Trapezoidal channel — normal, critical depth and a weir: 16-Civ-A5 Hydraulic Engineering, Undated paper, Q6

Question 6: Trapezoidal channel — normal, critical depth and a weir (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. Trapezoidal section, bottom width $b=1.5$ m, side slope $z=1$ (1:1, H:V), bank height 2 m.

Given data — Question 6
Quantity	Symbol	Value
Bottom width / side slope	$b,\ z$	1.5 m, 1:1
Bank height	—	2 m
Manning roughness / bed slope	$n,\ S_0$	0.020, 0.0002
Discharge	$Q$	10 m³/s

Find. (a) $y_n$, $Fr_n$; (b) $y_c$, $Fr_c$; (c) the flow regime at normal depth; (d) the effect of a downstream broad-crested weir on the surface profile.

Figure 6-1. Trapezoidal section (1.5 m base, 1:1 sides). At $S_0=0.0002$ the Manning normal depth (2.80 m) exceeds the 2 m bank height, so the reach overtops before reaching normal depth — see the callout.

Approach. Use the trapezoidal geometry $A=(b+zy)y$, $P=b+2y\sqrt{1+z^2}$, $T=b+2zy$. Normal depth solves Manning; critical depth solves $Q^2/g=A^3/T$. Compare to fix the regime; the specific-energy curve explains the weir.

Normal depth and Froude number (part a). Solving $Q=\tfrac1n A R^{2/3}S_0^{1/2}$, i.e. $A^{5/3}/P^{2/3}=Qn/S_0^{1/2}=10(0.020)/\sqrt{0.0002}=14.14$, by iteration gives $$\boxed{y_n=2.80\ \text{m}.}$$ Then $A=(1.5+2.80)2.80=12.0\ \text{m}^2$, $T=1.5+2(2.80)=7.10\ \text{m}$, hydraulic depth $D_h=A/T=1.69\ \text{m}$, $v=Q/A=0.832\ \text{m/s}$: $$Fr_n=\frac{v}{\sqrt{gD_h}}=\frac{0.832}{\sqrt{9.81(1.69)}}=\boxed{0.20.}$$
Critical depth and Froude number (part b). Critical flow satisfies $Q^2/g=A^3/T$, i.e. $A^3/T=10^2/9.81=10.19$. Iterating on $y$: $$\boxed{y_c=1.25\ \text{m},}$$ with $A=3.44\ \text{m}^2$, $T=4.00\ \text{m}$, $v_c=Q/A=2.90\ \text{m/s}$. By definition of critical flow, $$\boxed{Fr_c=1.00.}$$
Flow regime at normal depth (part c). Because $y_n=2.80\ \text{m}\gt y_c=1.25\ \text{m}$ (equivalently $S_0=0.0002\lt S_c=0.0056$, and $Fr_n=0.20\lt 1$), the bed is a mild slope and normal flow is subcritical. The channel is deep and tranquil, not shooting.
Broad-crested weir downstream (part d). A broad-crested weir raises the bed and acts as a control that forces the flow through critical depth at its crest, where specific energy is a minimum, $E_{\min}$ (the nose of the specific-energy curve, $E=y+q^2/2gy^2$). Approaching from the upstream mild, subcritical reach, the flow must gain enough head to pass the crest, so the water surface rises above normal depth — an M1 backwater extends upstream of the weir. Over the crest the depth drops to critical (velocity rises to $v_c$), the flow passes from the subcritical upper limb of the $E$–$y$ curve to the critical point; downstream of the crest it can accelerate onto the supercritical lower limb. In specific-energy terms, the weir converts part of the flow’s depth (potential) energy into velocity (kinetic) energy while holding the discharge fixed, and the crest sets the unique minimum-energy (critical) section that controls the upstream profile.

Check — normal depth exceeds the banks

At $S_0=0.0002$ the Manning normal depth (2.80 m) is above the stated 2 m bank height: at bankfull (2 m) the section conveys only about 4.9 m³/s, well short of 10 m³/s. In practice the reach would overtop its banks before reaching normal depth, so $y_n=2.80$ m is the theoretical value (trapezoid extended above the banks). The regime conclusion (mild, subcritical) is unaffected. Flagged per Exam Note 1.

Question 6 — results
Part	Quantity	Value
(a)	Normal depth / $Fr_n$	2.80 m / 0.20 (subcritical)
(b)	Critical depth / $Fr_c$	1.25 m / 1.00
(c)	Regime at normal depth	mild slope, subcritical
(d)	Broad-crested weir	M1 backwater upstream; critical depth over crest ($E_{\min}$)

Topic. Uniform, critical and gradually varied flow in a trapezoidal channel; control by a weir.

Key relations. $A=(b+zy)y$, $P=b+2y\sqrt{1+z^2}$, $T=b+2zy$; Manning $Q=\tfrac1n AR^{2/3}S_0^{1/2}$; critical $Q^2/g=A^3/T$; specific energy $E=y+Q^2/(2gA^2)$, minimum at critical.

Why this works. Normal depth is the uniform-flow balance of gravity and friction; critical depth is the geometry-only condition of minimum specific energy for a given $Q$. On a mild slope $y_n\gt y_c$, so uniform flow is subcritical and disturbances (weirs, constrictions) propagate upstream as backwater. A weir imposes critical flow at its crest — the one section where $E$ is minimised — and that control fixes the upstream M1 profile.

Common pitfall. Using the rectangular $y_c=(q^2/g)^{1/3}$ for a trapezoid — the top width $T$ varies with depth, so critical depth must be found from $Q^2/g=A^3/T$. Also, the hydraulic depth $D_h=A/T$ (not the flow depth $y$) belongs in the Froude number for a non-rectangular section.

Source. Chow, Open-Channel Hydraulics, §3–4, §6–9; Chin, Water-Resources Engineering, §5.2–5.4.

Question 6 of 6: Trapezoidal channel — normal, critical depth and a weir

Question 6: Trapezoidal channel — normal, critical depth and a weir (20 marks)