Design of Canals

Lacey's regime equations (basic equations) for a channel in regime:

$$V = \left(\frac{Q f^2}{140}\right)^{1/6}\qquad (1)$$ $$R = \frac{5V^2}{2f}\qquad (2)$$

where $V$ = mean velocity (m/s), $Q$ = discharge (m$^3$/s), $R$ = hydraulic mean radius (m), $f$ = silt factor ($f = 1.76\sqrt{d_{mm}}$), and the wetted perimeter $P$ in m.

Derivation of $P = 4.75\sqrt{Q}$

Step 1. Raise (1) to the 6th power, then take the square root:

$$V^6 = \frac{Q f^2}{140}\;\Rightarrow\; V^3 = f\sqrt{\frac{Q}{140}}\qquad (3)$$

Step 2. Continuity: $Q = A V$ and $A = P R$, so

$$Q = P\,R\,V$$

Step 3. Substitute $R$ from (2):

$$Q = P\left(\frac{5V^2}{2f}\right)V = \frac{5}{2f}\,P\,V^3$$

Step 4. Substitute $V^3$ from (3):

$$Q = \frac{5}{2f}\,P\,f\sqrt{\frac{Q}{140}} = \frac{5\,P\sqrt{Q}}{2\sqrt{140}}$$

Step 5. Solve for $P$ (note the silt factor $f$ cancels):

$$P = \frac{2\sqrt{140}}{5}\sqrt{Q} = 4.733\sqrt{Q}$$

Lacey rounded the coefficient to 4.75, so

$$\boxed{P = 4.75\sqrt{Q}}$$

Remarks: the wetted perimeter depends only on the discharge, not on the silt factor. Other Lacey relations follow: $A = Q/V$, $S = \dfrac{f^{5/3}}{3340\,Q^{1/6}}$.

Kennedy's silt theory (1895)

Kennedy observed Upper Bari Doab canals in Punjab (India) that were in regime. He stated that silt is kept in suspension by the vertical eddies generated at the bed by the flowing water. The upward component of these eddies balances the weight of the silt. Hence there is a critical velocity $V_0$ at which the channel neither silts nor scours:

$$V_0 = 0.55\,m\,D^{0.64}$$

where $D$ = depth of water (m) and $m$ = critical velocity ratio $V/V_0$ (depends on silt grade). The mean velocity $V$ is found with Kutter's (or Manning's) formula. A channel is designed by trial so that $V = V_0$; for a given $Q$ and $S$ the depth, width and velocity are fixed.

Lacey's regime theory (1929-1935)

Lacey said that a channel is in true regime only if it flows uniformly in unlimited incoherent alluvium of the same grade as it carries, and its section, slope and velocity all adjust themselves. Silt is held in suspension by the vertical components of eddies generated at all points of the wetted perimeter (bed and sides), and the section tends to be semi-elliptical. Using a silt factor $f = 1.76\sqrt{d_{mm}}$, he gave:

$$V = \left(\frac{Qf^2}{140}\right)^{1/6},\quad R = \frac{5V^2}{2f},\quad P = 4.75\sqrt{Q},\quad S = \frac{f^{5/3}}{3340\,Q^{1/6}}$$

For a given $Q$ and $f$ every dimension is fixed with no trial and error.

Comparison

Why Lacey's theory is superior

• It considers eddies from the entire wetted perimeter, which is closer to real flow.
• Silt size enters through $f$, so grain size is accounted for directly; Kennedy's $m$ is chosen by judgement.
• It gives the slope as an output, and treats $S$, $P$, $R$, $V$ as dependent on $Q$ and $f$ only, so the design is quick and unique.
• Does not need a roughness coefficient, which is hard to estimate.
• Defines regime clearly (initial and final regime, true regime, regime in the sense of sections) and also gives shape-dependence ($P \propto \sqrt Q$).
• Regime relations are based on many canal and river data of different countries, so they apply better outside Punjab.

Both are empirical and restricted to alluvial channels carrying fine silt; Lacey's equations give only the regime perimeter and do not indicate the section shape directly, which is the main criticism of his method.

Given: $Q = 10\ \text{m}^3/\text{s}$, $n = 0.0245$, $S = 0.0002$ (1 in 5000), $m = 1$, side slope 0.5H:1V.

Method (Kennedy's silt theory, trial and error)

1. Assume depth $D$.
2. Kennedy's critical velocity: $V_0 = 0.55\,m\,D^{0.64}$.
3. Area $A = Q/V_0$, bed width $B = (A - zD^2)/D$, perimeter $P = B + 2D\sqrt{1+z^2}$, $R = A/P$.
4. Find $V$ from Manning's formula, $V = \frac{1}{n}R^{2/3}S^{1/2}$.
5. Repeat until $V = V_0$.

Trials

The velocities $V$ and $V_0$ become equal at $D \approx 0.89$ m, so this is the required depth.

Final section

Adopt $D = 0.89$ m, $B = 21.6$ m.

$$\begin{aligned} A &= (21.6 + 0.5\times 0.89)\times 0.89 = 19.62\ \text{m}^2 \\ P &= 21.6 + 2\times 0.89\sqrt{1+0.5^2} = 23.59\ \text{m},\quad R = 0.832\ \text{m} \\ V &= 0.511\ \text{m/s},\quad V_0 = 0.55\times 1\times 0.89^{0.64} = 0.510\ \text{m/s} \\ Q &= AV = 19.62\times 0.511 = 10.02\ \text{m}^3/\text{s} \end{aligned}$$

Answer: Depth $D \approx 0.89$ m, bed width $B \approx 21.6$ m, side slope 0.5H:1V, bed slope 1 in 5000, velocity $\approx 0.51$ m/s (discharge check 10.0 m³/s, i.e. within about 0.2% of the design value).

Given: $Q = 40\ \text{m}^3/\text{s}$, $n = 0.018$, $S = 0.0002$ (1 in 5000), $m = 1.0$, side slope 0.5H:1V.

Method (Kennedy's silt theory, trial and error)

1. Assume depth $D$.
2. Kennedy's critical velocity: $V_0 = 0.55\,m\,D^{0.64}$.
3. Area $A = Q/V_0$, bed width $B = (A - zD^2)/D$, perimeter $P = B + 2D\sqrt{1+z^2}$, $R = A/P$.
4. Find $V$ from Manning's formula, $V = \frac{1}{n}R^{2/3}S^{1/2}$.
5. Repeat until $V = V_0$.

Trials

The velocities $V$ and $V_0$ become equal at $D \approx 3.78$ m, so this is the required depth.

Final section

Adopt $D = 3.78$ m, $B = 6.3$ m.

$$\begin{aligned} A &= (6.3 + 0.5\times 3.78)\times 3.78 = 30.96\ \text{m}^2 \\ P &= 6.3 + 2\times 3.78\sqrt{1+0.5^2} = 14.75\ \text{m},\quad R = 2.099\ \text{m} \\ V &= 1.288\ \text{m/s},\quad V_0 = 0.55\times 1.0\times 3.78^{0.64} = 1.288\ \text{m/s} \\ Q &= AV = 30.96\times 1.288 = 39.87\ \text{m}^3/\text{s} \end{aligned}$$

Answer: Depth $D \approx 3.78$ m, bed width $B \approx 6.3$ m, side slope 0.5H:1V, bed slope 1 in 5000, velocity $\approx 1.29$ m/s (discharge check 39.9 m³/s, i.e. within about 0.3% of the design value).

Given: $Q = 30\ \text{m}^3/\text{s}$, $S = 1$ in 3700, $n = 0.0225$, $\tau_p = 0.0035\ \text{kN/m}^2$, side slope 0.5H:1V. The single permissible value is taken for both bed and sides (sides carry only 0.75 of the bed stress, so they are safer).

Principle

A channel is stable when the tractive force exerted by the flowing water on the boundary does not exceed the permissible (critical) tractive force of the soil. For a wide channel the maximum tractive stress on the bed is $\tau_0 = \gamma D S$ and on the sides about $0.75\gamma D S$.

Depth from bed stability

$$D = \frac{\tau_p}{\gamma S} = \frac{0.0035}{9.81\times 0.000270} = 1.320\ \text{m}$$

Adopt $D = 1.32$ m.

Bed width from Manning's equation

$Q = \frac{1}{n}AR^{2/3}S^{1/2}$ with $A = (B + 0.5D)D$, $P = B + 2D\sqrt{1+0.5^2}$. Solving by trial for $D = 1.32$ m gives $B = 26.62$ m, say 26.6 m.

Check with $B = 26.6$ m:

$$\begin{aligned} A &= 35.98\ \text{m}^2,\ P = 29.55\ \text{m},\ R = 1.218\ \text{m} \\ V &= \frac{1}{0.0225}\times 1.218^{2/3}\times 0.000270^{1/2} = 0.833\ \text{m/s} \\ Q &= AV = 29.98\ \text{m}^3/\text{s} \end{aligned}$$

Stresses: bed $\tau_0 = 9.81\times 1.32\times 0.000270 = 0.00350\ \text{kN/m}^2$ and sides $0.75\tau_0 = 0.00262\ \text{kN/m}^2$, both not more than the permissible value (sides are safe).

Answer: Depth $D = 1.32$ m, bed width $B \approx 26.6$ m, side slope 0.5H:1V, bed slope 1 in 3700, velocity 0.83 m/s (discharge 30.0 m³/s).

Given: $Q = 45\ \text{m}^3/\text{s}$, $S = 1$ in 4500, $n = 0.0225$, $\tau_p = 0.0035\ \text{kN/m}^2$, side slope 0.5H:1V. The single permissible value is taken for both bed and sides (sides carry only 0.75 of the bed stress, so they are safer).

Principle

A channel is stable when the tractive force exerted by the flowing water on the boundary does not exceed the permissible (critical) tractive force of the soil. For a wide channel the maximum tractive stress on the bed is $\tau_0 = \gamma D S$ and on the sides about $0.75\gamma D S$.

Depth from bed stability

$$D = \frac{\tau_p}{\gamma S} = \frac{0.0035}{9.81\times 0.000222} = 1.606\ \text{m}$$

Adopt $D = 1.61$ m.

Bed width from Manning's equation

$Q = \frac{1}{n}AR^{2/3}S^{1/2}$ with $A = (B + 0.5D)D$, $P = B + 2D\sqrt{1+0.5^2}$. Solving by trial for $D = 1.61$ m gives $B = 31.64$ m, say 31.6 m.

Check with $B = 31.6$ m:

$$\begin{aligned} A &= 52.17\ \text{m}^2,\ P = 35.20\ \text{m},\ R = 1.482\ \text{m} \\ V &= \frac{1}{0.0225}\times 1.482^{2/3}\times 0.000222^{1/2} = 0.861\ \text{m/s} \\ Q &= AV = 44.93\ \text{m}^3/\text{s} \end{aligned}$$

Stresses: bed $\tau_0 = 9.81\times 1.61\times 0.000222 = 0.00351\ \text{kN/m}^2$ and sides $0.75\tau_0 = 0.00263\ \text{kN/m}^2$, both not more than the permissible value (sides are safe).

Answer: Depth $D = 1.61$ m, bed width $B \approx 31.6$ m, side slope 0.5H:1V, bed slope 1 in 4500, velocity 0.86 m/s (discharge 44.9 m³/s).

Given: $Q = 50\ \text{m}^3/\text{s}$, $S = 1$ in 5000, $n = 0.022$, $\tau_p = 0.0025\ \text{kN/m}^2$, side slope 0.5H:1V. The single permissible value is taken for both bed and sides (sides carry only 0.75 of the bed stress, so they are safer).

Principle

A channel is stable when the tractive force exerted by the flowing water on the boundary does not exceed the permissible (critical) tractive force of the soil. For a wide channel the maximum tractive stress on the bed is $\tau_0 = \gamma D S$ and on the sides about $0.75\gamma D S$.

Depth from bed stability

$$D = \frac{\tau_p}{\gamma S} = \frac{0.0025}{9.81\times 0.000200} = 1.274\ \text{m}$$

Adopt $D = 1.27$ m.

Bed width from Manning's equation

$Q = \frac{1}{n}AR^{2/3}S^{1/2}$ with $A = (B + 0.5D)D$, $P = B + 2D\sqrt{1+0.5^2}$. Solving by trial for $D = 1.27$ m gives $B = 53.01$ m, say 53.0 m.

Check with $B = 53.0$ m:

$$\begin{aligned} A &= 68.12\ \text{m}^2,\ P = 55.84\ \text{m},\ R = 1.220\ \text{m} \\ V &= \frac{1}{0.022}\times 1.220^{2/3}\times 0.000200^{1/2} = 0.734\ \text{m/s} \\ Q &= AV = 49.99\ \text{m}^3/\text{s} \end{aligned}$$

Stresses: bed $\tau_0 = 9.81\times 1.27\times 0.000200 = 0.00249\ \text{kN/m}^2$ and sides $0.75\tau_0 = 0.00187\ \text{kN/m}^2$, both not more than the permissible value (sides are safe).

Answer: Depth $D = 1.27$ m, bed width $B \approx 53.0$ m, side slope 0.5H:1V, bed slope 1 in 5000, velocity 0.73 m/s (discharge 50.0 m³/s).

Semi-theoretical approach

Uses a basic flow formula (Manning's or Chezy's) with empirical limits for non-silting, non-scouring velocity and a chosen $B/D$ ratio (from standards). The section is found from $Q = AV$ and $V = \frac{1}{n}R^{2/3}S^{1/2}$, with $V$ and $B/D$ fixed by experience, and the slope is checked against the allowable value.

Kennedy's approach

Based on the critical velocity concept, $V_0 = 0.55mD^{0.64}$ (vertical eddies from the bed). Kutter's formula gives $V$. Trial and error on $D$ gives the section for a known $S$.

Lacey's approach

Regime theory for alluvial channels with a silt factor $f$; $V$, $R$, $P$ and $S$ come directly from $Q$ and $f$.

Critical comparison

Semi-theoretical designs are the practical choice for canals where silt is not a problem (hill irrigation systems, non-alluvial soils). Kennedy and Lacey apply to alluvial soils, with Lacey's method preferred for new works.

Lacey's regime relations for a channel are

$$V = \left(\frac{Qf^2}{140}\right)^{1/6}, \qquad R = \frac{5V^2}{2f}$$

where $V$ = mean velocity (m/s), $Q$ = discharge (m³/s), $f$ = silt factor and $R$ = hydraulic mean radius (m).

Derivation

From the second equation, $V^2 = \dfrac{2fR}{5}$. From the first, squaring the sixth-root relation to a third-root relation:

$$V^2 = \left(\frac{Qf^2}{140}\right)^{1/3}$$

Equating:

$$\begin{aligned} \frac{2fR}{5} &= \left(\frac{Qf^2}{140}\right)^{1/3} \\ R &= \frac{5}{2f}\left(\frac{Qf^2}{140}\right)^{1/3} = \frac{5}{2}\cdot\frac{Q^{1/3}f^{2/3}}{140^{1/3}\,f} \\ &= \frac{2.5}{5.19}\left(\frac{Q}{f}\right)^{1/3} \end{aligned}$$ $$R \approx 0.48\left(\frac{Q}{f}\right)^{1/3}$$

(Many textbooks write the constant as 0.47 because of rounding of the constants 140 and 5/2 in the original regime relations.)

Result: the regime hydraulic mean radius increases with the cube root of the discharge and decreases with the cube root of the silt factor. For example, $Q = 20$ m³/s and $f = 1.5$ gives $R = 0.48\,(13.33)^{1/3} = 1.14$ m, which agrees with $R = 5V^2/2f$ computed directly.

For a very wide channel the wetted perimeter is almost equal to the width, so $A \approx PR$ and the discharge per unit width is $q = VR$ (this is the regime flow per metre width).

Lacey's regime equation connecting $R$, $V$ and $f$:

$$R = \frac{5V^2}{2f} \quad\Rightarrow\quad V = \sqrt{\frac{2fR}{5}}$$

Proof

$$\begin{aligned} q &= V R = \sqrt{\frac{2fR}{5}}\;R = \sqrt{\frac{2f}{5}}\;R^{3/2} \\ R^{3/2} &= q\sqrt{\frac{5}{2f}} \\ R^{3} &= \frac{5}{2}\cdot\frac{q^2}{f} \\ R &= \left(\frac{5}{2}\right)^{1/3}\left(\frac{q^2}{f}\right)^{1/3} = 1.357\left(\frac{q^2}{f}\right)^{1/3} \end{aligned}$$ $$R \approx 1.35\left(\frac{q^2}{f}\right)^{1/3}$$

where $R$ is the scour depth (m), $q$ the discharge per metre width (m³/s/m) and $f$ the silt factor. This is Lacey's scour depth formula used for the depth of cutoff walls, guide bunds and aprons ($D = R$ for normal scour, $2R$ for the nose of guide bunds).

Lacey's regime values

Given: $Q = 20\ \text{m}^3/\text{s}$, $f = 1.500$, side slope 0.5H:1V.

$$\begin{aligned} V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{20\times 1.500^2}{140}\right)^{1/6} = 0.828\ \text{m/s} \\ A &= \frac{Q}{V} = \frac{20}{0.828} = 24.16\ \text{m}^2 \\ R &= \frac{5V^2}{2f} = \frac{5\times 0.828^2}{2\times 1.500} = 1.142\ \text{m} \\ P &= 4.75\sqrt{Q} = 4.75\sqrt{20} = 21.24\ \text{m} \\ S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1.500^{5/3}}{3340\times 20^{1/6}} = 0.000357 = \frac{1}{2800} \end{aligned}$$

Trapezoidal section

For side slope $z = 0.5$: $A = (B + zD)D$ and $P = B + 2D\sqrt{1+z^2} = B + 2.2361D$.

Eliminating $B$:

$$\begin{aligned} 1.7361\,D^2 - 21.24\,D + 24.16 &= 0 \\ D &= \frac{21.24 - \sqrt{21.24^2 - 4\times 1.7361\times 24.16}}{2\times 1.7361} = 1.269\ \text{m} \\ B &= P - 2.2361D = 21.24 - 2.2361\times 1.269 = 18.40\ \text{m} \end{aligned}$$

(The other root of the quadratic gives a very deep, narrow section, so the smaller root is taken.)

Check: $A = (18.40 + 0.5\times 1.269)\times 1.269 = 24.16\ \text{m}^2$ against required 24.16 $\text{m}^2$.

Answer: Bed width $B \approx 18.4$ m, depth of water $D \approx 1.27$ m, side slope 0.5H:1V, bed slope $S = 1$ in 2800 (0.000357), velocity $V = 0.83$ m/s, hydraulic mean radius $R = 1.14$ m.

The printed data look like misprints, so these are assumed: depth $D = 2.5$ m (not 25 m) and $n = 0.016$ (not 0.16; typical for concrete lining). Side slope 1.5H:1V (angle with horizontal $\theta = 0.588$ rad $= 33.7^\circ$ since $\tan\theta = 1/1.5$), $B = 5$ m, $S = 1/1000$.

Carrying capacity (Manning's formula)

$$\begin{aligned} A &= (B + zD)D = (5 + 1.5\times 2.5)\times 2.5 = 21.875\ \text{m}^2 \\ P &= B + 2D\sqrt{1+z^2} = 5 + 2\times 2.5\times 1.8028 = 14.014\ \text{m} \\ R &= A/P = 1.5610\ \text{m} \\ V &= \frac{1}{n}R^{2/3}S^{1/2} = \frac{1}{0.016}\times 1.5610^{2/3}\times 0.001^{1/2} = 2.660\ \text{m/s} \\ Q &= AV = 21.875\times 2.660 = 58.18\ \text{m}^3/\text{s} \end{aligned}$$

Maximum carrying capacity $Q \approx 58.2$ m³/s (full supply depth 2.5 m).

Irrigable area

Water delivered during one Kor period of 10 days:

$$\text{Volume} = Q\,t = 58.18\times 10\times 86400 = 5.027e+07\ \text{m}^3$$

Field irrigation requirement (depth) $= 150$ mm $= 0.15$ m. Neglecting losses beyond the stated field requirement:

$$\text{Area} = \frac{\text{Volume}}{\text{Depth}} = \frac{5.027e+07}{0.15} = 3.351e+08\ \text{m}^2 = 33,510\ \text{ha}$$

Answer: $Q \approx 58.2$ m³/s; irrigable area $\approx 33,510$ ha (about 33,500 ha). (Velocity 2.66 m/s is within the safe limit for concrete lining.)

Lining an existing canal is justified when the yearly benefit exceeds the yearly cost of the lining. The analysis compares cost and benefit on an annual basis.

Annual cost of lining

$$C = (\text{Initial cost of lining})\times(i + \text{depreciation}) + \text{Extra maintenance}$$

• Initial cost: lining cost per metre $\times$ length (materials, labour, subgrade preparation, joints).
• Interest at the annual rate $i$ and depreciation (sinking fund) over the life of the lining (say 30-40 years for concrete).

Annual benefits

1. Saving of water by reduced seepage – seepage loss before lining minus after lining (often 70-80% saved). The value of saved water is its irrigation worth, i.e. the extra crop it can produce on additional area.
2. Increased command area from the saved water.
3. Reduction in the cost of maintenance, silt clearance and weed control.
4. Saving in land and structures: higher velocity allows a smaller cross-section; less excavation and embankment.
5. Prevention of waterlogging, and the cost of drainage avoided.
6. Reduced breaches and easier operation (higher permissible velocity, more reliable supply).

Procedure

1. Estimate unlined seepage loss $=$ (seepage rate, m³/s per million m² of wetted area) $\times P\,L$ using Punjab or Davis-Wilson formulas; estimate lined loss similarly.
2. Water saved $=$ difference, converted to hectares extra irrigation using duty $\Delta$ (ha per m³/s).
3. Annual benefit $=$ (extra area $\times$ net return per ha) $+$ maintenance savings $+$ other savings.
4. Compute annual cost of lining (above).
5. Benefit-cost ratio:

$$\text{B/C} = \frac{\text{Annual benefit}}{\text{Annual cost}} \ge 1$$

The lining is economical if $B/C > 1$. The break-even cost of lining per square metre is the one at which $B/C = 1$. The type of lining (concrete, brick, soil-cement, plastic membrane) is chosen as the cheapest one that gives acceptable benefit.

Given: $Q = 5\ \text{m}^3/\text{s}$, $n = 0.025$, $S = 0.0005$ (1 in 2000), $m = 1.1$, side slope 0.5H:1V. Bed slope $= 0.5$ m/km $= 0.0005$. Side slope is not given, so 0.5H:1V is assumed; Manning's formula is used with the given $n$.

Method (Kennedy's silt theory, trial and error)

1. Assume depth $D$.
2. Kennedy's critical velocity: $V_0 = 0.55\,m\,D^{0.64}$.
3. Area $A = Q/V_0$, bed width $B = (A - zD^2)/D$, perimeter $P = B + 2D\sqrt{1+z^2}$, $R = A/P$.
4. Find $V$ from Manning's formula, $V = \frac{1}{n}R^{2/3}S^{1/2}$.
5. Repeat until $V = V_0$.

Trials

The velocities $V$ and $V_0$ become equal at $D \approx 1.69$ m, so this is the required depth.

Final section

Adopt $D = 1.69$ m, $B = 2.7$ m.

$$\begin{aligned} A &= (2.7 + 0.5\times 1.69)\times 1.69 = 5.99\ \text{m}^2 \\ P &= 2.7 + 2\times 1.69\sqrt{1+0.5^2} = 6.48\ \text{m},\quad R = 0.925\ \text{m} \\ V &= 0.849\ \text{m/s},\quad V_0 = 0.55\times 1.1\times 1.69^{0.64} = 0.846\ \text{m/s} \\ Q &= AV = 5.99\times 0.849 = 5.09\ \text{m}^3/\text{s} \end{aligned}$$

Answer: Depth $D \approx 1.69$ m, bed width $B \approx 2.7$ m, side slope 0.5H:1V, bed slope 1 in 2000, velocity $\approx 0.85$ m/s (discharge check 5.1 m³/s, i.e. within about 1.7% of the design value).

Given: $Q = 7\ \text{m}^3/\text{s}$, $n = 0.025$, $S = 0.0003$ (1 in 3333), $m = 1.1$, side slope 0.5H:1V. Bed slope $= 0.3$ m/km $= 0.0003$. Side slope is not given, so 0.5H:1V is assumed; Manning's formula is used with the given $n$.

Method (Kennedy's silt theory, trial and error)

1. Assume depth $D$.
2. Kennedy's critical velocity: $V_0 = 0.55\,m\,D^{0.64}$.
3. Area $A = Q/V_0$, bed width $B = (A - zD^2)/D$, perimeter $P = B + 2D\sqrt{1+z^2}$, $R = A/P$.
4. Find $V$ from Manning's formula, $V = \frac{1}{n}R^{2/3}S^{1/2}$.
5. Repeat until $V = V_0$.

Trials

The velocities $V$ and $V_0$ become equal at $D \approx 1.18$ m, so this is the required depth.

Final section

Adopt $D = 1.18$ m, $B = 8.2$ m.

$$\begin{aligned} A &= (8.2 + 0.5\times 1.18)\times 1.18 = 10.37\ \text{m}^2 \\ P &= 8.2 + 2\times 1.18\sqrt{1+0.5^2} = 10.84\ \text{m},\quad R = 0.957\ \text{m} \\ V &= 0.673\ \text{m/s},\quad V_0 = 0.55\times 1.1\times 1.18^{0.64} = 0.673\ \text{m/s} \\ Q &= AV = 10.37\times 0.673 = 6.98\ \text{m}^3/\text{s} \end{aligned}$$

Answer: Depth $D \approx 1.18$ m, bed width $B \approx 8.2$ m, side slope 0.5H:1V, bed slope 1 in 3333, velocity $\approx 0.67$ m/s (discharge check 7.0 m³/s, i.e. within about 0.3% of the design value).

Given: $Q = 50\ \text{m}^3/\text{s}$, $S = 0.001000$ (1 in 1000), particle size $d = 50$ mm, $\phi = 36^\circ$. Side slope is not given, so 2H:1V is assumed (flatter than $\phi$). Side slope $2.00H:1V$ (26.6° with horizontal); $\gamma = 9.81$ kN/m³, specific gravity of particles 2.65. $n = \dfrac{d^{1/6}}{21.1} = \dfrac{0.05^{1/6}}{21.1} = 0.0288$ (Strickler).

Critical shear stress of the bed (Shields)

$$\tau_c = 0.047(\gamma_s - \gamma)d = 0.047\times(2.65-1)\times 9.81\times 0.05 = 38.04\ \text{N/m}^2$$

Depth limited by the bed

$$\gamma D S \le \tau_c \Rightarrow D_{bed} = \frac{\tau_c}{\gamma S} = \frac{0.03804}{9.81\times 0.001000} = 3.88\ \text{m}$$

Depth limited by the side slope

Reduction factor for the side:

$$K = \sqrt{1 - \frac{\sin^2\theta}{\sin^2\phi}} = \sqrt{1 - \frac{\sin^2 26.6^\circ}{\sin^2 36^\circ}} = 0.649$$

Maximum stress on the side is $0.75\gamma DS$, so:

$$D_{side} = \frac{K\tau_c}{0.75\gamma S} = \frac{0.649\times 0.03804}{0.75\times 9.81\times 0.001000} = 3.35\ \text{m}$$

The smaller value governs: stable depth $D = 3.35$ m (side controlled).

Bed width from Manning's equation

Using $D = 3.35$ m and $Q = \frac{1}{n}AR^{2/3}S^{1/2}$, trial gives $B = 2.58$ m, adopt $B = 2.6$ m. Check: $A = 31.16$ m², $R = 1.772$ m, $V = 1.610$ m/s, $Q = 50.15$ m³/s.

Answer: Stable depth $D \approx 3.35$ m, bed width $B \approx 2.6$ m, side slope 2.0H:1V.

This section is deep and narrow. Because every depth up to 3.35 m is stable, a wider and shallower section (larger $B$, smaller $D$) carrying the same discharge is equally stable and is usually preferred in practice.

Given: $Q = 20\ \text{m}^3/\text{s}$, $S = 0.000667$ (1 in 1500), particle size $d = 40$ mm, $\phi = 36^\circ$. Side slope is not given, so 2H:1V is assumed. Side slope $2.00H:1V$ (26.6° with horizontal); $\gamma = 9.81$ kN/m³, specific gravity of particles 2.65. $n = 0.0250$ (given).

Critical shear stress of the bed (Shields)

$$\tau_c = 0.047(\gamma_s - \gamma)d = 0.047\times(2.65-1)\times 9.81\times 0.04 = 30.43\ \text{N/m}^2$$

Depth limited by the bed

$$\gamma D S \le \tau_c \Rightarrow D_{bed} = \frac{\tau_c}{\gamma S} = \frac{0.03043}{9.81\times 0.000667} = 4.65\ \text{m}$$

Depth limited by the side slope

Reduction factor for the side:

$$K = \sqrt{1 - \frac{\sin^2\theta}{\sin^2\phi}} = \sqrt{1 - \frac{\sin^2 26.6^\circ}{\sin^2 36^\circ}} = 0.649$$

Maximum stress on the side is $0.75\gamma DS$, so:

$$D_{side} = \frac{K\tau_c}{0.75\gamma S} = \frac{0.649\times 0.03043}{0.75\times 9.81\times 0.000667} = 4.03\ \text{m}$$

The smaller value governs: stable depth $D = 4.03$ m (side controlled).

Section for the discharge

With $D = 4.03$ m the flow area needed for 20 m³/s is small compared with the section (even a very small bed width would carry more than 20 m³/s), so the stable depth is the upper limit of the depth. Any flow depth up to 4.03 m is stable. To pass the discharge choose a practical bed width $B = 8$ m; Manning's equation then gives a flow depth

$$D = 1.57\ \text{m}\ (< 4.03\ \text{m}),\quad A = 17.49\ \text{m}^2,\quad R = 1.165\ \text{m},\quad V = 1.143\ \text{m/s}$$

and the bed stress $\gamma DS = 10.3$ N/m² is below $\tau_c = 30.4$ N/m².

Answer: Maximum stable depth $= 4.03$ m. Adopted section: $B = 8$ m, $D = 1.57$ m, side slope 2.0H:1V, $V = 1.14$ m/s.

Given: $Q = 40\ \text{m}^3/\text{s}$, $S = 0.000833$ (1 in 1200), particle size $d = 50$ mm, $\phi = 37^\circ$. Unprotected banks, so the side slope governs; side slope assumed 2H:1V. Side slope $2.00H:1V$ (26.6° with horizontal); $\gamma = 9.81$ kN/m³, specific gravity of particles 2.65. $n = \dfrac{d^{1/6}}{21.1} = \dfrac{0.05^{1/6}}{21.1} = 0.0288$ (Strickler).

Critical shear stress of the bed (Shields)

$$\tau_c = 0.047(\gamma_s - \gamma)d = 0.047\times(2.65-1)\times 9.81\times 0.05 = 38.04\ \text{N/m}^2$$

Depth limited by the bed

$$\gamma D S \le \tau_c \Rightarrow D_{bed} = \frac{\tau_c}{\gamma S} = \frac{0.03804}{9.81\times 0.000833} = 4.65\ \text{m}$$

Depth limited by the side slope

Reduction factor for the side:

$$K = \sqrt{1 - \frac{\sin^2\theta}{\sin^2\phi}} = \sqrt{1 - \frac{\sin^2 26.6^\circ}{\sin^2 37^\circ}} = 0.669$$

Maximum stress on the side is $0.75\gamma DS$, so:

$$D_{side} = \frac{K\tau_c}{0.75\gamma S} = \frac{0.669\times 0.03804}{0.75\times 9.81\times 0.000833} = 4.15\ \text{m}$$

The smaller value governs: stable depth $D = 4.15$ m (side controlled).

Section for the discharge

With $D = 4.15$ m the flow area needed for 40 m³/s is small compared with the section (even a very small bed width would carry more than 40 m³/s), so the stable depth is the upper limit of the depth. Any flow depth up to 4.15 m is stable. To pass the discharge choose a practical bed width $B = 15$ m; Manning's equation then gives a flow depth

$$D = 1.73\ \text{m}\ (< 4.15\ \text{m}),\quad A = 31.83\ \text{m}^2,\quad R = 1.401\ \text{m},\quad V = 1.257\ \text{m/s}$$

and the bed stress $\gamma DS = 14.1$ N/m² is below $\tau_c = 38.0$ N/m².

Answer: Maximum stable depth $= 4.15$ m. Adopted section: $B = 15$ m, $D = 1.73$ m, side slope 2.0H:1V, $V = 1.26$ m/s.

Given: $Q = 32\ \text{m}^3/\text{s}$, $S = 0.000667$ (1 in 1500), particle size $d = 30$ mm, $\phi = 37^\circ$. Unprotected banks, so the side slope governs; side slope assumed 2H:1V. Side slope $2.00H:1V$ (26.6° with horizontal); $\gamma = 9.81$ kN/m³, specific gravity of particles 2.65. $n = \dfrac{d^{1/6}}{21.1} = \dfrac{0.03^{1/6}}{21.1} = 0.0264$ (Strickler).

Critical shear stress of the bed (Shields)

$$\tau_c = 0.047(\gamma_s - \gamma)d = 0.047\times(2.65-1)\times 9.81\times 0.03 = 22.82\ \text{N/m}^2$$

Depth limited by the bed

$$\gamma D S \le \tau_c \Rightarrow D_{bed} = \frac{\tau_c}{\gamma S} = \frac{0.02282}{9.81\times 0.000667} = 3.49\ \text{m}$$

Depth limited by the side slope

Reduction factor for the side:

$$K = \sqrt{1 - \frac{\sin^2\theta}{\sin^2\phi}} = \sqrt{1 - \frac{\sin^2 26.6^\circ}{\sin^2 37^\circ}} = 0.669$$

Maximum stress on the side is $0.75\gamma DS$, so:

$$D_{side} = \frac{K\tau_c}{0.75\gamma S} = \frac{0.669\times 0.02282}{0.75\times 9.81\times 0.000667} = 3.11\ \text{m}$$

The smaller value governs: stable depth $D = 3.11$ m (side controlled).

Bed width from Manning's equation

Using $D = 3.11$ m and $Q = \frac{1}{n}AR^{2/3}S^{1/2}$, trial gives $B = 1.58$ m, adopt $B = 1.6$ m. Check: $A = 24.32$ m², $R = 1.568$ m, $V = 1.319$ m/s, $Q = 32.08$ m³/s.

Answer: Stable depth $D \approx 3.11$ m, bed width $B \approx 1.6$ m, side slope 2.0H:1V.

This section is deep and narrow. Because every depth up to 3.11 m is stable, a wider and shallower section (larger $B$, smaller $D$) carrying the same discharge is equally stable and is usually preferred in practice.

Kennedy's theory (R.G. Kennedy, 1895, from the Upper Bari Doab canal, Punjab) is for the design of alluvial earthen channels that neither silt nor scour.

Salient features

• Silt is kept in suspension by the vertical components of eddies formed at the channel bed. The channel remains stable when the mean velocity equals the critical velocity $V_0$.
• Critical velocity: $V_0 = 0.55\,m\,D^{0.64}$, $D$ = depth (m), $m$ = critical velocity ratio $V/V_0$ ($m$ = 1.1 for coarse silt, 0.8 for very fine silt, 1.0 for normal).
• The mean velocity in the channel is found from Kutter's formula (or Manning's) with a rugosity coefficient $n$.
• For a given $Q$ and $S$ the section is found by trial and error so that $V = V_0$, with $A = Q/V$.
• Valid for stable channels in alluvium which carry fine silt, with trapezoidal section and side slopes taken from soil types.
• A tabular/graphical (Garret's diagram) solution can be used to reduce the work.

Limitations

1. Considers eddies from the bed only; sides ignored.
2. $m$ is chosen by judgement; there is no relation to the silt grade or the silt charge.
3. A single equation ($V_0$) is used for a design that actually involves several unknowns ($B, D, S$), so the depth and width are not uniquely determined unless $S$ or $B/D$ is assumed.
4. Requires Kutter's $n$, which is not well defined for silt-laden flow.
5. Based on the data of one region (Upper Bari Doab canals), not on fully regime channels.
6. Trial-and-error design; the bed width/depth relation $B/D$ is not controlled, and the ratio can become unreasonable.
7. Does not account for the silt load carried, so cannot be used where the canal carries heavy silt.

The semi-theoretical approach designs a canal using a standard flow formula together with empirical limits obtained from experience.

• Flow equation: Manning's, $V = \frac{1}{n}R^{2/3}S^{1/2}$, with $Q = AV$ (or Chezy's formula).
• Non-silting and non-scouring conditions are met by keeping $V$ between permissible limits: not less than about 0.3-0.6 m/s (to avoid silting and weed growth) and not more than the maximum non-scouring velocity of the boundary material.
• The ratio $B/D$ and side slopes are taken from standard tables of design practice for the given discharge range.
• The slope is chosen to suit the land slope, so the design is checked by trial and adjusted until the section carries $Q$ at the desired velocity.

It is used for canals which are not fully alluvial (hilly regions, lined canals, canals in rigid boundaries) where regime theories are not applicable. It is simple but depends on experience (empirical limits), unlike Kennedy's and Lacey's theories, which are based on silt transport concepts.