(i) Stage–discharge curves for streamflow prediction (6 marks)
A stage-discharge curve (rating curve) is generated by making a series of concurrent field measurements of river stage (water-surface elevation, read from a staff gauge or continuous recorder) and discharge (measured directly by current-meter or ADCP gauging across the cross-section) over a range of flow conditions, then fitting a smooth curve — typically a power-law relation $Q=a(h-h_0)^b$, where $h_0$ is the stage of zero flow — through the paired points. Once established, the rating curve converts a continuously recorded stage (cheap and automatic) into a continuous discharge record without having to gauge the river directly every day, which is how most streamflow records (including annual-maximum flood series such as the Red River data in Problem 1) are actually produced.
Two key assumptions that influence accuracy are: (1) a stable, unchanging channel control section — the rating assumes the stage-discharge relationship at the control (a natural riffle, constriction, or engineered structure) does not shift over time; scour, deposition, vegetation growth or ice effects change the control geometry and invalidate the existing rating, requiring periodic re-gauging to detect and correct for the drift; and (2) steady, single-valued (non-looped) flow — the simple rating assumes a given stage always corresponds to the same discharge, which holds for steady or slowly-varying flow but breaks down during rapidly rising/falling flood waves, where the same stage carries different discharge on the rising limb versus the falling limb (hysteresis/looped rating), so peak-flood discharges estimated from a simple rating carry additional uncertainty.
(ii) Trapezoidal channel: discharge and Reynolds number (6 marks)
Fig. 4 — Trapezoidal channel cross-section at normal depth (not to scale).
Given.
Quantity
Value
Normal depth, $y$
1.5 m
Base width, $b$
13 m
Side slope, H:V
1:4 → $z=0.25$
Bed slope, $S_0$
3% = 0.03
Manning's $n$ (grass-lined channel)
0.030 (Chow, short/mowed grass, range 0.027–0.035)
Find. Discharge $Q$ in m³/min; Reynolds number/flow regime.
Approach. Trapezoidal geometry gives area $A$, wetted perimeter $P$ and hydraulic radius $R$; Manning's equation gives velocity and hence $Q$; the open-channel Reynolds number then uses $R$ as the characteristic length.
(b) Reynolds number. Using the hydraulic radius as the open-channel characteristic length,
$$Re = \frac{VR}{\nu} = \frac{6.69\times1.247}{1.0\times10^{-6}} = \boxed{8.3\times10^{6}}$$
This is far above the open-channel turbulent threshold ($Re\gtrsim2000$), so the flow is turbulent.
Quantity
Value
Area, $A$
20.06 m²
Hydraulic radius, $R$
1.247 m
Velocity, $V$
6.69 m/s
Discharge, $Q$
8050 m³/min (134.2 m³/s)
Reynolds number, $Re$
$8.3\times10^6$ (turbulent)
Check: the 3% bed slope on this wide, steep-sided channel produces a normal velocity of 6.7 m/s, well beyond the maximum permissible velocity for a grass lining (typically about 1.2–2.4 m/s, depending on grass cover and slope, per standard erosion-control guidance) — the Manning arithmetic follows directly from the stated $S_0$, $n$ and geometry, but a real grass-lined channel at this velocity would fail by erosion and would need a riprap or engineered hard lining instead, or a flatter design grade.
(iii) Gradually varied flow over a bed rise — choked crossing (8 marks)
Fig. 5 — Specific-energy profile over the 0.7 m bed rise: the hump exceeds the maximum rise the upstream flow can clear without choking, so critical depth occurs at the crest and the upstream depth must back up above the stated 1.5 m.
Given. Same trapezoidal channel ($b=13\ \text{m}$, $z=0.25$) as part (ii); $Q=20\ \text{m}^3/\text{s}$; upstream depth $Y_1=1.5\ \text{m}$; bed rise $\Delta z=0.7\ \text{m}$ over the 30 m reach; frictional losses negligible.
Find. Downstream depth $Y_2$ at the raised section.
Approach. With no friction loss, total energy is conserved but the bed rises by $\Delta z$, so specific energy must fall by exactly $\Delta z$: $E_2=E_1-\Delta z$. Before solving $E_2=y_2+\dfrac{Q^2}{2gA(y_2)^2}$ for $y_2$, the critical depth $Y_c$ and minimum specific energy $E_{min}$ must be checked, because a bed rise larger than $E_1-E_{min}$ cannot physically be cleared at the given upstream depth — the flow chokes.
Critical depth and minimum specific energy. Solving $\dfrac{Q^2T}{gA^3}=1$ (with $T=b+2zy$ the top width) numerically for this section gives
$$Y_c \approx \boxed{0.620\ \text{m}} \qquad E_{min} = Y_c+\frac{Q^2}{2gA(Y_c)^2} = \boxed{0.927\ \text{m}}$$
Since $Y_1=1.5\ \text{m} > Y_c$, the approach flow is subcritical, as expected on this mild-graded channel.
Check whether the hump chokes the flow. The maximum bed rise the upstream flow can clear without forcing critical depth at the crest is
$$\Delta z_{max} = E_1-E_{min} = 1.551-0.927 = 0.624\ \text{m}$$
The stated rise $\Delta z = 0.7\ \text{m}$ exceeds $\Delta z_{max}=0.624\ \text{m}$, so the crossing is choked: the channel physically cannot pass $20\ \text{m}^3/\text{s}$ over a 0.7 m hump while holding the upstream depth at 1.5 m. At choking, the depth at the crest is forced to critical depth,
$$Y_2 = Y_c = \boxed{0.62\ \text{m}}$$
and (beyond what the question asks, but the necessary physical consequence) the upstream depth must rise above the stated 1.5 m — to a new depth satisfying $E_1' = E_{min}+\Delta z = 1.627\ \text{m}$ — producing a backwater curve upstream of the hump rather than the smooth subcritical draw-down a smaller rise would give.
