SlideShare a Scribd company logo
Exercise lecture 7 culverts
Downstream
Upstream
Cross-section
Dropwaterlevel
Length culvert
Length culvert is 50 m, cross-section 2 x 2 m, λ=0,022 and μ =0,6
Question 1
Calculate the discharge if the drop in water level is 1 m and the velocity downstream
and upstream is 0 m/s. Make a sketch of the H and h line, with numbers
Because the velocity upstream and downstream is 0 m/s, velocity head is 0, total head
is equal to pressure line, difference in water level is the same as difference in energy
level. Δy = ΔH. ΔH = 1 m.
2g
u
ξΔΗ
2
c
tottot ⋅=
m
O
A
R 50,0
2222
22
=
+++
⋅
==
44,01
1
2
=





−=
µ
ξi
55,0
5,04
50
022,0 =
⋅
⋅=⋅=
D
l
w λξ
1=uξ
994,1155,044,0)ξξξ( uwi =++=++=totξ
smu /17,3=
s
mQ
3
67,12=
Head losses at culvert:
m22,0
20
17,3
44,0
2g
u
ξΔΗ
22
culvert
i =⋅=⋅=
m28,0
20
17,3
55,0
2g
u
ξΔΗ
22
culvert
ffriction =⋅=⋅=
m50,0
20
17,3
1
2g
u
ξΔΗ
22
culvert
oo =⋅=⋅=
When I add al losses it should be 1 m, which is the total head loss ΔH
Velocity head culvert
m50,0
20
17,3
2
22
culvert
==
g
u
Due to contraction the Aera at the inlet of the culvert will be 0,6 x 4 =2,4 m2
So velocity will be 12,67/2,4=5,28 m/s.
This makes the velocity head at the contraction m39,1
20
28,5 2
=
Velocity head upstream culvert
m0
20
0
2
22
upstream
==
g
u
Velocity head downstream culvert
m0
20
0
2
22
downstream
==
g
u
Question 2
Calculate the discharge en velocity culvert if the velocity upstream is 1 m/s and
downstream is 2 m/s. Drop in waterlevel is 1 m. Make a sketch of the H and h line,
with numbers
The difference with question 1 is that the velocity upstream and downstream are not
0, so velocity head upstream and downstream are not 0, so total head upstream is not
equal to pressure line upstream, so total head downstream is not equal to pressure line
downstream.
Velocity head upstream m05,0
2
12
=
g
, downstream is m2,0
2
22
=
g
So ΔH=1 + 0,05 –0,2 =0,85 m.
The other numbers are the same as with question 1.
s
mHgAq tot
tot
v
3
2 68,1185,0204
994,1
1
2
1
=⋅⋅⋅=∆⋅⋅⋅=
ξ
sm
A
Q
u /92,2
4
68,11
===
Of course you also can use the formulas:
2g
v
ξΔΗ
2
c
tottot ⋅= And
For the sketch, you can use the same strategy as with question 1, only the velocity
now is different, 2,92 m/s. And the velocity head upstream(0,05m) and
downstream (0,2 m)are not zero
Head losses at culvert:
m19,0
20
92,2
44,0
2g
u
ξΔΗ
22
culvert
i =⋅=⋅=
m23,0
20
92,2
55,0
2g
u
ξΔΗ
22
culvert
ffriction =⋅=⋅=
m43,0
20
92,2
1
2g
u
ξΔΗ
22
culvert
oo =⋅=⋅=
When I add al losses it should be 0,85 m, which is the total head loss ΔH
Velocity head culvert
m43,0
20
92,2
2
22
culvert
==
g
u
Due to contraction the Aera at the inlet of the culvert will be 0,6 x 4 =2,4 m2
So velocity will be 11,68/2,4=4,87 m/s.
This makes the velocity head at the contraction m18,1
20
87,4 2
=
1 m
0,05 m
0,2 m
ΔH
duiker
Question 3
Suppose the water level downstream is 3 m above the bottom of the culvert, velocity
downstream and upstream is 0,5 m/s and the flow-rate is 10 m3
/s.
Calculate the water level upstream. Make a sketch of the H and h line, with numbers
De duiker staat geheel gevuld met water.
tot
tot
v HgAq ∆⋅⋅⋅= 2
1
2
ξ
Er is 1 onbekende, ΔH. In this case ΔH al other numbers
are given.
totH∆⋅⋅⋅= 204
994,1
1
10 oplossen geeft ΔH=0,62 m.
Dit is het energieverschil!!!! Niet het drukverschil (=waterstandsverschil)
Er moet nog rekening gehouden worden met de snelheidshoogte m0125,0
2
5,0 2
=
g
.
Bovenstrooms en benedenstrooms is de snelheidshoogte gelijk.
To transfer this to difference in water level the velocity head upstream and
downstream have to be taken into account!!!!!
Het waterstandsverschil wordt (water level difference will be)
Δy=0,0125 + 0,62 - 0,0125=0,62 m
De waterstand bovenstrooms is (water upstream) 3 + 0,62 =3,62 m tov de bodem.
Note: Because velocity downstream and upstream are equal, both velocity heads are
equal so difference in water level(Δy) is equal to difference in head (ΔH)
For the sketch you may use the same strategy as with question 2.
Velocity culvert is different, velocity upstream / downstream is different.
0,0125 m
m
0,0125 m
m
ΔH=0,62 m
Culvert
Δh=??
Question 4
We use same data as question 3. Suppose the calculated water level upstream is to
high. What possibilities do you have to lower the upstream water level, without
changing the dimensions of cross-section of the culvert.
Het energieverschil, en dus waterstandsverschil kan met een van de volgende
formules bepaald worden:
tot
tot
v HgAq ∆⋅⋅⋅= 2
1
2
ξ
of
2g
u
ξΔΗ
2
duiker
⋅= tot
If you look at the formula above, q, A and u do not change. The only number you can
change is totaalξ
totaalξ exists off
44,0=iξ Door de vorm aan te passen ( afronding met grote straal) is het mogelijk
de waarde te verlagen tot 0.
If you use a smooth shape, you can reduce the value to 0
55,0=⋅=
D
l
w λξ Hier valt niks aan te passen. Can’t change this,, we assume that we
don’t use another material
1=uξ Door de vorm aan te passen is het mogelijk de waarde te verlagen tot 0,3
By changing the shape (is somewhere in the PTT) you can reduce the value to 0,3
totaalξ wordt nu 0 + 0,55 + 0,3 = 0,85 ipv 1,992
totH∆⋅⋅⋅= 204
85,0
1
10 oplossen geeft ΔH=0,27 m. Aangezien de snelheidshoogte
beneden en bovenstrooms gelijk is (zie uitwerking opgave 3) is ook het
waterstandverschil 0, 27 m. Because velocity head downstream and upstream are
the same, difference in waterlevel is 0,27 m.
In dit geval is het dus mogelijk om het waterstandverschil meer dan te halveren, door
het aanpassen, optimaliseren van de instroom en uitstroomopening. So by optimizing
the shape you can reduce the head loss with about 50%.
Question 5
Suppose the discharge is3 m3
/s. Velocity upstream is 1 m/s, velocity downstream is
0,5 m/s. Reference line is the bottom of the culvert, water level downstream is 3 m
above reference, water level upstream is 3.5 m above reference. Calculate the
dimensions of the cross-section of the culvert. Make a sketch of the H (total head) and
h (pressure) line, with numbers
De duiker ligt geheel onder water.
Snelheidshoogte bovenstrooms is m05,0
2
12
=
g
, benedenstrooms is
m0125,0
2
5,0 2
=
g
ΔH=0,5 + 0,05 – 0,0125 =0,54 m. We gaan uit van de volgende formule
2g
u
ξΔΗ
2
duiker
⋅= tot en
A
Q
u = samen
2g
Q
ξΔΗ 2
2
duiker
⋅
⋅=
A
tot
waarbij :
44,0=iξ Zie opgave 1 deze blijft hetzelfde
1=uξ Zie opgave 1 deze blijft hetzelfde
RRD
l
w
275,0
4
50
022,0 =⋅=⋅= λξ
R
tot
275,0
44,1 +=ξ en
O
A
R =
Onbekenden zijn dus A (afmetingen van de duiker) en R.
De oplossing kan je vinden door te proberen. Daarbij is het zo dat er een soort
standaard afmetingen zijn bij duikers. Kijk bv op www.waco.nl bij afmetingen
duikers
The numbers which are missing are A and R. This is difficult to solve
mathematically , I would suggest to use the try and error method.
Eerste poging, doorsnede duiker is 2 x 2 m. (first attempt)
m
O
A
R 50,0
2222
22
=
+++
⋅
== 99,1
5,0
275,0
44,1 =+=totξ
m
A
tot 05,1
204
13
99,1
2g
Q
ξΔΗ 2
2
2
2
duiker
=
⋅
⋅=
⋅
⋅= >0,54 m voldoet niet
Tweede poging, doorsnede duiker is 3 x 2,5 m.
m
O
A
R 68,0
5,25,233
5,23
=
+++
⋅
== 84,1
68,0
275,0
44,1 =+=totξ
m
A
tot 28,0
205,7
13
84,1
2g
Q
ξΔΗ 2
2
2
2
duiker
=
⋅
⋅=
⋅
⋅= <0,54 m is wel erg ruim
0,5 m
0,05 m
0,0125m
ΔH
duiker
Derde poging, doorsnede duiker is 3 x 2 m.
m
O
A
R 6,0
2233
23
=
+++
⋅
== 90,1
6,0
275,0
44,1 =+=totξ
m
A
tot 45,0
206
13
90,1
2g
Q
ξΔΗ 2
2
2
2
duiker
=
⋅
⋅=
⋅
⋅= <0,54 m voldoet
De waterstandsverhoging tgv van de duiker 3 x 2 m wordt : 0,45 + 0,0125 – 0,05 = 0,
41 m, terwijl er een maximale stijging van 0,50 m was toegestaan.
Dit soort berekening kunnen heel goed in een spreadsheet uitgevoerd worden.

More Related Content

What's hot

Cu06997 lecture 9_open channel
Cu06997 lecture 9_open channelCu06997 lecture 9_open channel
Cu06997 lecture 9_open channel
Henk Massink
 
Cu06997 lecture 5_reynolds_and_r
Cu06997 lecture 5_reynolds_and_rCu06997 lecture 5_reynolds_and_r
Cu06997 lecture 5_reynolds_and_r
Henk Massink
 
Cu06997 lecture 9-10_exercises
Cu06997 lecture 9-10_exercisesCu06997 lecture 9-10_exercises
Cu06997 lecture 9-10_exercises
Henk Massink
 
Cu06997 lecture 2_answer
Cu06997 lecture 2_answerCu06997 lecture 2_answer
Cu06997 lecture 2_answer
Henk Massink
 
Cu06997 lecture 8_sewers
Cu06997 lecture 8_sewersCu06997 lecture 8_sewers
Cu06997 lecture 8_sewers
Henk Massink
 
Cu06997 lecture 12_sediment transport and back water
Cu06997 lecture 12_sediment transport and back waterCu06997 lecture 12_sediment transport and back water
Cu06997 lecture 12_sediment transport and back water
Henk Massink
 
Answers assignment 4 real fluids-fluid mechanics
Answers assignment 4 real fluids-fluid mechanicsAnswers assignment 4 real fluids-fluid mechanics
Answers assignment 4 real fluids-fluid mechanics
asghar123456
 
Answers assignment 3 integral methods-fluid mechanics
Answers assignment 3 integral methods-fluid mechanicsAnswers assignment 3 integral methods-fluid mechanics
Answers assignment 3 integral methods-fluid mechanics
asghar123456
 
Cu06997 exercise5
Cu06997 exercise5Cu06997 exercise5
Cu06997 exercise5
Henk Massink
 
assignment 1 properties of fluids-Fluid mechanics
assignment 1 properties of fluids-Fluid mechanicsassignment 1 properties of fluids-Fluid mechanics
assignment 1 properties of fluids-Fluid mechanics
asghar123456
 
Physics LO 4
Physics LO 4Physics LO 4
Physics LO 4
Victoria Purcell
 
Cu06997 lecture 10_froude
Cu06997 lecture 10_froudeCu06997 lecture 10_froude
Cu06997 lecture 10_froude
Henk Massink
 
Answers assignment 2 fluid statics-fluid mechanics
Answers assignment 2 fluid statics-fluid mechanicsAnswers assignment 2 fluid statics-fluid mechanics
Answers assignment 2 fluid statics-fluid mechanics
asghar123456
 
Solution manual for water resources engineering 3rd edition - david a. chin
Solution manual for water resources engineering 3rd edition - david a. chinSolution manual for water resources engineering 3rd edition - david a. chin
Solution manual for water resources engineering 3rd edition - david a. chin
Salehkhanovic
 
Solutions Manual for Water-Resources Engineering 3rd Edition by Chin
Solutions Manual for Water-Resources Engineering 3rd Edition by ChinSolutions Manual for Water-Resources Engineering 3rd Edition by Chin
Solutions Manual for Water-Resources Engineering 3rd Edition by Chin
MolinaLan
 
Local Energy (Head) Losses (Lecture notes 03)
Local Energy (Head) Losses (Lecture notes 03)Local Energy (Head) Losses (Lecture notes 03)
Local Energy (Head) Losses (Lecture notes 03)
Shekh Muhsen Uddin Ahmed
 
Flows under Pressure in Pipes (Lecture notes 02)
Flows under Pressure in Pipes  (Lecture notes 02)Flows under Pressure in Pipes  (Lecture notes 02)
Flows under Pressure in Pipes (Lecture notes 02)
Shekh Muhsen Uddin Ahmed
 
Chapter 11
Chapter 11Chapter 11
Chapter 11
Younes Sina
 
Unit 3 Fluid Static
Unit 3 Fluid StaticUnit 3 Fluid Static
Unit 3 Fluid Static
Malaysia
 
Solucionario de fluidos_white
Solucionario de fluidos_whiteSolucionario de fluidos_white
Solucionario de fluidos_white
jonathan
 

What's hot (20)

Cu06997 lecture 9_open channel
Cu06997 lecture 9_open channelCu06997 lecture 9_open channel
Cu06997 lecture 9_open channel
 
Cu06997 lecture 5_reynolds_and_r
Cu06997 lecture 5_reynolds_and_rCu06997 lecture 5_reynolds_and_r
Cu06997 lecture 5_reynolds_and_r
 
Cu06997 lecture 9-10_exercises
Cu06997 lecture 9-10_exercisesCu06997 lecture 9-10_exercises
Cu06997 lecture 9-10_exercises
 
Cu06997 lecture 2_answer
Cu06997 lecture 2_answerCu06997 lecture 2_answer
Cu06997 lecture 2_answer
 
Cu06997 lecture 8_sewers
Cu06997 lecture 8_sewersCu06997 lecture 8_sewers
Cu06997 lecture 8_sewers
 
Cu06997 lecture 12_sediment transport and back water
Cu06997 lecture 12_sediment transport and back waterCu06997 lecture 12_sediment transport and back water
Cu06997 lecture 12_sediment transport and back water
 
Answers assignment 4 real fluids-fluid mechanics
Answers assignment 4 real fluids-fluid mechanicsAnswers assignment 4 real fluids-fluid mechanics
Answers assignment 4 real fluids-fluid mechanics
 
Answers assignment 3 integral methods-fluid mechanics
Answers assignment 3 integral methods-fluid mechanicsAnswers assignment 3 integral methods-fluid mechanics
Answers assignment 3 integral methods-fluid mechanics
 
Cu06997 exercise5
Cu06997 exercise5Cu06997 exercise5
Cu06997 exercise5
 
assignment 1 properties of fluids-Fluid mechanics
assignment 1 properties of fluids-Fluid mechanicsassignment 1 properties of fluids-Fluid mechanics
assignment 1 properties of fluids-Fluid mechanics
 
Physics LO 4
Physics LO 4Physics LO 4
Physics LO 4
 
Cu06997 lecture 10_froude
Cu06997 lecture 10_froudeCu06997 lecture 10_froude
Cu06997 lecture 10_froude
 
Answers assignment 2 fluid statics-fluid mechanics
Answers assignment 2 fluid statics-fluid mechanicsAnswers assignment 2 fluid statics-fluid mechanics
Answers assignment 2 fluid statics-fluid mechanics
 
Solution manual for water resources engineering 3rd edition - david a. chin
Solution manual for water resources engineering 3rd edition - david a. chinSolution manual for water resources engineering 3rd edition - david a. chin
Solution manual for water resources engineering 3rd edition - david a. chin
 
Solutions Manual for Water-Resources Engineering 3rd Edition by Chin
Solutions Manual for Water-Resources Engineering 3rd Edition by ChinSolutions Manual for Water-Resources Engineering 3rd Edition by Chin
Solutions Manual for Water-Resources Engineering 3rd Edition by Chin
 
Local Energy (Head) Losses (Lecture notes 03)
Local Energy (Head) Losses (Lecture notes 03)Local Energy (Head) Losses (Lecture notes 03)
Local Energy (Head) Losses (Lecture notes 03)
 
Flows under Pressure in Pipes (Lecture notes 02)
Flows under Pressure in Pipes  (Lecture notes 02)Flows under Pressure in Pipes  (Lecture notes 02)
Flows under Pressure in Pipes (Lecture notes 02)
 
Chapter 11
Chapter 11Chapter 11
Chapter 11
 
Unit 3 Fluid Static
Unit 3 Fluid StaticUnit 3 Fluid Static
Unit 3 Fluid Static
 
Solucionario de fluidos_white
Solucionario de fluidos_whiteSolucionario de fluidos_white
Solucionario de fluidos_white
 

Similar to Cu06997 assignment 6 2014_answer

14.pdf
14.pdf14.pdf
Solucionário Introdução à mecânica dos Fluidos - Chapter 01
Solucionário Introdução à mecânica dos Fluidos - Chapter 01Solucionário Introdução à mecânica dos Fluidos - Chapter 01
Solucionário Introdução à mecânica dos Fluidos - Chapter 01
Rodolpho Montavoni
 
Momentum equation.pdf
 Momentum equation.pdf Momentum equation.pdf
Momentum equation.pdf
Dr. Ezzat Elsayed Gomaa
 
Problema 5
Problema 5Problema 5
Problema 5
yeisyynojos
 
Solutions fox
Solutions   foxSolutions   fox
Solutions fox
Arthur Lott Bezerra
 
Jamuna Oil Comany Ltd recruitment question & ans (5th july 2018)
Jamuna Oil Comany Ltd recruitment question & ans (5th july 2018)Jamuna Oil Comany Ltd recruitment question & ans (5th july 2018)
Jamuna Oil Comany Ltd recruitment question & ans (5th july 2018)
Alamin Md
 
Hidraulica ejercicios
Hidraulica ejercicios Hidraulica ejercicios
Hidraulica ejercicios
FiorbelaGutierrez
 
Answers assignment 5 open channel hydraulics-fluid mechanics
Answers assignment 5 open channel hydraulics-fluid mechanicsAnswers assignment 5 open channel hydraulics-fluid mechanics
Answers assignment 5 open channel hydraulics-fluid mechanics
asghar123456
 
Hardycross method
Hardycross methodHardycross method
Hardycross method
Muhammad Nouman
 
Inclass ass
Inclass assInclass ass
Inclass ass
dr walid
 
Presentation 4 ce 801 OCF by Rabindraa ranjan Saha
Presentation 4 ce 801 OCF by Rabindraa ranjan SahaPresentation 4 ce 801 OCF by Rabindraa ranjan Saha
Presentation 4 ce 801 OCF by Rabindraa ranjan Saha
World University of Bangladesh
 
Flow visualization
Flow visualizationFlow visualization
Flow visualization
Wolkite University
 
Introduction to basic principles of fluid mechanics
Introduction to basic principles of fluid mechanicsIntroduction to basic principles of fluid mechanics
Introduction to basic principles of fluid mechanics
Alwin Johnnie DoraiRaj
 
groundwater flood routing presentationhazard.pptx
groundwater flood routing presentationhazard.pptxgroundwater flood routing presentationhazard.pptx
groundwater flood routing presentationhazard.pptx
MDShohag54
 
@Pipeflow
@Pipeflow@Pipeflow
pipe lines lec 2.pptx
pipe lines lec 2.pptxpipe lines lec 2.pptx
pipe lines lec 2.pptx
amirashraf61
 
006
006006
BASIC EQUATIONS FOR ONE-DIMENSIONAL FLOW (Chapter 04)
BASIC EQUATIONS FOR ONE-DIMENSIONAL FLOW (Chapter 04)BASIC EQUATIONS FOR ONE-DIMENSIONAL FLOW (Chapter 04)
BASIC EQUATIONS FOR ONE-DIMENSIONAL FLOW (Chapter 04)
Shekh Muhsen Uddin Ahmed
 
Steady flow energy eq....by Bilal Ashraf
Steady flow energy eq....by Bilal AshrafSteady flow energy eq....by Bilal Ashraf
Steady flow energy eq....by Bilal Ashraf
Bilal Ashraf
 
Solutions Manual for Foundations Of MEMS 2nd Edition by Chang Liu
Solutions Manual for Foundations Of MEMS 2nd Edition by Chang LiuSolutions Manual for Foundations Of MEMS 2nd Edition by Chang Liu
Solutions Manual for Foundations Of MEMS 2nd Edition by Chang Liu
HildaLa
 

Similar to Cu06997 assignment 6 2014_answer (20)

14.pdf
14.pdf14.pdf
14.pdf
 
Solucionário Introdução à mecânica dos Fluidos - Chapter 01
Solucionário Introdução à mecânica dos Fluidos - Chapter 01Solucionário Introdução à mecânica dos Fluidos - Chapter 01
Solucionário Introdução à mecânica dos Fluidos - Chapter 01
 
Momentum equation.pdf
 Momentum equation.pdf Momentum equation.pdf
Momentum equation.pdf
 
Problema 5
Problema 5Problema 5
Problema 5
 
Solutions fox
Solutions   foxSolutions   fox
Solutions fox
 
Jamuna Oil Comany Ltd recruitment question & ans (5th july 2018)
Jamuna Oil Comany Ltd recruitment question & ans (5th july 2018)Jamuna Oil Comany Ltd recruitment question & ans (5th july 2018)
Jamuna Oil Comany Ltd recruitment question & ans (5th july 2018)
 
Hidraulica ejercicios
Hidraulica ejercicios Hidraulica ejercicios
Hidraulica ejercicios
 
Answers assignment 5 open channel hydraulics-fluid mechanics
Answers assignment 5 open channel hydraulics-fluid mechanicsAnswers assignment 5 open channel hydraulics-fluid mechanics
Answers assignment 5 open channel hydraulics-fluid mechanics
 
Hardycross method
Hardycross methodHardycross method
Hardycross method
 
Inclass ass
Inclass assInclass ass
Inclass ass
 
Presentation 4 ce 801 OCF by Rabindraa ranjan Saha
Presentation 4 ce 801 OCF by Rabindraa ranjan SahaPresentation 4 ce 801 OCF by Rabindraa ranjan Saha
Presentation 4 ce 801 OCF by Rabindraa ranjan Saha
 
Flow visualization
Flow visualizationFlow visualization
Flow visualization
 
Introduction to basic principles of fluid mechanics
Introduction to basic principles of fluid mechanicsIntroduction to basic principles of fluid mechanics
Introduction to basic principles of fluid mechanics
 
groundwater flood routing presentationhazard.pptx
groundwater flood routing presentationhazard.pptxgroundwater flood routing presentationhazard.pptx
groundwater flood routing presentationhazard.pptx
 
@Pipeflow
@Pipeflow@Pipeflow
@Pipeflow
 
pipe lines lec 2.pptx
pipe lines lec 2.pptxpipe lines lec 2.pptx
pipe lines lec 2.pptx
 
006
006006
006
 
BASIC EQUATIONS FOR ONE-DIMENSIONAL FLOW (Chapter 04)
BASIC EQUATIONS FOR ONE-DIMENSIONAL FLOW (Chapter 04)BASIC EQUATIONS FOR ONE-DIMENSIONAL FLOW (Chapter 04)
BASIC EQUATIONS FOR ONE-DIMENSIONAL FLOW (Chapter 04)
 
Steady flow energy eq....by Bilal Ashraf
Steady flow energy eq....by Bilal AshrafSteady flow energy eq....by Bilal Ashraf
Steady flow energy eq....by Bilal Ashraf
 
Solutions Manual for Foundations Of MEMS 2nd Edition by Chang Liu
Solutions Manual for Foundations Of MEMS 2nd Edition by Chang LiuSolutions Manual for Foundations Of MEMS 2nd Edition by Chang Liu
Solutions Manual for Foundations Of MEMS 2nd Edition by Chang Liu
 

More from Henk Massink

Cu07821 ppt9 recapitulation
Cu07821 ppt9 recapitulationCu07821 ppt9 recapitulation
Cu07821 ppt9 recapitulation
Henk Massink
 
Gastcollege mli
Gastcollege mliGastcollege mli
Gastcollege mli
Henk Massink
 
Cu07821 10management and maintenance2015
Cu07821 10management and maintenance2015Cu07821 10management and maintenance2015
Cu07821 10management and maintenance2015
Henk Massink
 
Cu07821 9 zoning plan2015
Cu07821 9 zoning plan2015Cu07821 9 zoning plan2015
Cu07821 9 zoning plan2015
Henk Massink
 
Cu07821 8 weirs
Cu07821 8 weirsCu07821 8 weirs
Cu07821 8 weirs
Henk Massink
 
Cu07821 7 culverts new
Cu07821 7 culverts newCu07821 7 culverts new
Cu07821 7 culverts new
Henk Massink
 
Cu07821 6 pumping stations_update
Cu07821 6 pumping stations_updateCu07821 6 pumping stations_update
Cu07821 6 pumping stations_update
Henk Massink
 
Cu07821 5 drainage
Cu07821 5 drainageCu07821 5 drainage
Cu07821 5 drainage
Henk Massink
 
Cu07821 4 soil
Cu07821 4 soilCu07821 4 soil
Cu07821 4 soil
Henk Massink
 
Cu07821 3 precipitation and evapotranspiration
Cu07821 3  precipitation and evapotranspirationCu07821 3  precipitation and evapotranspiration
Cu07821 3 precipitation and evapotranspiration
Henk Massink
 
Cu07821 2 help
Cu07821 2 helpCu07821 2 help
Cu07821 2 help
Henk Massink
 
Cu07821 1 intro_1415
Cu07821 1 intro_1415Cu07821 1 intro_1415
Cu07821 1 intro_1415
Henk Massink
 
Research portfolio delta_academy_s2_2014_2015
Research portfolio delta_academy_s2_2014_2015Research portfolio delta_academy_s2_2014_2015
Research portfolio delta_academy_s2_2014_2015
Henk Massink
 
Research portfolio da arc 2014-2015 s1
Research portfolio da arc  2014-2015 s1Research portfolio da arc  2014-2015 s1
Research portfolio da arc 2014-2015 s1
Henk Massink
 
Research portfolios1 2013_2014 jan july 2014
Research portfolios1 2013_2014 jan july 2014Research portfolios1 2013_2014 jan july 2014
Research portfolios1 2013_2014 jan july 2014
Henk Massink
 
Presentatie AET voor scholieren 15-11-2013
Presentatie AET voor scholieren 15-11-2013Presentatie AET voor scholieren 15-11-2013
Presentatie AET voor scholieren 15-11-2013
Henk Massink
 
Vision group1(5)
Vision group1(5)Vision group1(5)
Vision group1(5)
Henk Massink
 
Final presentation spain quattro
Final presentation spain quattroFinal presentation spain quattro
Final presentation spain quattro
Henk Massink
 
Final presentation group 3
Final presentation group 3Final presentation group 3
Final presentation group 3
Henk Massink
 

More from Henk Massink (20)

Cu07821 ppt9 recapitulation
Cu07821 ppt9 recapitulationCu07821 ppt9 recapitulation
Cu07821 ppt9 recapitulation
 
Gastcollege mli
Gastcollege mliGastcollege mli
Gastcollege mli
 
Cu07821 10management and maintenance2015
Cu07821 10management and maintenance2015Cu07821 10management and maintenance2015
Cu07821 10management and maintenance2015
 
Cu07821 9 zoning plan2015
Cu07821 9 zoning plan2015Cu07821 9 zoning plan2015
Cu07821 9 zoning plan2015
 
Cu07821 8 weirs
Cu07821 8 weirsCu07821 8 weirs
Cu07821 8 weirs
 
Cu07821 7 culverts new
Cu07821 7 culverts newCu07821 7 culverts new
Cu07821 7 culverts new
 
Cu07821 6 pumping stations_update
Cu07821 6 pumping stations_updateCu07821 6 pumping stations_update
Cu07821 6 pumping stations_update
 
Cu07821 5 drainage
Cu07821 5 drainageCu07821 5 drainage
Cu07821 5 drainage
 
Cu07821 4 soil
Cu07821 4 soilCu07821 4 soil
Cu07821 4 soil
 
Cu07821 3 precipitation and evapotranspiration
Cu07821 3  precipitation and evapotranspirationCu07821 3  precipitation and evapotranspiration
Cu07821 3 precipitation and evapotranspiration
 
Cu07821 2 help
Cu07821 2 helpCu07821 2 help
Cu07821 2 help
 
Cu07821 1 intro_1415
Cu07821 1 intro_1415Cu07821 1 intro_1415
Cu07821 1 intro_1415
 
Research portfolio delta_academy_s2_2014_2015
Research portfolio delta_academy_s2_2014_2015Research portfolio delta_academy_s2_2014_2015
Research portfolio delta_academy_s2_2014_2015
 
Research portfolio da arc 2014-2015 s1
Research portfolio da arc  2014-2015 s1Research portfolio da arc  2014-2015 s1
Research portfolio da arc 2014-2015 s1
 
Jacobapolder
JacobapolderJacobapolder
Jacobapolder
 
Research portfolios1 2013_2014 jan july 2014
Research portfolios1 2013_2014 jan july 2014Research portfolios1 2013_2014 jan july 2014
Research portfolios1 2013_2014 jan july 2014
 
Presentatie AET voor scholieren 15-11-2013
Presentatie AET voor scholieren 15-11-2013Presentatie AET voor scholieren 15-11-2013
Presentatie AET voor scholieren 15-11-2013
 
Vision group1(5)
Vision group1(5)Vision group1(5)
Vision group1(5)
 
Final presentation spain quattro
Final presentation spain quattroFinal presentation spain quattro
Final presentation spain quattro
 
Final presentation group 3
Final presentation group 3Final presentation group 3
Final presentation group 3
 

Recently uploaded

Pengantar Penggunaan Flutter - Dart programming language1.pptx
Pengantar Penggunaan Flutter - Dart programming language1.pptxPengantar Penggunaan Flutter - Dart programming language1.pptx
Pengantar Penggunaan Flutter - Dart programming language1.pptx
Fajar Baskoro
 
Jemison, MacLaughlin, and Majumder "Broadening Pathways for Editors and Authors"
Jemison, MacLaughlin, and Majumder "Broadening Pathways for Editors and Authors"Jemison, MacLaughlin, and Majumder "Broadening Pathways for Editors and Authors"
Jemison, MacLaughlin, and Majumder "Broadening Pathways for Editors and Authors"
National Information Standards Organization (NISO)
 
BBR 2024 Summer Sessions Interview Training
BBR  2024 Summer Sessions Interview TrainingBBR  2024 Summer Sessions Interview Training
BBR 2024 Summer Sessions Interview Training
Katrina Pritchard
 
C1 Rubenstein AP HuG xxxxxxxxxxxxxx.pptx
C1 Rubenstein AP HuG xxxxxxxxxxxxxx.pptxC1 Rubenstein AP HuG xxxxxxxxxxxxxx.pptx
C1 Rubenstein AP HuG xxxxxxxxxxxxxx.pptx
mulvey2
 
Beyond Degrees - Empowering the Workforce in the Context of Skills-First.pptx
Beyond Degrees - Empowering the Workforce in the Context of Skills-First.pptxBeyond Degrees - Empowering the Workforce in the Context of Skills-First.pptx
Beyond Degrees - Empowering the Workforce in the Context of Skills-First.pptx
EduSkills OECD
 
Electric Fetus - Record Store Scavenger Hunt
Electric Fetus - Record Store Scavenger HuntElectric Fetus - Record Store Scavenger Hunt
Electric Fetus - Record Store Scavenger Hunt
RamseyBerglund
 
How to Setup Warehouse & Location in Odoo 17 Inventory
How to Setup Warehouse & Location in Odoo 17 InventoryHow to Setup Warehouse & Location in Odoo 17 Inventory
How to Setup Warehouse & Location in Odoo 17 Inventory
Celine George
 
Mule event processing models | MuleSoft Mysore Meetup #47
Mule event processing models | MuleSoft Mysore Meetup #47Mule event processing models | MuleSoft Mysore Meetup #47
Mule event processing models | MuleSoft Mysore Meetup #47
MysoreMuleSoftMeetup
 
Bonku-Babus-Friend by Sathyajith Ray (9)
Bonku-Babus-Friend by Sathyajith Ray  (9)Bonku-Babus-Friend by Sathyajith Ray  (9)
Bonku-Babus-Friend by Sathyajith Ray (9)
nitinpv4ai
 
Temple of Asclepius in Thrace. Excavation results
Temple of Asclepius in Thrace. Excavation resultsTemple of Asclepius in Thrace. Excavation results
Temple of Asclepius in Thrace. Excavation results
Krassimira Luka
 
Chapter wise All Notes of First year Basic Civil Engineering.pptx
Chapter wise All Notes of First year Basic Civil Engineering.pptxChapter wise All Notes of First year Basic Civil Engineering.pptx
Chapter wise All Notes of First year Basic Civil Engineering.pptx
Denish Jangid
 
Nutrition Inc FY 2024, 4 - Hour Training
Nutrition Inc FY 2024, 4 - Hour TrainingNutrition Inc FY 2024, 4 - Hour Training
Nutrition Inc FY 2024, 4 - Hour Training
melliereed
 
math operations ued in python and all used
math operations ued in python and all usedmath operations ued in python and all used
math operations ued in python and all used
ssuser13ffe4
 
مصحف القراءات العشر أعد أحرف الخلاف سمير بسيوني.pdf
مصحف القراءات العشر   أعد أحرف الخلاف سمير بسيوني.pdfمصحف القراءات العشر   أعد أحرف الخلاف سمير بسيوني.pdf
مصحف القراءات العشر أعد أحرف الخلاف سمير بسيوني.pdf
سمير بسيوني
 
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
PECB
 
Stack Memory Organization of 8086 Microprocessor
Stack Memory Organization of 8086 MicroprocessorStack Memory Organization of 8086 Microprocessor
Stack Memory Organization of 8086 Microprocessor
JomonJoseph58
 
Gender and Mental Health - Counselling and Family Therapy Applications and In...
Gender and Mental Health - Counselling and Family Therapy Applications and In...Gender and Mental Health - Counselling and Family Therapy Applications and In...
Gender and Mental Health - Counselling and Family Therapy Applications and In...
PsychoTech Services
 
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
GeorgeMilliken2
 
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
Nguyen Thanh Tu Collection
 
SWOT analysis in the project Keeping the Memory @live.pptx
SWOT analysis in the project Keeping the Memory @live.pptxSWOT analysis in the project Keeping the Memory @live.pptx
SWOT analysis in the project Keeping the Memory @live.pptx
zuzanka
 

Recently uploaded (20)

Pengantar Penggunaan Flutter - Dart programming language1.pptx
Pengantar Penggunaan Flutter - Dart programming language1.pptxPengantar Penggunaan Flutter - Dart programming language1.pptx
Pengantar Penggunaan Flutter - Dart programming language1.pptx
 
Jemison, MacLaughlin, and Majumder "Broadening Pathways for Editors and Authors"
Jemison, MacLaughlin, and Majumder "Broadening Pathways for Editors and Authors"Jemison, MacLaughlin, and Majumder "Broadening Pathways for Editors and Authors"
Jemison, MacLaughlin, and Majumder "Broadening Pathways for Editors and Authors"
 
BBR 2024 Summer Sessions Interview Training
BBR  2024 Summer Sessions Interview TrainingBBR  2024 Summer Sessions Interview Training
BBR 2024 Summer Sessions Interview Training
 
C1 Rubenstein AP HuG xxxxxxxxxxxxxx.pptx
C1 Rubenstein AP HuG xxxxxxxxxxxxxx.pptxC1 Rubenstein AP HuG xxxxxxxxxxxxxx.pptx
C1 Rubenstein AP HuG xxxxxxxxxxxxxx.pptx
 
Beyond Degrees - Empowering the Workforce in the Context of Skills-First.pptx
Beyond Degrees - Empowering the Workforce in the Context of Skills-First.pptxBeyond Degrees - Empowering the Workforce in the Context of Skills-First.pptx
Beyond Degrees - Empowering the Workforce in the Context of Skills-First.pptx
 
Electric Fetus - Record Store Scavenger Hunt
Electric Fetus - Record Store Scavenger HuntElectric Fetus - Record Store Scavenger Hunt
Electric Fetus - Record Store Scavenger Hunt
 
How to Setup Warehouse & Location in Odoo 17 Inventory
How to Setup Warehouse & Location in Odoo 17 InventoryHow to Setup Warehouse & Location in Odoo 17 Inventory
How to Setup Warehouse & Location in Odoo 17 Inventory
 
Mule event processing models | MuleSoft Mysore Meetup #47
Mule event processing models | MuleSoft Mysore Meetup #47Mule event processing models | MuleSoft Mysore Meetup #47
Mule event processing models | MuleSoft Mysore Meetup #47
 
Bonku-Babus-Friend by Sathyajith Ray (9)
Bonku-Babus-Friend by Sathyajith Ray  (9)Bonku-Babus-Friend by Sathyajith Ray  (9)
Bonku-Babus-Friend by Sathyajith Ray (9)
 
Temple of Asclepius in Thrace. Excavation results
Temple of Asclepius in Thrace. Excavation resultsTemple of Asclepius in Thrace. Excavation results
Temple of Asclepius in Thrace. Excavation results
 
Chapter wise All Notes of First year Basic Civil Engineering.pptx
Chapter wise All Notes of First year Basic Civil Engineering.pptxChapter wise All Notes of First year Basic Civil Engineering.pptx
Chapter wise All Notes of First year Basic Civil Engineering.pptx
 
Nutrition Inc FY 2024, 4 - Hour Training
Nutrition Inc FY 2024, 4 - Hour TrainingNutrition Inc FY 2024, 4 - Hour Training
Nutrition Inc FY 2024, 4 - Hour Training
 
math operations ued in python and all used
math operations ued in python and all usedmath operations ued in python and all used
math operations ued in python and all used
 
مصحف القراءات العشر أعد أحرف الخلاف سمير بسيوني.pdf
مصحف القراءات العشر   أعد أحرف الخلاف سمير بسيوني.pdfمصحف القراءات العشر   أعد أحرف الخلاف سمير بسيوني.pdf
مصحف القراءات العشر أعد أحرف الخلاف سمير بسيوني.pdf
 
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
 
Stack Memory Organization of 8086 Microprocessor
Stack Memory Organization of 8086 MicroprocessorStack Memory Organization of 8086 Microprocessor
Stack Memory Organization of 8086 Microprocessor
 
Gender and Mental Health - Counselling and Family Therapy Applications and In...
Gender and Mental Health - Counselling and Family Therapy Applications and In...Gender and Mental Health - Counselling and Family Therapy Applications and In...
Gender and Mental Health - Counselling and Family Therapy Applications and In...
 
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
 
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
 
SWOT analysis in the project Keeping the Memory @live.pptx
SWOT analysis in the project Keeping the Memory @live.pptxSWOT analysis in the project Keeping the Memory @live.pptx
SWOT analysis in the project Keeping the Memory @live.pptx
 

Cu06997 assignment 6 2014_answer

  • 1. Exercise lecture 7 culverts Downstream Upstream Cross-section Dropwaterlevel Length culvert Length culvert is 50 m, cross-section 2 x 2 m, λ=0,022 and μ =0,6 Question 1 Calculate the discharge if the drop in water level is 1 m and the velocity downstream and upstream is 0 m/s. Make a sketch of the H and h line, with numbers
  • 2. Because the velocity upstream and downstream is 0 m/s, velocity head is 0, total head is equal to pressure line, difference in water level is the same as difference in energy level. Δy = ΔH. ΔH = 1 m. 2g u ξΔΗ 2 c tottot ⋅= m O A R 50,0 2222 22 = +++ ⋅ == 44,01 1 2 =      −= µ ξi 55,0 5,04 50 022,0 = ⋅ ⋅=⋅= D l w λξ 1=uξ 994,1155,044,0)ξξξ( uwi =++=++=totξ smu /17,3= s mQ 3 67,12=
  • 3. Head losses at culvert: m22,0 20 17,3 44,0 2g u ξΔΗ 22 culvert i =⋅=⋅= m28,0 20 17,3 55,0 2g u ξΔΗ 22 culvert ffriction =⋅=⋅= m50,0 20 17,3 1 2g u ξΔΗ 22 culvert oo =⋅=⋅= When I add al losses it should be 1 m, which is the total head loss ΔH Velocity head culvert m50,0 20 17,3 2 22 culvert == g u Due to contraction the Aera at the inlet of the culvert will be 0,6 x 4 =2,4 m2 So velocity will be 12,67/2,4=5,28 m/s. This makes the velocity head at the contraction m39,1 20 28,5 2 = Velocity head upstream culvert m0 20 0 2 22 upstream == g u Velocity head downstream culvert m0 20 0 2 22 downstream == g u
  • 4.
  • 5. Question 2 Calculate the discharge en velocity culvert if the velocity upstream is 1 m/s and downstream is 2 m/s. Drop in waterlevel is 1 m. Make a sketch of the H and h line, with numbers The difference with question 1 is that the velocity upstream and downstream are not 0, so velocity head upstream and downstream are not 0, so total head upstream is not equal to pressure line upstream, so total head downstream is not equal to pressure line downstream. Velocity head upstream m05,0 2 12 = g , downstream is m2,0 2 22 = g So ΔH=1 + 0,05 –0,2 =0,85 m. The other numbers are the same as with question 1. s mHgAq tot tot v 3 2 68,1185,0204 994,1 1 2 1 =⋅⋅⋅=∆⋅⋅⋅= ξ sm A Q u /92,2 4 68,11 === Of course you also can use the formulas: 2g v ξΔΗ 2 c tottot ⋅= And For the sketch, you can use the same strategy as with question 1, only the velocity now is different, 2,92 m/s. And the velocity head upstream(0,05m) and downstream (0,2 m)are not zero Head losses at culvert: m19,0 20 92,2 44,0 2g u ξΔΗ 22 culvert i =⋅=⋅= m23,0 20 92,2 55,0 2g u ξΔΗ 22 culvert ffriction =⋅=⋅= m43,0 20 92,2 1 2g u ξΔΗ 22 culvert oo =⋅=⋅= When I add al losses it should be 0,85 m, which is the total head loss ΔH Velocity head culvert m43,0 20 92,2 2 22 culvert == g u Due to contraction the Aera at the inlet of the culvert will be 0,6 x 4 =2,4 m2 So velocity will be 11,68/2,4=4,87 m/s. This makes the velocity head at the contraction m18,1 20 87,4 2 = 1 m 0,05 m 0,2 m ΔH duiker
  • 6. Question 3 Suppose the water level downstream is 3 m above the bottom of the culvert, velocity downstream and upstream is 0,5 m/s and the flow-rate is 10 m3 /s. Calculate the water level upstream. Make a sketch of the H and h line, with numbers De duiker staat geheel gevuld met water. tot tot v HgAq ∆⋅⋅⋅= 2 1 2 ξ Er is 1 onbekende, ΔH. In this case ΔH al other numbers are given. totH∆⋅⋅⋅= 204 994,1 1 10 oplossen geeft ΔH=0,62 m. Dit is het energieverschil!!!! Niet het drukverschil (=waterstandsverschil) Er moet nog rekening gehouden worden met de snelheidshoogte m0125,0 2 5,0 2 = g . Bovenstrooms en benedenstrooms is de snelheidshoogte gelijk. To transfer this to difference in water level the velocity head upstream and downstream have to be taken into account!!!!! Het waterstandsverschil wordt (water level difference will be) Δy=0,0125 + 0,62 - 0,0125=0,62 m De waterstand bovenstrooms is (water upstream) 3 + 0,62 =3,62 m tov de bodem. Note: Because velocity downstream and upstream are equal, both velocity heads are equal so difference in water level(Δy) is equal to difference in head (ΔH) For the sketch you may use the same strategy as with question 2. Velocity culvert is different, velocity upstream / downstream is different. 0,0125 m m 0,0125 m m ΔH=0,62 m Culvert Δh=??
  • 7. Question 4 We use same data as question 3. Suppose the calculated water level upstream is to high. What possibilities do you have to lower the upstream water level, without changing the dimensions of cross-section of the culvert. Het energieverschil, en dus waterstandsverschil kan met een van de volgende formules bepaald worden: tot tot v HgAq ∆⋅⋅⋅= 2 1 2 ξ of 2g u ξΔΗ 2 duiker ⋅= tot If you look at the formula above, q, A and u do not change. The only number you can change is totaalξ totaalξ exists off 44,0=iξ Door de vorm aan te passen ( afronding met grote straal) is het mogelijk de waarde te verlagen tot 0. If you use a smooth shape, you can reduce the value to 0 55,0=⋅= D l w λξ Hier valt niks aan te passen. Can’t change this,, we assume that we don’t use another material 1=uξ Door de vorm aan te passen is het mogelijk de waarde te verlagen tot 0,3 By changing the shape (is somewhere in the PTT) you can reduce the value to 0,3 totaalξ wordt nu 0 + 0,55 + 0,3 = 0,85 ipv 1,992 totH∆⋅⋅⋅= 204 85,0 1 10 oplossen geeft ΔH=0,27 m. Aangezien de snelheidshoogte beneden en bovenstrooms gelijk is (zie uitwerking opgave 3) is ook het waterstandverschil 0, 27 m. Because velocity head downstream and upstream are the same, difference in waterlevel is 0,27 m. In dit geval is het dus mogelijk om het waterstandverschil meer dan te halveren, door het aanpassen, optimaliseren van de instroom en uitstroomopening. So by optimizing the shape you can reduce the head loss with about 50%.
  • 8.
  • 9. Question 5 Suppose the discharge is3 m3 /s. Velocity upstream is 1 m/s, velocity downstream is 0,5 m/s. Reference line is the bottom of the culvert, water level downstream is 3 m above reference, water level upstream is 3.5 m above reference. Calculate the dimensions of the cross-section of the culvert. Make a sketch of the H (total head) and h (pressure) line, with numbers De duiker ligt geheel onder water. Snelheidshoogte bovenstrooms is m05,0 2 12 = g , benedenstrooms is m0125,0 2 5,0 2 = g ΔH=0,5 + 0,05 – 0,0125 =0,54 m. We gaan uit van de volgende formule 2g u ξΔΗ 2 duiker ⋅= tot en A Q u = samen 2g Q ξΔΗ 2 2 duiker ⋅ ⋅= A tot waarbij : 44,0=iξ Zie opgave 1 deze blijft hetzelfde 1=uξ Zie opgave 1 deze blijft hetzelfde RRD l w 275,0 4 50 022,0 =⋅=⋅= λξ R tot 275,0 44,1 +=ξ en O A R = Onbekenden zijn dus A (afmetingen van de duiker) en R. De oplossing kan je vinden door te proberen. Daarbij is het zo dat er een soort standaard afmetingen zijn bij duikers. Kijk bv op www.waco.nl bij afmetingen duikers The numbers which are missing are A and R. This is difficult to solve mathematically , I would suggest to use the try and error method. Eerste poging, doorsnede duiker is 2 x 2 m. (first attempt) m O A R 50,0 2222 22 = +++ ⋅ == 99,1 5,0 275,0 44,1 =+=totξ m A tot 05,1 204 13 99,1 2g Q ξΔΗ 2 2 2 2 duiker = ⋅ ⋅= ⋅ ⋅= >0,54 m voldoet niet Tweede poging, doorsnede duiker is 3 x 2,5 m. m O A R 68,0 5,25,233 5,23 = +++ ⋅ == 84,1 68,0 275,0 44,1 =+=totξ m A tot 28,0 205,7 13 84,1 2g Q ξΔΗ 2 2 2 2 duiker = ⋅ ⋅= ⋅ ⋅= <0,54 m is wel erg ruim 0,5 m 0,05 m 0,0125m ΔH duiker
  • 10. Derde poging, doorsnede duiker is 3 x 2 m. m O A R 6,0 2233 23 = +++ ⋅ == 90,1 6,0 275,0 44,1 =+=totξ m A tot 45,0 206 13 90,1 2g Q ξΔΗ 2 2 2 2 duiker = ⋅ ⋅= ⋅ ⋅= <0,54 m voldoet De waterstandsverhoging tgv van de duiker 3 x 2 m wordt : 0,45 + 0,0125 – 0,05 = 0, 41 m, terwijl er een maximale stijging van 0,50 m was toegestaan. Dit soort berekening kunnen heel goed in een spreadsheet uitgevoerd worden.