Question 7 of 7

Problem 7 (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, Reynolds number, hydraulic radius (8 marks)

Given.

Find. (a) $Q$; (b) $Re$ and flow type; (c) $R_h$.

Approach. Compute the trapezoidal section's area and wetted perimeter from $b$, $y$ and $z$, evaluate $R_h$, apply Manning's equation for $V$ and $Q$, then classify the flow with $Re=VR_h/\nu$.

1. Section properties. $$A = (b+zy)y = (12+0.25\times3)(3) = \boxed{38.25\ \text{m}^2}$$ $$P = b+2y\sqrt{1+z^2} = 12+2(3)\sqrt{1+0.25^2} = 18.18\ \text{m} \qquad R_h=\frac{A}{P}=\frac{38.25}{18.18}=\boxed{2.10\ \text{m}}$$
2. Manning velocity and discharge. $$V = \frac{1}{n}R_h^{2/3}S_0^{1/2} = \frac{1}{0.013}(2.10)^{2/3}(0.02)^{1/2} = 17.9\ \text{m/s}$$ $$Q = VA = (17.9)(38.25) = \boxed{683\ \text{m}^3/\text{s}}$$
3. Reynolds number and flow regime. $$Re = \frac{VR_h}{\nu} = \frac{(17.9)(2.10)}{1.004\times10^{-6}} = \boxed{3.74\times10^{7}}$$ $Re\gg4000$, so the flow is turbulent (open-channel flow is turbulent in essentially all practical engineering cases; this is not a borderline result).

(ii) Supercritical, critical and subcritical flow — the Froude number (6 marks)

The Froude number compares inertial (flow) forces to gravitational forces and, for a channel of top width $T$, is defined as

where $D_h=A/T$ is the hydraulic depth. Physically, $F_r$ compares the flow velocity $V$ to the celerity $c=\sqrt{gD_h}$ of a small gravity (surface) wave in that channel. Subcritical flow ($F_r\lt1$) occurs when $V\lt c$: the flow is relatively slow and deep, disturbances (a wave, a control downstream) can propagate upstream faster than the flow moves, so downstream conditions control the flow profile. Critical flow ($F_r=1$) occurs when $V=c$ exactly: this is the depth of minimum specific energy for a given discharge, and a surface disturbance is exactly stationary relative to the bed. Supercritical flow ($F_r\gt1$) occurs when $V\gt c$: the flow is fast and shallow, disturbances cannot propagate upstream at all (they are swept downstream), so the flow profile is controlled entirely from upstream. This is why a hydraulic jump (a transition from supercritical to subcritical flow, as often occurs downstream of a sluice gate) is an abrupt, energy-dissipating discontinuity rather than a smooth transition — information cannot travel upstream through the supercritical zone to smooth it out.

(iii) Discharge under a sluice gate from the energy equation (6 marks)

Given.

Find. Discharge $Q$.

Approach. A sluice gate is a frictionless, horizontal-bed transition, so total energy head is conserved between the upstream and downstream sections; write $V_1=Q/(by_1)$ and $V_2=Q/(by_2)$ and solve the resulting single equation in $Q$.

1. Apply the energy equation with $z_1=z_2$. $$y_1+\frac{V_1^2}{2g} = y_2+\frac{V_2^2}{2g} \;\Rightarrow\; y_1-y_2 = \frac{Q^2}{2gb^2}\left(\frac{1}{y_2^2}-\frac{1}{y_1^2}\right)$$
2. Solve for $Q$. $$Q^2 = \frac{2gb^2(y_1-y_2)}{1/y_2^2-1/y_1^2} = \frac{2(9.81)(3)^2(5-1)}{1/1^2-1/5^2} = \frac{706.3}{0.96} = 735.8$$ $$Q = \boxed{27.1\ \text{m}^3/\text{s}}$$
3. Check. $V_1=Q/(by_1)=27.1/15=1.81\ \text{m/s}$, $V_2=Q/(by_2)=27.1/3=9.04\ \text{m/s}$; total head $E_1=y_1+V_1^2/2g=5.167\ \text{m}$ equals $E_2=y_2+V_2^2/2g=5.167\ \text{m}$, confirming energy is conserved as required.