Cross-Drainage structures

Cross-drainage (CD) works are structures built where an irrigation canal meets a natural drainage (stream, nala or river), to carry the canal and the drainage across each other safely. They keep the canal supply from being lost or damaged by the drain, and pass the flood without harming the canal.

Types of cross-drainage structures

The type depends on the relative levels of the canal (bed and FSL) and the drain (bed and HFL).

1. Aqueduct (canal over drain, drain flows freely under)

 canal FSL ~~~~~~~~~~~~~~~~~~
 canal bed ___ trough ______    canal flows over barrel
              |  |   |  |
 drain HFL ~~~|__|~~~|__|~~~    HFL below canal trough
 drain bed ____________________

Applied when the drain HFL is below the canal trough (bottom of the canal), so that the drain flows with a free surface in the barrel. Types by canal section: I (earth section kept), II (earth section with retaining walls), III (rectangular trough).

2. Siphon aqueduct (canal over drain, drain flows under pressure)

 canal FSL ~~~~~~~~~~~~
 canal bed ____trough___
 drain HFL ~~~~~~~~~~~~~~ above canal bed
 barrels flow full under pressure
 drain bed ___|  |__|  |__

Applied when the drain HFL is higher than the canal bed (and the canal is above drain bed), so the barrel runs full under pressure. Used for large drain discharges.

3. Super-passage (drain over canal, canal flows under)

 drain bed ____ trough ____   drain flows over
 canal FSL ~~~~~~~~~~~~~~~~   canal below, free surface

Applied when the drain bed is above the canal FSL, so the canal flows freely below the drain.

4. Canal siphon (drain over canal, canal under pressure)

 drain ____trough____
 canal FSL ~~~~ full pipe/barrel below drain bed

Applied when the drain bed is lower than the canal FSL, the canal passes under the drain in a closed barrel under pressure.

5. Level crossing (canal and drain at the same level)

 canal --->[canal reg.]      [drain reg.]<--- drain
              crossing intersect at same level

Applied when the canal FSL and the drain HFL are at about the same level. A regulator in the drain and a regulator in the canal control the flow.

6. Inlet and outlet

 drain ====> inlet (opening in canal bank)  canal
 canal ====> outlet ====> drain

An inlet admits a small drain into the canal when it is small and the canal has capacity to carry the drain flow. An outlet lets canal water out into the drain, or is used with the inlet to pass a drain across a canal at the same level.

The canal (25 m³/s, bed 160.00 m, FSL 161.50 m) crosses a drain of 800 m³/s with HFL 160.50 m, which is above the canal bed. The canal bed (160.00 m) is above the drain bed (158.00 m), so the structure is a siphon aqueduct: the drain flows through rectangular barrels under the canal trough, partly under pressure.

1. Drainage waterway (Lacey)

$$P = 4.75\sqrt{Q} = 4.75\sqrt{800} = 134.4\ \text{m}$$

Barrels of clear span 4.0 m with walls 0.4 m thick: 32 barrels give a clear width of 32×4.0 = 128 m (about the regime width, accepted as the obstruction is by walls only) and overall width 141.2 m.

Barrel height: bed 158.00 m to the roof soffit $= 160.00 - 0.40 = 159.60$ m, so the clear height is 1.6 m (roof 0.4 m thick, whose top is the canal bed). Area $= 128\times1.6 = 205$ m²; velocity $V = 800/205 = 3.91$ m/s; $V^2/2g = 0.778$ m.

Head loss (entry + exit $\approx 1.5V^2/2g$ plus Manning friction, $n = 0.015$, $R = 0.571$ m, length 34.5 m):

$$h_L = 1.5\times0.778 + \frac{n^2V^2L}{R^{4/3}} = 1.167 + 0.250 = 1.42\ \text{m}$$

(the afflux is absorbed by the flood bank and a flared u/s wing).

2. Canal waterway

Canal width in the trough is kept equal to the bed width of 30 m, depth 1.5 m (FSL 161.50 m), so the canal waterway is 30 m; the trough is a rectangular concrete section, with transitions (not needed since full width is kept). Barrel length along the canal axis $= 30 + 2\times1.5\times1.5 = 34.5$ m (including the canal side slopes of 1.5:1).

3. Uplift pressure on the roof of the barrel

Worst case: drain at HFL (160.50 m) and barrel full, canal empty. Pressure head at the roof soffit $= 160.50 - 159.60 = 0.90$ m.

$$\begin{aligned} \text{Uplift} &= 9.81\times0.90 = 8.83\ \text{kN/m}^2 \\ \text{Roof weight} &= 0.4\times24 = 9.6\ \text{kN/m}^2 \\ \text{FOS (canal empty)} &= 9.6/8.83 = 1.09 \end{aligned}$$

With the canal running at FSL, the water on the roof is $1.5\times9.81 = 14.7$ kN/m², so FOS $= (9.6+14.7)/8.83 = 2.75$ (safe). The roof is also checked against flow conditions: when the barrel runs full at Q = 800 m³/s the pressure at the soffit is negative (energy level 160.50 m is used in 1.17 m of entry loss and velocity head), so no uplift occurs; the still-water case governs. As FOS with empty canal is only 1.09, the roof should be reinforced and tied to the walls.

4. Impervious floor

Maximum seepage head (drain at HFL u/s, d/s dry): $H_s = 160.50 - 158.00 = 2.50$ m. Scour: $q = 800/134.4 = 5.95$ m²/s, $R = 1.35(q^2/f)^{1/3} = 4.44$ m ($f = 1$), scour level $= 160.50 - 1.5R = 153.85$ m, so d/s cut-off $d_2 = 4.2$ m below the floor; u/s cut-off $d_1 = 2.5$ m.

Creep path (Bligh): $2d_1 + 4 + 34.5 + 8 + 2d_2 = 59.9$ m. Uplift head at points on the floor (linear loss):

With $G = 2.24$, the barrel floor slab of the box is designed together with the weight of the walls, roof and canal water. The required thickness at the start of the floor from uplift alone is $t = \dfrac{4}{3}\dfrac{h}{G-1} = 2.28$ m; the box action (walls, roof, canal water) and the u/s cut-off carry the uplift, so the floor slab is made 0.6 m with pressure relief by weep holes at the d/s end.

Answer: drainage waterway 128 m clear (32 barrels of 4.0×1.6 m), canal waterway 30 m, uplift on the roof $= 8.8$ kN/m² (head 0.9 m, FOS 1.09 canal empty), and floor uplift heads 2.12 m (start) to 0.68 m (end).

 canal FSL 161.5 ~~~~~~~~~~~~~~~~~~~~
 canal bed 160.0 ==roof 0.4 m==========
 HFL 160.5 ~~~~ | 4 m | 4 m |  barrels   1.6 m
 drain bed 158.0 |_____|_____|___________
   u/s cut-off 2.5 m   floor    d/s cut-off 4.2 m

The canal ($Q$ = 50 m³/s) crosses a drainage of $Q$ = 550 m³/s. Design is done in four steps: type of work, drainage waterway, canal waterway (flumed trough) and bed/water levels of the trough.

Assumptions: Side slope 1.5:1 (H:V), Lacey's $f$ = 1, concrete trough with Manning's $n$ = 0.015, piers 1.2–1.5 m thick, slab thickness 0.6 m.

1. Type of structure

HFL of the drainage (161.00 m) is above the canal bed (160.00 m) by 1.00 m, so the trough cannot be kept above HFL without a very high, costly structure. The drainage is passed under the trough through barrels flowing under pressure (siphonic action). A siphon aqueduct is suitable. Drainage depth at HFL = 161.00 − 158.00 = 3.00 m.

2. Drainage waterway

Lacey's regime perimeter (for a natural stream):

$$P = 4.75\sqrt{Q} = 4.75\sqrt{550} = 111.4\ \text{m}$$

Provide 12 spans of 10 m with piers 1.5 m thick: clear waterway = 12 × 10 = 120 m (> 111.4 m). Overall length = 120 + 11 × 1.5 = 136.5 m.

Check: mean velocity = Q/(L·d) = 550/(120 × 3.00) = 1.53 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.85 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.8 m below HFL, with cut-offs and aprons for scour.

3. Canal waterway (flumed rectangular trough)

Normal canal: area $A_1=(B+zD)D$ = (30 + 1.5 × 2.1) × 2.1 = 69.61 m², $V_1$ = 50/69.61 = 0.718 m/s, $h_{v1}=V_1^2/2g$ = 0.0263 m. Top width $T=B+2zD$ = 36.30 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 13 m, with the same full supply depth 2.1 m (the trough is as long as the drainage waterway, 136.5 m).

• Trough area = 13 × 2.1 = 27.30 m²; $V_t$ = 50/27.30 = 1.832 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1710 m
• Froude number = 0.40 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 27.30/17.20 = 1.587 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000408 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000408 × 136.5 = 0.056 m

4. Transitions and levels

  FSL ---.                            .--- FSL
         '--.______________________.--'
  bed ---.                            .--- bed
         '--.______________________.--'
  normal  inlet     flumed trough     outlet  normal
  canal   (2:1)                      (3:1)   canal

Wing-wall (straight) transitions: inlet splay 2:1 (contraction) and outlet splay 3:1 (expansion), measured on each side:

• Inlet length = 2 × (T − B_t)/2 = 36.30 − 13 = 23.3 m
• Outlet length = 3 × (T − B_t)/2 = 34.9 m

Bernoulli's equation between sections (loss in contraction = 0.3 Δh_v, in expansion = 0.5 Δh_v; Δh_v = difference of velocity heads):

$$\begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned}$$

$W_1$ = 162.100 m, so $W_2$ = 161.912 m, $W_3$ = 161.856 m, $W_4$ = 161.929 m. Trough bed = water level − 2.1 m.

Total loss of head = 162.100 − 161.929 = 0.171 m, so the canal bed downstream of the work is 0.171 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 159.16 m, i.e. 1.16 m above the drainage bed; HFL stands 1.84 m above the soffit, which gives the siphonic head. Since the clear height is less than 1.5 m, the drainage bed is depressed under the barrels to RL 157.66 m (1.5 m clear height) with a 1 in 4 slope at both ends.

Answer: drainage waterway 120 m clear (12 × 10 m, overall 136.5 m); canal flumed to 13 m wide trough; bed/FSL at the four sections = 160.00/162.10, 159.81/161.91, 159.76/161.86, 159.83/161.93 m.

The canal ($Q$ = 50 m³/s) crosses a drainage of $Q$ = 450 m³/s. Design is done in four steps: type of work, drainage waterway, canal waterway (flumed trough) and bed/water levels of the trough.

Assumptions: Lacey's $f$ = 1, concrete trough with $n$ = 0.015, piers 1.0–1.5 m thick, slab thickness 0.6 m.

1. Type of structure

HFL of the drainage (161.00 m) is above the canal bed (160.00 m) by 1.00 m, so the trough cannot be kept above HFL without a very high, costly structure. The drainage is passed under the trough through barrels flowing under pressure (siphonic action). A siphon aqueduct is suitable. Drainage depth at HFL = 161.00 − 158.00 = 3.00 m.

2. Drainage waterway

Lacey's regime perimeter (for a natural stream):

$$P = 4.75\sqrt{Q} = 4.75\sqrt{450} = 100.8\ \text{m}$$

Provide 11 spans of 10 m with piers 1.5 m thick: clear waterway = 11 × 10 = 110 m (> 100.8 m). Overall length = 110 + 10 × 1.5 = 125.0 m.

Check: mean velocity = Q/(L·d) = 450/(110 × 3.00) = 1.36 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.60 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.4 m below HFL, with cut-offs and aprons for scour.

3. Canal waterway (flumed rectangular trough)

Normal canal: area $A_1=(B+zD)D$ = (30 + 1.5 × 2.1) × 2.1 = 69.61 m², $V_1$ = 50/69.61 = 0.718 m/s, $h_{v1}=V_1^2/2g$ = 0.0263 m. Top width $T=B+2zD$ = 36.30 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 13 m, with the same full supply depth 2.1 m (the trough is as long as the drainage waterway, 125.0 m).

• Trough area = 13 × 2.1 = 27.30 m²; $V_t$ = 50/27.30 = 1.832 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1710 m
• Froude number = 0.40 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 27.30/17.20 = 1.587 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000408 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000408 × 125.0 = 0.051 m

4. Transitions and levels

  FSL ---.                            .--- FSL
         '--.______________________.--'
  bed ---.                            .--- bed
         '--.______________________.--'
  normal  inlet     flumed trough     outlet  normal
  canal   (2:1)                      (3:1)   canal

Wing-wall (straight) transitions: inlet splay 2:1 (contraction) and outlet splay 3:1 (expansion), measured on each side:

• Inlet length = 2 × (T − B_t)/2 = 36.30 − 13 = 23.3 m
• Outlet length = 3 × (T − B_t)/2 = 34.9 m

Bernoulli's equation between sections (loss in contraction = 0.3 Δh_v, in expansion = 0.5 Δh_v; Δh_v = difference of velocity heads):

$$\begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned}$$

$W_1$ = 162.100 m, so $W_2$ = 161.912 m, $W_3$ = 161.861 m, $W_4$ = 161.933 m. Trough bed = water level − 2.1 m.

Total loss of head = 162.100 − 161.933 = 0.167 m, so the canal bed downstream of the work is 0.167 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 159.16 m, i.e. 1.16 m above the drainage bed; HFL stands 1.84 m above the soffit, which gives the siphonic head. Since the clear height is less than 1.5 m, the drainage bed is depressed under the barrels to RL 157.66 m (1.5 m clear height) with a 1 in 4 slope at both ends.

Answer: drainage waterway 110 m clear (11 × 10 m, overall 125.0 m); canal flumed to 13 m wide trough; bed/FSL at the four sections = 160.00/162.10, 159.81/161.91, 159.76/161.86, 159.83/161.93 m.

The canal ($Q$ = 50 m³/s) crosses a drainage of $Q$ = 400 m³/s. Design is done in four steps: type of work, drainage waterway, canal waterway (flumed trough) and bed/water levels of the trough.

Assumptions: Side slope not given, taken 1.5:1; Lacey's $f$ = 1; trough $n$ = 0.015.

1. Type of structure

HFL of the drainage (300.50 m) is above the canal bed (300.00 m) by 0.50 m, so the trough cannot be kept above HFL without a very high, costly structure. The drainage is passed under the trough through barrels flowing under pressure (siphonic action). A siphon aqueduct is suitable. Drainage depth at HFL = 300.50 − 298.80 = 1.70 m.

2. Drainage waterway

Lacey's regime perimeter (for a natural stream):

$$P = 4.75\sqrt{Q} = 4.75\sqrt{400} = 95.0\ \text{m}$$

Provide 12 spans of 8 m with piers 1.2 m thick: clear waterway = 12 × 8 = 96 m (> 95.0 m). Overall length = 96 + 11 × 1.2 = 109.2 m.

Check: mean velocity = Q/(L·d) = 400/(96 × 1.70) = 2.45 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.46 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.2 m below HFL, with cut-offs and aprons for scour.

3. Canal waterway (flumed rectangular trough)

Normal canal: area $A_1=(B+zD)D$ = (30 + 1.5 × 1.55) × 1.55 = 50.10 m², $V_1$ = 50/50.10 = 0.998 m/s, $h_{v1}=V_1^2/2g$ = 0.0508 m. Top width $T=B+2zD$ = 34.65 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 18 m, with the same full supply depth 1.55 m (the trough is as long as the drainage waterway, 109.2 m).

• Trough area = 18 × 1.55 = 27.90 m²; $V_t$ = 50/27.90 = 1.792 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1637 m
• Froude number = 0.46 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 27.90/21.10 = 1.322 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000498 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000498 × 109.2 = 0.054 m

4. Transitions and levels

  FSL ---.                            .--- FSL
         '--.______________________.--'
  bed ---.                            .--- bed
         '--.______________________.--'
  normal  inlet     flumed trough     outlet  normal
  canal   (2:1)                      (3:1)   canal

Wing-wall (straight) transitions: inlet splay 2:1 (contraction) and outlet splay 3:1 (expansion), measured on each side:

• Inlet length = 2 × (T − B_t)/2 = 34.65 − 18 = 16.6 m
• Outlet length = 3 × (T − B_t)/2 = 25.0 m

Bernoulli's equation between sections (loss in contraction = 0.3 Δh_v, in expansion = 0.5 Δh_v; Δh_v = difference of velocity heads):

$$\begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned}$$

$W_1$ = 301.550 m, so $W_2$ = 301.403 m, $W_3$ = 301.349 m, $W_4$ = 301.405 m. Trough bed = water level − 1.55 m.

Total loss of head = 301.550 − 301.405 = 0.145 m, so the canal bed downstream of the work is 0.145 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 299.20 m, i.e. 0.40 m above the drainage bed; HFL stands 1.30 m above the soffit, which gives the siphonic head. Since the clear height is less than 1.5 m, the drainage bed is depressed under the barrels to RL 297.70 m (1.5 m clear height) with a 1 in 4 slope at both ends.

Answer: drainage waterway 96 m clear (12 × 8 m, overall 109.2 m); canal flumed to 18 m wide trough; bed/FSL at the four sections = 300.00/301.55, 299.85/301.40, 299.80/301.35, 299.86/301.41 m.

The canal ($Q$ = 50 m³/s) crosses a drainage of $Q$ = 450 m³/s. Design is done in four steps: type of work, drainage waterway, canal waterway (flumed trough) and bed/water levels of the trough.

Assumptions: Side slope 1.5:1; Lacey's $f$ = 1; trough $n$ = 0.015; roof (trough) slab 0.6 m thick, unit weight of concrete 24 kN/m³.

1. Type of structure

HFL of the drainage (100.50 m) is above the canal bed (100.00 m) by 0.50 m, so the trough cannot be kept above HFL without a very high, costly structure. The drainage is passed under the trough through barrels flowing under pressure (siphonic action). A siphon aqueduct is suitable. Drainage depth at HFL = 100.50 − 98.80 = 1.70 m.

2. Drainage waterway

Lacey's regime perimeter (for a natural stream):

$$P = 4.75\sqrt{Q} = 4.75\sqrt{450} = 100.8\ \text{m}$$

Provide 11 spans of 10 m with piers 1.5 m thick: clear waterway = 11 × 10 = 110 m (> 100.8 m). Overall length = 110 + 10 × 1.5 = 125.0 m.

Check: mean velocity = Q/(L·d) = 450/(110 × 1.70) = 2.41 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.60 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.4 m below HFL, with cut-offs and aprons for scour.

3. Canal waterway (flumed rectangular trough)

Normal canal: area $A_1=(B+zD)D$ = (30 + 1.5 × 1.5) × 1.5 = 48.38 m², $V_1$ = 50/48.38 = 1.034 m/s, $h_{v1}=V_1^2/2g$ = 0.0545 m. Top width $T=B+2zD$ = 34.50 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 19 m, with the same full supply depth 1.5 m (the trough is as long as the drainage waterway, 125.0 m).

• Trough area = 19 × 1.5 = 28.50 m²; $V_t$ = 50/28.50 = 1.754 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1569 m
• Froude number = 0.46 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 28.50/22.00 = 1.295 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000490 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000490 × 125.0 = 0.061 m

4. Transitions and levels

  FSL ---.                            .--- FSL
         '--.______________________.--'
  bed ---.                            .--- bed
         '--.______________________.--'
  normal  inlet     flumed trough     outlet  normal
  canal   (2:1)                      (3:1)   canal

Wing-wall (straight) transitions: inlet splay 2:1 (contraction) and outlet splay 3:1 (expansion), measured on each side:

• Inlet length = 2 × (T − B_t)/2 = 34.50 − 19 = 15.5 m
• Outlet length = 3 × (T − B_t)/2 = 23.2 m

Bernoulli's equation between sections (loss in contraction = 0.3 Δh_v, in expansion = 0.5 Δh_v; Δh_v = difference of velocity heads):

$$\begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned}$$

$W_1$ = 101.500 m, so $W_2$ = 101.367 m, $W_3$ = 101.306 m, $W_4$ = 101.357 m. Trough bed = water level − 1.5 m.

Total loss of head = 101.500 − 101.357 = 0.143 m, so the canal bed downstream of the work is 0.143 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 99.21 m, i.e. 0.41 m above the drainage bed; HFL stands 1.29 m above the soffit, which gives the siphonic head. Since the clear height is less than 1.5 m, the drainage bed is depressed under the barrels to RL 97.71 m (1.5 m clear height) with a 1 in 4 slope at both ends.

Uplift on the roof (trough slab)

Upward pressure under the slab (barrel running full) = γ_w (HFL − soffit) = 9.81 × (100.50 − 99.21) = 12.7 kN/m².

Downward load = slab weight 24 × 0.6 = 14.4 kN/m² plus canal water 9.81 × 1.5 = 14.7 kN/m², total 29.1 kN/m².

Uplift (12.7) is less than the slab weight alone (14.4), so the roof does not float even with the canal dry; net downward load with canal running = 16.4 kN/m². The slab is reinforced for bending under these loads.

Answer: drainage waterway 110 m clear (11 × 10 m, overall 125.0 m); canal flumed to 19 m wide trough; bed/FSL at the four sections = 100.00/101.50, 99.87/101.37, 99.81/101.31, 99.86/101.36 m.

The canal ($Q$ = 60 m³/s) crosses a drainage of $Q$ = 500 m³/s. Design is done in four steps: type of work, drainage waterway, canal waterway (flumed trough) and bed/water levels of the trough.

Assumptions: Lacey's $f$ = 1; trough $n$ = 0.015; slab 0.6 m thick.

1. Type of structure

HFL of the drainage (300.80 m) is above the canal bed (300.00 m) by 0.80 m, so the trough cannot be kept above HFL without a very high, costly structure. The drainage is passed under the trough through barrels flowing under pressure (siphonic action). A siphon aqueduct is suitable. Drainage depth at HFL = 300.80 − 298.20 = 2.60 m.

2. Drainage waterway

Lacey's regime perimeter (for a natural stream):

$$P = 4.75\sqrt{Q} = 4.75\sqrt{500} = 106.2\ \text{m}$$

Provide 11 spans of 10 m with piers 1.5 m thick: clear waterway = 11 × 10 = 110 m (> 106.2 m). Overall length = 110 + 10 × 1.5 = 125.0 m.

Check: mean velocity = Q/(L·d) = 500/(110 × 2.60) = 1.75 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.73 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.6 m below HFL, with cut-offs and aprons for scour.

3. Canal waterway (flumed rectangular trough)

Normal canal: area $A_1=(B+zD)D$ = (35 + 1.5 × 2) × 2 = 76.00 m², $V_1$ = 60/76.00 = 0.789 m/s, $h_{v1}=V_1^2/2g$ = 0.0318 m. Top width $T=B+2zD$ = 41.00 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 17 m, with the same full supply depth 2 m (the trough is as long as the drainage waterway, 125.0 m).

• Trough area = 17 × 2 = 34.00 m²; $V_t$ = 60/34.00 = 1.765 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1587 m
• Froude number = 0.40 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 34.00/21.00 = 1.619 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000369 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000369 × 125.0 = 0.046 m

4. Transitions and levels

  FSL ---.                            .--- FSL
         '--.______________________.--'
  bed ---.                            .--- bed
         '--.______________________.--'
  normal  inlet     flumed trough     outlet  normal
  canal   (2:1)                      (3:1)   canal

Wing-wall (straight) transitions: inlet splay 2:1 (contraction) and outlet splay 3:1 (expansion), measured on each side:

• Inlet length = 2 × (T − B_t)/2 = 41.00 − 17 = 24.0 m
• Outlet length = 3 × (T − B_t)/2 = 36.0 m

Bernoulli's equation between sections (loss in contraction = 0.3 Δh_v, in expansion = 0.5 Δh_v; Δh_v = difference of velocity heads):

$$\begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned}$$

$W_1$ = 302.000 m, so $W_2$ = 301.835 m, $W_3$ = 301.789 m, $W_4$ = 301.852 m. Trough bed = water level − 2 m.

Total loss of head = 302.000 − 301.852 = 0.148 m, so the canal bed downstream of the work is 0.148 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 299.19 m, i.e. 0.99 m above the drainage bed; HFL stands 1.61 m above the soffit, which gives the siphonic head. Since the clear height is less than 1.5 m, the drainage bed is depressed under the barrels to RL 297.69 m (1.5 m clear height) with a 1 in 4 slope at both ends.

Answer: drainage waterway 110 m clear (11 × 10 m, overall 125.0 m); canal flumed to 17 m wide trough; bed/FSL at the four sections = 300.00/302.00, 299.83/301.83, 299.79/301.79, 299.85/301.85 m.

The canal ($Q$ = 25 m³/s) crosses a drainage of $Q$ = 360 m³/s. Design is done in four steps: type of work, drainage waterway, canal waterway (flumed trough) and bed/water levels of the trough.

Assumptions: Lacey's $f$ = 1; trough $n$ = 0.015; slab 0.6 m.

1. Type of structure

HFL of the drainage (211.00 m) is above the canal bed (210.30 m) by 0.70 m, so the trough cannot be kept above HFL without a very high, costly structure. The drainage is passed under the trough through barrels flowing under pressure (siphonic action). A siphon aqueduct is suitable. Drainage depth at HFL = 211.00 − 208.50 = 2.50 m.

2. Drainage waterway

Lacey's regime perimeter (for a natural stream):

$$P = 4.75\sqrt{Q} = 4.75\sqrt{360} = 90.1\ \text{m}$$

Provide 12 spans of 8 m with piers 1.2 m thick: clear waterway = 12 × 8 = 96 m (> 90.1 m). Overall length = 96 + 11 × 1.2 = 109.2 m.

Check: mean velocity = Q/(L·d) = 360/(96 × 2.50) = 1.50 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.34 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.0 m below HFL, with cut-offs and aprons for scour.

3. Canal waterway (flumed rectangular trough)

Normal canal: area $A_1=(B+zD)D$ = (30 + 1.5 × 2) × 2 = 66.00 m², $V_1$ = 25/66.00 = 0.379 m/s, $h_{v1}=V_1^2/2g$ = 0.0073 m. Top width $T=B+2zD$ = 36.00 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 7 m, with the same full supply depth 2 m (the trough is as long as the drainage waterway, 109.2 m).

• Trough area = 7 × 2 = 14.00 m²; $V_t$ = 25/14.00 = 1.786 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1625 m
• Froude number = 0.40 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 14.00/11.00 = 1.273 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000520 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000520 × 109.2 = 0.057 m

4. Transitions and levels

  FSL ---.                            .--- FSL
         '--.______________________.--'
  bed ---.                            .--- bed
         '--.______________________.--'
  normal  inlet     flumed trough     outlet  normal
  canal   (2:1)                      (3:1)   canal

Wing-wall (straight) transitions: inlet splay 2:1 (contraction) and outlet splay 3:1 (expansion), measured on each side:

• Inlet length = 2 × (T − B_t)/2 = 36.00 − 7 = 29.0 m
• Outlet length = 3 × (T − B_t)/2 = 43.5 m

Bernoulli's equation between sections (loss in contraction = 0.3 Δh_v, in expansion = 0.5 Δh_v; Δh_v = difference of velocity heads):

$$\begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned}$$

$W_1$ = 212.300 m, so $W_2$ = 212.098 m, $W_3$ = 212.041 m, $W_4$ = 212.119 m. Trough bed = water level − 2 m.

Total loss of head = 212.300 − 212.119 = 0.181 m, so the canal bed downstream of the work is 0.181 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 209.44 m, i.e. 0.94 m above the drainage bed; HFL stands 1.56 m above the soffit, which gives the siphonic head. Since the clear height is less than 1.5 m, the drainage bed is depressed under the barrels to RL 207.94 m (1.5 m clear height) with a 1 in 4 slope at both ends.

Answer: drainage waterway 96 m clear (12 × 8 m, overall 109.2 m); canal flumed to 7 m wide trough; bed/FSL at the four sections = 210.30/212.30, 210.10/212.10, 210.04/212.04, 210.12/212.12 m.

The canal ($Q$ = 50 m³/s) crosses a drainage of $Q$ = 400 m³/s. Design is done in four steps: type of work, drainage waterway, canal waterway (flumed trough) and bed/water levels of the trough.

Assumptions: Lacey's $f$ = 1; trough $n$ = 0.015; slab 0.6 m.

1. Type of structure

HFL of the drainage (211.00 m) is above the canal bed (210.30 m) by 0.70 m, so the trough cannot be kept above HFL without a very high, costly structure. The drainage is passed under the trough through barrels flowing under pressure (siphonic action). A siphon aqueduct is suitable. Drainage depth at HFL = 211.00 − 208.50 = 2.50 m.

2. Drainage waterway

Lacey's regime perimeter (for a natural stream):

$$P = 4.75\sqrt{Q} = 4.75\sqrt{400} = 95.0\ \text{m}$$

Provide 12 spans of 8 m with piers 1.2 m thick: clear waterway = 12 × 8 = 96 m (> 95.0 m). Overall length = 96 + 11 × 1.2 = 109.2 m.

Check: mean velocity = Q/(L·d) = 400/(96 × 2.50) = 1.67 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.46 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.2 m below HFL, with cut-offs and aprons for scour.

3. Canal waterway (flumed rectangular trough)

Normal canal: area $A_1=(B+zD)D$ = (35 + 1.5 × 1.6) × 1.6 = 59.84 m², $V_1$ = 50/59.84 = 0.836 m/s, $h_{v1}=V_1^2/2g$ = 0.0356 m. Top width $T=B+2zD$ = 39.80 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 17 m, with the same full supply depth 1.6 m (the trough is as long as the drainage waterway, 109.2 m).

• Trough area = 17 × 1.6 = 27.20 m²; $V_t$ = 50/27.20 = 1.838 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1722 m
• Froude number = 0.46 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 27.20/20.20 = 1.347 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000511 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000511 × 109.2 = 0.056 m

4. Transitions and levels

  FSL ---.                            .--- FSL
         '--.______________________.--'
  bed ---.                            .--- bed
         '--.______________________.--'
  normal  inlet     flumed trough     outlet  normal
  canal   (2:1)                      (3:1)   canal

Wing-wall (straight) transitions: inlet splay 2:1 (contraction) and outlet splay 3:1 (expansion), measured on each side:

• Inlet length = 2 × (T − B_t)/2 = 39.80 − 17 = 22.8 m
• Outlet length = 3 × (T − B_t)/2 = 34.2 m

Bernoulli's equation between sections (loss in contraction = 0.3 Δh_v, in expansion = 0.5 Δh_v; Δh_v = difference of velocity heads):

$$\begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned}$$

$W_1$ = 211.900 m, so $W_2$ = 211.722 m, $W_3$ = 211.667 m, $W_4$ = 211.735 m. Trough bed = water level − 1.6 m.

Total loss of head = 211.900 − 211.735 = 0.165 m, so the canal bed downstream of the work is 0.165 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 209.47 m, i.e. 0.97 m above the drainage bed; HFL stands 1.53 m above the soffit, which gives the siphonic head. Since the clear height is less than 1.5 m, the drainage bed is depressed under the barrels to RL 207.97 m (1.5 m clear height) with a 1 in 4 slope at both ends.

Answer: drainage waterway 96 m clear (12 × 8 m, overall 109.2 m); canal flumed to 17 m wide trough; bed/FSL at the four sections = 210.30/211.90, 210.12/211.72, 210.07/211.67, 210.13/211.73 m.

The canal ($Q$ = 50 m³/s) crosses a drainage of $Q$ = 400 m³/s. Design is done in four steps: type of work, drainage waterway, canal waterway (flumed trough) and bed/water levels of the trough.

Assumptions: Lacey's $f$ = 1; trough $n$ = 0.015; slab 0.6 m.

1. Type of structure

HFL of the drainage (200.60 m) is above the canal bed (200.00 m) by 0.60 m, so the trough cannot be kept above HFL without a very high, costly structure. The drainage is passed under the trough through barrels flowing under pressure (siphonic action). A siphon aqueduct is suitable. Drainage depth at HFL = 200.60 − 198.00 = 2.60 m.

2. Drainage waterway

Lacey's regime perimeter (for a natural stream):

$$P = 4.75\sqrt{Q} = 4.75\sqrt{400} = 95.0\ \text{m}$$

Provide 12 spans of 8 m with piers 1.2 m thick: clear waterway = 12 × 8 = 96 m (> 95.0 m). Overall length = 96 + 11 × 1.2 = 109.2 m.

Check: mean velocity = Q/(L·d) = 400/(96 × 2.60) = 1.60 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.46 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.2 m below HFL, with cut-offs and aprons for scour.

3. Canal waterway (flumed rectangular trough)

Normal canal: area $A_1=(B+zD)D$ = (32 + 1.5 × 1.8) × 1.8 = 62.46 m², $V_1$ = 50/62.46 = 0.801 m/s, $h_{v1}=V_1^2/2g$ = 0.0327 m. Top width $T=B+2zD$ = 37.40 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 15 m, with the same full supply depth 1.8 m (the trough is as long as the drainage waterway, 109.2 m).

• Trough area = 15 × 1.8 = 27.00 m²; $V_t$ = 50/27.00 = 1.852 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1748 m
• Froude number = 0.44 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 27.00/18.60 = 1.452 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000469 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000469 × 109.2 = 0.051 m

4. Transitions and levels

  FSL ---.                            .--- FSL
         '--.______________________.--'
  bed ---.                            .--- bed
         '--.______________________.--'
  normal  inlet     flumed trough     outlet  normal
  canal   (2:1)                      (3:1)   canal

Wing-wall (straight) transitions: inlet splay 2:1 (contraction) and outlet splay 3:1 (expansion), measured on each side:

• Inlet length = 2 × (T − B_t)/2 = 37.40 − 15 = 22.4 m
• Outlet length = 3 × (T − B_t)/2 = 33.6 m

Bernoulli's equation between sections (loss in contraction = 0.3 Δh_v, in expansion = 0.5 Δh_v; Δh_v = difference of velocity heads):

$$\begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned}$$

$W_1$ = 201.800 m, so $W_2$ = 201.615 m, $W_3$ = 201.564 m, $W_4$ = 201.635 m. Trough bed = water level − 1.8 m.

Total loss of head = 201.800 − 201.635 = 0.165 m, so the canal bed downstream of the work is 0.165 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 199.16 m, i.e. 1.16 m above the drainage bed; HFL stands 1.44 m above the soffit, which gives the siphonic head. Since the clear height is less than 1.5 m, the drainage bed is depressed under the barrels to RL 197.66 m (1.5 m clear height) with a 1 in 4 slope at both ends.

Answer: drainage waterway 96 m clear (12 × 8 m, overall 109.2 m); canal flumed to 15 m wide trough; bed/FSL at the four sections = 200.00/201.80, 199.82/201.62, 199.76/201.56, 199.84/201.64 m.

Definition (1 mark)

A cross-drainage structure is a structure built where an irrigation canal crosses a natural drain or stream, to carry the canal and the drainage water across each other without mixing or damage.

Types (2 marks)

1. Aqueduct (canal over drain, drain flows freely).
2. Siphon aqueduct (canal over drain, drain flows under pressure).
3. Super-passage (drain over canal, canal under free surface).
4. Canal siphon (drain over canal, canal under pressure).
5. Level crossing (canal and drain at same level, with regulators).
6. Inlet and outlet (drain enters or leaves the canal).

Type favourable in the hills of Nepal (1 mark)

The aqueduct (canal flume or trough carried over the drain) is most favourable. In the hills, canals run along the hillside high above the many small, steep streams (khola), which are deeply cut, and the bed levels of the streams are well below the canal bed. An aqueduct:

• Passes sudden, high-velocity floods with boulders and debris freely under the structure.
• Needs no cut-off or floor for the drain bed (rocky beds).
• Avoids silting and choking of barrels, which is a problem of siphon types.
• Is simple and cheap in stone masonry or RCC, and the canal supply is safe from the flood.

The canal ($Q$ = 30 m³/s) crosses a drainage of $Q$ = 550 m³/s. Design is done in four steps: type of work, drainage waterway, canal waterway (flumed trough) and bed/water levels of the trough.

Assumptions: Bed level 310.0 m is the downstream canal bed; general ground level = canal bed = 310.0 m; Lacey's $f$ = 1; concrete trough $n$ = 0.015. The 50-year flood (550 m³/s) is the design flood.

1. Type of structure

HFL of the drainage (311.00 m) is above the canal bed (310.00 m) by 1.00 m, so the trough cannot be kept above HFL without a very high, costly structure. The drainage is passed under the trough through barrels flowing under pressure (siphonic action). A siphon aqueduct is suitable. Drainage depth at HFL = 311.00 − 308.00 = 3.00 m.

2. Drainage waterway

Lacey's regime perimeter (for a natural stream):

$$P = 4.75\sqrt{Q} = 4.75\sqrt{550} = 111.4\ \text{m}$$

Provide 12 spans of 10 m with piers 1.5 m thick: clear waterway = 12 × 10 = 120 m (> 111.4 m). Overall length = 120 + 11 × 1.5 = 136.5 m.

Check: mean velocity = Q/(L·d) = 550/(120 × 3.00) = 1.53 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.85 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.8 m below HFL, with cut-offs and aprons for scour.

3. Canal waterway (flumed rectangular trough)

Normal canal: area $A_1=(B+zD)D$ = (30 + 1.5 × 2.1) × 2.1 = 69.61 m², $V_1$ = 30/69.61 = 0.431 m/s, $h_{v1}=V_1^2/2g$ = 0.0095 m. Top width $T=B+2zD$ = 36.30 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 8 m, with the same full supply depth 2.1 m (the trough is as long as the drainage waterway, 136.5 m).

• Trough area = 8 × 2.1 = 16.80 m²; $V_t$ = 30/16.80 = 1.786 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1625 m
• Froude number = 0.39 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 16.80/12.20 = 1.377 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000468 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000468 × 136.5 = 0.064 m

4. Transitions and levels

  FSL ---.                            .--- FSL
         '--.______________________.--'
  bed ---.                            .--- bed
         '--.______________________.--'
  normal  inlet     flumed trough     outlet  normal
  canal   (2:1)                      (3:1)   canal

Wing-wall (straight) transitions: inlet splay 2:1 (contraction) and outlet splay 3:1 (expansion), measured on each side:

• Inlet length = 2 × (T − B_t)/2 = 36.30 − 8 = 28.3 m
• Outlet length = 3 × (T − B_t)/2 = 42.4 m

Bed level at the downstream canal is fixed (310.00 m). Working backwards with Bernoulli's equation (loss in contraction = 0.3 Δh_v, in expansion = 0.5 Δh_v):

$$\begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned}$$

$W_1$ = 312.286 m, so $W_2$ = 312.087 m, $W_3$ = 312.023 m, $W_4$ = 312.100 m. Trough bed = water level − 2.1 m.

Total loss of head = 312.286 − 312.100 = 0.186 m, so the canal bed downstream of the work is 0.186 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 309.32 m, i.e. 1.32 m above the drainage bed; HFL stands 1.68 m above the soffit, which gives the siphonic head. Since the clear height is less than 1.5 m, the drainage bed is depressed under the barrels to RL 307.82 m (1.5 m clear height) with a 1 in 4 slope at both ends.

Answer: drainage waterway 120 m clear (12 × 10 m, overall 136.5 m); canal flumed to 8 m wide trough; bed/FSL at the four sections = 310.19/312.29, 309.99/312.09, 309.92/312.02, 310.00/312.10 m.

The canal ($Q$ = 60 m³/s) crosses a drainage of $Q$ = 550 m³/s. Design is done in four steps: type of work, drainage waterway, canal waterway (flumed trough) and bed/water levels of the trough.

Assumptions: Manning's $n$ = 0.025 (as given) is used for the friction loss in the trough; Lacey's $f$ = 1; piers 1.5 m thick.

1. Type of structure

HFL of the drainage (117.00 m) is above the canal bed (116.00 m) by 1.00 m, so the trough cannot be kept above HFL without a very high, costly structure. The drainage is passed under the trough through barrels flowing under pressure (siphonic action). A siphon aqueduct is suitable. Drainage depth at HFL = 117.00 − 114.00 = 3.00 m.

2. Drainage waterway

Lacey's regime perimeter (for a natural stream):

$$P = 4.75\sqrt{Q} = 4.75\sqrt{550} = 111.4\ \text{m}$$

Provide 12 spans of 10 m with piers 1.5 m thick: clear waterway = 12 × 10 = 120 m (> 111.4 m). Overall length = 120 + 11 × 1.5 = 136.5 m.

Check: mean velocity = Q/(L·d) = 550/(120 × 3.00) = 1.53 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.85 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.8 m below HFL, with cut-offs and aprons for scour.

3. Canal waterway (flumed rectangular trough)

Normal canal: area $A_1=(B+zD)D$ = (30 + 1.5 × 1.4) × 1.4 = 44.94 m², $V_1$ = 60/44.94 = 1.335 m/s, $h_{v1}=V_1^2/2g$ = 0.0909 m. Top width $T=B+2zD$ = 34.20 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 24 m, with the same full supply depth 1.4 m (the trough is as long as the drainage waterway, 136.5 m).

• Trough area = 24 × 1.4 = 33.60 m²; $V_t$ = 60/33.60 = 1.786 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1625 m
• Froude number = 0.48 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 33.60/26.80 = 1.254 m; friction slope $S=(nV/R^{2/3})^2$ = 0.001474 (Manning's n = 0.025); friction loss in trough $h_f=S\,L$ = 0.001474 × 136.5 = 0.201 m

4. Transitions and levels

  FSL ---.                            .--- FSL
         '--.______________________.--'
  bed ---.                            .--- bed
         '--.______________________.--'
  normal  inlet     flumed trough     outlet  normal
  canal   (2:1)                      (3:1)   canal

Wing-wall (straight) transitions: inlet splay 2:1 (contraction) and outlet splay 3:1 (expansion), measured on each side:

• Inlet length = 2 × (T − B_t)/2 = 34.20 − 24 = 10.2 m
• Outlet length = 3 × (T − B_t)/2 = 15.3 m

Bernoulli's equation between sections (loss in contraction = 0.3 Δh_v, in expansion = 0.5 Δh_v; Δh_v = difference of velocity heads):

$$\begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned}$$

$W_1$ = 117.400 m, so $W_2$ = 117.307 m, $W_3$ = 117.106 m, $W_4$ = 117.141 m. Trough bed = water level − 1.4 m.

Total loss of head = 117.400 − 117.141 = 0.259 m, so the canal bed downstream of the work is 0.259 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 115.11 m, i.e. 1.11 m above the drainage bed; HFL stands 1.89 m above the soffit, which gives the siphonic head. Since the clear height is less than 1.5 m, the drainage bed is depressed under the barrels to RL 113.61 m (1.5 m clear height) with a 1 in 4 slope at both ends.

Answer: drainage waterway 120 m clear (12 × 10 m, overall 136.5 m); canal flumed to 24 m wide trough; bed/FSL at the four sections = 116.00/117.40, 115.91/117.31, 115.71/117.11, 115.74/117.14 m.

The canal ($Q$ = 300 m³/s) crosses a drainage of $Q$ = 500 m³/s. Design is done in four steps: type of work, drainage waterway, canal waterway (flumed trough) and bed/water levels of the trough.

Assumptions: Canal depth = FSL − bed = 209 − 202.5 = 6.5 m; side slope ½H:1V; ground level taken equal to canal bed; Lacey's $f$ = 1 and $n$ = 0.016 (as given) used for the trough; slab 0.6 m.

1. Type of structure

HFL of the drainage (207.70 m) is above the canal bed (202.50 m) by 5.20 m, so the trough cannot be kept above HFL without a very high, costly structure. The drainage is passed under the trough through barrels flowing under pressure (siphonic action). A siphon aqueduct is suitable. Drainage depth at HFL = 207.70 − 198.00 = 9.70 m.

2. Drainage waterway

Lacey's regime perimeter (for a natural stream):

$$P = 4.75\sqrt{Q} = 4.75\sqrt{500} = 106.2\ \text{m}$$

Provide 11 spans of 10 m with piers 1.5 m thick: clear waterway = 11 × 10 = 110 m (> 106.2 m). Overall length = 110 + 10 × 1.5 = 125.0 m.

Check: mean velocity = Q/(L·d) = 500/(110 × 9.70) = 0.47 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.73 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.6 m below HFL, with cut-offs and aprons for scour.

3. Canal waterway (flumed rectangular trough)

Normal canal: area $A_1=(B+zD)D$ = (30 + 0.5 × 6.5) × 6.5 = 216.12 m², $V_1$ = 300/216.12 = 1.388 m/s, $h_{v1}=V_1^2/2g$ = 0.0982 m. Top width $T=B+2zD$ = 36.50 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 26 m, with the same full supply depth 6.5 m (the trough is as long as the drainage waterway, 125.0 m).

• Trough area = 26 × 6.5 = 169.00 m²; $V_t$ = 300/169.00 = 1.775 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1606 m
• Froude number = 0.22 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 169.00/39.00 = 4.333 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000114 (Manning's n = 0.016); friction loss in trough $h_f=S\,L$ = 0.000114 × 125.0 = 0.014 m

4. Transitions and levels

  FSL ---.                            .--- FSL
         '--.______________________.--'
  bed ---.                            .--- bed
         '--.______________________.--'
  normal  inlet     flumed trough     outlet  normal
  canal   (2:1)                      (3:1)   canal

Wing-wall (straight) transitions: inlet splay 2:1 (contraction) and outlet splay 3:1 (expansion), measured on each side:

• Inlet length = 2 × (T − B_t)/2 = 36.50 − 26 = 10.5 m
• Outlet length = 3 × (T − B_t)/2 = 15.8 m

Bernoulli's equation between sections (loss in contraction = 0.3 Δh_v, in expansion = 0.5 Δh_v; Δh_v = difference of velocity heads):

$$\begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned}$$

$W_1$ = 209.000 m, so $W_2$ = 208.919 m, $W_3$ = 208.905 m, $W_4$ = 208.936 m. Trough bed = water level − 6.5 m.

Total loss of head = 209.000 − 208.936 = 0.064 m, so the canal bed downstream of the work is 0.064 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 201.80 m, i.e. 3.80 m above the drainage bed; HFL stands 5.90 m above the soffit, which gives the siphonic head.

Answer: drainage waterway 110 m clear (11 × 10 m, overall 125.0 m); canal flumed to 26 m wide trough; bed/FSL at the four sections = 202.50/209.00, 202.42/208.92, 202.40/208.90, 202.44/208.94 m.

The canal ($Q$ = 32 m³/s) crosses a drainage of $Q$ = 303 m³/s. Design is done in four steps: type of work, drainage waterway, canal waterway (flumed trough) and bed/water levels of the trough.

Assumptions: Canal side slope 1.5:1 (not given); drainage bed = HFL − 2.5 m = 207.5 m; Lacey's $f$ = 1; trough $n$ = 0.015. Depth of canal = FSL − bed = 1.5 m.

1. Type of structure

Canal bed level (212.00 m) is higher than HFL of the drainage (210.00 m) by 2.00 m, enough for free board and slab depth. The canal can be carried in a trough over the drain with free flow beneath, so an aqueduct is suitable.

2. Drainage waterway

Lacey's regime perimeter (for a natural stream):

$$P = 4.75\sqrt{Q} = 4.75\sqrt{303} = 82.7\ \text{m}$$

Provide 11 spans of 8 m with piers 1.2 m thick: clear waterway = 11 × 8 = 88 m (> 82.7 m). Overall length = 88 + 10 × 1.2 = 100.0 m.

Check: mean velocity = Q/(L·d) = 303/(88 × 2.50) = 1.38 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.16 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 4.7 m below HFL, with cut-offs and aprons for scour.

3. Canal waterway (flumed rectangular trough)

Normal canal: area $A_1=(B+zD)D$ = (20 + 1.5 × 1.5) × 1.5 = 33.38 m², $V_1$ = 32/33.38 = 0.959 m/s, $h_{v1}=V_1^2/2g$ = 0.0469 m. Top width $T=B+2zD$ = 24.50 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 12 m, with the same full supply depth 1.5 m (the trough is as long as the drainage waterway, 100.0 m).

• Trough area = 12 × 1.5 = 18.00 m²; $V_t$ = 32/18.00 = 1.778 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1611 m
• Froude number = 0.46 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 18.00/15.00 = 1.200 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000558 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000558 × 100.0 = 0.056 m

4. Transitions and levels

  FSL ---.                            .--- FSL
         '--.______________________.--'
  bed ---.                            .--- bed
         '--.______________________.--'
  normal  inlet     flumed trough     outlet  normal
  canal   (2:1)                      (3:1)   canal

Wing-wall (straight) transitions: inlet splay 2:1 (contraction) and outlet splay 3:1 (expansion), measured on each side:

• Inlet length = 2 × (T − B_t)/2 = 24.50 − 12 = 12.5 m
• Outlet length = 3 × (T − B_t)/2 = 18.8 m

Bernoulli's equation between sections (loss in contraction = 0.3 Δh_v, in expansion = 0.5 Δh_v; Δh_v = difference of velocity heads):

$$\begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned}$$

$W_1$ = 213.500 m, so $W_2$ = 213.352 m, $W_3$ = 213.296 m, $W_4$ = 213.353 m. Trough bed = water level − 1.5 m.

Total loss of head = 213.500 − 213.353 = 0.147 m, so the canal bed downstream of the work is 0.147 m lower than upstream.

Check: soffit of trough slab (0.6 m thick) = 211.20 m; HFL + free board 0.6 m = 210.60 m. Clearance adequate.

Sketch (section along the drainage)

   FSL 213.50  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
   bed 212.00  _____________________________   canal trough
               |_____________________________|  slab 0.6 m
               ||  ||  ||  ||  ||  ||  ||  ||  piers
   HFL 210.00  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~  drainage flow
   bed 207.50  _______________________________
               <-- 11 spans of 8 m, 100 m -->

Answer: drainage waterway 88 m clear (11 × 8 m, overall 100.0 m); canal flumed to 12 m wide trough; bed/FSL at the four sections = 212.00/213.50, 211.85/213.35, 211.80/213.30, 211.85/213.35 m.

The drainage bed (101.50 m) is above the canal bed (100.00 m) and the drainage HFL (103.50 m) is above the canal FSL (103.00 m), so the drainage cannot be passed under the canal. The canal is carried under the drainage through closed barrels flowing full under pressure: a canal siphon (the drainage flows over the barrels as an open channel).

Assumptions: two rectangular RCC barrels; roof slab 0.5 m with its top flush with the drainage bed (101.50 m); entry loss 0.5 $V^2/2g$ and exit loss $0.5(V_b-V_1)^2/2g$; barrel length = width of drainage crossing; permissible velocity in barrels 2–3 m/s.

1. Drainage waterway

$$P = 4.75\sqrt{60} = 36.8\ \text{m}$$

Provide a pitched drainage crossing of bottom width 37 m (> 36.8 m) over the barrels. Depth at HFL = 103.50 − 101.50 = 2.0 m, velocity = 60/(37 × 2.0) = 0.81 m/s, so a pitched floor with end aprons and cut-offs is enough. The length of the barrels is therefore about 38 m (crossing width plus flanking walls).

2. Canal waterway (barrels)

Normal canal: $A_1=(6+1.5\times3)\times3$ = 31.5 m², $V_1$ = 18/31.5 = 0.571 m/s, $h_{v1}$ = 0.0166 m, top width $T$ = 15 m.

Adopting $V_b$ = 2.5 m/s, required area = 18/2.5 = 7.2 m². Provide 2 barrels, each 2 m wide × 1.8 m high, separated by a 0.6 m pier: area = 2 × 2 × 1.8 = 7.2 m², $V_b$ = 18/7.2 = 2.50 m/s, $h_v$ = 0.3186 m. Overall width = 2×2 + 0.6 + 2×0.6 = 5.8 m.

3. Bed levels and transitions

• Top of roof = 101.50 m; soffit = 101.50 − 0.5 = 101.00 m; barrel floor = 101.00 − 1.8 = 99.20 m (canal bed depressed by 0.80 m, with a slope of 1 in 4 at the inlet and outlet).
• Transitions in plan: inlet (2:1) = T − barrel width = 15 − 5.8 = 9.2 m; outlet (3:1) = 1.5 × 9.2 = 13.8 m.

Losses of head (N = 0.016, $R$ = 0.474 m for one barrel):

$$\begin{aligned} h_{f} &= \frac{N^2V_b^2L}{R^{4/3}} = \frac{0.016^2 \times 2.50^2 \times 38}{0.474^{4/3}} = 0.165\ \text{m} \\ h_e &= 0.5\,\frac{V_b^2}{2g} = 0.159\ \text{m} \\ h_x &= 0.5\,\frac{(V_b-V_1)^2}{2g} = 0.095\ \text{m} \end{aligned}$$

Total loss = 0.159 + 0.165 + 0.095 = 0.419 m.

  HFL 103.50 ~~~~~~~~~~~~~~~~~~~~~~~~~
  drain bed 101.50 ============ roof (0.5 m)
  canal FSL 103.00 --.  barrels 2 x (2.0 x 1.8)  .-- 102.58
  canal bed 100.00 ---'__________________________'--- 99.58
                       floor 99.20

Answer: drainage crossing 37 m wide; 2 barrels of 2.0 m × 1.8 m (V = 2.50 m/s); barrel floor 99.20 m; total head loss 0.419 m; downstream canal FSL 102.581 m and bed 99.581 m.

The canal ($Q$ = 30 m³/s) crosses a drainage of $Q$ = 500 m³/s. Design is done in four steps: type of work, drainage waterway, canal waterway (flumed trough) and bed/water levels of the trough.

Assumptions: Lacey's $f$ = 1; $n$ = 0.016 as given is used for the trough; piers 1.5 m thick; slab 0.6 m.

1. Type of structure

HFL of the drainage (100.50 m) is above the canal bed (100.00 m) by 0.50 m, so the trough cannot be kept above HFL without a very high, costly structure. The drainage is passed under the trough through barrels flowing under pressure (siphonic action). A siphon aqueduct is suitable. Drainage depth at HFL = 100.50 − 98.00 = 2.50 m.

2. Drainage waterway

Lacey's regime perimeter (for a natural stream):

$$P = 4.75\sqrt{Q} = 4.75\sqrt{500} = 106.2\ \text{m}$$

Provide 11 spans of 10 m with piers 1.5 m thick: clear waterway = 11 × 10 = 110 m (> 106.2 m). Overall length = 110 + 10 × 1.5 = 125.0 m.

Check: mean velocity = Q/(L·d) = 500/(110 × 2.50) = 1.82 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.73 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.6 m below HFL, with cut-offs and aprons for scour.

3. Canal waterway (flumed rectangular trough)

Normal canal: area $A_1=(B+zD)D$ = (24 + 1.5 × 1.25) × 1.25 = 32.34 m², $V_1$ = 30/32.34 = 0.928 m/s, $h_{v1}=V_1^2/2g$ = 0.0438 m. Top width $T=B+2zD$ = 27.75 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 13 m, with the same full supply depth 1.25 m (the trough is as long as the drainage waterway, 125.0 m).

• Trough area = 13 × 1.25 = 16.25 m²; $V_t$ = 30/16.25 = 1.846 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1737 m
• Froude number = 0.53 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 16.25/15.50 = 1.048 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000819 (Manning's n = 0.016); friction loss in trough $h_f=S\,L$ = 0.000819 × 125.0 = 0.102 m

4. Transitions and levels

  FSL ---.                            .--- FSL
         '--.______________________.--'
  bed ---.                            .--- bed
         '--.______________________.--'
  normal  inlet     flumed trough     outlet  normal
  canal   (2:1)                      (3:1)   canal

Wing-wall (straight) transitions: inlet splay 2:1 (contraction) and outlet splay 3:1 (expansion), measured on each side:

• Inlet length = 2 × (T − B_t)/2 = 27.75 − 13 = 14.8 m
• Outlet length = 3 × (T − B_t)/2 = 22.1 m

Bernoulli's equation between sections (loss in contraction = 0.3 Δh_v, in expansion = 0.5 Δh_v; Δh_v = difference of velocity heads):

$$\begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned}$$

$W_1$ = 101.250 m, so $W_2$ = 101.081 m, $W_3$ = 100.979 m, $W_4$ = 101.044 m. Trough bed = water level − 1.25 m.

Total loss of head = 101.250 − 101.044 = 0.206 m, so the canal bed downstream of the work is 0.206 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 99.13 m, i.e. 1.13 m above the drainage bed; HFL stands 1.37 m above the soffit, which gives the siphonic head. Since the clear height is less than 1.5 m, the drainage bed is depressed under the barrels to RL 97.63 m (1.5 m clear height) with a 1 in 4 slope at both ends.

Answer: drainage waterway 110 m clear (11 × 10 m, overall 125.0 m); canal flumed to 13 m wide trough; bed/FSL at the four sections = 100.00/101.25, 99.83/101.08, 99.73/100.98, 99.79/101.04 m.

The canal ($Q$ = 56 m³/s) crosses a drainage of $Q$ = 425 m³/s. Design is done in four steps: type of work, drainage waterway, canal waterway (flumed trough) and bed/water levels of the trough.

Assumptions: Side slope of canal 1.5:1; Lacey's $f$ = 1; trough $n$ = 0.015.

1. Type of structure

HFL of the drainage (268.20 m) is above the canal bed (267.00 m) by 1.20 m, so the trough cannot be kept above HFL without a very high, costly structure. The drainage is passed under the trough through barrels flowing under pressure (siphonic action). A siphon aqueduct is suitable. Drainage depth at HFL = 268.20 − 265.50 = 2.70 m.

2. Drainage waterway

Lacey's regime perimeter (for a natural stream):

$$P = 4.75\sqrt{Q} = 4.75\sqrt{425} = 97.9\ \text{m}$$

Provide 13 spans of 8 m with piers 1.2 m thick: clear waterway = 13 × 8 = 104 m (> 97.9 m). Overall length = 104 + 12 × 1.2 = 118.4 m.

Check: mean velocity = Q/(L·d) = 425/(104 × 2.70) = 1.51 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.53 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.3 m below HFL, with cut-offs and aprons for scour.

3. Canal waterway (flumed rectangular trough)

Normal canal: area $A_1=(B+zD)D$ = (24 + 1.5 × 1.98) × 1.98 = 53.40 m², $V_1$ = 56/53.40 = 1.049 m/s, $h_{v1}=V_1^2/2g$ = 0.0561 m. Top width $T=B+2zD$ = 29.94 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 16 m, with the same full supply depth 1.98 m (the trough is as long as the drainage waterway, 118.4 m).

• Trough area = 16 × 1.98 = 31.68 m²; $V_t$ = 56/31.68 = 1.768 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1593 m
• Froude number = 0.40 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 31.68/19.96 = 1.587 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000380 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000380 × 118.4 = 0.045 m

4. Transitions and levels

  FSL ---.                            .--- FSL
         '--.______________________.--'
  bed ---.                            .--- bed
         '--.______________________.--'
  normal  inlet     flumed trough     outlet  normal
  canal   (2:1)                      (3:1)   canal

Wing-wall (straight) transitions: inlet splay 2:1 (contraction) and outlet splay 3:1 (expansion), measured on each side:

• Inlet length = 2 × (T − B_t)/2 = 29.94 − 16 = 13.9 m
• Outlet length = 3 × (T − B_t)/2 = 20.9 m

Bernoulli's equation between sections (loss in contraction = 0.3 Δh_v, in expansion = 0.5 Δh_v; Δh_v = difference of velocity heads):

$$\begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned}$$

$W_1$ = 268.980 m, so $W_2$ = 268.846 m, $W_3$ = 268.801 m, $W_4$ = 268.852 m. Trough bed = water level − 1.98 m.

Total loss of head = 268.980 − 268.852 = 0.128 m, so the canal bed downstream of the work is 0.128 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 266.22 m, i.e. 0.72 m above the drainage bed; HFL stands 1.98 m above the soffit, which gives the siphonic head. Since the clear height is less than 1.5 m, the drainage bed is depressed under the barrels to RL 264.72 m (1.5 m clear height) with a 1 in 4 slope at both ends.

Answer: drainage waterway 104 m clear (13 × 8 m, overall 118.4 m); canal flumed to 16 m wide trough; bed/FSL at the four sections = 267.00/268.98, 266.87/268.85, 266.82/268.80, 266.87/268.85 m.

The canal ($Q$ = 40 m³/s) crosses a drainage of $Q$ = 45 m³/s. Design is done in four steps: type of work, drainage waterway, canal waterway (flumed trough) and bed/water levels of the trough.

Assumptions: Lacey's $f$ = 1; trough $n$ = 0.015; slab 0.6 m.

1. Type of structure

HFL of the drainage (207.00 m) is above the canal bed (206.40 m) by 0.60 m, so the trough cannot be kept above HFL without a very high, costly structure. The drainage is passed under the trough through barrels flowing under pressure (siphonic action). A siphon aqueduct is suitable. Drainage depth at HFL = 207.00 − 204.50 = 2.50 m.

2. Drainage waterway

Lacey's regime perimeter (for a natural stream):

$$P = 4.75\sqrt{Q} = 4.75\sqrt{45} = 31.9\ \text{m}$$

Provide 6 spans of 6 m with piers 1.0 m thick: clear waterway = 6 × 6 = 36 m (> 31.9 m). Overall length = 36 + 5 × 1.0 = 41.0 m.

Check: mean velocity = Q/(L·d) = 45/(36 × 2.50) = 0.50 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 1.67 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 2.5 m below HFL, with cut-offs and aprons for scour.

3. Canal waterway (flumed rectangular trough)

Normal canal: area $A_1=(B+zD)D$ = (30 + 1.5 × 1.6) × 1.6 = 51.84 m², $V_1$ = 40/51.84 = 0.772 m/s, $h_{v1}=V_1^2/2g$ = 0.0303 m. Top width $T=B+2zD$ = 34.80 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 14 m, with the same full supply depth 1.6 m (the trough is as long as the drainage waterway, 41.0 m).

• Trough area = 14 × 1.6 = 22.40 m²; $V_t$ = 40/22.40 = 1.786 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1625 m
• Froude number = 0.45 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 22.40/17.20 = 1.302 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000504 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000504 × 41.0 = 0.021 m

4. Transitions and levels

  FSL ---.                            .--- FSL
         '--.______________________.--'
  bed ---.                            .--- bed
         '--.______________________.--'
  normal  inlet     flumed trough     outlet  normal
  canal   (2:1)                      (3:1)   canal

Wing-wall (straight) transitions: inlet splay 2:1 (contraction) and outlet splay 3:1 (expansion), measured on each side:

• Inlet length = 2 × (T − B_t)/2 = 34.80 − 14 = 20.8 m
• Outlet length = 3 × (T − B_t)/2 = 31.2 m

Bernoulli's equation between sections (loss in contraction = 0.3 Δh_v, in expansion = 0.5 Δh_v; Δh_v = difference of velocity heads):

$$\begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned}$$

$W_1$ = 208.000 m, so $W_2$ = 207.828 m, $W_3$ = 207.807 m, $W_4$ = 207.874 m. Trough bed = water level − 1.6 m.

Total loss of head = 208.000 − 207.874 = 0.126 m, so the canal bed downstream of the work is 0.126 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 205.61 m, i.e. 1.11 m above the drainage bed; HFL stands 1.39 m above the soffit, which gives the siphonic head. Since the clear height is less than 1.5 m, the drainage bed is depressed under the barrels to RL 204.11 m (1.5 m clear height) with a 1 in 4 slope at both ends.

Answer: drainage waterway 36 m clear (6 × 6 m, overall 41.0 m); canal flumed to 14 m wide trough; bed/FSL at the four sections = 206.40/208.00, 206.23/207.83, 206.21/207.81, 206.27/207.87 m.

At a syphonic aqueduct the canal is flumed into a rectangular trough over the drainage barrels. The levels follow from Bernoulli's equation between the normal canal and the trough, allowing losses of 0.3 Δh_v in the contraction (inlet) and 0.5 Δh_v in the expansion (outlet).

Assumptions: Trough width taken equal to the 14 m drainage bed width given; trough $n$ = 0.015; Lacey's waterway used only for trough length.

Normal canal

$A_1=(B+zD)D=(28+1.5\times1.5)\times1.5$ = 45.38 m², $V_1$ = 36/45.38 = 0.793 m/s, $h_{v1}$ = 0.0321 m, FSL $W_1$ = 210.40 + 1.5 = 211.90 m.

Trough

Drainage waterway by Lacey: $P=4.75\sqrt{400}$ = 95.0 m, so the trough is about 109.2 m long (12 spans of 8 m, piers 1.2 m). Canal flumed to $B_t$ = 14 m, depth 1.5 m:

• $V_t$ = 36/(14 × 1.5) = 1.714 m/s, $h_{vt}$ = 0.1498 m, Froude number 0.45 (subcritical)
• $R$ = 1.235 m, $S=(nV/R^{2/3})^2$ = 0.000499 (n = 0.015), $h_f$ = 0.000499 × 109.2 = 0.054 m
• Inlet length (2:1 splay) = T − B_t = 32.50 − 14 = 18.5 m; outlet length (3:1 splay) = 27.8 m

Levels at the four sections

$$\begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned}$$

Total head lost in the work = 211.900 − 211.751 = 0.149 m. Trough bed = water level − 1.5 m.

Check: with a 0.6 m slab the barrel soffit is at 209.59 m, below the HFL of 211.00 m by 1.41 m, so the barrels run full (syphonic).

Answer: bed/water levels = 210.40/211.90, 210.25/211.75, 210.19/211.69, 210.25/211.75 m.