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

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%.

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.