Question 7 of 7: Open-Channel Uniform Flow, Gradually Varied Flow, and Manning's Equation

Problem 7: Open-Channel Uniform Flow, Gradually Varied Flow, and Manning's Equation (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) Trapezoidal Channel — Discharge and Reynolds Number

Given. A fine-sand-lined trapezoidal channel in uniform flow.

Find. (a) discharge $Q$; (b) Reynolds number $Re$ and flow regime.

Approach. Compute the trapezoidal geometry ($A$, $P$, $R$), apply Manning's equation for $V$ and $Q$, then classify the flow from $Re=V(4R)/\nu$.

1. Geometry. $$A=(b+zy)y=(12+0.25\times3)\times3=\boxed{38.25\ \text{m}^2}$$ $$P=b+2y\sqrt{1+z^2}=12+2(3)\sqrt{1+0.25^2}=18.18\ \text{m},\qquad R=\frac{A}{P}=\boxed{2.10\ \text{m}}$$
2. (a) Manning velocity and discharge. $$V=\frac{1}{n}R^{2/3}S_0^{1/2}=\frac{1}{0.020}(2.10)^{2/3}(0.02)^{1/2}=\boxed{11.6\ \text{m/s}}$$ $$Q=VA=11.6\times38.25=\boxed{444\ \text{m}^3/\text{s}}$$
3. (b) Reynolds number. $$Re=\frac{V(4R)}{\nu}=\frac{11.6\times(4\times2.10)}{1.0\times10^{-6}}=\boxed{9.8\times10^{7}}$$ Since $Re\gg2000$, the flow is turbulent.

(ii) Gradually Varied Flow over a 0.6 m Bed Rise

Given. The same trapezoidal channel ($b=12$ m, $z=0.25$) now carries $Q=15\ \text{m}^3/\text{s}$ at $Y_1=1.5$ m; 15 m downstream the bed rises $\Delta z=0.6$ m; friction losses are negligible.

Find. The flow depth $Y_2$ at the raised section.

Approach. With no friction loss, specific energy relative to each local bed is conserved except for the bed rise itself: $E_2=E_1-\Delta z$. Compute $E_1$ from $Y_1$, confirm the flow is subcritical (so the physical root is the subcritical branch of $E(Y_2)=E_2$), then solve for $Y_2$.

1. Upstream specific energy. $A_1=(12+0.25\times1.5)(1.5)=18.56\ \text{m}^2$, $V_1=Q/A_1=15/18.56=0.808\ \text{m/s}$: $$E_1=Y_1+\frac{V_1^2}{2g}=1.5+\frac{0.808^2}{2(9.81)}=\boxed{1.53\ \text{m}}$$
2. Energy at the raised section. $$E_2=E_1-\Delta z=1.53-0.6=\boxed{0.933\ \text{m}}$$
3. Critical depth check. Solving $Fr=1$ for this channel gives $y_c=0.540$ m with $E_{\min}=0.807$ m. Since $E_2=0.933\ \text{m}>E_{\min}$, the hump does not choke the flow, and since $Y_1=1.5\ \text{m}>y_c$ the upstream flow is subcritical, so $Y_2$ is the subcritical (upper) root of $E(Y_2)=Y_2+\dfrac{Q^2}{2gA(Y_2)^2}=E_2$.
4. Solve for $Y_2$ (numerically, subcritical branch $y_c

(iii) Manning's Equation — Terms and Application to a Triangular Channel

Manning's equation, $V=\dfrac{1}{n}R^{2/3}S^{1/2}$ (SI units), with $Q=VA$, relates mean velocity to channel roughness and geometry. $n$ is Manning's roughness coefficient, describing the retarding effect of the channel boundary material and surface irregularity. $R$ is the hydraulic radius, $R=A/P$ (flow cross-sectional area divided by wetted perimeter) — a measure of how efficiently the cross-section conveys flow relative to the boundary friction it exposes. $S$ is the bed (or energy-grade-line) slope, equal to the bed slope for steady uniform flow. $V$ is the resulting mean cross-sectional velocity.

For a triangular channel with side slope $z$ (H:V $=z$:1) at flow depth $y$: area $A=zy^2$, wetted perimeter $P=2y\sqrt{1+z^2}$, so hydraulic radius $R=\dfrac{zy^2}{2y\sqrt{1+z^2}}=\dfrac{zy}{2\sqrt{1+z^2}}$. Substituting into Manning's equation gives the mean velocity directly in terms of $y$, $z$, $n$ and $S$: $$V=\frac{1}{n}\left(\frac{zy}{2\sqrt{1+z^2}}\right)^{2/3}S^{1/2}$$ and the discharge is then $$Q=VA=\frac{1}{n}\left(\frac{zy}{2\sqrt{1+z^2}}\right)^{2/3}S^{1/2}\cdot zy^2$$ — a single closed-form expression in the flow depth $y$ once $z$, $n$ and $S$ are fixed, evaluated directly for $Q$ at a known depth, or solved (typically by trial/numerically, since $y$ appears to a non-integer combined power) for the normal depth that carries a specified design flow.