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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
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