Hydrology

HYDROLOGICAL CALCULATIONS

Design discharge• Para 4 of substructure code –

Hydrological design investigations• Estimation of design discharge :

Para 4.2.2 of Substructure code says 50 year return period.

Bridge design : use of Q

• Waterway :• HFL• Depth of scour :• Clearance• Afflux• Free Board• River training works

QCPw 811.1

31

2

473.0

f

QD

101524.0

88.17

22

a

AX

Vh

FLOOD ESTIMATION METHODS

1. Statistical analysis (flood frequency approach)- when sufficient data is available (probabilty distribution curves e.g.Gumbel/gamma)

2. UNIT hydrograph approach : (not for catchments more than 2500 sq km ) when limited or no records are available. Use of flood estimation reports

3. For catchments having area < 25 sqkm : Modified rational formula – RDSO report RBF-16

Earlier methods were based on empirical formulae (Dickens,Ryves, Inglis , Nawab Ali Jung etc) and had no concept of frequency or recurrence interval concept

SYNTHETIC UNIT HYDROGRAPH (SUH)

Use of flood estimation reports : The country has been divided into 7 hydro meteorological zones and 26 subzones .Flood estimation reports have been published. (a joint work of CWC, RDSO, IMD, and MOST)

UNIT HYDROGRAPH Concepts

The UH of a drainage basin is defined as a hydrograph of direct runoff (DSRO) resulting from one unit of effective rainfall which is uniformly distributed over the basin at a uniform rate during the specified period of time known as unit time or unit duration. The unit quantity of effective rainfall is generally taken as 1cm and the outflow hydrograph is expressed by the discharge ordinates. The unit duration may be 1 hour, 2 hour, 3 hours or so depending upon the size of the catchment and storm characteristics. However, the unit duration cannot be more than the time of concentration tc, which is the time that is taken by the water from the furthest point of the catchment to reach the outlet.

UNIT HYDROGRAPH -DEFINITION

1. Effective rainfall is uniformly distributed over the basin, that is, if there are ‘N’ rain gauges spread uniformly over the basin, then all the gauges should record almost same amount of rainfall during the specified time.

2. Effective rainfall is constant over the catchment during the unit time. 3. The direct runoff hydrograph for a given effective rainfall for a catchment is

always the same irrespective of when it occurs. Hence, any previous rainfall event is not considered. This antecedent precipitation is otherwise important because of its effect on soil-infiltration rate, depressional and detention storage, and hence, on the resultant hydrograph.

4. The ordinates of the unit hydrograph are directly proportional to the effective rainfall hyetograph ordinate. Hence, if a 6-h unit hydrograph due to 1 cm rainfall is given, then a 6-h hydrograph due to 2 cm rainfall would just mean doubling the unit hydrograph ordinates. Hence, the base of the resulting hydrograph (from the start or rise up to the time when discharge becomes zero) also remains the same

ASSUMPTIONS

Flood or discharge at the point of interest depends

upon the catchments characteristics and rainfall

characteristics

Q

QQ

Q

Q

Q

Q

Q

Q

= Discharge

Typical river system

Shape, size and slope of the catchment

• A catchment that is shaped in the form of a pear, with the narrow end towards the upstream and the broader end nearer the catchment outlet (Figure 1a) shall have a hydrograph that is fast rising and has a rather concentrated high peak

A catchment with the same area but shaped with its narrow end towards the outlet (oblong shape) has a hydrograph that is slow rising and with a somewhat lower peak (Figure 2) for the same amount of rainfall.

Shape, size and slope of the catchment…

Loss rateInterception , evaporation, transpiration, evapo-transpiration, infiltration, watershed leakage etc

Catchment characteristics are reflected in the UH of the typical

catchment

Rainfall characteristics

• Rainfall intensity• Rainfall or storm duration• Areal distribution of rainfall• Time distribution of rainfall• Direction of storm w.r.t.

catchment

Rainfall characteristics

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

Time (hrs.)

Pre

cip

itat

ion

(in

ches

)

Uniform loss rate of 0.2 inches per hour.

Rainfall excess

Hyetograph

Rainfall or storm duration

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

Time (hrs.)

Exc

ess

Pre

c. (

inch

es)

Small amounts of excess precipitation at beginning and end may

be omitted.

Derived unit hydrograph is the result of approximately 6 hours

of excess precipitation.

The concept of Isochrones might be helpful for explaining the effect of the duration of a uniform rainfall on the shape of hydrograph. Isochrones are imaginary lines across the catchment (see Figure 5) from where water particles traveling downward take the same time to reach the catchment outlet. Version

If the rainfall event starts at time zero, then the hydrograph at the catchment outlet will go on rising and after a time‘Δt’, the flow from the isochrone I would have reached the catchment outlet. Thus, after a gap of time Δt, all the area A1 contributes to the outflow hydrograph. Continuing in this fashion, it can be concluded that after a lapse of time ‘4Δt’, all the catchment area would be contributing to the catchment outflow, provided the rain continues to fall for atleast up to a time 4Δt. If rainfall continues further, then the hydrograph would not increase further and thus would reach a plateau.

TIME AND AREAL DISTRIBUTION OF RAINFALL

DIRECT RUNOFF HYDROGRAPH – CAN BE DERIVED BY SUMMATION OF INDIVIDUAL HYDROGRAPHS

Finding catchment – use of TOPO sheets

STREAMS

1. Blue lines

2. pattern of the contour lines V patterns. ‘V’ always points u/s

On flood plains, drainage lines are less well defined and the contour lines flatten out. Run-off flows across the whole of the floodplain rather than in a well-defined channel.

Identifying features on toposheets

Identifying ridge line and streams (valley)

• Contour lines in valleys form 'V' shapes (typically narrowing and steep valleys), which can spread to wide 'U' shapes (typically broad, shallow valleys) - but the effect is the same - the closed end points to higher ground (see Fig. 2).

• Ridge contours can be confused with valley contours as they, too, form 'U' shapes - the difference is that the closed end of the 'U' points to lower ground. (See Fig.3)

One of the easiest landscapes to visualize on a topographic map is an isolated hill. If this hill is more or less circular the map will show it as a series of more or less concentric circles . A contour is a line that joins points of equal elevationContour interval is the vertical distance between contour lines

IDENTIFYING AN ISOLATED HILL/SADDLE

Very often the closed ends of two 'U's' can be seen pointing toward each other. This is a sure indication of a saddle between two areas of higher ground.

Contour Patterns…• Contour lines close together

show steep slopes• Contour lines far apart show

gentle slopes• Contour lines evenly spaced

show uniform slope• If the spacing decreases when

going from high to low, the slope is convex

• If the spacing increases when going from high to low, the slope is concave

After identifying the stream we locate highest point of each stream/tributary. A general rule of thumb is that topographic lines always point upstream. In the Figure for example, the direction of stream flow is from point A to point B. Ultimately, we reach the highest point upstream. This is the head of the watershed, beyond which the land slopes away into another watershed or catchment. If we join all of these high points around the stream we have the watershed boundary. (High points are generally hill tops, ridge lines, or saddles).

Delineating catchment on toposheets

Delineating catchment on topo sheet

Catchment Parameters

AREA OF CATCHMENT -A

LENGTH OF LONGEST STREAM : L

CG OF THE CATCHMENT

LENGTH OF STREAM FROM NEAREST TO CG TO THE BRIDGE SITE – Lc

EQ. OR STASTICAL STREAM SLOPE

LLc

1

2

3

45

FLOOD ESTIMATION STEP BY STEP

1. Calculate catchment parameters 2. Calculate SUH parameters using the

equations (equations given in the subzone reports)

3. Develop SUH4. Obtain effective rainfall excess of design

duration 5. Apply effective rainfall excess on the SUH to

obtain the flood hydrograph or peak flood.

STEP-1 –Catchment parameters1. Find catchment area A-in sqkm – planimeter, graph by counting

squares, autocad2. Find length of the longest stream L in KM3. Find CG of the area- by hanging a cardboard of the shape of catchment

area and joining the line4. Length of the stream from the bridge site near to CG. Take the same

stream which was taken for step-2 5. Calculation of equivalent stream slope S- using analytical method or

graphical method as given in the flood estimation report.

Is observed to be an important parameter of the catchment and related with the basin lag tpS

LLc

Determining stream slope (equivalent or statistical slope)

1 2 3 4 5

e0 e1e2

e3e4

e5e6

D0DatumD1

D2D3 D4

D5 D6

l1 l2 l3 l4 l5 l6

n

L

ii liDDS

1

12

Time - Hrs

Q-c

umec

s

50W

75W

50RW

75RW

pq

BT

45.0

253.0

S

LLXt c

p

842.0968.1 pp tq

018.150 3.2 pqW

035.175 581.0 pqW

078.150 954.0 pR qW

035.175 581.0 pR qW

9.0572.4 pB tT

STEP-2–Calculating UH parameters

UH equations for subzone-3(f)

Calculation of UH parameters

Parameter

Area of the

catchment A

Length of the stream

from farthest point of

the catchment

to the bridge L

Length of the

longest strem

from the point

nearest to CG to the bridge Lc

Equivalent slope m/km

qp = 0.9178*(L/s)-

0.4313

tp = 1.5607*(qp)-

1.0814

W50=1.9251(qp)-

1.0896

W75=1.0189(qp)-1.0443

WR50=0.5788(qp)-

1.1072

WR75=0.3469(qp

)-1.0538

TB = 7.3801(tp)0.734

3

tm=tp+tr/2

Qp=qp*A

Design storm

duration Td

=1.1*tp

R50 : 50 year 24

hrs point

rainfall

R50(Td) : 50

year td hrs

point rainfall

R50(Td) : 50

year td hrs

areal rainfall

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Br No sqkm km km m/kmcumecs/sq km

hrs hrs hrs hrs hrs hrs hrs cumecs hrs mm mm mm

119 11.3 6.75 3.25 8.22 0.999 1.56 1.93 1.02 0.58 0.35 10.24 2.06 11.26 1.72 420 159.6 156.41

125 3.7 4.3 2.25 10.78 1.364 1.12 1.37 0.74 0.41 0.25 8.00 1.62 5.06 1.23 440 154 154

135 &136

17.6 7.2 4.7 2.49 0.581 2.81 3.48 1.80 1.06 0.62 15.76 3.31 10.23 3.09 440 211.2 202.75

166 10.6 5.75 2.9 3.03 0.696 2.31 2.86 1.49 0.86 0.51 13.64 2.81 7.40 2.54 440 193.6 183.92

169 232.8 19.55 10.1 3.26 0.424 3.95 4.91 2.50 1.50 0.86 20.23 4.45 98.68 4.34 440 237.6 197.21

173 24.6 14.15 7.9 7.77 0.709 2.27 2.80 1.46 0.85 0.50 13.45 2.77 17.41 2.49 440 198 186.12

318 2.2 2.2 1.1 11.36 1.863 0.80 0.98 0.53 0.29 0.18 6.24 1.30 4.02 0.88 480 144 144

358+359

1352.0 81 40 2.36 0.200 8.91 11.13 5.48 3.44 1.89 36.77 9.41 270.09 9.80 520 379.6 273.31

400+401

2072.0 160.25 82.25 2.59 0.155 11.73 14.69 7.15 4.56 2.48 45.00 12.23 320.92 12.90 520 416 312

411 23.4 8.55 4.05 1.46 0.428 3.91 4.86 2.47 1.48 0.85 20.09 4.41 10.02 4.30 380 173.3 166.37

428 36.9 14.15 7.75 1.12 0.307 5.60 6.97 3.50 2.14 1.20 26.15 6.10 11.33 6.16 380 200 192

17 46.1 20 11 1.01 0.254 6.88 8.58 4.27 2.64 1.47 30.42 7.38 11.68 7.57 380 220 211.2

Drawing 1 Hr UH….

Parameter

X Cordinate of pea

k

Cordinates of

b1

Cordinates of

b2

Cordinates of c1

Cordinates of c2

Br No Xp X1 Y1 X2 Y2 X3 Y3 X4 Y4

119 2.06 1.48 5.63 3.41 5.63 1.71 8.44 2.73 8.44

125 1.62 1.21 2.53 2.58 2.53 1.37 3.80 2.10 3.80

135 &136 3.31 2.25 5.12 5.73 5.12 2.69 7.68 4.49 7.68

166 2.81 1.94 3.70 4.80 3.70 2.30 5.55 3.79 5.55

169 4.45 2.95 49.34 7.86 49.34 3.59 74.01 6.09 74.01

173 2.77 1.92 8.70 4.72 8.70 2.27 13.06 3.73 13.06

318 1.30 1.01 2.01 1.98 2.01 1.12 3.02 1.65 3.02

358+359 9.41 5.96 135.0 17.10 135.05 7.51 202.57 12.99 202.57

400+401 12.23 7.66 160.4 22.35 160.46 9.75 240.69 16.90 240.69

75W

Time - Hrs

Q-c

umec

s

50W

50RW

75RW

pq

BT

75W

Xp

X1 X2

X3 X4

Measure hourly UH ordinatesBr No 169 –Virar –SuratTime Hr UH

0 01 52 173 404 855 956 757 648 519 43

10 3511 2912 2413 2014 1715 1416 1217 818 619 420 221 0

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 220

10

20

30

40

50

60

70

80

90

100

Br 169 catchment : 1 Hr UH

STEP-4 -Rainfall Characteristicsi. Design storm duration TD = 1.1 x tp- it gives

maximum flood peakii. Then to find TD hr rainfall - find 24 hr point

rainfall from isopluvial maps (if not available directly) and multiply with the conversion factor to get TD hr point rainfall.

iii. Multiply by areal reduction factor (ARF) given in the relevant sub-zone report to get TD hr areal rainfall

iv. Use time distribution graphs /tables to obtain rainfall depth (cumulative) for each 1 hr interval.

v. then find rainfall increments by subtraction of successive cumulative values

vi. Obtain 1 hr effective rainfall increment by subtracting the loss rate (given for each subzone)

Rainfall characteristics

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

Time (hrs.)

Pre

cip

itat

ion

(in

ches

)

Uniform loss rate of 0.2 inches per hour.

Rainfall excess or 1 hr -effective rainfall

Rainfall Hyetograph

Example : 4-Hr duration rainfall in sub-zone 5(a) Br No 169 in VR-ST

• A = 233 sq km• L = 19.55 KM• Lc : 10.1 KM• Slope : 3.26 m/ KM• R50 24 Hrs rainfall : 400 mm• R50 – 4 hrs rainfall : 180 mm• Loss rate : 0.19 cm/hr• Base flow : 0.15 cumecs/sq KM

– rainfall dist in 4 hrs : 57%,81%,94%,100% This comes to 13.28 , 18.87,21.9,23.3cm

– Hence rainfall excess : 13.28, (18.87-13.28)=5.59, (21.9-18.87)=3.03, (23.3-21.9)=1.4

– Effective rainfall excess : (13.28-0.19) , (5.59-0.19), (3.03-0.19), (1.4-0.19) : 13.09 5.4 2.84 1.21

STEP-5: (i) Critical sequence of 1-hr effective areal rainfall excess

Br No 169 –Virar –SuratTime Hr UH RE Q

0 0    1 5    2 17    3 40    4 85 5.4 459.175 95 13.09 1243.656 75 2.84 212.937 64 1.21 77.318 51   1993.19 43 B/FLOW 35.00

10 35   2028.111 29    12 24    13 20    14 17    15 14    16 12    17 8    18 6    19 4    20 2    21 0    

Arrange 1 –hr effective areal rainfall against 1 hr UG ordinates such that max value against max ordinate and next lower value against next lower ordinate and so on… this can be done easily in Microsoft excel worksheet. Multiply the UH ordinate by the above corresponding values and add them to get DSRO. Add base flow to get Q

STEP-V-ii :Obtaining flood hydrograph

Reverse the sequence of 1 –hr effective areal rainfall and obtain ordinate for each hydrograph by multiplying the corresponding UHG ordinates and adding all such hydrographs (ordinates) to obtain flood hydrograph (ordinates)

Time UH ord 1 Hr effective rainfall cm

Total D.S.R.

O

Base flow

Total flow

cumecsHr Q 1.21 2.84 13.09 5.40

0 0 0.0       0 35 351 5 6.0 0.0     6 35 412 17 20.5 14.2 0.0   35 35 703 40 48.3 48.3 65.5 0.0 162 35 1974 85 102.7 113.6 222.5 27.0 466 35 5015 95 114.8 241.3 523.6 91.8 972 35 10076 75 90.6 269.7 1112.7 216.1 1689 35 17247 64 77.3 212.9 1243.6 459.2 1993 35 20288 51 61.6 181.7 981.8 513.2 1738 35 17739 43 51.9 144.8 837.8 405.2 1440 35 1475

10 35 42.3 122.1 667.6 345.7 1178 35 121311 29 35.0 99.4 562.9 275.5 973 35 100812 24 29.0 82.3 458.2 232.3 802 35 83713 20 24.2 68.1 379.6 189.1 661 35 69614 17 20.5 56.8 314.2 156.7 548 35 58315 14 16.9 48.3 261.8 129.6 457 35 49216 12 14.5 39.7 222.5 108.0 385 35 42017 8 9.7 34.1 183.3 91.8 319 35 35418 6 7.2 22.7 157.1 75.6 263 35 29819 4 4.8 17.0 104.7 64.8 191 35 22620 2 2.4 11.4 78.5 43.2 136 35 17121 0 0.0 5.7 52.4 32.4 90 35 125      0.0 26.2 21.6 48 35 83        0.0 10.8 11 35 46          0.0 0 35 35

0 2 4 6 8 10 12 14 16 18 20 22 240

500

1000

1500

2000

2500

Alternative method : Simplified equations- RDSO report TM 50

• Developed for some sub-zones and recommended for preliminary checks /surveys or temporary works

e.g. for 3(f) :

45.0

388.0

st

cd

S

LLt

2548.0

18945.0

)(5050

c

stt

LL

SKARQ

d

Flood estimation for small catchments : area < 25 sq km –

RDSO report RBF-16

• Uniform rainfall characteristics• tc is very small therefore we need UH of very

small unit duration. Such data od small intervals are generally not available.

Modified rational formula RBF-16 suggests to use modified rational formula which incorporate s recurrence interval concept over the normal rational formula

ACIQ 5050 278.0C= runoff coefficientA : catchment area in sq KMI50 : 50 year rainfall intensity

mm/hr = R50(tc)/tc

Runoff coefficient• Depends upon nature of soil, soil cover and

location of catchment :

R = 50 year 24 hrs rainfall (cm) F : Areal reduction factor K : 0.249 to 0.498 depends on soil type and

location

2.0).( FRKC

Runoff coefficient……

In absence of no description of catchment is available then

Where : 50 year tc hour point rainfall and tc is given

by

For normal situations H : Elevation of farthest point above the elevation of the bridge

For steep slopes in upper reaches and flat slope in lower reaches, here Sl is average slope from source to site in %

1.0179.0537.0 ctRC

)(50:ct

RR

345.03

H

Ltc

2.01.0 .

.618.0

l

cSA

Lt

Sample calculations – Virar -Surat

Bridge No

50 year 24 hrs

rainfall R50(24)

(Ref Map 6.1 p53 of

TM 50)

Area reduction

factor F (Table 6.2 p 42 TM 50)

50 year 1

hrs rainfall R50(1)

Length of the

stream from

farthest point of

the catchment to the bridge L

Area of the catchment

A

Diff of elevation of bed at farthest point of

catchment and the bridge H

tc = (L3/H)0.3

45

C = 0.332*(R.F)0.2

tc hr ratio R50(tc)/R5

0(24)

1 hr ratio R50(tc)/R50(24)

coefficient K

= (10)/(1

1)

R50(1)=0.38*R50(24

)

R50(tc) = K * R50(1)

I50=R50

(tc)/tc

Q50=0.278*C*I50*A

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Br No cm   cm km sqkm m hrs         mm mm mm/hrs cumecs

119 42 0.7 15.96 6.75 11.2675 297 1.01 0.65 0.425 0.38 1.12 159.6 178.50 176.36 360.65

125 44 0.71 16.72 4.3 3.7125 98 0.93 0.66 0.4 0.38 1.05 167.2 176.00 189.17 129.01

135 &136 44 0.68 16.72 7.2 17.63 19 2.79 0.66 0.58 0.38 1.53 167.2 255.20 91.35 293.32

166 44 0.7 16.72 5.75 10.63 58 1.51 0.66 0.48 0.38 1.26 167.2 211.20 140.22 273.05

169 44 0.68 16.72 19.55 232.83 104 4.37 0.66 0.58 0.38 1.53 167.2 255.20 58.40 2476.48

173 44 0.68 16.72 14.15 24.565 456 1.88 0.66 0.53 0.38 1.39 167.2 233.20 124.18 555.56

318 40 0.72 15.2 2.2 2.16 40 0.63 0.65 0.33 0.38 0.87 152 132.00 208.39 81.36

358+359 52 0.68 19.76 81 1352 649 10.12 0.68 0.58 0.38 1.53 197.6 301.60 29.81 7589.90

400+401 52 0.68 19.76 160.25 2072 1095 17.11 0.68 0.58 0.38 1.53 197.6 301.60 17.62 6876.04

411 38 0.68 14.44 8.55 23.425 10 4.16 0.64 0.58 0.38 1.53 144.4 220.40 52.92 219.25

428 38 0.68 14.44 14.15 36.935 15 6.10 0.64 0.58 0.38 1.53 144.4 220.40 36.14 236.05

17 38 0.68 14.44 20 46.05 20 7.90 0.64 0.58 0.38 1.53 144.4 220.40 27.89 227.18

Deciding HFL • H.F.L. from enquiry : observed HFL• Calculated HFL : using design discharges Q By

area velocity relationship : velocity by manning’s formula -trial and error

• Effect of afflux • Effect of constraint of waterway at D/S

21

321SR

nV

EFFECT OF AFFLUX

EFFECT OF VARIATION IN VELOCITY

For larger streams depth can be calculated for each compartment

THANKS