07-Str-B11 · May 2016

Question 6 of 6: Rectangular Channel — Normal Depth, Critical Depth and the Specific-Energy Diagram

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, May 2016 — 07-Str-B11, Hydraulic Engineering. Three hours, closed book; one 8.5 × 11 inch aid sheet (both sides may be used) and any non-communicating calculator are permitted. The paper prints six questions of equal value, and a candidate completes any five of them; where a question has parts, the parts also carry equal value. All six are worked below, so the set is useful whichever five a reader chooses.

Reference texts: Mays, L.W., Water Resources Engineering (3rd ed., Wiley) — Hazen-Williams pipe hydraulics, branched and looped network analysis, control valves and extended-period simulation; Chin, D.A., Water-Resources Engineering (3rd ed., Pearson) — North-American distribution practice, maximum-day-plus-fire-flow loading and service-pressure criteria; Chow, V.T., Open-Channel Hydraulics (McGraw-Hill, 1959) — uniform flow, Manning’s equation, specific energy and channel transitions; Henderson, F.M., Open Channel Flow (Macmillan) — critical flow, controls and surface-profile classification; White, F.M., Fluid Mechanics (7th ed., McGraw-Hill) — the pipe force balance, the friction-factor definition and laminar film flow.

Notation and conventions used throughout. The paper supplies the SI Hazen-Williams form $Q = 0.278\,C\,D^{2.63}S^{0.54}$ with $Q$ in m³/s, $D$ in metres and $S = h_f/L$; it is used exactly as printed. Writing $K = 0.278\,C\,D^{2.63}$ inverts it to the head-loss form $h_f = L\,(Q/K)^{1.852}$, the exponent $n = 1.852$ being the value the paper itself quotes in its loop-correction note. Following the cover-page Note 6, local losses and velocity head are neglected, so the hydraulic grade line (HGL) and the energy grade line coincide and “pressure head at a node” means $p/\gamma = \mathrm{HGL} - z$. Water is taken as $\rho = 1000$ kg/m³ with $\nu = 1.31\times10^{-6}$ m²/s (the cover page prints the exponent without its minus sign, which is an obvious typesetting slip — $1.31\times10^{6}$ m²/s is not a physical viscosity).

Source note — the Question 4 schematic is not reproduced in the examination paper. Question 4 refers to “the figure below”, but no figure is printed between Question 4 and Question 5 on page 3 of the paper, and page 4 opens with the continuation of Question 5. The network is therefore assumed to be a single loop: a source reservoir feeding node N1 through pipe P1 and node N2 through pipe P2, with pipe P3 closing the loop between N1 and N2. That topology is adopted here and is stated explicitly in the Given.

Question 6: Rectangular Channel — Normal Depth, Critical Depth and the Specific-Energy Diagram (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. A prismatic rectangular channel of fixed width and slope, terminating at a broad-crested weir that forces critical flow.

Given data — Question 6
Quantity	Symbol	Value
Discharge	$Q$	3.0 m³/s
Channel width	$b$	11 m
Side height	—	2 m
Manning’s roughness	$n$	0.013
Longitudinal slope	$S_0$	0.001

Find. The normal depth from Manning’s equation, the critical depth from the discharge and width alone, the flow regime that comparison of the two implies, and a specific-energy diagram showing how the flow moves from that regime to critical at the weir.

Approach. Normal depth is the depth at which Manning’s equation returns the given discharge on the given slope, and is found by iteration because the hydraulic radius depends on it. Critical depth for a rectangular section is explicit in the unit discharge. Comparing the two classifies the slope and hence the regime, and the specific-energy curve then displays the transition from the sub-critical branch to the nose of the curve.

Part (a) — set up Manning’s equation for the rectangular section. With $A = by$ and wetted perimeter $P = b + 2y$, $$Q = \frac{1}{n}A R^{2/3}S_0^{1/2},\ \ R = \frac{A}{P} = \frac{11y}{11 + 2y}$$ Substituting the data and grouping the constants, $$3.0 = \frac{1}{0.013}\,(11y)\left(\frac{11y}{11+2y}\right)^{2/3}(0.001)^{1/2} = 2.4325\,(11y)\left(\frac{11y}{11+2y}\right)^{2/3}$$ so the depth must satisfy $(11y)\,[11y/(11+2y)]^{2/3} = 1.2333$.
Part (a) — iterate to the normal depth. Trying $y = 0.30$ m gives $A = 3.30$ m², $R = 0.2845$ m and a left-hand side of 1.4275, which is too large; $y = 0.27$ m gives 1.2015, which is too small. Interpolating and refining converges on $$\boxed{y_n = 0.2743\ \text{m}}$$ with $A = 3.0175$ m², $P = 11.549$ m, $R = 0.26129$ m and $R^{2/3} = 0.40871$. Back-substituting, $Q = (1/0.013)(3.0175)(0.40871)(0.031623) = 3.000$ m³/s, which closes the check. The mean velocity is $V = 3.0/3.0175 = 0.994$ m/s and the flow occupies only 0.27 m of the 2 m side height, so the section is nowhere near its capacity.
Part (b) — critical depth from the unit discharge. For a rectangular channel the critical condition $Fr = 1$ reduces to an explicit formula in the discharge per unit width. With $$q = \frac{Q}{b} = \frac{3.0}{11} = 0.27273\ \text{m}^2\text{/s}$$ $$y_c = \left(\frac{q^{2}}{g}\right)^{1/3} = \left(\frac{0.27273^{2}}{9.81}\right)^{1/3} = (0.0075821)^{1/3}$$ $$\boxed{y_c = 0.1965\ \text{m}}$$ At that depth the velocity is $V_c = q/y_c = 1.388$ m/s and the Froude number is exactly 1.000, which confirms the value. The corresponding minimum specific energy is $E_{min} = 1.5\,y_c = 0.2947$ m.
Part (c) — classify the regime well upstream of the weir. Well upstream the weir’s influence has died away and the flow has reached normal depth, so the comparison is between $y_n$ and $y_c$: $$y_n = 0.2743\ \text{m}\ \gt\ y_c = 0.1965\ \text{m}$$ Equivalently, the Froude number at normal depth is $$Fr_n = \frac{V_n}{\sqrt{g\,y_n}} = \frac{0.9942}{\sqrt{9.81 \times 0.2743}} = 0.606\ \lt\ 1$$ $$\boxed{\text{Flow well upstream of the weir is SUB-CRITICAL}}$$ The same conclusion follows from the slope: the critical slope for this discharge and roughness is $S_c = 0.00299$, so the channel’s 0.001 is a mild slope and normal flow on it is necessarily sub-critical. Because the flow is sub-critical, the weir is a downstream control and its influence propagates upstream, which is exactly why a broad-crested weir works as a measuring structure here.
Part (d) — read the transition off the specific-energy diagram. Specific energy is the head measured above the channel bed, $$E = y + \frac{V^{2}}{2g} = y + \frac{q^{2}}{2gy^{2}}$$ which for a fixed $q$ is a two-branched curve: the upper, sub-critical branch approaches the line $E = y$ as depth grows, and the lower, super-critical branch rises steeply as depth falls. The two meet at the nose, where $E$ is a minimum and the flow is critical. At normal depth, $$E_n = 0.2743 + \frac{0.9942^{2}}{2 \times 9.81} = 0.2743 + 0.0504 = 0.3247\ \text{m}$$ against $E_{min} = 0.2947$ m at critical. The channel therefore approaches the weir with 0.030 m of specific energy in hand.

Figure 6.1 — Specific-energy curve for q = 0.2727 m²/s per metre of width. The flow arrives from the left along the upper (sub-critical) branch at normal depth, and as it nears the weir the depth falls and the specific energy is drawn down along the red path to the nose of the curve, where critical depth occurs on the weir crest. The passage from the sub-critical branch to the critical point is the M2 drawdown curve.

(d) What the diagram shows. The red path on Figure 6.1 is the answer part (d) asks for. Moving from the blue point at $(E_n,\,y_n) = (0.325,\,0.274)$ to the red point at $(E_{min},\,y_c) = (0.295,\,0.196)$, the depth falls and the specific energy falls with it, because the mild-slope bed is no longer able to supply as much energy as friction removes once the drawdown starts. In backwater-classification terms this is an M2 profile: it lies in zone 2 of a mild slope, between $y_c$ and $y_n$, and it always ends at critical depth at a free overfall or a broad-crested control. A candidate should also note what the diagram forbids: the flow cannot continue past the nose, because there is no point on the curve with $E \lt E_{min}$, and any attempt to force more discharge through the section than the available energy permits raises the upstream water level instead, which is the phenomenon of choking.

Final results — Question 6
Quantity	Result
(a) Normal depth	0.2743 m
(a) Area / hydraulic radius / mean velocity at normal depth	3.0175 m² / 0.2613 m / 0.994 m/s
(b) Unit discharge	0.27273 m²/s
(b) Critical depth at the weir	0.1965 m
(b) Minimum specific energy	0.2947 m
(c) Froude number at normal depth	0.606
(c) Regime well upstream	Sub-critical (mild slope; $S_c = 0.00299$)
(d) Specific energy at normal depth	0.3247 m
(d) Transition shown on the diagram	M2 drawdown, upper branch to the nose (see Figure 6.1)

Topic. Uniform flow by Manning’s equation, critical flow in a rectangular channel, slope classification, and the specific-energy diagram as a picture of a channel control.

Key relations. Manning $Q = (1/n)A R^{2/3}S_0^{1/2}$ with $R = A/P$; rectangular critical depth $y_c = (q^2/g)^{1/3}$ with $q = Q/b$; $E_{min} = 1.5\,y_c$; the Froude number $Fr = V/\sqrt{gy}$; and specific energy $E = y + q^2/(2gy^2)$.

Why this works. Normal depth and critical depth answer two entirely different questions, which is why comparing them classifies the flow. Normal depth is a balance between gravity and friction and therefore depends on slope and roughness; critical depth is a property of the discharge and the cross-section alone and knows nothing about either. When normal depth exceeds critical depth the channel is mild and uniform flow on it is sub-critical, so disturbances travel upstream and controls are located downstream; when it is less, the channel is steep and control passes upstream. The specific-energy curve makes the same statement geometrically: at a fixed discharge, every depth has an energy, the function has a single minimum, and the two depths that share any energy above that minimum are the alternate depths, one on each branch. A broad-crested weir works precisely by forcing the flow to the minimum, because the depth over the crest is then fixed by the discharge alone and a single measurement of head gives the flow.

Common pitfall. Three recur. Using the top width rather than the wetted perimeter in $R = A/P$ is the most common, and it under-states friction and normal depth; the perimeter here is $b + 2y$, not $b$. Applying $y_c = (q^2/g)^{1/3}$ to a non-rectangular section is the second, since that formula is a rectangular special case of the general $Q^2 T/(gA^3) = 1$. The third is confusing specific energy with total energy: specific energy is measured from the channel bed and therefore falls along a mild-slope drawdown even though the total energy line falls monotonically everywhere.

Source. Chow, Open-Channel Hydraulics, §3.2, §5.4 and §9 (specific energy, critical flow, and the classification of surface profiles); Henderson, Open Channel Flow, Ch. 2 and Ch. 4; Chin, Water-Resources Engineering 3e, Ch. 3.

← Question 5 Back to the paper →

Quantity	Symbol	Value
Discharge	\(Q\)	3.0 m³/s
Channel width	\(b\)	11 m
Side height	—	2 m
Manning’s roughness	\(n\)	0.013
Longitudinal slope	\(S_0\)	0.001