Question 5 of 5: Slug Tests — Method Comparison, Bouwer-Rice (Unconfined), and Cooper-Bredehoeft-Papadopulos (Confined)

Question 5: Slug Tests — Method Comparison, Bouwer-Rice (Unconfined), and Cooper-Bredehoeft-Papadopulos (Confined) (equal value)

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.

Approach. Part (a) is a qualitative comparison. Part (b) applies the Bouwer-Rice method (explicit effective radius of influence given, appropriate for an unconfined aquifer). Part (c) applies the Cooper-Bredehoeft-Papadopulos (CBP) type-curve method for a confined aquifer, reading the transmissivity and storativity directly off the stated match point.

Slug tests vs. pump tests (part a). Advantages of a slug test: (1) it requires no pump, discharge measurement, or water disposal — a single well and a fast, inexpensive field procedure (displace the water level instantaneously, then log its recovery); (2) it can be run in low-yield or very tight formations where a sustained pumping rate could not even be maintained. Disadvantages: (1) it interrogates only a small volume of aquifer immediately around the well (its radius of influence is small compared with a pump test), so it is far more sensitive to near-well skin effects, well development, and heterogeneity than a regional average; (2) it yields hydraulic conductivity/transmissivity reliably but is far less reliable for storativity, and cannot directly reveal aquifer BOUNDARIES (recharge or barrier boundaries, leaky aquitards) the way an extended pump test with multiple observation wells can.

Given. (b) Unconfined-aquifer slug test: casing radius $r_c=7.5$ cm, effective well radius (incl. gravel pack) $r_w=10$ cm, screen length $L_e=2$ m, initial displacement $y_0=0.7$ m, displacement after $t=15$ s is $y_1=0.05$ m, radius of influence $R_e=85$ m. (c) Confined-aquifer slug test: $r_c=r_w=5$ cm, match point $t_1=15$ s at $Tt/r_c^2=1$, matched type curve $\log\alpha=-4$ (the paper's $\mu$ is this same type-curve parameter, written $\alpha$ below).

Find. (b) hydraulic conductivity $K$ of the unconfined aquifer. (c) transmissivity $T$ and storativity $S$ of the confined aquifer.

1. Part (b) — Bouwer-Rice conductivity. $$K=\frac{r_c^2\ln(R_e/r_w)}{2L_et}\ln\!\left(\frac{y_0}{y_1}\right)=\frac{(7.5)^2\ln(8500/10)}{2(200)(15)}\ln\!\left(\frac{0.7}{0.05}\right)=\boxed{0.167\ \text{cm/s}\ (1.67\times10^{-3}\ \text{m/s})}.$$
2. Part (c) — CBP transmissivity from the match point. The matched dimensionless time is $\beta=Tt/r_c^2=1$ at $t_1=15$ s, so: $$T=\frac{r_c^2\beta}{t_1}=\frac{(0.05)^2(1)}{15}=\boxed{1.67\times10^{-4}\ \text{m}^2\text{/s}}.$$
3. Storativity from the matched curve parameter. The CBP type-curve family is indexed by $\alpha=r_w^2S/r_c^2$; the matched curve has $\log\alpha=-4\Rightarrow\alpha=10^{-4}$. Since $r_w=r_c=5$ cm here, $\alpha=S$ directly: $$S=\alpha\,\frac{r_c^2}{r_w^2}=10^{-4}(1)=\boxed{1.0\times10^{-4}}.$$