Falling-Head Permeameter on a Three-Layer Sample: 18-Env-A3 Geotechnical and Hydrogeological Engineering, December 2016, Q4

Question 4: Falling-Head Permeameter on a Three-Layer Sample (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. A falling-head test on a sample built of three layers stacked along the flow direction.

Given data
Quantity	Symbol	Value
Sample length / diameter	$L,\,D$	120 mm / 80 mm
Standpipe diameter	$d$	4 mm
Initial / final head	$h_0,\,h_1$	1100 mm / 420 mm
Layer thickness (1,2,3)	$L_1,L_2,L_3$	20, 60, 40 mm
Layer permeability (1,2,3)	$k_1,k_2,k_3$	3×10^-3, 5×10^-4, 17×10^-4 mm/s

Find. (a) the time for the head to fall from 1100 mm to 420 mm; (b) the average (effective) permeability of the three-layer sample.

Figure 3 (schematic) — falling-head permeameter with the sample built of three layers in series along the flow path.

Approach. The layers are stacked one above the other in the direction of flow (20+60+40 = 120 mm = the full sample length), so the same flow rate passes through each in series — the effective permeability is the thickness-weighted harmonic mean. That effective $k$ then substitutes directly into the standard falling-head time formula.

Part (b) — average permeability (needed first). For flow in series across layers of equal cross-section, $$\begin{aligned} k_{avg}&=\frac{L}{\sum L_i/k_i}\\ &=\frac{120}{20/(3\times10^{-3})+60/(5\times10^{-4})+40/(17\times10^{-4})}\\ &=\boxed{7.99\times10^{-4}\ \text{mm/s}}\ \big(8.0\times10^{-5}\ \text{cm/s}\big). \end{aligned}$$ Note this average is dragged down close to the LOWEST individual $k_2=5\times10^{-4}$ mm/s — typical of series (perpendicular-to-bedding) flow, where the least-permeable layer dominates.
Part (a) — time for the head to fall. With sample area $A=\tfrac{\pi}{4}(80)^2=5027\ \text{mm}^2$ and standpipe area $a=\tfrac{\pi}{4}(4)^2=12.57\ \text{mm}^2$, the falling-head relation $$t=\frac{aL}{Ak_{avg}}\ln\!\frac{h_0}{h_1}=\frac{12.57\times120}{5027\times7.99\times10^{-4}}\ln\!\frac{1100}{420}=\boxed{362\ \text{s}}\ (6.03\ \text{min}).$$

Quantity	Value
(b) Average permeability, $k_{avg}$	7.99×10^-4 mm/s
(a) Time for head to fall 1100→420 mm	362 s (6.03 min)

Concept & Theory

Topic. Falling-head permeability testing and equivalent permeability of layered soil.

Key relations. $k=\dfrac{aL}{At}\ln\dfrac{h_0}{h_1}$ (falling head); series (flow-perpendicular) layering: $k_{avg}=L/\sum(L_i/k_i)$; parallel (flow-parallel) layering would instead use the thickness-weighted arithmetic mean.

Why this works. With the layers stacked along the flow path, continuity forces the SAME discharge (and hence the same head loss gradient scaled by each layer's own $k_i$) through every layer in turn — exactly analogous to resistors in series, which is why the harmonic-mean form appears.

Common pitfall. Using the arithmetic (thickness-weighted) mean of $k_1,k_2,k_3$ instead of the harmonic mean — that would badly overstate $k_{avg}$ here (≈1.4×10^-3 mm/s vs the correct 8.0×10^-4 mm/s) because it ignores that the least-permeable layer bottlenecks the whole series path.

Source. Das, Principles of Geotechnical Engineering, permeability chapter.

Question 4 of 6: Falling-Head Permeameter on a Three-Layer Sample

Question 4: Falling-Head Permeameter on a Three-Layer Sample (20 marks)