river engineering flood design

@ McGraw-Hill Education

1

7.11 - DESIGN FLOOD For economic consideration, it’s not practical to

design the hydraulic structure for safety purpose

against maximum possible flood all the times

Culverts / storm drainage can be designed for less severe

flood

This section show the procedure to select the

flood magnitude for the design of hydraulic

structure

3 Types

Spillway design flood

Standard project flood (SPF)

Probable maximum flood (PMF)


2

• Spillway design flood

Design flood specific for designing spillway of storage

structure

This term is used to donate maximum discharge that can

passed in hydraulic structure without causing damage or

threat to the stability of structure

• Standard project flood (SPF)

Flood that result from combination of meteorological &

hydrological factors

Extreme rare combination of factors are excluded

Used in situation where failure would cause less damage

40% - 60% of PMF for same drainage basin


3

•Probable maximum flood (PMF)

Extreme flood that physically possible from severe

combination,

including rare combination of meteorological &

hydrological combination

Used in situation where failure of structure cause life

and severe damage

Table 7.8

Guideline for selecting design flood

Adopted by CWC India


4


5

INDIAN STANDARD GUIDELINE

FOR DESIGN OF FLOOD FOR

DAMS

Classification of dams

According to size (table 7.9(a))

By using hydraulic heads & gross storage

• Hydraulic head define as

Difference between maximum water level on

upstream & normal annual average flood level

downstream

• For safety of dam

The inflow design flood (IDF) is considered

Given by Table 7.9 (b)


6


7

7.12 – DESIGN STORM

To estimate the design flood by using unit

hydrograph

• Computation of design storm

By hydrometeorologist by using meteorological data

Various method use which depend on availability of

reliable data and expertise

2 method

Storm-producing probable maximum

precipitation (PMP) derived from PMF

Standard project storm (SPS) for SPF calculation


8

Below show procedure from India

1. Find the critical rainfall.

This will be the basin lag if the flood peak is of

interest

If flood volume is of interest, find duration of the

longest storm experienced in the basin

2. Find the past major storm

DAD analysis is performed and the enveloping

curve represent max depth-duration relation for

the study is obtained


9

3. From enveloping curve

Find the rainfall depth for convenient time

intervals (eg. 6 hr)

Increment to be arranged in order to find critical

sequence which produce max flood peak when

applied to unit hydrograph

Critical sequence find by

trail & error

Increment of precipitation arranged by

I. Maximum rainfall increment is against

maximum unit hydrograph ordinate

II. Second highest rainfall increment is against

the second largest unit hydrograph ordinate

III. The sequence of rainfall increment

arranged from I & II is reversed ( eg 7.8)


10

4. Combine design storm & abstraction most

conducive to high runoff

Low initial loss & lowest infiltration rate to get to

hyetograph of rainfall excess to operate on unit

hydrograph


11

Eg 7. 10

The ordinates of cumulative rainfall from

the enveloping maximum depth-duration

curve of a basin are given below. Also given

are the ordinates of a 6-h unit hydrograph.

Design the critical sequence of rainfall

excesses by taking the ø index to be 0.15

cm/h


12


13

Given data

Column 1,2 & 4

• Column 2

Row 2 – row 1 = 15 – 0 = 15

Row 3 – row 2 = 24.1 – 15 = 9.1

• Column 5

Max data in col. 3 is pair with col.4

• Column 6

Reversed of column 5

• Column 7

Given data infiltration loss = 0.15 cm/h x 6h

= 0.9 cm/6h

• Column 8

Column 6 – column 7


14

7.13 – RISK, RELIABILITY AND SAFETY FACTORS

Design of hydraulic structure involve uncertainty &

such involve risk of failure

Risk, Rˉ is the probability of occurrence of an

event (x≥xt) at least once over n successive year

Rˉ = 1 – (probability of non-occurrence of the

event x ≥ xt in n years )

P = probability P (x ≥ xt) = 1/T

T = return period

• Reliability, Re = 1 - Rˉ = (1 – 1/T)^n

Rˉ = 1 – ( 1 – P ) ^ n ( equ 7.29)

= 1 – ( 1 - 1/ T) ^ n


15

Safety factor

Due to uncertainty in water resource development project such as during constructional of the building or from economic causes

A safety factor with parameter M is apply to the development project

Parameter, M

Includes items such as flood discharge magnitude, max. river stage, reservoir capacity and free board

Safety margin is the difference between Cam & Chm (Cam – Chm)

Actual values of parameter

M adopted in the

Safety factor = (SF)m = design of projects . = Cam

( for parameter M) Values of parameters M Chm

obtained from

hydrological consideration


16

Eg. 7.11 A bridge has an expected life of 25 years and is

designed for flood magnitude of return period 100

years

a) what is the risk of this hydrological design

b) if a 10% risk is acceptable, what return period

will have to be adopted

• Solution :

a) Given n = 25 years

T = 100 years

Using equation 7.29


17

.

R = 1 – ( 1 - 1/ T) ^ n

= 1 – ( 1 – 1/100) ^ 25

= 0.222

Hence, inbuilt risk design is 22.2 %

.

b) Given R = 10% = 0.10

0.10 = 1 – ( 1 – 1/T)^25

T = 238 years ≈ 240 years

Hence, for 10 % risk, the bridge must be design with

240 years of return period


18

Eg. 7. 12 Analysis of annual flood series of a river yielded a

sample mean of 1000 m^3/s and standard deviation

of 500m^3/s. estimate the design flood of a

structure on this river to provide 90% assurance

that the structure will not fail in the next 50 years.

Use Gumbel method and assume samples size is

large


19

Solution:

Given

xˉ = 1000 m^3/s

σn-1 = 500 m^3/s

Reliability, Re = 90% = 0.9

n = 50 year

We need to find xt

xt = xˉ + Kσn-1

yt = - (ln x ln( T /( T -1) )

K = (yt – 0.588) / 1.2825


20

Re = ( 1 – 1/T)^n

0.90 = (1 – 1/T)^50

T = 0.998 = 475 years

yt = - (ln x ln( T /( T -1) )

= - (ln x ln ( 475 / (475 – 1))

= 6.16226

K = (yt – 0.588) / 1.2825

= 4.355

xt = xˉ + Kσn-1 = 3177 m^3/s


21

Eg 7.13 Annual flood data of the river Narmada at

Garudeshwar covering the period of 1948 to 1979

yield for annual flood discharge a mean of 29,600

m^3/s & standard deviation of 14,860 m^3/s. For a

proposed bridge on this rover this site it is decided

to have an acceptable risk of 10% in its expected

life of 50 years

a) estimate flood discharge by Gumbel Method

b) if the actual flood values in this design is

125,000m^3/s, what are the safety factor and safety

margin relating to the maximum flood discharge


22

Solution

Given

Rˉ= 10% = 0.1

n = 50 year

N = 32 years (1948 – 1979)

Hence R ˉ = 1 – ( 1 - 1/ T) ^ n

0.1 = 1 – ( 1 - 1/ T) ^ 50

T =475 years

From tables 7.3 & 7.4

y ˉn =0.5380

S n = 1.1193


23

yt = - (ln x ln( T /( T -1) )

= - (ln x ln ( 475 / (475 – 1))

= 6.16226

K = (yt – y ˉn ) / Sn

= 5.0248

xt = xˉt + Kσn-1 = 29600 + ( 5.0248 x 14860)

= 104568 ≈ 105000m^3/s

Safety factor, (SF)flood = 125000/105000 = 1.19

Safety margin = 12500 - 105000 = 20000m^3/s