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July 2002

Road Drainage Design Manual Chapter 7: Worked Examples

7Chapter 7

WorkedExamples

Chapter 8References

Chapter 7Worked Examples

Chapter 6Maintenance and Remediation

Chapter 5Erosion and Sediment Control

Chapter 4Design

Chapter 3Hydrology and Design Criteria

Chapter 2Site Assessment

Chapter 1Overview

Manual Contents

July 2002


7

Table of Contents

Introduction 7-1

7.1 Open Channel Flow 7-1

7.2 Hydrology - Calculation of the Design Flood Discharge 7-3

7.3 Procedure for the Selection of Culvert Size 7-5

7.4 Gully Inlets on Grade 7-7

7.5 Aquaplaning 7-9

7.6 Extended Outlet Protection for Culverts 7-9

7.7 Energy Dissipators for Supercritical Flow 7-12

7.8 Scour Hole Dimensions 7-16

7.9 Tidal Range at Site 7-17

7.10 Floodway Calculations 7-19

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Introduction

Worked examples of some of the more commonlyused design calculations are given in this chapter.

It has been considered more appropriate to giveother worked examples immediately followingthe description of design principles in otherchapters. For example, substantial calculations areshown in the Appendix to the UndergroundDrainage Systems Chapter.

Worked examples in this Chapter are:

7.1 Open Channel flow

7.2 Hydrology - Calculation of the DesignFlood Discharge

7.3 Procedure for the Selection of Culvert Size

7.4 Gully Inlets on Grade

7.5 Aquaplaning

7.6 Extended Outlet Protection for Culverts

7.7 Energy Dissipators for Supercritical Flow

7.8 Scour Hole Dimensions

7.9 Tidal Range at Site

7.10 Floodway Calculations

7.1 Open Channel Flow

Preliminary

Check that a reasonable effort has been made toensure that the channel cross section and slope hasbeen adequately defined, as indicated by thefollowing.

For a bridge size catchment (nominally greaterthan 5 km² in area with a major channel), bed and

debris/water gradients and changes in crosssections should have been surveyed for a distanceof at least 500 metres both upstream anddownstream of the job site.

For smaller culvert structures to calculate tail-water levels or for small open channels, shorterdistances upstream and downstream down to anabsolute minimum of 100 metres should similarlybe surveyed.

For the larger catchments, field inspections inaccordance with the Department’s Form 2759,Field Report - Bridge Waterways should havebeen carried out including interviews with localresidents about observed or reported floods at ornear the site.

For all jobs either a site investigation should havetaken place or adequate photographs are availableto define the roughness of the bed and banks ofthe open channel and areas of any overflow.

Worked Example

For the conditions listed below and illustrated inthe diagram determine the depth of flow for adischarge of 42.5 m³/s.

The direction of flow in the main channel is at askew of 10° to the normal to the road centreline.Allow for the same skew in any overflow.

Figure 7.1

(a) The average slope over a section of thestream for 200 m both upstream anddownstream of the cross section is 0.001m/m.

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Chapter 7

Worked Examples

(b) The stream is fairly uniform in cross sectionand alignment over this 400 m section and islined mainly with thick grass about 200 mmhigh.

(c) The flood debris levels from a recent floodare known to be at the top of the banks.

Step 1

Figure 7.2

Select an appropriate value of vegetal retardancefor the stream from Table 4.5 (viz. C), determinea value of “n” for the stream flow to debris levelfrom Figure F19C. Then, assuming a trapezoidalchannel shape as illustrated, calculate the flow todebris level (height 22.0 m) using Manning’sformula (for more irregular shaped sections, aplanimeter is necessary to determine cross-sectional areas).

Depth of flow (debris level - bed level) = 2 m

A = 20 m²

n = 0.03

Q = A.cos100.V= 20 x 0.985 x l.24

Q = 24.4 m³/s

Step 2

For the higher design discharge of 42.5 m³/s,water will overtop the banks. As the flood plainswill flow at a shallower depth, for the same grass

cover as in the main channel, the roughnesscoefficient will be higher up to a certain limit.Also for shallower flow, the hydraulic radius willbe smaller and so will be the velocity, for the samehydraulic gradient.

Divide the flow path into three sections as shownin Figure 7.3 to allow for the above differences.

Figure 7.3

Step 3

For rating curves (discharge versus height), thearea and wetted perimeter need to be calculatedfor various flood heights for discharge calcula-tions. In this example, it is required to find theflood level corresponding to a total discharge of42.5 m³/s.

For a water to water boundary such as thatbetween each overflow and the main channel, halfthe depth of the water at such boundaries may beadded to each wetted perimeter as illustratedbelow.

Assume a flood height 22.5 m, giving a depth offlow of 0.5 m in each overflow.

Left OverflowA = 6 x 0.5

= 3 m²

Wetted perimeter (WP) = 0.5 + 6 + 0.5/2= 6.75 m

R = 3/6.75= 0.444 m

n = 0.075 (Figure F19C)

S = 0.001 m/m

V = 0.25 m/s (from Manning’s formula)

Q = 0.74 m³/s (A.cos100.V)

(Manning’s formula)

(Section 4.3.6)m/s 1.24 n

SR V0.50.667

=

=

1.27

ABCD) perimeter, tted(area / we 15.820

R

=

=

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Main ChannelA = 27.5 m²

WP = 0.25 + 5.39 +5 + 5.39 + 0.25= 16.28 m

R = 1.69

n = 0.03

S = 0.001 m/m

V = 1.50 m/s (Manning’s formula)

Q = A.cos100.V = 27.5 x 0.985 x 1.50

Q = 40.63 m³/s

Right OverflowQ = 0.74 m³/s as per left overflow

Total DischargeTotal Discharge = Qleft overflow + Qmain channel +

Qright overflow= 0.74 + 40.63 + 0.74= 42.11 m³/s

Therefore, the design flood of 42.5 m³/s is atheight 22.5 m, approx.

7.2 Hydrology - Calculationof the Design FloodDischarge

This example gives calculations for the estimationof the 50 year ARI rainfall runoff from acatchment. The runoff is conventionally taken asthe 50 year ARI design flood discharge as it isonly in very rare cases that the average recurrenceintervals (ARI) of the runoff and discharge varyfor the larger floods in Queensland.

The flood height corresponding to the design flooddischarge is easily found as the height on the ratingcurve giving this discharge. A rating curve is a plotof discharge versus flood height obtained by openchannel calculations shown in Section 4.3.6.

The Modified Friend Formula for time ofconcentration is used as the catchment is largerthan 5 square kilometres.

The Rational method is used for the designdischarge, but other procedures for a catchmentthis size are not shown.

Given:

It is required to calculate the 50 year ARI flooddischarge for a crossing on a road between tworelatively large towns in western Queensland

From 1:100 000 contour maps:

Catchment area (M) = 118.2 km²

Length of catchment (L) = 26.2 km

Weighted average slope (H) = 74 m in 26200m= 0.28 %

The slope is fairly uniform.

From field inspection:

Maximum reported flood, Ht 294.7 m (February,1992 - from local resident) Manning’s roughnesscoefficients at site:

n = 0.06 (left overflow)

n = 0.055 (main channel)

n = 0.065 (right overflow)

The stream cross-section is shown approximatelyto scale in Figure 7.4. The flow direction is squareto the road centreline (zero skew).

Figure 7.4

Step 1

Assume a flood level as approximating theexpected design flood level.

This initial level is usually either

• the maximum reported/recorded flood level;

• at or just over the main stream bank; or

• berm or intermediate bank level, if there is alarge cross-section for a small catchment.

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Note: Further calculations will give a design flooddischarge and a design flood level correspondingto this discharge. However, if this design floodlevel is more than about 0.5 m different to theoriginally assumed flood level, go back to Step 1,assume another flood level and repeat the processdescribed below until the difference is smaller.

In this example select the maximum recordedflood level, Ht 294.70 m. The average hydraulicradius at this level is Rs.

Step 2

Find the average hydraulic radius at the site,Rs for a flood at Ht 294.70 m.

From open channel calculations, for the 3sections, with S = 0.0017 m/m in all sections:

A1= 67.6 m²; WP1 = 113.5 m; R1 = 0.60m;n1 = 0.065; V1 = 0.48 m/s; q1 = 32.6 m³/s

A2 = 53.3 m²; WP2 = 21.4 m; R2 = 2.49 m;n2 = 0.055; V2 = 1.36 m/s; q2 = 72.6 m³/s

A3 = 28.2 m²; WP3 = 73.1 m; R3 = 0.39 m;n3 = 0.065; V3 = 0.33 m/s; q3 = 9.4 m³/s

Qtot = 32.6 + 72.6 + 9.4∴Qtot = 114.6 m³/s

∴Rs = 1.78 m

Step 3

Calculate the time of concentration using theModified Friend Formula.

It should be noted that as the time of concentrationis for zero (start of rainfall) to maximum flow in acreek, an adjustment to the hydraulic radius at thesite and possible adjustments to the roughnesscoefficient are made.

where

Tc = time of concentration (h)

ch = Chezy’s No. = R0.167 / n

R = adopted average hydraulic radius= 0.75Rs where the fall of the stream from the

top of the catchment to the site is fairlyuniform

= 0.65 Rs where there are significant lengthsof contrasting steep and flat sections of thestream slope

n = average coefficient of roughness of themain channel and overflows over the wholestream length. It may be the same or 0.005to 0.01 higher than n at the site.

L, M and H are as defined in “Given” above.

Adopt R= 0.75Rs = 0.75 x 1.78 = 1.34m

∴ ch = 17.49

∴ Tc =13.1 hours

Step 4

Calculate the average rainfall intensity for thetime of concentration for the design storm.

i.e. for this example, the average rainfall intensityfor a storm of 13.1 hours duration and an ARI of50 years is required.

The RAIN2 Program from the Hydraulics Sectionof Main Roads and available in most Districts isused. Other programs are available which also useparameters from “Australian Rainfall and Runoff”(IEAust, 1987).

The RAIN2 Program which is interactive requeststhe following input data. The actual Map numberand value of parameter from ARR is also shown.

4.01.0c 28.0x2.118x49.172.26x5.8T =

06.034.1ch

167.0=

4.01.0c HchML5.8T =

6.11439.0x4.949.2x6.7260.0x6.32Rs

++=

tot

332211s Q

RqRqRqR ++=

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2i1 (2 year, 1 hr log-normal rainfall intensity) =37.5 mm/h - Map 1.4



50il (50 year, 1 hr log-normal rainfall intensity) =69 mm/h - Map 4.4

50i12 (50 year, 12 hr log-normal rainfall intensity)= 10.8 mm/h - Map 5.4

50i72 (50 year 72 hr log-normal rainfall intensity)= 3 mm/h - Map 6.4

SKEWNESS FACTOR (G) = 0.21 (Map 7.c)

GEOGRAPHICAL FACTOR F2 = 4.28 (Map 8)

GEOGRAPHICAL FACTOR F50 = 16.6 (Map 9)

From the program, average rainfall intensity =10.61 mm/hr for a 13.1 hr storm of 50 year ARI.

Step 5

Calculate the 50 Year ARI design flooddischarge.

The Rational Method (See Section 3.5.2) states:

Q = 0.28 C.I.A

where

Q = discharge (m³/s)

0.28 is a conversion factor to ensure units areconsistent for A in km²

C = runoff coefficient (dimensionless)

I = average rainfall intensity (mm/h)

A = area of catchment (km²)

From Table 3.5, the runoff coefficient is firstcalculated for the catchment which has thefollowing characteristics:

(i) Rainfall intensity = 10.61 mm/h

(ii) Rolling, with slopes 1.5 - 4%

(iii) Well defined system of small watercourses

(iv) Open forest or grassed land

∴ C=0.50

Therefore

Q = 0.28 x 0.50 x 10.61 x 118.2

∴Q=175.6m³/s

This corresponds to a calculated flood level, Ht295.04 m at the site. As the initial assumed floodHt 294.70 m is less than 0.5 m different, there isno need to repeat the calculation procedure with ahigher initial flood level.

Adopt Q50 = 175.6 m³/s at flood Ht 295.04 m.

7.3 Procedure for theSelection of Culvert Size

(Refer to Section 4.2.2.)

Step 1

List the design data:

(a) Design discharge Q50 = 19.3 m³/s

(b) Allowable outlet velocity (with standardoutlet protection).Va = 1.8 m/s

(c) Flood level in natural channel F.L. = 31.8 m

(d) Invert level of channel at outlet IL0 = 30.0 m

(e) Slope of culvert in metres per metre.S0 = 0.01 m/m

(f) Allowable headwater depth in metres.HWa = 2.0 m

(g) Mean velocity = 0.55 m/s. Maximumvelocity = 0.61 m/s.

From Manning’s formula, maximumvelocity is in the main channel and the meanvelocity is over the total section includingthe overflow.

100401000C +++=

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Note: These velocities are used inconjunction with field observations of scourin the natural channel. If low velocitiesproduce scour, then either the allowabledesign velocity in (b) should be reduced orconsideration be given to additional outletprotection. However if high velocities occurin the natural channel with little or no scour,the allowable design velocity in (b) may beincreased.

(h) Length = 12.0 in.

Figure 4.7 is recommended as suitable to recordcalculations and to facilitate checking.

Step 2

Determine the first trial culvert size:

(a) From the equation A = Q/V area ofwaterway: A = 19.3/1.8 m² = 10.72 m²

(b) Select a culvert size where the soffit wouldbe above or just below the tailwater toutilise the full waterway opening, ifpractical.

Depth of Tailwater = Flood level - Invertlevel of natural channel = 31.8 - 30.0 = 1.8

Assuming a height of 1.5 m and selecting astandard size R.C. slab deck culvert:

Try 3/2400 x 1500 mm culvert (Area 10.80 m²)

Step 3

Find the headwater depth for the trial culvert

(a) Assume INLET CONTROL

Determine the discharge (Q) per cell

i.e.Q = 19.3 / 3 = 6.43 m³/s

Find the ratio of discharge to width Q/B =2.68 m³/s per metre), then using thenomograph in Figure 4.8A determine theHW/D ratio.

HW/D = 0.91

HW= 0.91 x 1.5= 1.37 m

This headwater is satisfactory as it is lessthan the allowable given in the design data,item (f) of Step 1.

(b) Assume OUTLET CONTROL

Since the tailwater level is above the culvertsoffit at the outlet, the HW can be calculatedfrom the equation:

HW = H + ho - LSo

ho = TW = 1.8 m

LSo = 12 x 0.01 = 0.12

From the nomograph in Figure F8E andusing ke = 0.4 from Table 4.2, for area ofbox = 2.4 x 1.5 = 3.60 m²,

H = 0.26 m

Therefore,

HW = 0.26 + 1.8 - 0.12 = 1.94 m

(c) As the HW for outlet control is higher thanthat for inlet control, outlet control is thegoverning factor; and being less than theallowable height of 2.0 m is acceptable.

Step 4

Try a culvert of another type or shape, if acomparison of alternative design costs is to bemade and determine size and headwater by theabove procedure.

Step 5

Compute the velocity through the 3/2400 x 1500RCBC.

Since outlet control governs, and the tailwater isabove the soffit of the culvert, the full waterwayarea is used.

Outlet velocity = Q/Ao = 19.3/10.8 = 1.79 m/s

Step 6

Make selection and record all relevant data onFigure F7 or similar form.

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7.4 Gully Inlets on Grade

Design charts from Sections 4.4.3 and 4.4.4 areused in this example.

Figure 7.5

For the conditions listed below and illustrated inFigure 7.5 determine the number, types andspacings of the gully inlets required.

(a) 1000 metres of 1.8% grade to be drained

(b) rainfall intensity = 200 mm/h

(c) Manning’s “n”, channel = 0.012; shoulder =0.014

(d) allowable spread of water to be checked forpedestrian safety

(e) channel terminates at a bridge abutment,hence zero bypass

(f) use grate or combination inlets.

Step 1

Determine the allowable spread. This is to be thesmaller of

• Allowable width of spread of water on the road= 4.9 m (from Figure F32A), or

• Width arising from the allowable depth ofwater at the kerb for pedestrian safety. It will befound, by trial and error, from Figure F34, thatthe allowable depth of water at the kerb = 138mm, giving a spread width of 2960 mm asillustrated in Figure 7.6.

Adopt maximum allowable spread = 2.960 m.

The following procedure was adopted todetermine the allowable depth of water at the kerbfor pedestrian safety.

• Assume a depth of water at the kerb, dg.

• Calculate the dimension BC (= 100.5 mm) andthen the area ABCD.

• Calculate the discharge in ABCD from FigureF34 and Note 3.

Calculate the velocity in ABCD from velocity =discharge / area to give Vave.

Calculate the product dgVave.

Repeat the procedure as required until dgVave <0.4 m²/s.

It will be found that the maximum allowabledepth of water at the kerb is 138 mm, givingdgVave = 0.39 m²/s. The resultant allowable spreadis 2960 mm.

Figure 7.6

Step 2

Calculate the flow corresponding to the maximumspread.

Using Figure F34 and Note 4, the flow for a 1.8%grade and a spread of 2.960 m will be found to be0.35 m³/s ( = 350 litres/sec).

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Step 3

Calculate the inlet capacities, allowing forblockage.

Assume initially concrete gullies, roadway type,kerb in line (Dwg No 1312). Then from the BCCType A Gully Inlet Capacity Charts, removing theblockage factor, the full capacities for an approachflow of 350 litres/sec (using the road crossfall1:30 as the nearest to the actual 1: 25) are:

With S lintel, captured flow = 188 litres/sec

With M lintel, captured flow = 216 litres/sec

With L lintel, captured flow = 228 litres/sec

From Table 4.11, allowing for blockage, thepercentage of theoretical capacity allowed is 80%.Therefore, the design capacities are:

S lintel, design capture flow = 0.8 x 188 = 150litres/sec

M lintel, design capture flow = 0.8 x 216 = 173litres/sec

L lintel, design capture flow = 0.8 x 228 = 182litres/sec

Step 4

Locate the inlets.

(a) First, place an inlet at the point where theallowable spread occurs.

Using Figure F31

For Q = 0.35 m³/s (from Step 2)

and W = 11.1 m(given)

the length of contributing road L = 625 m

Selecting the concrete gully with the Llintel, from Step 3,

Inlet capacity = 0.182 m³/s and

Bypass = gutter flow - inlet capacity= 0.35 - 0.182 = 0.168 m³/s

(b) The second inlet will have to take the bypassfrom the first inlet plus flow from thesection of roadway between the first andsecond inlets.

Placing the second inlet where the flow hasbuilt up to the allowable spread, then:

flow from roadway

= flow at allowable spread - bypass frominlet 1

= 0.35 - 0.168

= 0.182 m³/s

Using Figure F31

For Q = 0.182m³/s

and W = 11.1 (given)

the length of contributing road

L2 = 300 m

Place a concrete gully with the L lintel withinlet capacity = 0.182 m³/s as the previousone. Also as in inlet 1, the bypass flow willbe 0.168 m³/s.

(c) The third inlet in the sag (end of grade) willtake a total flow

= flow from 75 m roadway + bypass frominlet 2

= 0.043 + 0.168

= 0.211 m³/s

Select a gully inlet to capture 0.211 m³/s atthe sag. From the BCC Capacity Charts, theType A concrete gully with an S lintel, withdepth of water 95 mm at the undepressed lipof channel will capture this flow. The depthof water at the kerb face will be 132.5 mmwhich is acceptable < 138 mm allowed fromStep 1.

List the final selections as shown in Figure 7.7.

Figure 7.7

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7.5 Aquaplaning

The following example shows a check foraquaplaning in the transition area between reversesuperelevations, one of the main areas of potentialaquaplaning because of the small to zerocrossfalls.

Reference should be made to Section 4.4.4.2 ofthe Manual.

GivenCheck for aquaplaning.

Contours at 0.1 m intervals have been drawnusing a computer program.

Step 1

Find the longer drainage path lengths forchecking. List other parameters, calculated orspecified as follows for the longest drainage path.

Design speed = 100 km/h

Rainfall intensity (I) = 50 mm/h

Water film depth (D) to be -2.5 mm desirable maximum4 mm maximum

Pavement texture depth (T) = 0.6 mm (densegraded asphalt)

Drainage path length (L) = 104 m

Pavement crossfall 0% - 3%

Longitudinal grade = 1.79%

Slope of drainage path (S) = 1.87%

Step 2

Calculate the water film depth (D) for the longestdrainage path.

Therefore,

D = 4.94 mm

This is unacceptable as it is > 4 mm.

Step 3

Increase the changes in crossfall by the maximumrate of rotation given in the Road Planning andDesign Manual. This changes the longest lengthof drainage path and its slope to give:

Length of drainage path = 53 m

Slope of drainage path = 2.08%

Crossfall = 0% to 3%

Other parameters remain the same, therefore,

D = 3.37 mm

This is <4 mm and is acceptable.

Step 4

Check other drainage paths.

7.6 Extended OutletProtection for Culverts

The following procedure calculates the distancestandard culvert outlet protection has to beextended when design velocities higher thannormally acceptable are used.

For velocities in excess of 5 m/s, the use of energydissipators as in Section 4.5.2 should beconsidered. Also, the design of a stabilised scourhole downstream may be an option.

Figure F59 illustrates the theory for the cal-culations for extended outlet protection.

Flow effectively diverges from a culvert outlet atan angle such that

F31tan =α

6.008.2

50x53x6.0x103.0D 42.0

59.043.011.0−=

6.087.1

50x104x6.0x103.0D 42.0

59.043.011.0−=

TS

ILT103.0D 42.0

59.043.011.0−=

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v = velocity of flow at the outlet of the culvertbarrel (m/s)

g = acceleration due to gravity (9.8 m/s/s)

d = tailwater depth at the outlet of the culvertbarrel (m)

It is required to extend the outlet protection so thatat the end of the protection, the velocity of flowdoes not cause erosion.

Preliminary

The performance of existing culverts in similarnatural surface conditions should first be assessed.Normally this would be an assessment of scouringcaused by existing culverts in the immediatelocality of the new design and the design velocityfor those culverts.

If records of the original design velocity are notfound, assume that the existing culverts weredesigned for 1.8 - 2.4 m/s flowing full, which usedto be the standard practice.

At all times the design velocity mentioned in thisexample, is located at the outlet of the culvert.

Step 1

Adopt a design velocity for the new culvert.

In this example, afflux upstream is not a problemand culverts are expensive so that fewer culvertsdesigned for a higher than average velocity needto be considered.

Because of divergence of flow leaving a culvertbarrel, at the end of standard culvert protection,the actual velocity will be about half the designvelocity for most existing culverts. The number ofcells in a culvert will give a variation to thisvelocity.

Therefore, if a design velocity of 4.0 m/s is usedinstead of say the 2.4 m/s for existing culvertswith satisfactory performance in similar naturalsurface conditions at the outlet, the allowable

calculated velocity at the end of the extendedoutlet is still 1.2 m/s (half of 2.4) to avoidscouring downstream of the protection.

Adopt a design velocity of 4.0 m/s at the outlet ofthe culvert barrel and a velocity of 1.2 m/s at theend of the extended outlet protection.

Step 2

List the design data:

(a) 1/1200 x 900 RCBC

(b) Outlet velocity = 4.0 m/s at depth of 800mm.

(c) Extended outlet protection required withtheoretical velocity at the end of theprotection not to exceed 1.2 m/s.

(d) The dimensions of the outlet protection,ABCDEF in Figure 7.9 are to be defined.

5.0)gd(v=where F, the Froude Number

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Figure 7.8

Figure 7.9

Step 3

Find the width BE at the end of the protectionwhere the velocity will not cause erosion.

Assume that the depth (d) of the water remainsconstant as it flows downstream from the culvertbarrel, increasing its width at a uniform rate.

Then, asDischarge = area x velocity

= width x depth x velocityand is constant, then

WidthBE x d x VBE = WidthAF x d x VAF

Therefore,

Substituting,

BE= 4m

As CD = 1.2 m, then BC = DE = 1.4 m

Step 4

Find the distance AC to which the outletprotection has to be extended.

For water exiting the culvert at 4 m/s at a depth of800 mm, from Figure 7.8, α = 13.10.

Therefore,

AC = 6.0 m

Therefore, the outlet protection (gabions orequivalent) is to extend 6.0 m from the outlet ofthe barrel. The side boundaries are to be anextension of the wingwalls or from the calculatedflare angle if it falls outside the extension of thewingwalls.

7.7 Energy Dissipators forSupercritical Flow

A concrete channel 1.5 m wide has a design flowof 6.86 m³/s at a depth of 0.5 m and velocity 9.15m/s. The channel has a slope of 6% (0.06 m/m)and discharges into a wide, relatively flat existingchannel.

To minimise any scouring at the downstreamchannel, it is necessary to dissipate the energyfrom the flows from the concrete channel, bycreating a hydraulic jump by either

A. A horizontal apron with unchanged channelwidth;

B. A simple drop structure with widenedchannel and lowered bed; or

C. A stilling basin with concrete blocks andend sills.

Option C is considered to be impractical for thisjob in a relatively remote part of Queensland, andcalculations are not shown in this example.However, indicative lengths of some standardUSBR Basins from the Bureau of Reclamation(1964) are given.

233.04.1

1.13tanBCAC 0 ==

ACBCtan =α

2.14x2.1

VV2.1BEBE

AF ==

BE

AF

AF

BEVV

WidthWidth =

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A. Hydraulic Jump on a Horizontal Apronwith Unchanged Channel Width

Figure 7.10

Figure 7.10 shows the required hydraulic jumpwith water depth, D1 becoming D2 after the jump.

It is the simplest of all energy dissipators, but italso has the longest length of of apron, L1 to beprotected. The sides also have to be protected.

In this example,

V1 = 9.15 m/s

D1 = 0.5 m

It is required to find L1.

Step 1

First, calculate the Froude No.,

Froude No.,

Therefore, F = 4.13

Step 2

Find the sequent depth, D2.

From Figure 7.11 (Bureau of Reclamation, 1964)

Therefore,

D2 = 5.36 D1 = 5.36 x 0.5

D2 = 2.68 m

Figure 7.11

Step 3

Find the length of horizontal apron required toinduce the hydraulic jump.

From Figure 7.12 (Bureau of Reclamation, 1964)

Figure 7.12

Therefore, L1 = 15.44 m

For this design, a horizontal apron, 15.44 m longand protected against scouring is required.Because of turbulence, the sides of the channelalong the apron length have also to be protected toa height of D2 plus freeboard.

76.52D

L1 =

)1F81(21

DD 2

1

2 −+=

5.05.0 )5.0x8.9(15.9

)gD(VF ==

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B. Hydraulic Jump in a Simple Drop Basin

By both widening the channel and dropping thebed level, the length of apron required to inducethe hydraulic jump may be considerablyshortened. The design shown below has themaximum effective widening consistent with thenatural flare of the water entering the dropstructure.

With reference to Figure 7.13 (US Dept ofTransport, 1983):

Known:y0 = 0.5 m

V0 = 9.15 m/s

W0 = 1.5 m

S0 = 0.06

Z0 = 5.0 m say (height above a datum line of100.0 m say

Z3 = 4.52 m say (start of wide open channeldownstream)

Tw = 0.8 m say (calculated in open channel atoutlet to the drop basin) @ Ht = 105.32 m

ST = Ss = 0.5 say (0.33 or 0.5 suitable for thisdesign)

Z1 =Z2 = 3.5 m say (due to an assumed drop ofapprox. 3y0 at the end of the basin)

LT = 3.0 m (due to the 1.5 m drop)

L = 8.0 m say (about half of horizontal

apron in example A)

ST = Ss = 0.5 say (0.33 or 0.5 suitable for thisdesign)

Required to calculate:W1 = WB = basin width

y1 = depth of water at the bottom of the drop

y2 = sequent depth after hydraulic jump is formed

Acceptable Design:The design will be acceptable when the watersurface level at the end of the basin (top of y2) isbelow the tailwater level at this point. Thetailwater is required to be higher to induce thehydraulic jump.

The design procedure is:

Step 1

Locate the basin where the flow downstream ofthe basin is sub-critical or where a simple tran-sition as in Example 7.6 will lead to non scouring(usually sub-critical) flow. i.e. locate the basin ator near the end of the 6% slope. (Sub-critical flowoccurs when the Froude No. is < 1).

Step 2

Assume an initial basin shape for testing. Assumea length, L = 8.0 m (about half the length of thehorizontal basin required from example A.

• Assume a drop of 1.5 m at the front of the basin(equal to a drop of 3y0 to the floor of the basin).

• Assume a height DATUM = 100.00 m.

• Calculate Z0, Z1, Z2 etc.

Therefore, Z1 = Z2 = 3.5 m

LT = 3.0 m (due to the 1.5 m drop)

Step 3

Calculate the Froude No. at the start of the basin.

F0 = 4.13

Step 4

Calculate the maximum effective width, W1 in thebasin. This is due to the natural flaring of thewater as it leaves the upstream channel.

W1 = 2.04 m

13.4x315.03x25.1W

2

1++=

0

2TT

0B1 F31SL2

WWW+

+==

5.05.00

0 )5.0x8.9(14.9

)gd(VF ==

7-14 July 2002


7

Step 5

Find the water depth, y1, at the start of thehorizontal apron.

For this basin,

Q = y1WB [2g (Z0 - Z1+ y0 - y1) + V02] 0.5

The discharge Q = 6.86 m³/s, and y1 is found bytrial and error, starting from values about 0.6y0,until the correct discharge is obtained.

After various trials, say y1 = 0.31 m,

Q = 0.31 x 2.04 [2 x 9.8(5 - 3.5 + 0.5 - 0.31) +9.152] 0.5 = 6.84 m³/s

Adopt y1 = 0.31 m

Step 6

Find the velocity V1 corresponding to y1 and theFroude No. at the start of the horizontal apron.

V1 = Q1/A1 = 6.86 / 2.04 x 0.31 = 10.85 m/s

Therefore,

F1 = 6.22

Step 7

Find the sequent depth, y2 after the hydraulicjump has formed, and the water surface height.

The hydraulic jump equation is

where C1 = 1.0 for the hydraulic jump

y2 = 2.58 m

The water surface height at the end of the basin is,therefore, 100 + Z2 + y2

= 100 + 3.5 + 2.58

= 106.08 m

This is above the tailwater, TW height of105.32m. Therefore the design is unacceptable asthe hydraulic jump will not form.

2)122.6x81(31.0x1y

2

2−+=

2)1F81(yC

y2

1112

−+=

5.05.01

1 )31.0x8.9(85.10

)gd(VV ==

July 2002 7-15


7

Figure 7.13

Step 8

Repeat Steps 1 to 7 until an acceptable depth isobtained at y2. It is obvious the floor of the basinhas to be lowered to achieve this.

Final trial giving acceptable results gave:

Lower basin floor 2.25 m from edge of channel,giving:

Z0 = 5.0 m, Z1 = Z2 = 2.75 m

LT = 4.5 m

W0 = 1.5 m (unchanged)

W1 = WB = 2.52 m (calc.)

y1 = 0.258 m (calc.)

F1 = 7.24 (calc.)

y2 = 2.52 m (calc.)

Height of water surface of y2 @ 105.27 m

Tailwater height @ 105.32 > 105.27

Therefore, the design is acceptable.

Step 9

Check the length of the basin and the outlet levels,to ensure the initial total length assumed of 8.0 mis approximately correct.

The length of the horizontal apron,

LB = 2.52 m

From geometry,

Z3 - Z2 = 104.52 - 102.75 m = 1.77m

Therefore, Ls = 1.77 x 2 = 3.52 m

Total length = LT + LB + Ls

= 4.5 + 2.52 + 3.52

= 10. 54 m

This is longer than the assumed length.

Check that the tailwater water just past the end ofthe basin is still higher than the water level at theend of the basin.

If so, adopt the design.

If not, the basin may need to be moved further upthe slope and recalculated with a deeper drop atthe start.

Notes:

• The floor of the drop basin should be concreteprotected and the vertical sides also.

• A trapezoidal channel for convenience ofconstruction may require changes to the designprocedure.

• Approximate basin lengths of standard USBRBasins from graphs (Bureau of Reclamation,1964) are:

L = 9.8 m for Type II Basins with chute blocksand dentated end sill.

L = 6.2 m for Type III Basins with chute blocks,baffle piers and solid end sill.

7.8 Scour Hole Dimensions

The following example calculates scour holedimensions at existing culverts, or at new culvertswhere outlet protection is limited and notdesigned for the higher than average velocitiesthrough the culverts.

When scour holes have reached the calculateddimensions, only minimal protection is requiredas shown in Appendix 7.8.A to stop furthererosion.

The method is taken from the Ministry of Worksand Development, NZ (1978) based on testing onsand beds. In practice for other than sand at theoutlet, one-half to one-third of the calculateddimensions were observed in field inspections,hence adopted dimensions reflect thisproportion.

Given:

A 1/1500 x 900 RCBC has been designed to flowfull with an outlet velocity = 5.0 m/s and outletprotection extending only 1350 mm downstream.

The natural surface material downstream is sandyclay.

76.076.01

2B 24.7

52.2x5.4F

y5.4L ==

7-16 July 2002


7

Calculate the expected scour hole dimensions.

Step 1

Write down the following equation fromSection 4.5.1.6 and calculate the values Ye, Q,and Q/Ye2.5.

where

A = wetted area of culvert at outlet (m²)

Q = discharge (m³/s)

t = period of peak discharge (minutes). Use 30minutes if time not known.

α, β, θ, γ are coefficients and indices listed inTable 4.21.

Ye = 0.82

Q = AV = 1.5 x 0.9 x 5

Q = 6.75 m³/s

Step 2

Calculate the dimensions and volume of thescour hole.

Depth, hs = 0.95 (0.82)1.0 (11.09)0.375 (30)0.10

= 2.70 m

Adopt hs = 1.40 m (approx. 0.5 hs calculatedfor sand)

Width, Ws = 0.67 (0.82)1.0(11.09)0.915(30)0.15

= 9.89m

Adopt Ws = 4.5 m (approx. 0.5 Ws calculatedfor sand)

Length, Ls = 4.34 (0.82)1.0(11.09)0.71(30)0.125

= 35.9m

Adopt Ls = 18.0 m (approx 0.5 Ls calculatedfor sand)

Volume,Vs = 0.79 (0.82)3.0(11.09)2.0(30)0.375

= 191.8 m³

Adopt Vs = 95.9 m³ (approx. 0.5 Vs calculatedfor sand)

The wetted area A can be obtained from:

(1) Inlet control - generally normal depth(Figures 11.4A) but can be full depth in veryhigh tail water or flow under pressure

(2) Outlet control - tail water depth or criticaldepth (whichever is greater) when outlet isnot submerged; or D if outlet is submerged.

For the concrete channel, discussed in Section 7.7(i.e. 1.5 m wide with a flow depth 0.5 m and flowrate of 6.86 m³/s) the results would be:

Adopting dimensions 50% of those calculated,

Depth, hs = 1.4 m; Width, Ws = 6.2 m;

Length, Ls = 19.1 m and Volume, Vs = 179 m³.

7.9 Tidal Range at Site

At a job site some kilometres from the mouth of atidal stream, it may be necessary to calculate tidalranges at the site for construction reasons or toestimate clearance for boats under an existingbridge.

Tidal ranges and heights will become increasinglydifferent to those at the mouth of a creek, thefurther the distance upstream. The time to fill andempty tidal compartments, sand bars at the mouthor mounds of sand or bed material from erosionand deposition processes in the creek channelobstructing flows are just some of the reasonswhy tide levels will be different.

The example given here is an approximatecalculation and the more cycles of successive highand low tides measured at the job site, the moreaccurate the assessment.

09.11Y

Q5.2

e=

5.25.2e 82.0

75.6Y

Q =

5.05.0e )

29.0x5.1()

2A(Y ==

2/1e )

2A(Y =

θβγα= )t()Y

Q()Y(Dimension 5.2e

e

July 2002 7-17


7

Particularly where the tidal range is small, specialcare should be taken to check that “tide” levels atthe job site have not been affected by rainfallrunoff in the catchment.

Given:

It is required to calculate tidal ranges at a job site10 km from the mouth of a creek entering theCoral Sea just north of Mackay. In particular, theHAT (Highest Astronomical Tide) height isrequired at the job site.

Successive high - low - high heights have beenmeasured and a typical tidal range is that on the 2November, 1998. The nearest tide gauge(Queensland Transport) is that at Mackay OuterHarbour and the nearest high and low tide levelson that day are shown for comparison below.

Time Tide Heights (m)Job Site Mackay Outer Harbour

9.00 2.0529.09 5.52 (high tide)10.00 2.52211.00 2.48212.00 1.42213.00 0.43214.00 -0.03815.27 0.40 (low tide)

Step 1

Compare the tidal ranges at the site and at thegauge.

Difference between high and low tide (tidal range)at the job site on 2/11/98,

Tidal Range, TRsite = 2.522 - (-0.038) = 2.56 m

At Gauge, TRgauge = 5.54 - 0.40 = 5.14 m

Step 2

Calculate the Highest Astronomical Tide(HAT) height at the job site.

At Mackay Outer Harbour, HAT = 6.41 m

Height above 2/11/98 morning high tide =

= 6.41 - 5.54 = 0.87m

At job site, height HAT is above the 2/11/98morning high tide = 0.87 x Tide Ratio = 0.87 x0.50 m = 0.44 m

Therefore, calculated HAT at job site = 2.522 +0.44 m = 2.96 m

Adopt calculated HAT height 2.96 m.

Note: It is again emphasised that the moremeasurements of low to high or high to low tidesat the job site, the more accurate the calculationthe tide ratio and predicted tide levels become.

Step 3

Calculate the Mean High Water Springs(MHWS) and Mean Low water Springs(MLWS) heights at the job site.

At Mackay Outer Harbour,

MHWS = 5.28 m

MLWS = 0.72 m

Difference in spring tides = 5.28 - 0.72 m = 4.56 m

Equivalent difference in spring tides at job site =4.56 x Tide Ratio = 4.56 x 0.50 = 2.28 m

Next, find the MHWS tide height at the job site ina similar procedure as that for HAT.

At Mackay Outer harbour, height MHWS isbelow 2/11/98 high tide = 5.54 - 5.28 = 0.26 m

Therefore, at job site, height MHWS is below2/11/98 high tide = 0.26 x Tide Ratio = 0.26 x 0.50= 0.13m

Therefore, MHWS height at job site = 2.522 -0.13 = 2.39 m

Therefore, MLWS height at job site = 2.39 -2.28 m = 0.11 m

Adopt at job site, MHWS Ht 2.39 m and MLWSHt 0.11 m.

50.014.556.2

TRTR

gauge

site ==Tide Ratio,

7-18 July 2002


7

7.10 Floodway Calculations

The following example illustrates principles offloodway design described in Section 4.2.3.Reference should be made to this Section fordesign curves and to Section 4.2.2.

This example initially fixes the allowable depth ofwater over the road i.e. the headwater, and theorder of calculations is different to that shown inSection 4.2.2.

However, both Case A and Case B calculations inthe Section are also carried out, Case A beingcritical to check the allowable velocity throughthe culverts and Case B to calculate the afflux forfloods larger than the design flood.

Required:

A floodway with 20 year trafficability is requiredto be designed over a natural open channelapproximately trapezoidal in shape.

The floodway would be approximately 90 to100m long and for cost reasons, road batters willbe grass covered only for protection against scour.

Preliminary Considerations:

• Because the batters will be protected by grassonly, culverts will be required to build up thetailwater to not more than 300 mm below theedge of the downstream shoulder whenovertopping of the road first occurs. Allowingfor crossfall, there will be a head of 450 mmand a velocity of about 2.30-2.45 m/s throughthe culverts if this minimum tailwater isadopted. Is this acceptable on this job? (Thiswill be answered in the example).

• For 20 year ARI trafficability, the floodwaylevel should be at the 20 year unrestricted floodlevel to allow the maximum amount of waterover the road and save on culvert and overallcosts.

In general it is only in very long floodwayswith very little velocity in the open channeland/or where costly protection is unavoidable,that increasing culvert requirements by raisingthe road, thus decreasing the flow over the road

(to the extreme of a flood free road) may reducethe overall cost of the job.

Step 1

List all relevant criteria:Required standard: Trafficable in a 20 year ARIflood.

Time of closure: Maximum of 1 day in a 50 yearARI flood. (Calculated hydrograph shows thismaximum only a matter of hours - not includedhere).

Batter protection: Grass.

Width of floodway: 10 m.

Road crossfall: 3 %

Step 2

Calculate the rating curve for the unrestrictedchannel.

From open channel hydraulic calculations similarto those shown in Section 7.1, key results are:

Q50 = 162 m³/s @ Ht 322.76 m

Q20 = 130.4 m³/s @ Ht 322.58 m,

V = 0.68 m/s

Q10 = 108.1 m³/s @ Ht 322.44 m

Q = 70 m³/s @ Ht 322.13 m

Step 3

Adopt a road level and calculate the maximumallowable depth of water over the road.

Adopt the road level at the unrestricted 20 yearARI flood level and show the cross-sectionaldetails in Figure 7.14.

Figure 7.14

July 2002 7-19


7

From Section 4.2.3.5, closure to traffic occurswhen

Maximum allowable h, depth of water over theroad, occurs when H = 0.300 m

Therefore,

h = 0.28 m

Step 4

Calculate the discharge over the road andthrough the culverts in a 20 year ARI flood.

The total discharge over the road and through theculverts must equal the discharge in theunrestricted channel downstream with flow atheight 322.58 m.

i.e. Qtot = QR + QC = 130.4 m³/s

Flow over the road,

QR = CfLH1.5 (Section 4.2.3.3 for “free” flow,tailwater not above crown level of road)

where Cf = coefficient of discharge

Cf = 1.674 (From Figure F14,H/1 =.30/10 = 0.03 <0.15)

L = length of floodway = 94 m

H = 0.30 m

Therefore

QR = 1.674 x 94 x 0.31.5

QR = 25.9 m³/s

Discharge through culverts,

QC = Qtot - QR = 130.4 - 25.9

QC = 104.5 m³/s

Therefore, culverts are required to take a dis-charge of 104.5 m³/s operating under a head of0.30 m and outlet control for this design condition(say Case 1).

Step 5

Check for culvert requirements when the floodis at the point of overtopping the road (Case 2).

“Preliminary considerations” at the start of thisexample discussed batter protection, the need tobuild up the tailwater and a culvert velocity of2.30 - 2.45 m/s from a head of 450 mm.

These considerations apply to this example andthe velocity is acceptable. The discharge of70m³/s required through the culverts correspondsto a tailwater Ht 322.13 m (0.45 m below thecrown of the road).

Step 6

Detailed culvert design

Case 1 requires culverts to take a discharge of104.5 m³/s operating under a head of 0.3 m.

Case 2 requires culverts to take 70 m³/s from ahead of 450 mm.

Obviously more culverts will be required fromCase 1. Proceed with the design of culverts to take104.5 m³/s.

Height of culvert opening, D, from

D = crown level of road - crossfall - minimum fillabove culvert - thickness of deck slab - invertlevel

D = 322.58 - 5 x 0.03 - 0.100 - 0.180 - 320.55

D = 1.60 m.

For culvert design:

Outlet control with H = 0.30 m

Tailwater at Ht 322.58 m, HW at Ht 322.88 m

Invert at Ht 320.50 m say

From culvert design procedure in Chapter 11,illustrated in the example in Section 7.3,

13 / 2700 x 1500 RCBC have a capacity of104m³/s with Outlet Velocity = 1.98 m/s.

Adopt 13 / 2700 x 1500 RCBC and FloodwayHt 322.58 m and Length 94 m.

8.9x268.0300.0

g2VHh

22−=−=

mm300g2

VhH2>+=

7-20 July 2002


7

Step 7

If required, calculate the afflux in larger floods(e.g. 50 year ARI flood) with procedure shown inSection 4.2.3.3 (B. At Peak of the Flood).

July 2002 7-21


7

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