Exercise lecture 7 culverts 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
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Exercise lecture 7 culverts
Length culvert is 50 m, cross-section 2 x 2 m, λ=0,022 and μ =0,6
Question 1Calculate 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.
Head losses at culvert:
When I add al losses it should be 1 m, which is the total head loss ΔH
Velocity head culvert
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
Velocity head upstream culvert
Velocity head downstream culvert
Question 2Calculate 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 numbersThe 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 , downstream is
So ΔH=1 + 0,05 –0,2 =0,85 m.
The other numbers are the same as with question 1.
Of course you also can use the formulas:
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 zeroHead losses at culvert:
When I add al losses it should be 0,85 m, which is the total head loss ΔHVelocity head culvert
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
1 m
0,05 m
0,2 m
ΔH
duiker
Question 3Suppose 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.
Er is 1 onbekende, ΔH. In this case ΔH al other numbers are
given.
oplossen geeft ΔH=0,62 m.
Dit is het energieverschil!!!! Niet het drukverschil (=waterstandsverschil)
Er moet nog rekening gehouden worden met de snelheidshoogte .
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 mDe 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 4We 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:
of
If you look at the formula above, q, A and u do not change. The only number you can change is
exists off
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
Hier valt niks aan te passen. Can’t change this,, we assume that we
don’t use another material
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
wordt nu 0 + 0,55 + 0,3 = 0,85 ipv 1,992
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 5Suppose 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 , benedenstrooms is
ΔH=0,5 + 0,05 – 0,0125 =0,54 m. We gaan uit van de volgende formule
en samen
waarbij : Zie opgave 1 deze blijft hetzelfde
Zie opgave 1 deze blijft hetzelfde
en
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)