RDDM Chapter7 Examples

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July 2002

Road Drainage Design Manual Chapter 7: Worked Examples

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Chapter 7Worked

Examples


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Chapter 8References

Chapter 7Worked Examples

Chapter 6

Maintenance and Remediation

Chapter 5Erosion and Sediment Control

Chapter 4

Design

Chapter 3Hydrology and Design Criteria

Chapter 2Site Assessment

Chapter 1Overview

Manual Contents

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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 other

chapters. 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 has

been adequately defined, as indicated by the

following.

For a bridge size catchment (nominally greater than 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, shorter distances upstream and downstream down to anabsolute minimum of 100 metres should similarly

be surveyed.

For the larger catchments, field inspections inaccordance with the Department’s Form 2759,Field Report - Bridge Waterways should have

been carried out including interviews with localresidents about observed or reported floods at or near 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 of the 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


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(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, a

planimeter is necessary to determine cross-sectional areas).

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

A = 20 m²

n = 0.03

Q = A.cos10 0.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 will

be smaller and so will be the velocity, for the same

hydraulic 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 the

flood level corresponding to a total discharge of 42.5 m³/s.

For a water to water boundary such as that between each overflow and the main channel, half the depth of the water at such boundaries may beadded to each wetted perimeter as illustrated

below.

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

Left Overflow

A = 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.cos10 0.V)

(Manning’s formula)

(Section 4.3.6)m/s1.24 n

SR V

0.50.667

=

=

1.27

ABCD) perimeter,tted(area / we 15.8

20 R

=

=

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Main Channel

A = 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.cos10 0.V = 27.5 x 0.985 x 1.50

Q = 40.63 m³/s

Right Overflow

Q = 0.74 m³/s as per left overflowTotal Discharge

Total Discharge = Q left overflow + Q main 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 of concentration is used as the catchment is larger than 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 the

originally 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 R s.

Step 2

Find the average hydraulic radius at the site,

R s 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; WP 1 = 113.5 m; R 1 = 0.60m;n1 = 0.065; V 1 = 0.48 m/s; q 1 = 32.6 m³/s

A2 = 53.3 m²; WP 2 = 21.4 m; R 2 = 2.49 m;n2 = 0.055; V 2 = 1.36 m/s; q 2 = 72.6 m³/s

A3 = 28.2 m²; WP 3 = 73.1 m; R 3 = 0.39 m;n3 = 0.065; V 3 = 0.33 m/s; q 3 = 9.4 m³/s

Q tot = 32.6 + 72.6 + 9.4∴ Q tot = 114.6 m³/s

∴ R s = 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. = R 0.167 / n

R = adopted average hydraulic radius

= 0.75R s where the fall of the stream from thetop of the catchment to the site is fairlyuniform

= 0.65 R s 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= 0.75R s = 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 intensity

for a storm of 13.1 hours duration and an ARI of 50 years is required.

The RAIN2 Program from the Hydraulics Sectionof Main Roads and available in most Districts isused. Other programs are available which also use

parameters from “Australian Rainfall and Runoff”(IEAust, 1987).

The RAIN2 Program which is interactive requeststhe following input data. The actual Map number

and value of parameter from ARR is also shown.

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

06.034.1

ch167.0

=

4.01.0c HchM

L5.8T =

6.11439.0x4.949.2x6.7260.0x6.32

R s++

=

tot

332211s Q

R qR qR qR

++=

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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 Q 50 = 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 IL 0 = 30.0 m

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

(f) Allowable headwater depth in metres.

HW a = 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.

100401000

C +++

=

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Note: These velocities are used inconjunction with field observations of scour in the natural channel. If low velocities

produce scour, then either the allowable

design velocity in (b) should be reduced or consideration be given to additional outlet

protection. However if high velocities occur in 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 of waterway: A = 19.3/1.8 m² = 10.72 m²

(b) Select a culvert size where the soffit would be above or just below the tailwater toutilise the full waterway opening, if

practical.

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 + h o - LS oho = TW = 1.8 m

LS o = 12 x 0.01 = 0.12

From the nomograph in Figure F8E andusing k e = 0.4 from Table 4.2, for area of

box = 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 be

made 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/A o = 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 for

pedestrian 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 of water 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 V ave .

Calculate the product d gVave .

Repeat the procedure as required until d gVave <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 for blockage.

Assume initially concrete gullies, roadway type,kerb in line (Dwg No 1312). Then from the BCCType A Gully Inlet Capacity Charts, removing the

blockage 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/secFrom Table 4.11, allowing for blockage, the

percentage 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 has built 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 will

be 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 for aquaplaning in the transition area between reverse

superelevations, one of the main areas of potentialaquaplaning because of the small to zerocrossfalls.

Reference should be made to Section 4.4.4.2 of the Manual.

Given

Check for aquaplaning.

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

Step 1

Find the longer drainage path lengths for checking. List other parameters, calculated or specified 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

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

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

d = tailwater depth at the outlet of the culvert barrel (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 similar natural 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 were

designed 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 problem

and culverts are expensive so that fewer culvertsdesigned for a higher than average velocity needto be considered.

Because of divergence of flow leaving a culvert barrel, at the end of standard culvert protection,the actual velocity will be about half the designvelocity for most existing culverts. The number of cells 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 of the 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 the

protection 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


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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 culvert

barrel, increasing its width at a uniform rate.

Then, asDischarge = area x velocity

= width x depth x velocityand is constant, then

Width BE x d x V BE = Width AF x d x V AF

Therefore,

Substituting,

BE= 4m

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

Step 4

Find the distance AC to which the outlet protection has to be extended.

For water exiting the culvert at 4 m/s at a depth of 800 mm, from Figure 7.8, α = 13.1 0.

Therefore,

AC = 6.0 m

Therefore, the outlet protection (gabions or equivalent) is to extend 6.0 m from the outlet of the 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 for Supercritical 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 this job 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

1.13tan

BCAC 0 ==

ACBCtan =α

2.14x2.1

VV2.1

BEBE

AF ==

BE

AF

AF

BE

VV

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, D 1 becoming D 2 after the jump.

It is the simplest of all energy dissipators, but italso has the longest length of of apron, L 1 to be protected. The sides also have to be protected.

In this example,

V1 = 9.15 m/s

D1 = 0.5 m

It is required to find L 1.

Step 1First, calculate the Froude No.,

Froude No.,

Therefore, F = 4.13

Step 2

Find the sequent depth, D 2.

From Figure 7.11 (Bureau of Reclamation, 1964)

Therefore,

D2 = 5.36 D 1 = 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, L 1 = 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 D 2 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 the bed 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 of Transport, 1983):

Known:

y0 = 0.5 m

V0 = 9.15 m/sW0 = 1.5 m

S0 = 0.06

Z0 = 5.0 m say (height above a datum line of 100.0 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 = S s = 0.5 say (0.33 or 0.5 suitable for thisdesign)

Z1 =Z 2 = 3.5 m say (due to an assumed drop of approx. 3y 0 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 = S s = 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 water surface level at the end of the basin (top of y 2) is

below the tailwater level at this point. The

tailwater is required to be higher to induce thehydraulic jump.

The design procedure is:

Step 1

Locate the basin where the flow downstream of the 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 3y 0 to the floor of the basin).

• Assume a height DATUM = 100.00 m.

• Calculate Z 0, Z 1, Z 2 etc.

Therefore, Z 1 = Z 2 = 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, W 1 in the basin. This is due to the natural flaring of thewater as it leaves the upstream channel.

W1

= 2.04 m

13.4x315.03x2

5.1W2

1+

+=

0

2TT

0B1 F3

1SL2WWW

++==

5.05.00

0)5.0x8.9(

14.9

)gd(

VF ==

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Step 5

Find the water depth, y 1, at the start of thehorizontal apron.

For this basin,

Q = y 1WB [2g (Z 0 - Z 1+ y 0 - y 1) + V 02] 0.5

The discharge Q = 6.86 m³/s, and y 1 is found bytrial and error, starting from values about 0.6y 0,until the correct discharge is obtained.

After various trials, say y 1 = 0.31 m,

Q = 0.31 x 2.04 [2 x 9.8(5 - 3.5 + 0.5 - 0.31) +9.15 2] 0.5 = 6.84 m³/s

Adopt y 1 = 0.31 m

Step 6

Find the velocity V 1 corresponding to y 1 and theFroude No. at the start of the horizontal apron.

V1 = Q 1/A1 = 6.86 / 2.04 x 0.31 = 10.85 m/s

Therefore,

F1 = 6.22

Step 7

Find the sequent depth, y 2 after the hydraulic jump has formed, and the water surface height.

The hydraulic jump equation is

where C 1 = 1.0 for the hydraulic jump

y2 = 2.58 m

The water surface height at the end of the basin is,therefore, 100 + Z 2 + y 2

= 100 + 3.5 + 2.58

= 106.08 m

This is above the tailwater, TW height of 105.32m. Therefore the design is unacceptable asthe hydraulic jump will not form.

2)122.6x81(31.0x1

y2

2−+

=

2

)1F81(yCy

2111

2

−+=

5.05.01

1)31.0x8.9(

85.10)gd(

VV ==

July 2002 7-15


7

Figure 7.13


20/25

Step 8

Repeat Steps 1 to 7 until an acceptable depth isobtained at y 2. 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, Z 1 = Z 2 = 2.75 m

LT = 4.5 m

W0 = 1.5 m (unchanged)

W1 = W B = 2.52 m (calc.)

y1 = 0.258 m (calc.)

F1 = 7.24 (calc.)

y2 = 2.52 m (calc.)

Height of water surface of y 2 @ 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 - Z 2 = 104.52 - 102.75 m = 1.77m

Therefore, L s = 1.77 x 2 = 3.52 m

Total length = L T + L B + L s= 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 of

the 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 concrete protected and the vertical sides also.

• A trapezoidal channel for convenience of construction may require changes to the design

procedure.

• Approximate basin lengths of standard USBR Basins 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 further erosion.

The method is taken from the Ministry of Worksand Development, NZ (1978) based on testing on

sand 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 outlet

protection extending only 1350 mm downstream.

The natural surface material downstream is sandyclay.

76.076.01

2B

24.7

52.2x5.4

F

y5.4L ==

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Calculate the expected scour hole dimensions.

Step 1

Write down the following equation fromSection 4.5.1.6 and calculate the values Y e, Q,and Q/Y e2.5 .

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, h s = 0.95 (0.82) 1.0 (11.09) 0.375 (30) 0.10

= 2.70 m

Adopt h s = 1.40 m (approx. 0.5 h s calculatedfor sand)

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

= 9.89m

Adopt W s = 4.5 m (approx. 0.5 W s calculatedfor sand)

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

= 35.9m

Adopt L s = 18.0 m (approx 0.5 L s calculatedfor sand)

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

= 191.8 m³

Adopt V s = 95.9 m³ (approx. 0.5 V s 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 very

high tail water or flow under pressure(2) Outlet control - tail water depth or critical

depth (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, h s = 1.4 m; Width, W s = 6.2 m;

Length, L s = 19.1 m and Volume, V s = 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 existing

bridge.

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

75.6

Y

Q=

5.05.0e )

29.0x5.1

()2A

(Y ==

2/1e )

2A

(Y =

θβ γ α= )t()

Y

Q()Y(Dimension 5.2

ee

July 2002 7-17


7


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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 2

November, 1998. The nearest tide gauge

(Queensland Transport) is that at Mackay Outer Harbour 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 2.052

9.09 5.52 (high tide)

10.00 2.522

11.00 2.482

12.00 1.422

13.00 0.432

14.00 -0.038

15.27 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, TR site = 2.522 - (-0.038) = 2.56 m

At Gauge, TR gauge = 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 tides

at 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 is below 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

TR TR

gauge

site ==Tide Ratio,

7-18 July 2002


7


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7.10 Floodway Calculations

The following example illustrates principles of floodway design described in Section 4.2.3.

Reference should be made to this Section for design curves and to Section 4.2.2.

This example initially fixes the allowable depth of water 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 through

the culverts and Case B to calculate the afflux for floods 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 will

be 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 through

the 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 water over the road and save on culvert and overallcosts.

In general it is only in very long floodwayswith very little velocity in the open channel

and/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 year ARI 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 similar to 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 year ARI flood level and show the cross-sectionaldetails in Figure 7.14.

Figure 7.14

July 2002 7-19


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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).


7

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RDDM Chapter7 Examples