0.525(1.8-0.8\times0.525)=0.724$. $q_o=5143\times0.724$. $\boxed{q_o=3723\ \text{STBO/day}}$ — a 66% increase over the pre-frac 2248 STBO/day at the same $P_{wf}$.

Quantity	Value
Post-frac flow efficiency, $FE_{after}$	0.944
Fold of increase, FOI $=FE_{after}/FE_{before}$	1.89
(b) $q_o$ at $P_{wf}=2000$ psig, post-frac	3723 STBO/day
Conclusion	Successful frac job