Question 6 of 6: Normal depth, critical depth and specific energy

Question 6: Normal depth, critical depth and specific energy (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.

Given. Rectangular channel: $Q=3.0\text{ m}^3/\text{s}$, width $b=11\text{ m}$, wall height 2.5 m, $n=0.015$, $S_0=0.002$. Unit discharge $q=Q/b=3.0/11=0.273\text{ m}^2/\text{s}$.

Find. (a) normal depth $y_n$; (b) critical depth $y_c$; (c) the flow regime well upstream; (d) the specific-energy diagram and the draw-down path to the weir.

Approach. Normal depth from Manning’s uniform-flow equation (iterate, since $R$ depends on $y$); critical depth from the rectangular-channel formula $y_c=(q^2/g)^{1/3}$; compare the two to classify the regime; then sketch $E=y+q^2/2gy^2$.

1. (a) Normal depth (Manning). $Q=\tfrac{1}{n}A R^{2/3}S_0^{1/2}$ with $A=by$, $P=b+2y$, $R=A/P$. Solving $$3.0=\frac{1}{0.015}(11y)\!\left(\frac{11y}{11+2y}\right)^{2/3}(0.002)^{1/2}\ \Rightarrow\ \boxed{y_n=0.242\ \text{m}.}$$
2. (b) Critical depth. For a rectangular section, $$y_c=\left(\frac{q^2}{g}\right)^{1/3}=\left(\frac{0.273^2}{9.81}\right)^{1/3}=\boxed{0.196\ \text{m}.}$$
3. (c) Regime well upstream. Since $y_n=0.242\ \text{m} \gt y_c=0.196\ \text{m}$, the uniform flow is sub-critical (a mild slope). Equivalently the Froude number $Fr=q/\!\left(y_n\sqrt{g\,y_n}\right)=0.73 \lt 1$, confirming sub-critical flow.
4. (d) Specific-energy path. On the specific-energy curve $E=y+\dfrac{q^2}{2gy^2}$, sub-critical flow sits on the upper limb at $y_n=0.242$ m ($E=0.307$ m). Approaching the broad-crested weir the flow accelerates and the depth is drawn down along the upper limb to the critical point at the nose of the curve, $y_c=0.196$ m, where the specific energy is a minimum, $E_{\min}=1.5\,y_c=0.295$ m. Figure 6 shows this draw-down.