24-Pet-B2 Oil and Gas Evaluation and Economics · December 2014
Question 6 of 7: Two-Rate Drawdown Test — True Skin and Non-Darcy Factor
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams December 2014, 98-Pet-B2, Natural Gas Engineering — 3 hours, closed book (Casio/Sharp approved calculators only), 7 questions of 20 marks each. NOTES item 5 states only the first five questions in the answer book are marked; all 7 are solved.
Reference texts: Katz et al., Handbook of Natural Gas Engineering; Lee & Wattenbarger, Gas Reservoir Engineering (SPE Textbook Series Vol. 5); Ahmed, Reservoir Engineering Handbook, 5th ed.; Mohitpour et al., Pipeline Design and Construction, 3rd ed. (ASME Press); McCain, The Properties of Petroleum Fluids, 3rd ed.
Question 6: Two-Rate Drawdown Test — True Skin and Non-Darcy Factor (20 marks)
Check: no permeability is given, so it is solved from each test’s own semilog slope. The chart legends read “q=20 MSCFD” and “q=40 MSCFD” and are taken as the 20,000 and 40,000 MSCFD tests named in the question. Each series has eight markers at $t=$ 0.23, 0.4, 0.6, 0.8, 1, 2, 4 and 6 hr. An earlier reading shifted four of the 20,000-MSCFD times and dropped both 1-hr markers, and its displayed skin arithmetic did not close; all values below are recomputed. The two slopes imply different permeabilities (9.6 vs 6.8 mD), so each test is analysed with its own $k$, as the single-test $S'$ formula is written. Forcing one averaged $k$ into both tests gives a slightly negative (nonphysical) $D$. $D$ itself is small and sensitive to the chart reading (reading $\psi_i$ as $4.95\times10^8$ instead of $5.00\times10^8$ roughly doubles it), but the true skin stays near $-1.9$ to $-2.2$ either way.
Given. $p_i=2500$ psia; $T=610^{\circ}$R; $h=50$ ft; $\phi=0.10$; $r_w=0.3$ ft; $\mu_i=0.02$ cp; $c_{ti}=0.0004$ psi$^{-1}$; two constant-rate drawdown tests at $q_1=20{,}000$ and $q_2=40{,}000$ MSCFD, digitized $\psi(t)$ ($\times10^6$ psia$^2$/cp) at $t=$ 0.23, 0.4, 0.6, 0.8, 1, 2, 4, 6 hr: $q_1$: 389.1, 383.2, 370.3, 366.7, 362.9, 349.6, 338.5, 331.9; $q_2$: 194.0, 166.7, 145.7, 129.2, 118.3, 78.8, 45.2, 30.5. Least-squares semilog lines ($R^2=0.993$ and 0.998): $q_1$: $\psi_{1hr}=3.632\times10^8$, $m_1=-4.143\times10^7$ psia$^2$/cp per log cycle; $q_2$: $\psi_{1hr}=1.184\times10^8$, $m_2=-1.182\times10^8$ psia$^2$/cp per log cycle. $\psi(p_i=2500)=5.000\times10^8$ psia$^2$/cp (endpoint of the Question 5 $\psi(p)$ function).
Find. True skin factor $S$ and non-Darcy coefficient $D$.
Approach. Get permeability from each test’s own slope ($m=1637qT/kh$), compute each test’s apparent (rate-dependent) skin $S'$ from its own semilog intercept and $k$, then solve the two simultaneous equations $S_i'=S+Dq_i$ for the rate-independent true skin $S$ and the non-Darcy coefficient $D$.
Permeability from each test. $k=1637qT/(|m|h)$. Test 1: $k_1=1637(20{,}000)(610)/(4.143\times10^7\times50)=\boxed{k_1=9.64\ \text{mD}}$. Test 2: $k_2=1637(40{,}000)(610)/(1.182\times10^8\times50)=\boxed{k_2=6.76\ \text{mD}}$. The charted data do not give one common $k$ (see the check note), so each test keeps its own value.
Apparent skin per test. $S'=1.151\left[\dfrac{\psi_i-\psi_{1hr}}{|m|}-\log_{10}\!\left(\dfrac{k}{\phi\mu_ic_{ti}r_w^2}\right)+3.23\right]$. Here $\phi\mu_ic_{ti}r_w^2=0.10\times0.02\times0.0004\times0.09=7.2\times10^{-8}$. Test 1: $(5.000-3.632)\times10^8/4.143\times10^7=3.302$ and $\log_{10}(9.64/7.2\times10^{-8})=8.127$, giving $S_1'=1.151(3.302-8.127+3.23)=\boxed{S_1'=-1.836}$. Test 2: $(5.000-1.184)\times10^8/1.182\times10^8=3.228$ and $\log_{10}(6.76/7.2\times10^{-8})=7.973$, giving $S_2'=1.151(3.228-7.973+3.23)=\boxed{S_2'=-1.743}$.
Solve for true skin and non-Darcy factor. $S_2'-S_1'=D(q_2-q_1)$: $D=\dfrac{-1.743-(-1.836)}{40{,}000-20{,}000}=\boxed{D=4.7\times10^{-6}\ \text{MSCFD}^{-1}}$. Then $S=S_1'-Dq_1=-1.836-(4.67\times10^{-6})(20{,}000)$: $\boxed{S=-1.93}$. Check on test 2: $-1.93+(4.67\times10^{-6})(40{,}000)=-1.743$. The well is mildly stimulated, and the non-Darcy term is small (only about $+0.19$ skin units even at 40,000 MSCFD).
Fig. 3 — Semilog pseudopressure vs. time for both drawdown-test rates (eight markers each, read from the printed figure); the two least-squares straight lines give each test’s slope $m$ and 1-hour intercept $\psi_{1hr}$.
