Non-uniform gradually varied flow (GVF)

Dynamic equation of GVF

Assumptions: steady flow; hydrostatic pressure distribution (small slope); the channel is prismatic; the energy loss is given by the uniform-flow formula using the local depth and velocity; $\alpha = 1$.

Take a short reach $dx$ of an open channel. The total head above a datum is

$$H = z + y + \frac{V^2}{2g}$$

where $z$ is the bed level, $y$ the depth and $V = Q/A$. The bed slope is $S_0 = -dz/dx$ and the energy slope is $S_f = -dH/dx$. Differentiating with respect to $x$:

$$\frac{dH}{dx} = \frac{dz}{dx} + \frac{dy}{dx} + \frac{d}{dx}\left(\frac{V^2}{2g}\right)$$ $$-S_f = -S_0 + \frac{dy}{dx} + \frac{d}{dy}\left(\frac{Q^2}{2gA^2}\right)\frac{dy}{dx}$$

Since $dA = T\,dy$:

$$\frac{d}{dy}\left(\frac{Q^2}{2gA^2}\right) = -\frac{Q^2}{gA^3}\frac{dA}{dy} = -\frac{Q^2 T}{gA^3} = -Fr^2$$

Therefore

$$-S_f = -S_0 + \frac{dy}{dx}\left(1 - Fr^2\right)$$ $$\boxed{\frac{dy}{dx} = \frac{S_0 - S_f}{1 - Fr^2}}\qquad Fr^2 = \frac{Q^2 T}{gA^3}$$

Physical meaning: the numerator is the difference between the bed slope and the slope of the energy line. If $S_0 > S_f$, depth tends to increase in the flow direction (for subcritical flow); the denominator decides the sign by flow regime ($1 - Fr^2 > 0$ subcritical, $< 0$ supercritical). When $y = y_n$, $S_f = S_0$ and $dy/dx = 0$ (uniform flow). When $y = y_c$, $Fr = 1$ and $dy/dx \to \infty$ (the surface meets the critical depth line steeply).

Wide rectangular form using Manning's equation

For a wide rectangular channel, $R \approx y$, $A = by$ and $q = Q/b$.

Friction slope from Manning's equation $V = \frac{1}{n}R^{2/3}S_f^{1/2}$:

$$S_f = \frac{n^2 V^2}{R^{4/3}} = \frac{n^2 q^2}{y^{10/3}}$$

At normal depth $S_f = S_0$ and $y = y_n$:

$$S_0 = \frac{n^2 q^2}{y_n^{10/3}} \;\Rightarrow\; \frac{S_f}{S_0} = \left(\frac{y_n}{y}\right)^{10/3}$$

Froude number: at critical depth $q^2 = g\,y_c^3$, so

$$Fr^2 = \frac{q^2}{g y^3} = \left(\frac{y_c}{y}\right)^3$$

Substituting into the dynamic equation:

$$\boxed{\frac{dy}{dx} = \frac{S_0\left[1 - \left(\dfrac{y_n}{y}\right)^{10/3}\right]}{1 - \left(\dfrac{y_c}{y}\right)^3}}$$

Terms: NDL = normal depth line ($y_n$); CDL = critical depth line ($y_c$). On a mild slope $y_n > y_c$ (NDL above CDL); on a steep slope $y_n < y_c$ (NDL below CDL). Subcritical flow is controlled from downstream; supercritical flow is controlled from upstream.

Step by step

1. Mild reach (upstream). The flow far upstream is uniform at $y_n$ (subcritical). At the break to the steep slope the flow must pass through critical depth. The profile is a drawdown, M2, falling from the NDL to the CDL at the break.
2. Steep reach, before the gate. Flow leaves the break at $y_c$ and falls towards the steep NDL as an S2 curve (supercritical, accelerating). The sluice gate is an artificial control: it holds the water behind it at a depth above $y_c$. Supercritical flow at $y_n$ cannot meet this deeper water smoothly, so a hydraulic jump forms and the deep water behind the gate is an S1 curve rising to the gate.
3. Steep reach, after the gate. The jet from the gate (vena contracta) is shallower than $y_n$, so the depth increases gradually towards the NDL as an S3 curve.
4. Steep to mild break. The supercritical flow at $y_n$ (steep) enters a mild slope where the uniform depth is the deeper $y_n$ (mild). It rises gradually as M3 and then a hydraulic jump raises it to the mild normal depth, after which flow is uniform. (If the mild tailwater is high enough, the jump moves up to the break and the M3 curve disappears.)

 MILD |          STEEP (gate)           | MILD
 -----+---- NDL1 ....................... NDL3 (high)
  \   |                       gate        _____
   \M2|  S2        J  S1       |G|  S3  J/ M3
 CDL..\._________  /''''''      |_|__   /
 NDL2      yn ..\_/......        yn_/_/

The figure is not available here, so the method is given first and then worked for the three arrangements that these figures normally show: (a) a steep reach joining a mild reach, (b) a mild reach with a sluice gate, (c) a mild reach ending in a free overfall.

Method

1. Draw the NDL and CDL on every reach: mild $y_n > y_c$, steep $y_n < y_c$, horizontal $y_n = \infty$, adverse $y_n$ does not exist.
2. Control points (where depth is known):
   • critical depth at a break from mild to steep, and at a free overfall (taken as $y_c$ slightly upstream of the brink);
   • the depth set by a sluice gate (vena contracta downstream, backwater depth upstream);
   • the depth at a reservoir or pool.
3. Subcritical flow is controlled from downstream; supercritical flow from upstream. Build the profile from each control point, using the family M1, M2, M3, S1, S2, S3, H2, H3.
4. Where a supercritical profile must join a subcritical one, put a hydraulic jump, with the supercritical depth and subcritical depth conjugate.

(a) Steep reach then mild reach

• Control point: uniform supercritical flow upstream at $y_n$ (steep); uniform subcritical flow far downstream at $y_n$ (mild).
• Profile: uniform flow on the steep reach, an M3 curve rising on the mild reach, then a hydraulic jump to the mild normal depth.

 STEEP                MILD
 ---- NDL  yn(steep) .  M3   J  ______ yn(mild)
 ____________________/'''' |-'
                       (jump)

(b) Mild reach with a sluice gate

• Control points: gate opening gives the depth $a C_c$ downstream (supercritical) and a deeper pool upstream; uniform subcritical flow far downstream.
• Upstream of the gate: M1 backwater curve (depth above $y_n$) rising towards the gate.
• Downstream of the gate: M3 curve from the vena contracta, ending with a hydraulic jump to the NDL.

        M1                  gate
 NDL ......___---'''''|G|          J  ___ NDL
                       |_|__ M3 __/|--'

(c) Mild reach ending at a free overfall

• Control point: depth $\approx y_c$ at the brink.
• Profile: an M2 drawdown curve from the NDL (asymptotic far upstream) falling to $y_c$ at the end. There is no jump.

 NDL ......___
              ''--.__  M2
 CDL - - - - - - - -  '--.   (free overfall)

Every profile is tangent to the NDL where it approaches uniform flow, meets the CDL steeply (except at a jump or overfall), and a jump is shown wherever supercritical flow meets subcritical flow of greater depth.

Mild slope

A channel slope is mild when the normal depth is greater than the critical depth, $y_n > y_c$ (equivalently $S_0 < S_c$). Uniform flow on it is subcritical.

Analysis

Use the dynamic equation in the wide-channel form

$$\frac{dy}{dx} = S_0\,\frac{1 - (y_n/y)^{10/3}}{1 - (y_c/y)^3}$$

with $y_n > y_c$. The NDL and CDL divide the flow into three zones.

Zone 1: $y > y_n > y_c$ (M1 curve)

Numerator $> 0$, denominator $> 0$, so $dy/dx > 0$ (depth increases downstream).

• Upstream end: $y \to y_n$, so $dy/dx \to 0$. The curve is asymptotic to the NDL.
• Downstream end: $y \to \infty$, so $dy/dx \to S_0$. The water surface becomes horizontal, as in a reservoir.
• Backwater curve, caused by dams or weirs.

Zone 2: $y_n > y > y_c$ (M2 curve)

Numerator $< 0$, denominator $> 0$, so $dy/dx < 0$ (depth falls downstream).

• Upstream end: $y \to y_n$, $dy/dx \to 0$: asymptotic to the NDL.
• Downstream end: $y \to y_c$, denominator $\to 0$, so $dy/dx \to -\infty$: the curve meets the CDL at a steep angle.
• Drawdown curve, as at a free overfall or a break to a steeper slope.

Zone 3: $y < y_c < y_n$ (M3 curve)

Numerator $< 0$, denominator $< 0$, so $dy/dx > 0$ (depth rises downstream).

• Upstream end: starts from a control at small depth (e.g. below a gate).
• Downstream end: $y \to y_c$, $dy/dx \to \infty$: the curve meets the CDL steeply, and in practice a hydraulic jump forms before it.

Steep slope

A channel slope is steep when the normal depth is smaller than the critical depth, $y_n < y_c$ (equivalently $S_0 > S_c$). Uniform flow on it is supercritical.

Analysis

$$\frac{dy}{dx} = S_0\,\frac{1 - (y_n/y)^{10/3}}{1 - (y_c/y)^3},\qquad y_c > y_n$$

Zone 1: $y > y_c > y_n$ (S1 curve)

Numerator $> 0$, denominator $> 0$, so $dy/dx > 0$.

• Upstream end: $y \to y_c$, $dy/dx \to \infty$. The curve meets the CDL at a steep angle, i.e. it begins at a hydraulic jump.
• Downstream end: $y \to \infty$, $dy/dx \to S_0$; the surface becomes horizontal.
• It is a backwater curve behind a dam, weir or sluice gate on a steep slope.

Zone 2: $y_c > y > y_n$ (S2 curve)

Numerator $> 0$, denominator $< 0$, so $dy/dx < 0$ (depth falls downstream).

• Upstream end: $y \to y_c$, $dy/dx \to -\infty$: steep departure from the CDL (e.g. from a break from mild to steep slope).
• Downstream end: $y \to y_n$, $dy/dx \to 0$: asymptotic to the NDL.

Zone 3: $y < y_n < y_c$ (S3 curve)

Numerator $< 0$, denominator $< 0$, so $dy/dx > 0$ (depth rises downstream).

• Upstream end: starts at a control at small depth (below a sluice gate).
• Downstream end: $y \to y_n$, $dy/dx \to 0$: asymptotic to the NDL.

Given: $b = 5$ m, $Q = 1.85\ \text{m}^3/\text{s}$, $S_0 = 0.00076$, $n = 0.0165$, $\alpha = 1.1$, $y = 0.25$ m at a section. Find the distance to $y = 0.40$ m with $\Delta y = 0.05$ m.

Depths

$$q = 0.37\ \text{m}^2/\text{s},\quad y_c = \left(\frac{\alpha q^2}{g}\right)^{1/3} = \left(\frac{1.1 \times 0.37^2}{9.81}\right)^{1/3} = 0.2485\ \text{m}$$

By Manning's equation (trial, $R = A/P$): $y_n = 0.431$ m.

Since $y_n > y_c$ the slope is mild. The depth 0.25 m satisfies $y_n > y > y_c$, so the profile is M2 (drawdown). The depth rises upstream, so 0.40 m is upstream of the 0.25 m section.

Direct step method

$$E = y + \alpha\frac{V^2}{2g},\quad S_f = \frac{n^2V^2}{R^{4/3}},\quad \Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f}$$ $$L = 4.8 + 23.3 + 77.4 \approx 105.6\ \text{m upstream}$$

The last step is close to $y_n$ where the profile flattens, so a finer calculation gives about 126 m.

Answer: M2 profile; the depth is 0.40 m about 106 m upstream of the section where it is 0.25 m.

Given: $b = 4$ m, $Q = 2.0\ \text{m}^3/\text{s}$, $S_0 = 0.00075$, $n = 0.016$, $\alpha = 1.1$, $y = 0.2$ m at a section. Find the distance to $y = 0.4$ m with $\Delta y = 0.1$ m.

Depths

$$q = 0.5\ \text{m}^2/\text{s},\quad y_c = \left(\frac{1.1 \times 0.5^2}{9.81}\right)^{1/3} = 0.304\ \text{m}$$

By Manning's equation (trial, $R = A/P$): $y_n = 0.525$ m.

The slope is mild ($y_n > y_c$) and $y = 0.2$ m is below $y_c$, so the profile is M3 (supercritical, depth rising downstream).

Graphical integration. From the dynamic equation,

$$\frac{dx}{dy} = \frac{1 - \alpha Q^2T/(gA^3)}{S_0 - S_f},\qquad x = \int_{y_1}^{y_2}\frac{dx}{dy}\,dy = \text{area under the } \frac{dx}{dy}\text{ vs } y \text{ curve}$$

Plot $dx/dy$ against $y$ and find the area under the curve.

• From 0.2 to 0.3 m, the trapezoidal area is $\tfrac{169.4 + 10.9}{2}(0.1) = 9.0$ m; a finer reading of the curve (extra point at 0.25 m, where $dx/dy = 115.8$) gives 10.7 m.
• At $y_c = 0.304$ m, $dx/dy$ changes sign (the denominator of the dynamic equation, $1 - Fr^2$, goes through zero). The M3 curve cannot cross the critical depth gradually; it ends in a hydraulic jump at or before $y_c$. Hence the depth 0.40 m cannot be reached by a continuous profile from 0.20 m, and the 0.3 to 0.4 m "area" has no physical meaning.

Answer: M3 profile. The depth rises from 0.20 m to about 0.30 m (critical) over about 10 m downstream; the curve then ends in a hydraulic jump, so 0.40 m is not reached by gradually varied flow.

On a critical slope, $S_0 = S_c$ and the normal depth equals the critical depth, $y_n = y_c$. The NDL and CDL coincide, so only two zones exist and zone 2 is absent.

$$\frac{dy}{dx} = S_0\,\frac{1 - (y_c/y)^{10/3}}{1 - (y_c/y)^3}$$

Zone 1: $y > y_n = y_c$ (C1 curve)

Numerator $> 0$ and denominator $> 0$, so $dy/dx > 0$ (depth rises downstream).

• Downstream end: $y \to \infty$, so $dy/dx \to S_0$: the surface becomes horizontal.
• Upstream end: $y \to y_c$. The numerator and denominator both tend to zero, and the limit (with $x = y_c/y \to 1$) is

$$\lim_{x\to 1}\frac{1 - x^{10/3}}{1 - x^3} = \frac{10/3}{3} = \frac{10}{9}$$

so $dy/dx \to 1.11\,S_0$ (with the Chezy-type exponent 3 it would be exactly $S_0$). The profile approaches the critical line at a finite, nearly horizontal angle (not vertically), and is unstable with surface undulations because $Fr \approx 1$.

Zone 3: $y < y_n = y_c$ (C3 curve)

Numerator $< 0$ and denominator $< 0$, so $dy/dx > 0$.

• Upstream end: starts from a control at small depth (e.g. below a gate).
• Downstream end: $y \to y_c$ with the same finite slope $\approx 1.11\,S_0$: it approaches the critical line at a gentle angle, without a jump.

Zone 2 would lie between $y_n$ and $y_c$, which are equal, so no C2 curve exists. Uniform flow at $y_n = y_c$ itself is unstable and is avoided in design.

The A3 curve occurs on an adverse slope ($S_0 < 0$) in zone 3, where $y < y_c$ (supercritical). Write the bed slope as $-|S_0|$ in the dynamic equation:

$$\frac{dy}{dx} = \frac{S_0 - S_f}{1 - Fr^2} = \frac{-(|S_0| + S_f)}{1 - Fr^2}$$

In zone 3, $Fr > 1$, so the denominator is negative and $dy/dx > 0$: the depth increases downstream.

(i) Near the critical depth line ($y \to y_c$)

$Fr \to 1$, so the denominator $1 - Fr^2 \to 0$ while the numerator $-(|S_0| + S_f)$ stays finite and non-zero. Hence

$$\frac{dy}{dx} \to \infty$$

The tangent to the curve is vertical, so the curve meets the $y_c$ line normally (at a right angle). (In practice a hydraulic jump forms before this point is reached.)

(ii) Near the channel bottom ($y \to 0$)

For a wide channel, with $S_f = n^2q^2/y^{10/3}$ and $Fr^2 = q^2/(gy^3)$, as $y \to 0$ both $S_f$ and $Fr^2$ become very large and the $|S_0|$ and $1$ terms are negligible:

$$\frac{dy}{dx} \approx \frac{-S_f}{-Fr^2} = \frac{n^2 q^2 / y^{10/3}}{q^2/(g y^3)} = g\,n^2\,y^{-1/3}$$ $$\text{As } y \to 0:\quad \frac{dy}{dx} \to \infty$$

So the tangent is again vertical, and the curve meets the channel bottom normally.

Hence the A3 profile starts vertically at the bed (or at a small depth just after a gate) and ends vertically at the CDL.

Controls: subcritical flow (mild reaches) is controlled from downstream; supercritical flow (steep reach) is controlled from upstream. The U.C. at the head of the steep reach is the critical depth at the break; the gate is the artificial control (A.C.) on the steep reach.

Profiles

1. First mild reach. Uniform subcritical flow at $y_n$ far upstream. The flow must pass through $y_c$ at the break, so the surface falls as an M2 drawdown curve towards the CDL at the break.
2. Steep reach, upstream of the gate. Flow leaves the break at $y_c$ and falls as S2, approaching $y_n$ (supercritical). The gate holds deeper water behind it, so a hydraulic jump forms on the steep reach and the depth continues to rise to the gate as S1.
3. Steep reach, downstream of the gate. The vena-contracta depth is smaller than $y_n$, so an S3 curve rises towards the NDL.
4. Second mild reach. Supercritical flow at $y_n$ (steep) meets uniform subcritical flow in the mild reach. It rises as M3 and then a hydraulic jump raises it to $y_n$ of the mild slope.

 MILD    |       STEEP        |  MILD
 ---NDL  |                    |
   \_M2  |  S2     J  S1 |G|  S3   M3  J
 CDL.\.__|___      /'''' | |_..__ __/|---- NDL
 yn(st)  |    '--yn.....  |_|    ----'

Given: wide rectangular channel, $q = 3.8\ \text{m}^2/\text{s}$, $n = 0.012$, $S_{01} = 1/120 = 0.00833$, $S_{02} = 1/1500 = 0.000667$. For a wide channel $R \approx y$.

Normal and critical depths

$$y_n = \left(\frac{nq}{\sqrt{S_0}}\right)^{3/5},\qquad y_c = \left(\frac{q^2}{g}\right)^{1/3} = \left(\frac{3.8^2}{9.81}\right)^{1/3} = 1.138\ \text{m}$$ $$y_{n1} = \left(\frac{0.012 \times 3.8}{\sqrt{0.00833}}\right)^{0.6} = 0.659\ \text{m}\ (< y_c: \text{steep})$$ $$y_{n2} = \left(\frac{0.012 \times 3.8}{\sqrt{0.000667}}\right)^{0.6} = 1.407\ \text{m}\ (> y_c: \text{mild})$$

Sketch

The upstream slope is steep (supercritical, $y_{n1} = 0.659$ m) and the downstream slope is mild (subcritical, $y_{n2} = 1.407$ m). Supercritical flow must change to subcritical through a hydraulic jump, with an M3 curve before it (see the jump location below).

 STEEP              |  MILD
 ---- CDL 1.138 - - | - - - - - - - - - - - -
                    |            J  _____ yn2=1.407
 ______ yn1=0.659 __|__M3_____/'''|

Does a jump occur, and where?

Sequent depth of $y_{n1}$:

$$Fr_1 = \frac{3.8/0.659}{\sqrt{9.81 \times 0.659}} = 2.27,\qquad y_2 = \frac{y_{n1}}{2}\left(\sqrt{1 + 8Fr_1^2} - 1\right) = 1.809\ \text{m}$$

Since $y_2 = 1.809\ \text{m} > y_{n2} = 1.407\ \text{m}$, the downstream depth is too low to force a jump at the break. A jump does occur, on the downstream (mild) slope. Flow follows an M3 curve from $y_{n1} = 0.659$ m upwards until its conjugate depth equals $y_{n2}$:

$$y_j:\ \frac{y_j}{2}\left(\sqrt{1 + 8\frac{q^2}{g y_j^3}} - 1\right) = 1.407\ \Rightarrow\ y_j = 0.905\ \text{m}$$

The jump takes the depth from 0.905 m to 1.407 m.

Length of the M3 profile (direct step, 2 steps, mild slope $S_0 = 1/1500$)

$$E = y + \frac{V^2}{2g},\quad S_f = \frac{n^2q^2}{y^{10/3}},\quad \Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f}$$

Steps: $y = 0.659 \to 0.782 \to 0.905$ m.

$$L = 62.7 + 57.8 = 120.5\ \text{m}$$

Answer: A jump occurs on the downstream (mild) slope, from 0.905 m to 1.407 m. The M3 profile from the break to the jump is about 120 m long.

Given: wide rectangular channel, $q = 3.12\ \text{m}^2/\text{s}$, $n = 0.02$, $S_{01} = 0.0002$, $S_{02} = 0.001$. Wide channel: $R \approx y$.

Critical and normal depths

$$y_c = \left(\frac{3.12^2}{9.81}\right)^{1/3} = 0.997\ \text{m}$$ $$y_{n1} = \left(\frac{0.02 \times 3.12}{\sqrt{0.0002}}\right)^{3/5} = 2.437\ \text{m},\qquad y_{n2} = \left(\frac{0.02 \times 3.12}{\sqrt{0.001}}\right)^{3/5} = 1.504\ \text{m}$$

Both $y_{n1}$ and $y_{n2}$ exceed $y_c$, so both slopes are mild, the second being steeper. Flow is subcritical throughout, so the control is downstream.

Classification

• Downstream slope ($S_{02}$): uniform flow at $y_{n2} = 1.504$ m.
• Upstream slope ($S_{01}$): the depth must fall from $y_{n1} = 2.437$ m (far upstream) to $y_{n2} = 1.504$ m at the break. This lies between $y_n$ and $y_c$, so the profile is an M2 (drawdown) curve.

Reading of the 0.4 m depth. A depth of 0.4 m is below $y_c = 0.997$ m, which cannot occur on a mild slope for this discharge (it would be an M3 curve and needs a gate or other supercritical control). The calculation below is therefore done for the actual M2 curve: the distance upstream of the break at which the depth has recovered to $0.98\,y_{n1} = 2.388$ m.

Direct step (3 steps, $S_0 = 0.0002$)

$$E = y + \frac{V^2}{2g},\quad S_f = \frac{n^2 q^2}{y^{10/3}},\quad \Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f}$$

The negative sign means the distances are measured upstream of the slope change.

$$\sum \Delta x = -397.6 - 1055.6 - 3682.9 = -5136\ \text{m}$$

Answer: Both reaches are mild (M2 profile on the flatter upstream slope, uniform flow on the steeper one). The depth rises from 1.50 m at the break to 2.39 m at about 5.1 km upstream. A depth of 0.4 m is below critical depth (0.997 m) and cannot occur in this flow without a control.

Given: $b = 4\ \text{m}$, $S_0 = 0.0008$, $n = 0.014$, $y_1 = 1.85\ \text{m}$ (upstream of gate), $y_2 = 0.35\ \text{m}$ (downstream of gate).

Discharge (energy equation across the gate, no loss):

$$y_1 + \frac{V_1^2}{2g} = y_2 + \frac{V_2^2}{2g} \;\Rightarrow\; Q = b\,y_1 y_2\sqrt{\frac{2g}{y_1 + y_2}} = 4(1.85)(0.35)\sqrt{\frac{19.62}{2.2}} = 7.735\ \text{m}^3/\text{s}$$

Critical and normal depths ($q = 1.934\ \text{m}^2/\text{s}$):

$$y_c = \left(\frac{q^2}{g}\right)^{1/3} = 0.725\ \text{m},\qquad y_n = 1.171\ \text{m}\ (\text{from Manning's equation, } R = A/P)$$

Since $y_n > y_c$, the slope is mild.

i) Profiles

• Upstream of the gate: $y = 1.85\ \text{m} > y_n$, so an M1 (backwater) curve, rising towards the gate.
• Downstream of the gate: $y = 0.35\ \text{m} < y_c$, so an M3 curve rising from the vena contracta. The sequent depth of 0.35 m is 1.311 m, which is greater than $y_n = 1.171$ m, so the jump is not at the gate. The M3 curve continues until its conjugate depth equals $y_n$ (at $y_j = 0.411\ \text{m}$), where a hydraulic jump to $y_n$ forms.

        M1                     gate
  ...........___----''''''''''|G|
  yn=1.171 ..........         | |  M3  J     yn
                              |_|____/'|----------

ii) Lengths by direct step

$$E = y + \frac{V^2}{2g},\quad S_f = \frac{n^2V^2}{R^{4/3}},\quad \Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f}$$

Upstream (M1), $y$: 1.85 to 1.25 m in 3 steps of 0.20 m:

$$L_{up} = 348.3 + 420.7 + 681.3 \approx 1450\ \text{m (upstream of the gate)}$$

Downstream (M3), $y$: 0.35 to 0.411 m in 3 steps:

$$L_{down} = 5.46 + 5.39 + 5.30 \approx 16.1\ \text{m (downstream of the gate, up to the jump)}$$

Answer: M1 upstream (about 1450 m to within 7% of $y_n$) and M3 downstream (about 16 m, then a jump from 0.411 m to 1.171 m).

Given: $b = 7\ \text{m}$, $Q = 8\ \text{m}^3/\text{s}$ ($q = 1.143\ \text{m}^2/\text{s}$). Slope 1: $S_{01} = 0.00032$, $n_1 = 0.011$. Slope 2: $S_{02} = 0.0911$, $n_2 = 0.021$. Slope 3: $S_{03} = 0.0002$, with $y = 1.2$ m at its downstream end. $n_3$ is not given, so $n_3 = n_1 = 0.011$ is assumed.

Critical and normal depths

$$y_c = \left(\frac{1.143^2}{9.81}\right)^{1/3} = 0.511\ \text{m}$$

i) Longitudinal profiles

• Reach 1 (mild): flow upstream is uniform at 0.886 m. At the break to the steep slope the flow passes through critical depth, so the surface falls as an M2 curve from $y_n$ to $y_c$ at the end of the reach.
• Reach 2 (steep): starts at $y_c$ and falls as S2 to the uniform supercritical depth $0.224$ m.
• Reach 3 (mild): the depth at the downstream end is $1.2\ \text{m} > y_{n3}$, so a backwater curve M1 rises downstream from near $y_{n3} = 1.033$ m up to 1.2 m. Supercritical flow of 0.224 m has a sequent depth of $0.983$ m, which is less than the tailwater depth ($\geq 1.033$ m). The jump is therefore forced up to the toe of reach 2 (it forms at or just above the break).

 R1 (mild)     |  R2 (steep)       | R3 (mild)
 -- yn=0.886   |                   |            1.2 m
   \__ M2      |\ S2               |  M1  _____---
 yc  '-.. - - -|  '-._ yn=0.224 J  |__---'

ii) Length of the M2 profile in reach 1

The M2 curve approaches $y_n$ only asymptotically, so it is taken from $0.98\,y_{n1} = 0.868\ \text{m}$ to $y_c = 0.511$ m in 3 equal steps (the control is at the downstream end).

$$\Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f},\qquad S_f = \frac{n^2V^2}{R^{4/3}}$$ $$L = 748.6 + 171.2 + 30.7 \approx 950\ \text{m}$$

Answer: Profile sequence M2, S2, (jump at the toe of the steep slope), M1. The M2 profile in slope 1 is about 950 m long (measured from $0.98\,y_n$ to $y_c$, in the flow direction).

Given: $b = 20\ \text{m}$, $Q = 65\ \text{m}^3/\text{s}$ ($q = 3.25\ \text{m}^2/\text{s}$), $S_0 = 0.0001$, $n = 0.025$, $\alpha = 1.1$, $y = 2\ \text{m}$ at a section.

Critical and normal depth

$$y_c = \left(\frac{\alpha q^2}{g}\right)^{1/3} = \left(\frac{1.1 \times 3.25^2}{9.81}\right)^{1/3} = 1.058\ \text{m}$$

Manning's equation with $R = A/P$, solved by trial: $y_n = 4.024\ \text{m}$ ($A = 80.5\ \text{m}^2$, $R = 2.87$ m, $V = 0.81$ m/s).

Type of profile

$y_n > y > y_c$ on a gentle slope ($y_n > y_c$, so mild), therefore the profile is M2 (drawdown). For M2 the depth rises upstream, so the depth of 2.6 m lies upstream of the 2 m section.

Bresse's method

For a wide channel with constant Chezy $C$:

$$x = \frac{y_n}{S_0}\left[u - \left(1 - \left(\frac{y_c}{y_n}\right)^3\right)\phi(u)\right],\qquad u = \frac{y}{y_n}$$ $$\phi(u) = \frac{1}{6}\ln\frac{u^2 + u + 1}{(u-1)^2} + \frac{1}{\sqrt3}\tan^{-1}\frac{2u + 1}{\sqrt3}$$

Here $\dfrac{y_n}{S_0} = 40\,238\ \text{m}$ and $1 - (y_c/y_n)^3 = 1 - (0.2629)^3 = 0.9818$.

Distances relative to the 2 m section:

• 2.0 m to 2.3 m: $-12\,709 - (-12\,228) = -481$ m
• 2.3 m to 2.6 m: $-13\,521 - (-12\,709) = -812$ m

$$x_{2.0 \to 2.6} = -1292\ \text{m}$$

The negative sign means upstream. A direct-step check with exact geometry gives 1221 m, which agrees well.

Answer: M2 profile; the depth is 2.6 m about 1.29 km upstream of the section where the depth is 2 m.

M1 profile

On a mild slope ($y_n > y_c$), the M1 profile is the backwater curve in zone 1, where $y > y_n > y_c$. Its depth increases downstream. It is asymptotic to the normal depth line upstream and approaches a horizontal water surface downstream. It occurs upstream of dams, weirs and reservoirs. (The profile in the numerical part, upstream of a free outfall, is the drawdown curve M2, because the depth falls from $y_n$ to $y_c$.)

Numerical part

Given: $b = 10\ \text{m}$, $Q = 50\ \text{m}^3/\text{s}$, $S_0 = 0.0006$, free outfall. Concrete lining: $n = 0.014$ assumed.

Depths ($q = 5\ \text{m}^2/\text{s}$):

$$y_c = \left(\frac{5^2}{9.81}\right)^{1/3} = 1.366\ \text{m},\qquad y_n = 2.169\ \text{m}\ (\text{Manning, } R = A/P)$$

Since $y_n > y_c$ the slope is mild, and the profile is M2. The depth is $y_c$ at $4y_c = 5.46$ m upstream of the fall, which is the control section. The point asked about is 2000 m upstream of the fall, i.e. $2000 - 5.46 = 1994.5\ \text{m}$ upstream of the control.

Bresse's method (three depth steps)

$$x = \frac{y_n}{S_0}\left[u - \left(1 - \left(\frac{y_c}{y_n}\right)^3\right)\phi(u)\right],\quad u = \frac{y}{y_n}$$ $$\phi(u) = \frac{1}{6}\ln\frac{u^2 + u + 1}{(u-1)^2} + \frac{1}{\sqrt3}\tan^{-1}\frac{2u + 1}{\sqrt3}$$

$\dfrac{y_n}{S_0} = 3614.5\ \text{m}$ and $1 - (y_c/y_n)^3 = 1 - 0.6298^3 = 0.7502$.

Three equal depth steps from $y_c$ upwards (trial for the end depth):

The last row is the depth at which the distance equals 1994.5 m upstream. A fine numerical integration gives 2.135 m, which confirms it.

Answer: Depth of flow 2 km upstream of the fall $\approx 2.14\ \text{m}$ (very close to $y_n = 2.17$ m).

Given: $Q = 8.75\ \text{m}^3/\text{s}$, $b = 5\ \text{m}$, $S_0 = 1/5000 = 0.0002$, $n = 0.015$, depth at the dam $y = 2.5$ m.

Depths

$$q = 1.75\ \text{m}^2/\text{s},\quad y_c = \left(\frac{1.75^2}{9.81}\right)^{1/3} = 0.678\ \text{m}$$

By Manning's equation (trial, $R = A/P$): $y_n = 1.80\ \text{m}$.

Since $y = 2.5 > y_n > y_c$ on a mild slope, the profile is M1 (backwater). The depth falls going upstream towards $y_n$. We need the distance upstream to the point where $y = y_n + 0.1 = 1.90$ m.

Direct step method, 4 equal steps from 2.5 m to 1.90 m ($\Delta y = 0.15$ m):

$$E = y + \frac{V^2}{2g},\quad S_f = \frac{n^2V^2}{R^{4/3}},\quad \Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f}$$ $$L = 1323 + 1552 + 2003 + 3270 \approx 8150\ \text{m}$$

The negative sign means upstream. A finer integration gives about 8.2 km.

Answer: The depth is within 100 mm of the normal depth at about 8.1 km upstream of the dam (M1 backwater curve).