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Sun wind water earth life living legends for design
(AR1U010 Territory (design),AR0112 Civil engineering (calculations))
Prof.dr.ir. Taeke M. de Jong
Drs. M.J. Moens
Prof.dr.ir. C.M. Steenbergen
http://team.bk.tudelft.nl
Publish on your website:
AR1U010how you could take water, networks, traffic and civil works into account in your•earlier,•actual and•future work.
AR0112calculation and observations of streams in any
location and your design, check your
observations
As soon as you are ready with all subjects (Sun, Wind, Water, Earth, Life, Living, Traffic, Legends), send a message mailto:[email protected] referring your
web adress, student number and code AR1U010 or AR0112.
STREAMSWATER
TRAFFICNETWORKS
CIVIL WORKS
Total amount of water on Earth
1000 km3 salt fresh total m3/m2 mmatmosphere 12,9 12,9 0,025 25sea 1 338 000 1 338 000 2 624 2 624 021land, from w hich 12 957 35 004 47 960 94 94 057
snow and ice 24 364 24 364 48 47 782
subterranean 12 870 10 530 23 400 46 45 891
lakes 85,4 91 176,4 0,346 346
soil moisture 16,5 16,5 0,032 32
sw amps 2,1 2,1 0,004 4
life 1,1 1,1 0,002 2
total 1 350 957 35 004 1 385 960 2 718 2 718 079
Yearly gobal evaporation, precipitation and runoff
evaporation precipitation runoff evaporation precipitation runoff
sea 419 382 1157 1055land 69 106 37 467 717 250total 488 488 957 957
1000 km3/a mm/a
Global distribution of precipitation
European distribution of precipitation
Precipitation minus evaporation in The Netherlands
European river system
Soil types and average annual runoff
Simulating runoff
Distinguishing orders
1 2 3 4 51
10
100
1 103
Number( )Order
Length( )Order
Order
Theoretical orders of urban traffic infrastructure
km km km/km2
nominal mesh km/metropolis inclusive density exclusive mv/hdistrict roads 1 72000 2 1,33 1000city highways 3 24000 0,67 0,47 3000local highways 10 7200 0,2 0,13 10000regional highways 30 2400 0,07 0,05 30000national highways 100 720 0,02 0,02 100000
and so on nearly 3.00 2.00 total
Orders of dry and wet connections in a lattice
Opening up feather and tree like
Feather like Tree likedensity 29 sections 29 sectionsbifurcation ratio 18 2number of ‘orders’ 2 5
Wat’s efficient?
Feather like Tree likedensity 96 sections 98 sectionsbifurcation ratio 18 2number of ‘orders’ 2 6 or 9
Forms of deposit
Meandering and twining
Twining at R=100km,
meandering at R=30km
Deltas
Q by measurement
The velocity v of water can be measured on different vertical lines h with mutual distance b in a cross section of a river. You can multiply v x b x h and summon the outcomes in cross section A to get Q = (v*b*h).
Data from profile
0 5 104
2
0
hi
Bi
h .
0
1
3
3
1
0
mb .
0
2
2
2
2
2
m
Bi
= 0
i
x
bx
v .
0
1
2
3
2
1
m
sec
height h witdh b velocity v
Drainage subdivision
i ..0 5
ai
.bi
hi
..1
2b
ih
ih
i 1
A
i
ai
=A 16 m2
Qi
.vi
ai
Q
i
.vi
ai
=Q 36 m3 sec 1
vi
..0 m sec 1
..1 m sec 1
..2 m sec 1
..3 m sec 1
..2 m sec 1
..1 m sec 1
ai
.0 m2
.1 m2
.4 m2
.6 m2
.4 m2
.1 m2
Qi
..0 m3 sec 1
..1 m3 sec 1
..8 m3 sec 1
..18 m3 sec 1
..8 m3 sec 1
..1 m3 sec 1
Q on different water heights
0 100 200 300 4000
1
2
3
4
m3/sec
m
Mi
H( ),,a B Q
36
,.10 i Q
Q(height)
Q = 0,0003H8,7398
R2 = 0,9782
0
10
20
30
40
0 1 2 3 4
height in m
dra
ina
ge
in m
3/s
ec
Q = 0,0003H8,7398
R2 = 0,97820,1
1
10
100
1 10
height in m
dra
ina
ge
in m
3/s
ec
Normal representation Logarithmic representation
Hydrolic radius
0 5 100
2
4
f( )x
H
l1
r1
x
P
= l1
r1
j
Xj 1
Xj
2 Yj 1
Yj
2
A( )H .H r1
l1
dl1
r1
xf( )x
R( )HA( )H
P
Cross length (Natte omtrek) by Pythagoras:
Surface wet cross section:
AP
H
Hydrolic radius:
Method Chézy
The average velocity of water v = Q/A in m/sec is dependent on this hydrolic radius R, the roughness C it meets, and the slope of the river as drop of waterline s, in short v(C,R,s).
According to Chézy v(C,R,s)=CRs m/sec, and Q = Av = ACRs m3/sec.
Calculating C is the problem.
Method Strickler-Manning
Instead of v=CRs, Strickler-Manning used
v ..R
2
3 s
1
2
n
m
sec
Characteristics of bottom and slopes
from until
Concrete 0.010
0.013Gravel bed 0.02
00.03
0Natural streams:
Well maintained, straight 0.025
0.030Well maintained, winding 0.03
50.04
0Winding with vegetation 0.040
0.050Stones and vegetation 0.05
00.06
0River forelands:
Meadow
Agriculture
Shrubs
Tight shrubs
Tight forest
n
0.035
0.040
0.050
0.070
0.100
Method StevensInstead of v=CRs Stevens used v=cR considering Chézy’s Cs as a constant c to be calculated from local measurements.So, Q = Av = cAR m3/sec When we measure H and Q several times (H1, H2 …Hk and Q1, Q2 …
Qk), we can show different values of A(H)R(H) resulting from earlier
calculation as a straight line in the graph below.
0 1 2 3 4 5 6 7 8 9 10 11 1213 141516 1718 192021 222324 2526 272829 303132 3334350
10
20
30
40
.A( )H1 R( )H1.A( )Hk
R( )Hk
.A( )Hk
R( )Hk
H1 Q1
,H
k
m
Qk
m3
sec
A( )H .H r1
l1
dl1
r1
xf( )x
R( )HA( )H
P
Surface wet cross section:
Hydrolic radius:
Reading Q from H by Stevens
When we read today on our inspection walk a new water level H1 on the sounding rod of the profile concerned we can interpolate H1 between earlier measurements of H and read horizontally an estimated Q1 between the earlier corresponding values of Q to read Q from graph.
0 1 2 3 4 5 6 7 8 9 10 111213 141516 1718 192021 222324 2526 272829 303132 3334350
10
20
30
40
.A( )H1 R( )H1.A( )Hk
R( )Hk
.A( )Hk
R( )Hk
H1 Q1
,H
k
m
Qk
m3
sec
Hydrographs
River with continuous base discharge
River with periodical base discharge
Using drainage data
Duration line Dataset with peak discharges
Peak discharges
The peak discharge QT exceeded once in average T years (‘return
period’) is called ‘T-years discharge’.
The probability P of extreme values is called ‘extreme value distribution’.
The complementary probability P = 1 ‑ P’ discharge Q will exceed an observation (Q>X) is 1/T and the reverse P’ = 1 – P = 1 – 1/T. So, the ‘reduced variable’ y = -ln(-ln(1 – 1/T)).
P1
TP' 1 P 1
1
Te e
y
P( )y 1 e ey
T( )y1
( )exp( )exp( )y 1
Now we put in a graph:
and
Constructing Gumble I paper
T(y) and P(y) Logaritmically Gumbel I paper
Gumble I paper
Level and discharge regulators
Regulation principles
Retention in Rhine basin
Reservoirs
Storage
1
0
)(h
dhhA
When surface A varies with height h storage S is not proportional to height. By measuring surfaces on different heights A(h) you get an area-elevation curve. The storage on any height S(h) (capacity curve) is the sum of these layers or integral
Capacity calculation
You can simulate the working of a reservoir (‘operation study’) showing the cumulative sum of input minus output (inclusive evaporation and leakage). The graph is divided in intervals running from a peak to the next higher peak to start with the first peak. For every interval the difference between the first peak and its lowest level determines the required storage capacity of that interval. The highest value obtained this way is the required reservoir capacity.
Cumulative Rippl diagram
Avoiding floodings by reservoirsTo estimate the risk a reservoir can not store runoff long enough you need to know probability distributions of daily discharge.
Water management and hygiene
Strategies
Lowlands with spots of recognisable water management
Water managemant tasks in lowlands
05 Urban hydrology 06 Sewerage 07 Re-use of water 08 High tide management
09 Water management 10 Biological management 11 Wetlands 12 Water quality management
13 Bottom clearance 14 Law and organisation 15 Groundwater management 16 Natural purification
01 Water structuring 02 Saving water 03 Water supply and purificatien 04 Waste water management
Water management map
Overlay of observation points
Overlay of water supply
Need of drainage and flood control
Flooding of a canal in Delft Deep canal in Utrecht
Wet and dry functions
Area of lowlands with drainage and flood control problems
x1000 km2 1 crop 2 crops 3 crops TotalNorth America 170 210 30 400Centra America 20 190 210South America 60 290 1210 1560Europe 830 50 880Africa 300 1620 1920South Asia 10 460 580 1050North and Central Asia 1650 520 20 2190South-East Africa 530 530Australia 310 120 430
9170
Levels in lowland
Pumping stations in The Netherlands
Drainage by one to three pumping stations
A ‘row of windmills’ (‘molengang’)
One way sluice
The belt (‘boezem’) system of Delfland
Rising outside water levels and dropping ground levels
Polders
Distance between trenches
The necessary distance L between smallest ditches or drain pipes is determined by precipation q [m/24h], the maximally accepted height h [m] of ground water above drainage basis between drains and by soil characteristics. Soil is characterised by its permeability k [m/24h]. A simple formula is L=2(2Kh/q).
Soil permeability
Type of soilgravel
coarse sand with gravel 100 1000corse sand, frictured clay
in new polders10 100
middle fine sand 1 10very fine sand 0.2 1
sandy claypeat, heavy clayun-ripened clay 0.00001
Permeability k in m/24h>1000
0.10.01
Hooghoudt formula
A simple formula is L=2(2Kh/q). If we accept h=0.4m and several times per year precipitation is 0.008m/24h, supposing k=25m/24h the distance L between ditches is 100m. However, the permeability differs per soil layer. To calculate such differences more precise we need the Hooghoudt formula desribed by Ankum (2003).
Plot division in polders
Closed sluices
Uitwateringssluis Inlaatsluis
Ontlastsluis Keersluis
Open sluicesUitwateringssluis
IrrigatiesluisOntlastsluis
Inlaatsluis
Sluices
Ontlastsluis
Spuisluis Inundatiesluis
Damsluis
Weirs
Schotbalkstuw Schotbalkstuw met wegklapbare aanslagstijl
Naaldstuw Automatische klepstuw
Dakstuw Dubbele Stoneyschuif
Wielschuif rechtstreeks ondersteund door jukken
Wielschuif via losse stijlen ondersteund door jukken
Locks
Schutsluis Dubbelkerende schutsluis
Locks
Schutsluis Dubbelkerende schutsluis
Locks
Schutsluis Dubbelkerende schutsluis
Tweelingsluis Schachtsluis Driewegsluis
Sluis met verbrede kolk Bajonetsluis
LocksGekoppelde sluis
Entrance and exit constructions
Coastal protection
Delta project constructions