Falling-Head Permeameter Test: 18-Env-A3 Geotechnical and Hydrogeological Engineering, December 2018, Q5

Question 5: Falling-Head Permeameter Test (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.

Fig. Q5 — falling-head permeameter: standpipe (area $a$) feeds a soil sample of length $L$ and area $A$; the head $\Delta h$ in the standpipe decays exponentially as water drains through the sample.

Given.

Given data
Quantity	Symbol	Value
Sample length	$L$	5 cm = 0.05 m
Standpipe area	$a$	0.2 cm² = 0.2×10⁻⁴ m²
Sample cross-sectional area	$A$	30 cm² = 30×10⁻⁴ m²
Hydraulic conductivity	$K$	0.34 m/day
Head at start of part (a)	$\Delta h_0$	30 cm
Head at end of part (a)	$\Delta h_1$	15 cm
Additional drop, part (b)	—	5 cm (to $\Delta h_2=10$ cm)

Find. (a) time for $\Delta h$ to fall from 30 cm to 15 cm; (b) additional time for $\Delta h$ to fall from 15 cm to 10 cm.

Approach. Apply the standard falling-head time formula $t=\dfrac{aL}{AK}\ln\!\dfrac{\Delta h_i}{\Delta h_f}$ over each interval; the coefficient $aL/(AK)$ is the same for both parts, only the log ratio changes.

Falling-head coefficient. $$\frac{aL}{AK}=\frac{(0.2\times10^{-4})(0.05)}{(30\times10^{-4})(0.34)}=9.804\times10^{-4}\ \text{day}.$$
Part (a) — 30 cm → 15 cm. $$t_a=\frac{aL}{AK}\ln\frac{\Delta h_0}{\Delta h_1}=9.804\times10^{-4}\times\ln(2)=9.804\times10^{-4}\times0.6931\ \text{day} = \boxed{58.7\ \text{s}}.$$
Part (b) — a further 5 cm, 15 cm → 10 cm. Same coefficient, new interval: $$t_b=\frac{aL}{AK}\ln\frac{\Delta h_1}{\Delta h_2}=9.804\times10^{-4}\times\ln(1.5) = 9.804\times10^{-4}\times0.4055\ \text{day} = \boxed{34.3\ \text{s}}.$$

Check: $t_b \lt t_a$ despite covering a smaller absolute head drop (5 cm vs 15 cm) because falling-head decay is exponential in $\Delta h$, not linear — the log ratio $\ln(1.5)=0.405$ is smaller than $\ln(2)=0.693$, so the later interval (already closer to equilibrium) drains faster in relative terms. Both times converted from days to seconds by $\times86{,}400$.

Quantity	Value
(a) Time, 30→15 cm	58.7 s (9.80×10⁻⁴ day)
(b) Time, 15→10 cm	34.3 s (3.98×10⁻⁴ day)

Concept & Theory

Topic. Falling-head permeability test.

Key relations. $t=\dfrac{aL}{AK}\ln\dfrac{\Delta h_i}{\Delta h_f}$, derived from $a\,\dfrac{d(\Delta h)}{dt}=-KA\dfrac{\Delta h}{L}$.

Why this works. Darcy flow through the sample is proportional to the instantaneous head $\Delta h$, and that same flow drains the standpipe, giving a first-order ODE whose solution is a decaying exponential in $\Delta h(t)$ — hence the natural-log ratio of heads, not a linear head difference, sets the elapsed time.

Common pitfall. Using the constant-head formula ($Q=KiAt$, linear in $\Delta h$) instead of the exponential falling-head solution, and forgetting to keep $a$ and $A$ in consistent units with $K$ before substituting.

Source. Das, Principles of Geotechnical Engineering (9th ed.), Ch. 7 (laboratory permeability determination).

Question 5 of 6: Falling-Head Permeameter Test

Question 5: Falling-Head Permeameter Test (20 marks)