Rodrigo Proença de Oliveira
Hydrology, environment and water resources 2015 / 2016
Flood risk analysis
Flood cost
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Spatial distribution
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 3 17-11-2015
Flood losses
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 4
99 vitimas por ano
17-11-2015
Lisbon area, November 1967 110 mm / 5 hours: 700 deaths and 1000 loss their homes
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 5 17-11-2015
Lisbon area, November 1983
10 deaths e 200 families lost their homes
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 6 17-11-2015
Madeira, February 2010
42 deaths
7 IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 17-11-2015
Madeira, February 2010
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Madeira, February 2010
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Madeira, February 2010
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EARAM study (IST) results
17-11-2015
Lower Tagus flood, April 2013
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 11 17-11-2015
New Orleans, 2005 (Katrina hurricains)
1000 deaths, one million lost their homes
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 12 17-11-2015
Flood control
(flood risk reduction)
• Clearance of flood prone areas
/ Deslocação de pessoas e bens
de áreas com uma probabilidade
elevada de ocorrência de
inundações;
• River works to increase the
cross-section area / Melhoramento das secções dos
cursos de água para aumento da
sua capacidade de vazão;
• Building of dikes to increase
the cross-section area / Construções de diques para
aumento da secção de vazão;
• Use of reservoirs for flood
atenuation / Atribuição de um
volume de encaixe de cheias para
atenuação do hidrograma de
cheia;
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 13
Caudal afluente, Qa Caudal efluente, Qe
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Dikes in lower Tejo
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 14 17-11-2015
Mississipi 1927 130,000 homes lost; 700,000 people were displaced. 246
deaths; million 350 dollars of Property damage (5 billion dollars today)
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 15
Flood control act of 1928 and the Room for the river
project and the 2011 flood event
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 16
2011: • No deaths
• 21.000 homes and business impacted
• 2,8 billions USD losses
Scenario without the MRTSystem • 1,5 billions of homes and business impacted
• 238 billions USD losses
Holanda, Rhine
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 17 17-11-2015
Projecto Room for the river (Holanda)
• Dar espaço ao rio Reno / Room for
the river;
• Objectivos / Objectives:
– Reduzir o risco de cheias;
– Melhorar a qualidade da água.
• Metas / Targets:
– Até 2015: Assegurar uma
capacidade de escoamento de
16’000 m3/s;
– Até 2020: Reduzir os níveis
máximos de cheia em 70 cm.
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 18
Projecto Room for the river (Holanda)
19 IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015
• Medidas / Actions:
– Remoção de obstáculos / removal of flow obstacules;
– Rebaixamento e alargamento do leito / channel dredging ;
– Afastamento de diques / Dikes removals ;
– Remoção de polders / Polders removal;
– Melhoria do dique (casos pontuais) / Dikes improvement;
Custo:
2.1 biliões de euros.
17-11-2015
Exemplo: Room for the river (Holanda)
Plano base: 2015
20 IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015
Factors influencing flood risk
• Watershed area / Área
• Watershed shape / Forma
• Watershed orography / Relevo
• River network / Rede hidrográfica
• Soils / Solos
• Soil cover / Coberto vegetal
• Soil use / Uso da superficie
• Precipitation temporal distribution / Distribuição temporal da precipitação
• Temperature / Temperatura
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 22
Time for flow accumulation (concentration
time ands isochrones)
Tempo para acumulação do escoamento (tempo
de concentração / isócronas)
Infiltration
Infiltração
Factor Influences
Less important
Menos importante
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Flood hydrograph
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 23
Caudal, Q
Tempo, t
Escoamento total
Escoamento directo
Escoamento de base
Precipitação, P
Tempo de ascenção
Precipitação
útil
Intercepção, retenção
e infiltração
Tempo de decrescimento
Total flow
Direct flow
Base flow
Net precipitation
Interception, retention and infiltration
Rising time Decreasing time
Discharge
Q
Time, t
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Main parameters or variables that are usually
estimated
It depends on the purpose of the analysis: – Preliminary risk assessment / Avaliação preliminar do risco
de cheia
– Spillways design / Dimensionamento de descarregadores
– Bridge design / Dimensionamento de pontes
– Dikes design / Dimensionamento de diques
– Storm water drains design / Dimensionamento de colectores
– Design of flood attenuation storage volume / Dimensionamento de volumes de encaixe de cheia
– Flood damage evaluation studies / Avaliação da duração da
cheia e dos prejuizos causados
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 24
Time
Discharge
Flood volume Volume de cheia
Peak discharge Caudal de ponta
Flood peak / Caudal máximo
(ponta de cheia)
Flood peak / Caudal máximo
(ponta de cheia)
Flood volume / Volume de cheia
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Water balance simplification
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 25
Escoamento
superficial
(directo)
Escoamento
de base
Evapotranspiração
Intercepção
Retenção
Evaporação
Infiltração
Precipitação
Recarga
Escoamento
intermédio
Evaporação
Precipitação
P = H + E + Sp + Ss + Su + Ex – R
For a short time interval / Para um
intervalo de tempo curto:
E = 0 (evapotranspiration)
Su = 0 (aquíier contribution)
Ex – R = 0 (export-import)
Logo, H = P - Sp - Ss
Se Su <> 0
Logo, H = P - Sp - Ss + Su
Considered in a simplified way
Considerado de forma simplificada
Considered in a simplified way
Considerado de forma simplificada
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Opções para pequenas bacias hidrográficas
Assumptions for small watersheds analysis
Possible assumptions on hydrological fluxes during
a flood / Possíveis simplificações na análise dos fluxos
hidrológicos durante uma cheia:
• Evaporation and evapotranspiration: Often
ignored/ Evaporação + evapotranspiração: São
frequemente desprezadas na análise de situação de cheia
• Retention, interception and infiltration
(“losses”): A simple model is used / Retenção,
intercepção, infiltração (“perdas”) / Assume-se um modelo
simples:
• Intermediate flow: Often ignored / Escoamento
intermédio Frequentemente desprezado;
• Base flow: Often ignored or estimated using a
simple model independent from infiltration
estimate. / Escoamento de base: Por vezes ignorado ou
calculado com base num modelo simples e independente da
infiltração.
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 26 17-11-2015
Precipitação, P
Precipitação, P
Loss models: Constant infiltration / Taxa
constante;
Initial loss + Constant infiltration
/ Volume inicial + taxa constante;
Isochrones
Consider a rainfall event with a
duration equal to the double of the
time of concentration
Chuvada com uma duração da precipitação
útil igual ao dobro do tempo de
concentração da bacia hidrográfica
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 28
Time Tempo
Areas contributing to flow Áreas que contribuem para o
escoamento
0 0
1/3 x tc A
2/3 x tc A + B
tc A + B + C
4/3 x tc A + B + C
5/3 x tc A + B + C
2 x tc A + B + C
7/3 x tc B + C
8/3 x tc C
3 x tc 0
A
BC
Isócrona 1/3 x tc
Isócrona 2/3 x tc
Ponto mais
afastado
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Time of concentration
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 29
Assuming the precipitation falls uniformly over the
whole watershed (valid for small watersheds) Assumindo que a precipitação se distribui de forma
uniforme em toda a bacia hidrográfica (válido em
pequenas bacias).
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Discharge, Q
Time, t
Total flow
Direct flow
Base flow
Precipitation, P
Rising time
Net
precipitation
Interception,
infiltration
Falling time
Time of concentration
Discharge, Q
Tempo, t
Direct flow
Net precipitation, Pu
Time of concentration
Time of concentration
• Travel time from the most distant point to the watershed
outflow section / Tempo de percurso de uma gota de água entre o ponto
cinematicamente mais afastado da secção da bacia e a secção definidora da bacia;
• We assume it is an intrinsic characteristic of the
watershed; Propriedade intrínseca da bacia hidrográfica / Assume-se que é independente da precipitação, dependendo apenas das condições de
escoamento desde o ponto mais afastado até à secção final;
• Depends on the watershed characteristics / Na prática assume-se
que depende das características da bacia hidrográfica:
– Watershed area / Área da bacia (A);
– Length of the main watercourse / Comprimento do curso de água principal (L);
– Watershed average heigth / Altura média da bacia (hm);
– Slope of the main watercourse /Declive médio do curso de água principal (dm);
– Altitude difference along the main watercourse /Diferença de cotas do
talvegue do curso de água principal (H);
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 30 17-11-2015
Time of concentration estimation methods
• Analysis of observed hyetographs and hydrographs / Análise
dos hietogramas e hidrogramas observados;
• Cinematic computation based on distance travelled and
water velocity estimation / Cálculo cinemático: Cálculo do tempo de
escoamento do ponto cinematicamente mais afastado até à secção de referência,
aplicando por exemplo a fórmula de Manning-Strickler tendo em conta as condições de
escoamento ao longo das encostas e nos cursos de água de ordem crescente;
• Empirical formulas (for small watersheds) / Fórmulas empíricas
(para pequenas bacias hidrográficas):
– Giandotti:
– Temez:
– Kirpich (segundo Chow):
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 31
m
ch
LAt
8,0
5,14
76,0
25,03,0
m
cd
Lt
385,0
155,1
95,0H
Ltc
• tc (h) – Time of concentration
• A (km2) – Area of watershed
• L (km) – Length of main river
• hm (m) – Avg heigth of watersh
• dm (-) – Avg slope of main river
• H (m) – Altitude difference along main river
17-11-2015
Exercise
A net precipitation with intensity IA, IB, IC and ID occurred in each of the areas of
the represented watershed. The following table presents such distribution with I0
representing a reference intensity and tc the time of concentration of the
watershed. Determine the direct flow hydrograph resulting from such
precipitation in I0xA0 units. Em cada uma das zonas da bacia hidrográfica representada ocorreu precipitação útil de acordo com as
intensidades IA, IB, IC e ID que se apresentam no quadro seguinte, onde I0 representa uma intensidade de
referência e tc, o tempo de concentração da bacia hidrográfica. Determine o hidrograma do caudal do
escoamento directo resultante de tal precipitação, em unidades de I0 A0.
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 32
t/(tc/4) (-) 0 - 1 1 - 2 2 - 3 3 - 4 4 - 5
IA/I0 (-) 0.0 0.0 0.0 0.5 0.0
IB/I0 (-) 0.0 0.0 0.5 0.5 0.0
IC/I0 (-) 0.0 0.5 0.5 0.0 0.0
ID/I0 (-) 1.0 0.5 0.0 0.0 0.0
17-11-2015
Exercise
In particular river basin, with 290 km2, in prolonged periods without rainfall, the
flow rate in the reference section follows the following law / Em determinada bacia
hidrográfica, com 290 km2, em períodos prolongados sem precipitação, o caudal na secção
de referência segue com bastante aproximação uma lei de esgotamento do tipo.
Knowing that at 9 am of March 4 and March 13 the flow rate at the outlet of this
basin was respectively 10.9 and 7.3 m3/s and assuming that no precipitation
occurred between these two dates, estimate the volume of water that has passed
through the outlet from the latter date until full depletion of the basin. Express
this volume in water height evenly distributed over the basin / Sabendo que nessa
bacia hidrográfica, às 9 horas de 4 de Março e de 13 de Março o caudal era respectivamente
de 10,9 e 7,3 m3/s e admitindo que não ocorreu precipitação, estime o volume de água que
passou na secção de referência desde a última data até ao total esgotamento da bacia.
Exprima esse volume em altura de água uniformemente distribuída sobre a bacia.
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 33
)tt(k0
0eQQ
17-11-2015
Empirical formulas
• Iskowski (1886)
• Giandotti
• Soil Conservation Service
• Mockus
• Meyer
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 35
APmKQ Not a function of the return period
ct
h
AQ
pt
hk
A277,0Q
cc tt
hk
6,0
A277,0Q
ct2
Q – Peak flow (m3/s)
A – Watershed area (km2 )
h – Event precipitation (mm)
k – Peak factor in the range of 1 and 0,5
tp – Peak time = 0,5 D + 0,6 Tc
D – Duration of precipitation
Assumes a critical precipitation duration of
αACQ
Q – Peak flow (m3/s)
K – Coeficient dependent on soil type, land cover and orography
m – Coeficient dependent on the area of the watershed
P – Average annual precipitation (m)
A – Watershed area (km2)
Q – Peak flow (m3/s)
A – Watershed area (km2)
Tc – Time of concentration
– Coeficient dependent on the area of the watershed
h – Precipitation (mm)
Q – Peak discharge; (m3/s)
C – Coeficient dependent on the watershed characteristics and return period;
a –Coeficient dependent on the watershed characteristics (0.4 < a < 0.8)
A –Watershed area (km2)
17-11-2015
Rational formula
• C –Coeficient dependent on the watershed characteristics and return period / Coeficiente dependente das características da bacia (tipo de solo e uso do solo) e do período de retorno:
– C smaller for permeable soils and LU and larger watersheds / C menor para solos e uso de
solos mais permeáveis; bacias maiores;
– C greater for impervious soils and LU and higher return periods / C maior para solos e uso de
solos menos permeáveis; maiores períodos de retorno;
• i –Precipitation intensity / Intensidade de precipitação;
• A –Watershed area / Área da bacia;
• f - Majoration coeficient to consider non-uniform precipitation events / Factor de
majoração para considerar distribuições não uniformes da precipitação;
• n –Exponent of the rain depth-duration-frequency curve / Expoente da curva de
possibilidade udométrica;
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 36
nf 2
ntaP AiCQ
AiCfQ A
Precipitation rate, i
Peak
discharge
Q = C i A
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Linear reservoirs
Linear reservoirs:
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 37
𝒅𝑺
𝒅𝒕= 𝑰 − 𝑶
𝑶 = 𝒌 ∙ 𝑺
Inflow, I
Outflow, O
Storage, S
Outflow, O
Storage, S
𝒅𝑺
𝒅𝒕= 𝑰 − 𝒌 ∙ 𝑺
𝒊𝒇 𝑰 = 𝟎 𝒕𝒉𝒆𝒏 𝑺 = 𝑺𝒐 ∙ 𝒆−𝒌∙𝒕
S
t
Inflow, I
Outflow, O
Storage, S
Delay or lag
Attenuation
17-11-2015
Factor C
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 38 17-11-2015
Chow et al, 1988
Department of Water Affairs and
Forestry, Africa do Sul
What is the critical precipitation duration ?
Assuming that the precipitation falls uniformly
over the whole watershed (valid for small
watersheds):
• A small increase of the precipitation duration
leads to an increase in the area of the
watershed contributing to the flow at the
outlet;
• When the duration is higher than the time of
concentration there is an instant where the
whole watershed contributes to the flow at
the outlet;
• An increase of the duration of the
precipitation event leads to a reduction in
the average precipitation rate for a given
probability (return period);
• Therefore, precipitation events with
durations close to the time of concentrations
are the ones that lead to higher peak flows.
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 39 17-11-2015
Exercise
Use the rational formula to determine the peak flow associated with a
return period of 100 years from a watershed with 80 km2 and a time of
concentration of 2.5 h. Assume that 15% of the precipitation is lost to
infiltration and there is a peak flow attenuation through the watershed of
20%. Assume the following precipitation-duration curve, with P in mm and
t in min:
Utilize a fórmula racional para estimar o caudal de ponta de cheia associado a um
periodo de retorno de 100 anos, gerado numa bacia hidrográfica com 80 km2 de
área e 2,5 h de tempo de concentração. Assuma que 15% da precipitação é
perdida por infilitração e que há uma atenuação do caudal de ponta de cheia de
20%. Assuma que a linha de possibilidade udométrica para esse período de
retorno, com P em mm e t em min, é
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 40 17-11-2015
45,0t4P
45,0t4P
C = 0.8 x 0.85 = 0,68
D = tc = 2,5 h = 150 min
P = 4 x 150 0.45 = 38 mm
i = P / D = 15 mm/h
Q = C x i x A = 0,68 x 15 x 10-3 / 3600 x 80 x 106 = 230 m3/s
Exercise
Use the Temez formula and the rational formula to determine the peak
flow for T = 100 years of a rural watershed with 80 km2, where the
main river has 8 km and an average slope 0f 0.005. Assume the
following precipitation-duration curve for T=100 years, with P in mm
and t in min, is:
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 41
45,0t4P
17-11-2015
Exercise
Compute the design flow of an infra-structure whose functionality should
not be affected more 10 times in a century. The watershed of the infra-
structure has an area of 40 km2 and its main water course flows along 4 km
with an average steepness of 0.002. Use the Temez equation and the
rational formula with correcting factor, assuming a C coeficient of 0,5. The
watershed lies within a region with the following Precipitation Depth-
Duration-Frequency curve (P in mm and D in min).
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 42
D (h) T=10 T=50 T=100 T=500
a n a n a n a n
0,5 h – 6 h 37.837 0.398 50.233 0.368 55.673 0.360 67.615 0.344
6 h – 48 h 33.867 0.473 41.092 0.501 44.068 0.509 51.207 0.523
17-11-2015
Exercise
During a rainfall event, the flow capacity of a spillway, estimated in 120 m3/s,
was almost exceeded. The watershed of that spillway na área of 60 km2 and its
main waterscourse flows along 10 km, with an average steepness of 0.0008. The
watershed lies within a region with the following IDF curve (i em mm/h e D em
min). Compute the order of magnitude of the return period of the rainfall event,
using the rational formula and the Kirpich formula. Assume a C coeficient of 0,5.
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 43
t (min) T=10 T=50 T=100 T=500
a b a b a b a b
5 min – 30 min 567 -0,665 742 -0,652 815 -0,649 987 -0,643
30 min – 6 h 445 -0,602 668 -0.632 765 -0.640 992 -0.656
6 h – 48 h 293 -0,527 317 -0.499 329 -0.491 361 -0.477
17-11-2015
Exercise
A linear reservoir with a constant k = 15x10-6 s-1, has at a certain moment a
volume of water V0 = 105 m3. From that moment onwards the inflow is
represented by the hydrograph presented in the table is below, Determine the
hydrograph of the reservoir outflow (Q) from the starting time up to 18 h later
and plot the two hydrographs (I and Q).
Um reservatório linear, com uma constante k=15x10-6 s-1, apresenta em determinado
instante um volume de água V0=105 m3. Sabendo que a partir desse instante foi alimentado
de acordo com o hidrograma que se apresenta no quadro que se encontra abaixo,
determine o hidrograma do caudal saído do reservatório (Q) desde o instante inicial até 18
h depois e represente graficamente os dois hidrogramas (I e Q).
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 44
t (h) 0 1 2 3 4 5 6 7
I (m3/s) 0 14 20 30 46 28 12 0
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The concept of unit hydrograph
• Provides the relationship between the
flood hydrograph and the originating
precipitation / Relaciona a resposta da bacia
(hidrograma de cheia) com a precipitação que
lhe deu origem;
• We assume it is an intrinsic characteristic
of the watershed / It is the hydrograph
produced by a unit net precipitation that fall
during a certain duration;
• Assume-se que é uma propriedade
íntriseca da bacia, independente da
precipitação / É o hidrograma resultante de
uma chuvada de um valor unitário de
precipitação útil e com uma determinada
duração;
• Each unit hydrograph is associated with a
given duration / Está associado a uma
determinada duração da chuvada. IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 46 17-11-2015
Caudal, Q
Precipitação, P
1
How to apply the unit hydrograph (Additivity and proportionalities principles)
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 47 17-11-2015
0 1 2 3 4 5 6 7 8 Tempo / dt
q, u / umax
1
1
Precipitação útil (mm)
2
dt dt
Hidrograma unitário
para uma duração dt
0 1 2 3 4 5 6 7 8 Tempo / dt
q, u / umax
1
1
Precipitação útil (mm)
2
dt dt
Hidrograma unitário
para uma duração dt
Unit hydrograph for
a duration dt
Unit hydrograph for
a duration dt
Net precipitation (mm) Net precipitation (mm)
Application of the unit hydrograph
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 48 17-11-2015
0
.
8
447
34436
2433425
142332414
1322313
12212
111
q
Puuq
PuuPuuq
PuuPuuPuuq
PuuPuuPuuPuuq
PuuPuuPuuq
PuuPuuq
Puuq
4321 ,,, PuPuPuPu
0,,,,,0 543210 uuuuuu
0 1 2 3 4 5 6 7 8
Unit hydrograph characteristics
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 49 17-11-2015
Tc
Precipitation of 1 unit
ocurring in t
Unit hydrograph for
a duration of t
t
Corresponds to a precipitation
intensity of 1/t
The volume of the unit hydrograoh is equal
to 1 x A, where A is the watershed area.
When discretized in t
intervals, it has Tc/T values
that are different from 0.
mm
min
min
(m3/s)/mm
How to estimate a unit hydrograph
• The UH can be estimated from / O hidrograma
unitário pode ser estimado:
– Observed records from previous
precipitation events (hyetographs and
hydrographs) / A partir de registos observados
de precipitação (hietograma) e de caudal
(hidrograma) de um evento de cheia;
– From watershed characteristics using
synthetic unit hydrographs proposed by
different authors / A partir das características
da bacia – Hidrograma unitário sintético.
• Different authors have proposed synthetic unit
hydrographs / Vários autores avançaram com várias
propostas de hidrogramas unitários sintéticos:
– Giandotti
– Clark;
– Snyder;
– Soil Conservation Service;
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 51 17-11-2015
0 1 2 3 4 5 6 7 8 Tempo / dt
u / umax
1
1
Precipitação útil (mm)
(D + tc) / dt
ta / dt
D = dt
.
Giandotti unit hydrograph
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 52
Hidrograma de Giandotti
t
Q
ta tbtc tt
Qmax
Qmed
t
medt
APψQ
medmax QρQ
ct tγt
ca tρ
1ρt
cb tρ
ρ1)(γt
17-11-2015
A
(km2)
(-)
(-)
(-)
(-)
]0, 300] 10 4.0 0.50 1.25
]300, 500] 8 4.0 0.50 1.00
]500, 1000] 8 4.5 0.40 0.71
]1000, 8000] 6 5.0 0.30 0.36
]8000, 20000] 6 5.5 0.25 0.27
]20000, 70000] 6 6.0 0.20 0.20
Quintela aconselha que se adopte para A < 500 km2
= 6.5
= 4.0
= 0.50
O que corresponde a C = = 0.81
Snyder unit hydrograph
• Parâmetros:
– Lag (h), Tp
– Peak Coeficient, Cp
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 53
Mais informação em HEC-HMS Technical
Reference Manual, Cap. 6
Lag, Tp
Up
Caudal
Tempo
p
pp
T
C
A
U72,2
ccp TTT 75,05,0
17-11-2015
Clark unit hydrograph
• Parameters:
– Time of concentrarion (h), Tc
– Storage coeficient (h), R
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 54
More information in HEC-HMS Technical
Reference Manual, Cap. 6
Diagrama
tempo-área
Reservatório
fictício
tc
Área
Tempo de escoamento
para a secção de refência
17-11-2015
SCS unit hydrograph
• Parameters: Lag (h), Tlag
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 55
See more in HEC-HMS Technical
Reference Manual, Cap. 6
Lag, Tlag
Up
Caudal
Tempo
D
p
pT
AU 08.2
clag TT 6,0
lagp TD
T 2
17-11-2015
SCS triangular unit hydrograph
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 56
0
1
2
Pre
cip
ita
çã
o ú
til (m
m)
0
0.2
0.4
0.6
0.8
1
1.2
0 1 2 3
Tempo / ta
u /
um
ax
2
Dt0.6t ca
ab t3
5t
?Dttt bac a
maxt
A0.208u
ta tb 17-11-2015
Exercise
A continuous and constant net precipitation of 60 mm/h over a given watershed
generates the following hydrograph of direct flow. Please estimate:
– Time of concentration of the watershed.
– Area of the watershed.
– Unit hydrograph of the water for 0.25 h. Em determinada bacia hidrográfica, em resultado de uma precipitação útil com grande duração e
intensidade constante de 60 mm/h, obteve-se o hidrograma do escoamento directo indicado no quadro.
– Determine a área da bacia hidrográfica.
– Determine o hidrograma unitário para a duração de 0.25 h.
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 57
t (h) 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 …
Qd (m3/s) 0 60 150 280 320 340 350 350 …
17-11-2015
Exercise
A precipitation event with a duration of 1,5 h and net precipitation amounts of
5 mm, 10 mm and 3 mm in 3 successive 0.5h time periods over a given
watershed, generates the following hydrograph of direct flow. Please estimate:
– Time of concentration of the watershed.
– Unit hydrograph of the water for 0.5 h.
– Area of the watershed. Em determinada bacia hidrográfica, em resultado de precipitação útil que de 30 min em 30 min foi 5
mm, 10 mm e 3 mm, obteve-se o seguinte hidrograma do escoamento directo.
– Determine o tempo de concentração da bacia.
– Determine o hidrograma unitário para a duração de 0,5 h.
– Calcule a área da bacia, em km2.
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 58
t (h) 0,0 0,5 1,0 1,5 2,0 2,5 3,0
Qd (m3/s) 0 15 70 99 44 6 0
17-11-2015
Exercise
A precipitation event with a duration of 0,5 h and net precipitation amounts of
15 mm and 5 mm in 2 successive 0,25h time periods, generates the following
hydrograph of direct flow. Please estimate the unit hydrograph of the water for
0.5 h. Em determinada bacia hidrográfica, em resultado de uma precipitação útil de 15 mm em 15 min e de 5
mm nos 15 min seguintes, obteve-se o seguinte hidrograma do escoamento directo. Determine o
hidrograma unitário para a duração de 30 min.
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 59
t (h) 0.00 0.25 0.50 0.75 1.00 1.25 1.50
Qd (m3/s) 0 150 275 225 125 25 0
17-11-2015
Exercise
The table presents the unit hydrograph of a given watershed for unit hyetograph
with a duration of 0,5h. Please estimate:
– Time of concentration of the watershed.
– Area of the watershed.
– The direct hydrograph generated by an event with a net precipitation of
20 mm, 30 mm e 12 mm, in 3 successive intervals of 30 min.
– Unit hydrograph of the water for 1 h. Apresenta-se no quadro seguinte o hidrograma unitário de determinada bacia hidrográfica para a
precipitação útil com a duração de 0.5 h.
– Determine o tempo de concentração da bacia hidrográfica.
– Determine a área da bacia hidrográfica.
– Calcule o hidrograma do escoamento directo que resultaria na secção de referência dessa bacia
hidrográfica de uma precipitação útil de 20 mm, 30 mm e 12 mm, em intervalos sucessivos de 30
min.
– Calcule o HU para uma duração de 1 hora.
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 60
t (h) 0.0 0.5 1.0 1.5 2.0 2.5
u (m3/s/mm) 0 15 34 17 9 0
17-11-2015
What is the critical precipitation duration ?
Assuming that the precipitation falls uniformly
over the whole watershed (valid for small
watersheds):
• A small increase of the precipitation duration
leads to an increase in the area of the
watershed contributing to the flow at the
outlet;
• When the duration is higher than the time of
concentration there is an instant where the
whole watershed contributes to the flow at
the outlet;
• An increase of the duration of the
precipitation event leads to a reduction in
the average precipitation rate for a given
probability (return period);
• Therefore, precipitation events with
durations close to the time of concentrations
are the ones that lead to higher peak flows.
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 62 17-11-2015
Design hyetograph:
How to arrange the precipitation blocks?
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 63 17-11-2015
Q1=Pu1∙u1Q2=Pu2∙u1+Pu1∙u2
Q3=Pu3∙u1+Pu2∙u2+Pu1∙u3
….
• To maximize a given Qi:
– Multiply the largest u by the largest Pu;
– Multiply the 2nd largest u by the 2nd largest
Pu;
– ….
Precipitação útil (mm)
Tempo (h)
Precipitação útil (mm) Precipitação útil (mm) Precipitação útil (mm)
Tempo (h)Tempo (h)Tempo (h)
Caudal, Q
Tempo (h) Tempo (h) Tempo (h) Tempo (h)
Caudal, Q Caudal, Q Caudal, Q
Hietogramas com o mesmo volume útil Qual é a distribuição temporal que
conduz ao maior caudal de ponta?
Hidrogramas com o mesmo volume de cheia
What is the precipitation distribution patterns
that leads to the highest peak flow?
All hyetographs have the
same total ammount of
precipitation and this is the
same as the flood volume of
all hydrographs. Only the
time distribution changes.
Maximization of the flood peak
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 64 17-11-2015
nk
i
iproj
k
i
n
n
n
DaPPdtD
PPPdtaPdt
PPPdtaPdt
dtaPdt
11
121233123
112212
1
.........
33
22
n u Ordem u P Pu
0 0
1 u1 (5) P7 Pu7
2 u2 (2) P6 Pu6
3 u3 (1) P4 Pu4
4 u4 (3) P3 Pu3
5 u5 (4) P1 Pu1
6 u6 (6) P2 Pu2
7 u7 (7) P5 Pu5
8 0
0 1 2 3 4 5 6 7 8
u / umax
1
Precipitação útil (mm)
0 1 2 3 4 5 6 7 8
0 1 2 3 4 5 6 7 8
u / umax
1
Precipitação útil (mm)
0 1 2 3 4 5 6 7 8 The graphs are simetric around its
maximum value / Gráficos
simétricos em torno do seu máximo
Exercise
The unit hydrograph of a given watershed is indicated in the table. Assuming a
rainfall depth-duration-frequency curve for T=100 years equal to , with P in
mm and t in min, please compute:
• The hyetograph that maximizes the peak flow;
• The peak flow of resulting hyetograh;
• The peak flow that would result from a precipitation event with a uniform
precipitation rate. O hidrograma unitário de determinada bacia hidrográfica encontra-se representado no seguinte quadro.
Desprezando as perdas da precipitação e sabendo que a linha de possibilidade udométrica para o período
de retorno de 100 a na região é com P em mm e t em min, determine:
– a distribuição temporal da precipitação que maximiza o caudal de ponta de cheia para esse período de
retorno,
– o referido caudal máximo de ponta de cheia,
– o caudal de ponta de cheia que resultaria de uma precipitação com distribuição temporal uniforme.
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 65
5.05 tP
t (h) 0.0 0.5 1.0 1.5 2.0 2.5 3.0
u (m3/s/mm) 0 10 30 25 12 6 0
17-11-2015
5.05 tP
Unit hydrograph transformation
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 67 17-11-2015
One unit of net precipitation
during dt produces the UH for dt. Precipitação unitária com uma duração
de dt e hidrograma unitário para uma
duração de dt.
One can compute the flood
hydrograph produces by a rainfall
event discretized in dt. O hidrograma resultante de um
hietograma com um intervalo de
cálculo dt pode ser calculado a partir
de umhidrograma unitário para uma
duração de dt.
What should we do when the
precipitation is discretized in a
time interval distinct from dt? O que fazer quando o intervalode de
cálculo é diferente de dt? ?
0 1 2 3 4 5
u
0 1 2 3 4 5
Q
6 7 8
Estimation of unit hydrograph for a dt2 distinct from dt1
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 68 17-11-2015
Curva S
Curva S
UHs for dt1 Hidrogramas unitários para uma
precipitação unitária de duração
dt1
UHs for dt2 Hidrogramas unitários para uma
precipitação unitária de duração
dt2
Hydrograph from a continuous
long with an intensity of 1/dt1 Hidrograma resultante de uma
precipitação constante com
intensidade 1/dt1
Hydrograph from a continuous
long with an intensity of 1/dt2 Hidrograma resultante de uma
precipitação constante com
intensidade 1/dt2
(1) (2)
(3)
HU for dt1
Sum of
various HU
S curve for a precipitation
intensity of 1/dt1
S curve movement (translation)
by dt2 and subtraction of the
two S curves
HU for dt2
Multiplication by dt1/dt2
S curve for a precipitation
intensity of 1/dt2
S-curve shift and HU computation
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 69 17-11-2015
Ordinates of the UH for dt2
dt2
S’ curve for i=1/dt2
S’ curve for i=1/dt2 translated by dt2
Tempo Curva S Curva S' Tempo Curva S Curva S HU dt2
i=1/dt1 i=1/dt2 i=1/dt2 desfas. dt2
0.000 0 0.00 0.00 0.0 0.00 0
0.143 1 0 0.58 0.42 0.2 0.75 0.00 0.75
0.286 1 1 0 1.75 1.25 0.4 2.25 0.75 1.50
0.429 2 1 1 0 3.50 2.50 0.6 3.65 2.25 1.40
0.571 1 2 1 1 0 4.90 3.50 0.8 4.55 3.65 0.90
0.714 1 1 2 1 1 0 5.95 4.25 1.0 5.00 4.55 0.45
0.857 1 1 1 2 1 1 0 6.65 4.75 1.2 5.00 5.00 0.00
1.000 0 1 1 1 2 1 1 0 7.00 5.00 1.4 5.00 5.00 0.00
1.143 0 0 1 1 1 2 1 1 0 7.00 5.00 1.6 5.00 5.00 0.00
1.286 0 0 1 1 1 2 1 1 0 7.00 5.00 1.8 5.00 5.00 0.00
1.429 0 0 1 1 1 2 1 1 0 7.00 5.00 2.0 5.00 5.00 0.00
1.571 0 0 1 1 1 2 1 1 0 7.00 5.00 2.2 5.00 5.00 0.00
1.714 0 0 1 1 1 2 1 1 0 7.00 5.00 2.4 5.00 5.00 0.00
1.857 0 0 1 1 1 2 1 1 0 7.00 5.00 2.6 5.00 5.00 0.00
2.000 0 0 1 1 1 2 1 1 0 7.00 5.00 2.8 5.00 5.00 0.00
2.143 0 0 1 1 1 2 1 1 0 7.00 5.00 3.0 5.00 5.00 0.00
2.286 0 0 1 1 1 2 1 1 0 7.00 5.00 3.2 5.00 5.00 0.00
2.429 0 0 1 1 1 2 1 1 0 7.00 5.00 3.4 5.00 5.00 0.00
2.571 0 0 1 1 1 2 1 1 0 7.00 5.00 3.6 5.00 5.00 0.00
2.714 0 0 1 1 1 2 1 1 0 7.00 5.00
2.857 0 0 1 1 1 2 1 1 0 7.00 5.00
3.000 0 0 1 1 1 2 1 1 0 7.00 5.00
3.143 0 0 1 1 1 2 1 1 0 7.00 5.00
3.286 0 0 1 1 1 2 1 1 0 7.00 5.00
3.429 0 0 1 1 1 2 1 1 0 7.00 5.00
3.571 0 0 1 1 1 2 1 1 7.00 5.00
Hidrogramas unitários para dt1 desfasado de dt1
Estimation of unit hydrograph for a dt2 distinct from dt1
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 70 17-11-2015
Change of
discretization interval
Multiplication by
dt1/dt2
Translation by
dt2
HU for dt1
Sum
Exercise
Consider the S curve of a given watershed represented in the table. Determine:
– Time of concentration of the watershed
– Area of the watershed
– Unit hydrograph for a duration of 0,5h
O hidrograma em S de determinada bacia hidrográfica encontra-se representado no
seguinte quadro. Com base nesse hidrograma determine:
– o tempo de concentração da bacia hidrográfica;
– a área da bacia hidrográfica;
– o respectivo hidrograma unitário para a duração de 0,5 h.
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 71
t (h) 0,0 0,5 1,0 1,5 2,0 2,5
S (m3/s/mm) 0 2 10 20 36 36
17-11-2015
Exercise
The following direct flow hydrograph was generated from a long,
continuous, uniform precipitation event with a net rate of 600 mm/h.
Determine the unit hydrograph for 0.3 h.
Em determinada bacia hidrográfica, em resultado de uma precipitação útil com grande
duração e intensidade constante de 60 mm/h, obteve-se o hidrograma do escoamento
directo indicado no quadro. Determine o hidrograma unitário para a duração de 0.3 h.
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 72
t (h) 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 …
Qd (m3/s) 0 60 150 280 320 340 350 350 …
17-11-2015
Exercise
The following table shows the unit hydrograph of a given watershed for a
0,5h duration. Determine the unit hydrograph for 20 min.
Apresenta-se no quadro seguinte o hidrograma unitário de determinada bacia hidrográfica
para a precipitação útil com a duração de 0.5 h. Calcule o HU para uma duração de 20 min
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 73
t (h) 0.0 0.5 1.0 1.5 2.0 2.5
u (m3/s/mm) 0 15 34 17 9 0
17-11-2015
Reservoir storage zones
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015
Flood storage volume / pool
Volume de encaixe de cheias
Net/useful storage
Volume útil
Dead storage / Volume morto
NPA
NMC
Nme
Discharge Descarga
Turbined flow Volume turbinado
H
Evaporation Evaporação
Inflow Afluências
Uptake/abstraction Captação
Energy Energia
75
NMC –Maximum design storage level / Nível de maxima cheia
NPA – Full storage level or top of conservation pool / Nível de pleno armazenamento
Nme – Minimum operation level / Nível minimo de exploração
17-11-2015
Sizing of the flood control storage zone
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 76
17-11-2015
Encaixe de cheias
Volume útil
Volume morto
NPA
NMC
Nme
Descarga
Tempo Tempo
Qafluente
Qefluente
Volume “encaixado”
Redução do pico da cheia
Inflow
Outflow
Peak reduction
Controled flood volume
Net storage pool
Dead storage pool
Flood storage pool
In a reservoir the outflow is a
function of the store volume and of
the outflow control devices.
For example:
Inflow and outflow discharge
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 77
Inflow from the watershed, obtained from the
flood analysis.
23
_ 2 icali HgbcQe
17-11-2015
i Tempo(h) Qai Vi Hi Qei atrib Qei calc Abs(Qei_atrib-Qei_calc)
0 0 0 0 0 0 0 0
1 dt Qai 1 x x x x x
2 2dt Qai 2 x x x x x
3 3dt Qai 3 x x x x x
4 4dt Qai 4 x x x x x
5 5dt Qai 5 x x x x x
… … … … … … … …
Outflow discharge computation
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 78 17-11-2015
alb
ii
A
VH
dtQeQeQaQa
VV iiiiii
22
111
23
_ 2 icali HgbcQe
Valores
inicialmente
arbitrários
Goal seek: Abs(..)=0
bHgbcQaH 23
max 23
Outflow discharge computation
• For each time step:
• We need to automate this computation !
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 79 17-11-2015
alb
ii
A
VH 1
1
dtQeQeQaQa
VV iiiiii
22
111
23
12ˆ1
iHgbceQi
1iQearbitrar
??01,0ˆ11
ieQQei
1
ˆ1
ieQQei
Stop S
N
Goal seek global via VB
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 80 17-11-2015
Col Qei_atrib Col Abs(…)
Goal seek aplica-se de t=1 a ….
Existing mathematical models
HEC – Hydrological Engineering Center (USACE).
• Hydrologic model:
– Input: Hietogram;
– Output: Hidrogram;
– e.g.: HEC-HMS (ex HEC1);
• Hydraulic model:
– Input: Hidrogram;
– Output: h, U, flooded areas;
– e.g.: HEC-RAS (ex HEC2);
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 17-11-2015
Modelação
hidrológica
Modelação
hidráulica
Análise de registos
históricos
82
Hydrologic model
IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 83 17-11-2015
Hydraulic model
17-11-2015 IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015 84
Hydraulic model - results
17-11-2015 IST: Hydrology, environment and water resources 2015/16 © Rodrigo Proença de Oliveira, 2015
1748.12*1699.07* 1650.01*
1600.96*
1404.74*
1355.69*
1306.63*
1257.58*
1208.53
1179.97
1134.45
1086.27*
1038.10*
989.934*
941.763*
893.591*
845.419*
797.247*
749.076*
700.904*
652.732*
604.560*
556.389*
508.217*
460.045*
411.873*
363.702*
Novas seccoes_ApresntacaoNov2003 Plan: Plan 06 13/11/2003 4:52:18 PMGeom: novasseccoes_testecoordinates Flow: steady1
Legend
WS PF 1
WS PF 2
WS PF 3
Ground
Bank Sta
Ground
1846.23*
1797.17*
1748.12*
1699.07*1650.01*
1600.96*
1551.90*
1502.85*
1453.8*
1404.74*
1355.69*
1306.63*
1257.58*
1208.53
1179.97
1134.45
1086.27*
Novas seccoes_ApresntacaoNov2003 Plan: Plan 06 13/11/2003 4:52:18 PMGeom: novasseccoes_testecoordinates Flow: steady1
Legend
WS PF 1
WS PF 2
WS PF 3
Ground
Bank Sta
Ground
4115.34
4078.38
4024.89
4009.73
4000.74
Novas seccoes Plan: Plan 06 10/10/2003 3:17:13 PMGeom: novasseccoes_testecoordinates Flow: steady1
Legend
WS Max WS
Ground
Bank Sta
85