Falling-Head Permeability Test: 18-Env-A3 Geotechnical and Hydrogeological Engineering, December 2017, Q6

Question 6: Falling-Head Permeability 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.

Given.

Given data
Quantity	Symbol	Value
Specimen diameter / length	$D_{spec}$ / $L$	100 mm / 20 mm
Standpipe inside diameter	$d_{pipe}$	5 mm
Bath (tailwater) elevation above counter	—	100 mm
Standpipe elevation at $t=0$	—	500 mm above counter ($\Delta h_0=400\ \text{mm}$)
Standpipe elevation at $t=24\ \text{h}$	—	250 mm above counter ($\Delta h_1=150\ \text{mm}$)

Find. (a) hydraulic conductivity, with a comment on reasonableness; (b) standpipe elevation at $t=4$ h and $t=12$ h.

Approach. Apply the falling-head formula to the head loss measured ABOVE the constant bath level, then use the same exponential head-decay relation to evaluate the head (and hence elevation) at the two intermediate times.

Fig. Q6 — falling-head apparatus; heads $\Delta h_0$, $\Delta h(t)$, $\Delta h_1$ are measured above the constant bath level.

Areas and head loss. $a=\dfrac{\pi}{4}(0.5\ \text{cm})^2=0.1963\ \text{cm}^2$ (standpipe); $A=\dfrac{\pi}{4}(10\ \text{cm})^2=78.54\ \text{cm}^2$ (specimen); heads measured ABOVE the bath: $h_1=50-10=40\ \text{cm}$ at $t=0$, $h_2=25-10=15\ \text{cm}$ at $t=24\ \text{h}=86{,}400\ \text{s}$.
Part (a) — hydraulic conductivity. $$k=\frac{aL}{At}\ln\Big(\frac{h_1}{h_2}\Big)=\frac{0.1963\times2.0}{78.54\times86{,}400}\ln\Big(\frac{40}{15}\Big)=\boxed{5.68\times10^{-8}\ \text{cm/s}}.$$ This falls squarely in the typical clay range ($10^{-6}$ to $10^{-9}$ cm/s), so the value IS reasonable for a clay specimen.
Part (b) — head at intermediate times. The falling-head solution is an exponential decay of the head above the bath, $h(t)=h_1\exp\!\big[-\tfrac{kA}{aL}t\big]$; evaluating at $t=4$ and $12$ h (and confirming it reproduces $h_2=15\ \text{cm}$ at 24 h): $$\begin{aligned} h(4\ \text{h})&=34.0\ \text{cm}\ \Rightarrow\ \boxed{439.7\ \text{mm above the counter}} \\ h(12\ \text{h})&=24.5\ \text{cm}\ \Rightarrow\ \boxed{344.9\ \text{mm above the counter}} \end{aligned}$$

Check: elevations are the head above the bath PLUS the constant 100 mm bath level (e.g. $340\ \text{mm}+100\ \text{mm}=439.7$ mm reported to one decimal); the $t=24$ h check reproduces the given 250 mm exactly, confirming the fitted $k$.

Quantity	Value
(a) Hydraulic conductivity, $k$	$5.68\times10^{-8}$ cm/s (reasonable for clay)
(b) Standpipe elevation at $t=4$ h	439.7 mm
(b) Standpipe elevation at $t=12$ h	344.9 mm

Concept & Theory

Topic. Falling-head permeameter test on a low-permeability specimen.

Key relations. $k=\dfrac{aL}{At}\ln\big(\dfrac{h_1}{h_2}\big)$; head decays exponentially, $h(t)=h_1\exp\!\big[-\tfrac{kA}{aL}t\big]$.

Why this works. The falling-head test keeps a constant-head test running instant-by-instant, with the driving head itself decaying as the standpipe empties into the (constant-level) specimen outflow; the outflow rate through the specimen at any instant equals the rate the standpipe level is falling, which integrates to the logarithmic (exponential-decay) formula.

Common pitfall. Measuring the head from the LABORATORY COUNTER instead of from the constant bath level (the bath, not the counter, is the datum the flow actually drains to), and reporting $k$ without a reasonableness check against typical soil-type ranges — the question explicitly asks for that comment.

Source. Das, Principles of Geotechnical Engineering (9th ed.), Ch. 7 — Permeability.

Question 6 of 6: Falling-Head Permeability Test

Question 6: Falling-Head Permeability Test (20 marks)