Question 5 of 7: Stage-Discharge, Trapezoidal Channel Flow and Specific Energy Over a Bed Rise

Problem 5: Stage-Discharge, Trapezoidal Channel Flow and Specific Energy Over a Bed Rise (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.

(i) The Stage-Discharge Approach

The stage-discharge relationship (rating curve) converts an easily and continuously measured quantity — water surface elevation, or stage, $h$, recorded by a stream gauge — into discharge $Q$, which cannot be measured continuously in practice. It is derived by periodically measuring discharge directly (current-meter velocity-area gauging, or an ADCP survey) across a range of observed stages, then fitting a curve, typically a power law $Q=a(h-h_0)^b$, where $h_0$ is the stage at zero flow; once fitted, the rating curve lets the continuously recorded stage record be converted into a continuous discharge (streamflow) record without needing to gauge every day.

Two key parameters affecting the confidence of a streamflow prediction built this way over a 25-year period: (1) channel/control-section stability — if the gauging cross-section shifts (sediment deposition/scour, vegetation growth, ice effects, or a new beaver dam) the stage-discharge relationship itself shifts, so a rating built from historical gaugings silently drifts out of date unless it is periodically re-verified with fresh gaugings; and (2) extrapolation beyond the gauged range — ratings are built from gaugings taken at low-to-moderate flows (high flows are hard and hazardous to gauge directly), so predicting the rare, high-stage flood of a 25-year event usually requires extrapolating the rating curve beyond its calibrated range, which is a significant source of uncertainty in the resulting streamflow estimate.

(ii) Trapezoidal Channel: Discharge and Reynolds Number

Given.

Find. (a) $Q$ in m³/min; (b) $Re$ and flow type.

Approach. Compute the trapezoidal geometry ($A$, wetted perimeter $P$, hydraulic radius $R=A/P$), select a Manning's n appropriate to rock lining, apply Manning's equation for $Q$, then classify the flow with $Re=VR/\nu$.

1. Geometry. $$A=(b+zy)y=(10+\tfrac13\times2)\times2=\boxed{21.33\ \text{m}^2},$$ $$P=b+2y\sqrt{1+z^2}=10+2(2)\sqrt{1+\tfrac19}=\boxed{14.22\ \text{m}},\qquad R=\frac{A}{P}=\boxed{1.50\ \text{m}}.$$
2. Manning's equation. $$Q=\frac{1}{n}AR^{2/3}S_0^{1/2}=\frac{1}{0.035}(21.33)(1.50)^{2/3}(0.04)^{1/2}=\boxed{159.8\ \text{m}^3/\text{s}}=\boxed{9587\ \text{m}^3/\text{min}}.$$
3. Velocity and Reynolds number. $$V=\frac{Q}{A}=\frac{159.8}{21.33}=7.49\ \text{m/s},\qquad Re=\frac{VR}{\nu}=\frac{7.49\times1.50}{1.0\times10^{-6}}=\boxed{1.1\times10^{7}}.$$ $Re\gg4000\Rightarrow$ fully turbulent open-channel flow (open-channel flow is turbulent in essentially all practical municipal/civil applications).

(iii) Depth of Flow Over the Bed Rise (Specific Energy)

Given. Same channel ($b=10$ m, $z=1/3$); $Q=25$ m³/s; upstream normal depth $Y_1=2$ m; bed rise $\Delta z=0.5$ m over the 20 m reach; frictional losses negligible.

Find. $Y_2$, the flow depth 20 m downstream where the bed has risen $\Delta z=0.5$ m.

Approach. With friction losses negligible, apply conservation of specific energy referenced to each section's own bed: since the bed itself rises by $\Delta z$, $E_1=\Delta z+E_2$ (total head measured from a common datum is conserved). First check whether flow is sub- or supercritical (Froude number), then solve $E_2=Y_2+Q^2/(2gA(Y_2)^2)$ for $Y_2$ on the appropriate (subcritical or supercritical) branch, and confirm the bump does not choke the flow ($E_2\ge E_{c,\min}$, the critical specific energy).

1. Upstream velocity, Froude number and specific energy. $$A_1=(b+zY_1)Y_1=21.33\ \text{m}^2\ \text{(as in part ii)},\qquad V_1=\frac{Q}{A_1}=\frac{25}{21.33}=1.17\ \text{m/s}.$$ Top width $T_1=b+2zY_1=11.33$ m, so $Fr_1=V_1/\sqrt{g A_1/T_1}=1.17/\sqrt{9.81\times1.88}=0.27<1$: subcritical flow. $$E_1=Y_1+\frac{V_1^2}{2g}=2+\frac{1.17^2}{2(9.81)}=\boxed{2.07\ \text{m}}.$$
2. Specific energy after the rise. Since the datum-referenced total head is conserved and the bed itself is $\Delta z$ higher, $$E_2=E_1-\Delta z=2.07-0.50=\boxed{1.57\ \text{m}}.$$
3. Check for choking (critical depth). Solving $Q^2T_c/(gA_c^3)=1$ for this trapezoid gives critical depth $y_c\approx0.85$ m and minimum specific energy $E_{c,\min}\approx1.27$ m. Since $E_2=1.57\ \text{m} > E_{c,\min}=1.27\ \text{m}$, the bump does not choke the flow, so a subcritical solution for $Y_2$ exists.
4. Solve for $Y_2$ on the subcritical branch. Solving $E_2=Y_2+\dfrac{Q^2}{2g\,[(b+zY_2)Y_2]^2}=1.57$ m numerically for the root $y_cdecreases ($Y_2