Find. (a) the seepage $q$ per metre of wall; (b) the pressure head at point 5.
Figure 2 (schematic) — single-row sheet pile. Total head loss $H=100.0-93.0=7$ m. The equipotentials cross the pile at points 1–12; the pile tip (point 7) is the symmetry mid-drop, confirming $N_d=10$ equal drops of 0.7 m. Point 5 lies on the upstream face at el. 87.4 m.
Approach. Establish the head loss and the field counts, apply the flow-net discharge formula, then get the pressure head at point 5 from its total head (drop count) minus its elevation head.
Head loss. Water stands at el. 100.0 upstream and discharges at the ground surface (el. 93.0) downstream:
$$H = 100.0-93.0 = 7.0\ \text{m}.$$
Field counts and the drop. The equipotentials cross the sheet pile at the numbered points; going down the upstream face (2–6), around the tip (7) and up the downstream face (8–12) gives ten equal intervals, so $N_d=10$. The pile tip is the fifth drop, at the mid-head 96.5 m — the symmetry check that confirms the count. The net has $N_f=3$ flow channels. Each drop is
$$\Delta h=\frac{H}{N_d}=\frac{7.0}{10}=0.7\ \text{m}.$$
Seepage per metre of wall. Using the flow-net discharge equation with $k=2.2\times10^{-5}$ m/s,
$$q=k\,H\,\frac{N_f}{N_d}=(2.2\times10^{-5})(7.0)\frac{3}{10}=\boxed{4.62\times10^{-5}\ \text{m}^3/\text{s per m}}$$
$$=4.62\times10^{-5}\times86\,400\approx 4.0\ \text{m}^3/\text{day per m}.$$
Total head at point 5. Point 5 is the third equipotential down the upstream face ($n=3$), so three drops have been lost from the upstream head of 100.0 m:
$$h_5 = 100.0-3(0.7)=97.9\ \text{m}.$$
Pressure head at point 5. Point 5 sits at elevation 87.4 m. Pressure head = total head − elevation head:
$$h_{p,5}=h_5-z_5=97.9-87.4=\boxed{10.5\ \text{m}}.$$
The corresponding pore pressure is $u_5=\gamma_w h_{p,5}=9.81(10.5)\approx103\ \text{kPa}.$
Q5 — seepage results
Quantity
Value
Head loss $H$
7.0 m
Drop per field $\Delta h$
0.7 m ($N_d=10$)
Seepage $q$ per m
$4.62\times10^{-5}$ m$^3$/s ≈ 4.0 m$^3$/day
Pressure head at point 5
10.5 m ($u\approx103$ kPa)
Check: $N_f$ and $N_d$ are read from a hand-drawn net and carry a ±½-field tolerance (Exam Note 3). The numbered crossing points make $N_d=10$ unambiguous (tip = mid-drop); $N_f=3$ is taken from the three downstream flow channels. A count of $N_f=3$–4 changes $q$ by about ±30% but does not affect the point-5 pressure head, which depends only on the drop count.
