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1. 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. 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. 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. 4. 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
5. 5. 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=??
6. 6. 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%.
7. 7. 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
8. 8. 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.