Question 5 of 7

Problem 5 (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) Stage-discharge approach: derivation, use, and 25-year prediction parameters (7 marks)

Derivation and use. The stage-discharge (rating curve) approach relies on the fact that, at a stable, well-defined channel control (a natural constriction, riffle, or engineered weir/flume), discharge is a single-valued, repeatable function of stage: $Q=f(h)$. The relationship is established empirically by making periodic direct discharge measurements (current-meter or ADCP gauging) simultaneously with stage readings across a wide range of flows, then fitting a smooth curve (commonly a power-law form $Q=a(h-h_0)^b$, where $h_0$ is the stage of zero flow) through the paired $(h,Q)$ observations. Once established, the rating curve converts the CONTINUOUSLY recorded stage (from an automatic stage recorder, cheap and simple to operate continuously) into a continuous discharge record WITHOUT requiring a discharge measurement to be made at every time step — discharge measurement itself is comparatively labour-intensive and cannot be automated as easily as stage, which is exactly why the stage-discharge relationship is the practical bridge between continuously measurable stage and the discharge record actually needed for design and water-balance work.

Two key parameters affecting a 25-year streamflow prediction. (1) Channel control stability (bed/bank change over time). A rating curve is only valid as long as the control section's geometry is unchanged; channel aggradation/degradation, vegetation growth, ice effects, or a major flood that reshapes the control all shift the true $Q=f(h)$ relationship, so a rating curve established today can be significantly wrong 25 years later unless it is periodically re-verified with fresh discharge measurements (rating shifts are the dominant source of long-term streamflow-record error). (2) Land-use and climate change in the contributing watershed. Even with a perfectly stable control, the STAGE record itself (and hence the predicted discharge) reflects the watershed's runoff-generation behaviour at the time of measurement; over a 25-year horizon, urbanization (reduced infiltration, higher peak/lower baseflow), forestry/land clearing, and shifting precipitation patterns under climate change all alter the underlying rainfall-runoff relationship, so a stage-discharge-based prediction extrapolated 25 years forward implicitly assumes watershed conditions the gauge record may no longer represent unless it is explicitly adjusted for anticipated land-use/climate trends.

(ii) Trapezoidal channel: uniform-flow discharge and Reynolds number (6 marks)

Given.

Find. (a) Discharge $Q$; (b) Reynolds number and flow type.

Approach. Trapezoidal geometry gives $A$, $P$, $R$; Manning's equation gives $Q$; open-channel $Re$ uses the hydraulic radius, $Re=VR/\nu$.

1. Part (a) — geometry and discharge.$$A=(b+zy)y=(8+0.25\times2.5)(2.5)=\boxed{21.56\ \text{m}^2}\qquad P=b+2y\sqrt{1+z^2}=8+2(2.5)\sqrt{1.0625}=13.15\ \text{m}\qquad R=\frac{A}{P}=1.639\ \text{m}$$$$Q=\frac{1}{n}AR^{2/3}S_o^{1/2}=\frac{1}{0.013}(21.56)(1.639)^{2/3}(0.03)^{1/2}=\boxed{399.4\ \text{m}^3/\text{s}}=23964\ \text{m}^3/\text{min}$$
2. Part (b) — Reynolds number.$$V=\frac{Q}{A}=\frac{399.4}{21.56}=18.52\ \text{m/s}\qquad Re=\frac{VR}{\nu}=\frac{(18.52)(1.639)}{1.0\times10^{-6}}=\boxed{3.036\times10^{7}}$$Since open-channel flow is turbulent for $Re\gtrsim12{,}500$ (hydraulic-radius Reynolds number), this flow is strongly turbulent.

(iii) Specific energy over a bed rise, Q = 50 m³/s (7 marks)

Given. $Q=50\ \text{m}^3/\text{s}$, $Y_1=2.5$ m, bed rise $\Delta z=0.6$ m, same trapezoidal section ($b=8$ m, $z=0.25$); frictional losses negligible.

Find. Depth $Y_2$ at the raised section.

Approach. With no friction loss, specific energy referenced to the local bed falls by exactly $\Delta z$ across the rise, $E_2=E_1-\Delta z$. Before solving for $Y_2$, the crossing must be checked against the minimum specific energy (critical-flow) condition — if $\Delta z$ exceeds the available margin $E_1-E_{\min}$, no subcritical $Y_2$ exists and the crossing is choked.

1. Approach conditions at Section 1.$$A_1=(8+0.25\times2.5)(2.5)=21.56\ \text{m}^2\qquad V_1=\frac{50}{21.56}=2.319\ \text{m/s}\qquad E_1=Y_1+\frac{V_1^2}{2g}=2.5+0.274=\boxed{2.774\ \text{m}}$$$$\text{Froude number: } Fr_1=\frac{Q}{A_1\sqrt{gA_1/T_1}}=0.485<1\ \ (\text{subcritical})$$
2. Check whether the bump chokes the flow. Solving $Q^2T/(gA^3)=1$ numerically for the critical depth of this trapezoidal section gives $Y_c=1.559$ m, with minimum specific energy $E_{\min}=2.304$ m. The available margin is only$$\Delta z_{\max}=E_1-E_{\min}=2.774-2.304=0.470\ \text{m}$$Since the actual rise $\Delta z=0.6\ \text{m}>\Delta z_{\max}=0.470\ \text{m}$, the crossing is choked — no subcritical depth satisfies $E_2=E_1-\Delta z$ at this upstream depth.
3. Depth at the raised section. A choked crossing is forced through critical depth at its highest point (the flow cannot carry more energy deficit than $E_1-E_{\min}$, so the free surface adjusts until the crest itself runs at critical depth):$$\boxed{Y_2=Y_c=1.559\ \text{m}}$$
4. What actually happens upstream (afflux). Since $Y_1=2.5$ m cannot supply enough specific energy to clear a 0.6 m rise, water backs up until the approach section itself carries the higher specific energy $E_1'=E_{\min}+\Delta z=2.304+0.6=2.904$ m. Solving $E_1'=Y_1'+Q^2/(2gA(Y_1')^2)$ on the subcritical branch:$$Y_1'=\boxed{2.665\ \text{m}}$$i.e., the stated $Y_1=2.5$ m upstream depth is not sustainable once the 0.6 m bump is in place — the channel backs up (afflux) to about 2.665 m upstream of the rise before flow can pass over the crest at critical depth.