Falling-Head Hydraulic Conductivity Test: 22-Agric-A2 Soil Physics and Mechanics, May 2017, Q2

Question 2: Falling-Head Hydraulic Conductivity 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.

Quantity	Value
Specimen diameter, D	97 mm
Specimen length, L	20 mm
Standpipe inside diameter, d	6.0 mm
Constant bath (tailwater) level	120 mm above counter top
Standpipe level, start → end	510 mm → 261 mm above counter top
Elapsed time, t	46 hours

Find. The hydraulic conductivity K (a), and the further time needed for the standpipe to fall to 200 mm above the counter top (b).

Approach. The bath (tailwater) level is held constant by the overflow, so the head actually driving flow through the specimen at any instant is the standpipe level minus the bath level, not the standpipe level alone; apply the standard falling-head formula to that effective head, then reuse the same K for a second falling-head interval starting where the first one ended.

Cross-sectional areas. $$a=\frac{\pi}{4}d^2=\frac{\pi}{4}(6.0)^2=28.27\ \text{mm}^2, \qquad A=\frac{\pi}{4}D^2=\frac{\pi}{4}(97)^2=7390\ \text{mm}^2$$
Effective heads (relative to the constant bath level). $$h_0=510-120=390\ \text{mm}, \qquad h_1=261-120=141\ \text{mm}$$
a) Hydraulic conductivity. With t = 46 h = 165\,600 s, $$K=\frac{aL}{At}\ln\frac{h_0}{h_1}=\frac{(28.27)(20)}{(7390)(165600)}\ln\frac{390}{141} =\boxed{4.70\times10^{-10}\ \text{m/s}}$$ This falls squarely inside the textbook range for a clay (≈10^-12–10^-9 m/s), so the result is reasonable for the stated soil.
b) Further drop to 200 mm. The second interval starts where the first left off (h′₀ = 141 mm) and continues to h′₁ = 200−120 = 80 mm, using the SAME K just measured: $$t_2=\frac{aL}{AK}\ln\frac{h'_0}{h'_1}=\frac{(28.27)(20)}{(7390)(4.70\times10^{-7}\,\text{mm/s})} \ln\frac{141}{80}=92\,247\ \text{s}=\boxed{1537\ \text{min}\ (\approx25.6\ \text{h})}$$

Quantity	Value
Hydraulic conductivity, K	4.70×10^-10 m/s
Reasonable for a clay?	Yes — within 10^-12–10^-9 m/s
Further time to reach 200 mm, t₂	≈ 1537 min (25.6 h)

Concept & Theory

Topic: Falling-head permeability testing of low-K (clay) specimens.

Key relations: K = (aL)/(At) ln(h₀/h₁), with h measured between the standpipe level and the constant tailwater/bath level.

Why this works: Continuity requires the volume leaving the standpipe, −a dh/dt, to equal the Darcy flow through the specimen, K A (h/L); separating variables and integrating from h₀ to h₁ over the elapsed time gives the logarithmic formula used above. Because h decays exponentially with time, equal further drops always take progressively longer — consistent with t₂ here covering a smaller absolute drop (141→80 mm) in a shorter time than the original (390→141 mm) took, once the shrinking driving head is accounted for.

Common pitfall: Measuring h as the raw standpipe reading above the counter top instead of relative to the constant bath level — the bath is the downstream (tailwater) boundary, not zero head, so omitting it overstates K by a sizeable factor here (390 vs. 510 mm at the start alone is a ≈30% difference).

Source: B.M. Das, Principles of Geotechnical Engineering, 9th ed., Ch. 7 (permeability); R.F. Craig, Craig's Soil Mechanics, 9th ed., Ch. 2.

Question 2 of 6: Falling-Head Hydraulic Conductivity Test

Question 2: Falling-Head Hydraulic Conductivity Test (20 marks)