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# Computation exam

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### Computation exam

1. 1. Question 1In the picture you see a lock gate with at the left side freshwater and at the right sidesaltwater. Suppose the gate is free to turn, in which direction will the gate turn, to theleft or to the right? Make your argument based on a calculation.3,9m3,8 mfresh saltLock gate, width = 4 , Sluisdeur, breedte = 4mDruk onderin door zoetwater = 1000 x 10 x 3,9 =39.000 N/m2Gemiddelde druk = 19.500 N/m2Kracht door zoet = 3,9 x4 x 19.500 = 304.200 NDruk onderin door zoutwater = 1025 x 10 x 3,8 =38.950 N/m2Gemiddelde druk = 19.475 N/m2Kracht door zout = 3,8 x 4 x 19.475 = 296.020 NDe deur draait naar rechts𝑦 =𝑝𝜌 ∙ 𝑔p =FA
2. 2. Question 2An open channel has a width at the bottom of 3 m, a slope on both sides of 1:2. Waterdepth is 1,3 m.The length of the channel is 4 km. Manning’s roughness coefficient n is0,023 s/m1/3. At the end of the channel (downstream) a short crested weir is situated.(Width of the weir is 4,5 m, runoff coefficient m=1,8 m1/2/s). The water depthdownstream of the weir is 0,8 m The discharge is 5,1 m3/s. The discharge does notchange over the length of the channel.a. Calculate the bed slope of the channel. What is the difference in bottom heightbetween beginning and end of the channel.b. Calculate froudes number. Indicate if the flow in the channel is sub-critical orsuper-critical. As width you can use the average width of the channel. Give acomputation.c. Calculate the level of the crest related to the bottom level of the channel. Assumethe weir has a free flow.d. Check if the weir has a free flow. Give a motivation.Gaan uit van een evenwichtssituatieA = (3+2*1,3) x 1,3 =7,28 m2O= 3 + 2 x (1,32+ 2.62)1/2= 8,81 mR= A/O= 7,28 / 8,81 = 0,83 mu=Q/A = 5,1 / 7,28 = 0,70 m/s𝑉 =𝑅23 ∙ 𝑆𝑓12𝑛n=0,023 invullen geeft S=3,323 x 10-4of 1:3009Afschot over 4 km is 1.33 m
3. 3. Question 2An open channel has a width at the bottom of 3 m, a slope on both sides of 1:2. Waterdepth is 1,3 m.The length of the channel is 4 km. Manning’s roughness coefficient n is0,023 s/m1/3. At the end of the channel (downstream) a short crested weir is situated.(Width of the weir is 4,5 m, runoff coefficient m=1,8 m1/2/s). The water depthdownstream of the weir is 0,8 m The discharge is 5,1 m3/s. The discharge does notchange over the length of the channel.a. Calculate the bed slope of the channel. What is the difference in bottom heightbetween beginning and end of the channel.b. Calculate froudes number. Indicate if the flow in the channel is sub-critical orsuper-critical. As width you can use the average width of the channel. Give acomputation.c. Calculate the level of the crest related to the bottom level of the channel. Assumethe weir has a free flow.d. Check if the weir has a free flow. Give a motivation.Onderdeel bGemiddelde breedte = 3 + 2*1,3 = 5.6 m322bgqy vcy kritisch = 0,44 m, Vc = 2,08 m/s𝐹𝑟 =𝑉𝑔 ∙ 𝑦𝑐2=𝑉𝑉𝑐=0,72,08= 0,34Fr < 1, kortom we hebben te maken met stromend water
4. 4. Question 2An open channel has a width at the bottom of 3 m, a slope on both sides of 1:2. Waterdepth is 1,3 m.The length of the channel is 4 km. Manning’s roughness coefficient n is0,023 s/m1/3. At the end of the channel (downstream) a short crested weir is situated.(Width of the weir is 4,5 m, runoff coefficient m=1,8 m1/2/s). The water depthdownstream of the weir is 0,8 m The discharge is 5,1 m3/s. The discharge does notchange over the length of the channel.a. Calculate the bed slope of the channel. What is the difference in bottom heightbetween beginning and end of the channel.b. Calculate froudes number. Indicate if the flow in the channel is sub-critical orsuper-critical. As width you can use the average width of the channel. Give acomputation.c. Calculate the level of the crest related to the bottom level of the channel. Assumethe weir has a free flow.d. Check if the weir has a free flow. Give a motivation.Onderdeel c23Hbmqv Overstortende straal = 0,74 mSnelheidshoogte voor de overstort = v2/2g=0,72/20 = 0,0245mHoogte stuw vanaf de bodem = 1,3 +0,0245 – 0,74 = 0,58 mOnderdeel dHh 323 H=0,74 2/3 H = 0, 49 h3 0,8 -0,58 = 0,22 m
5. 5. Question 3In a channel the culvert above is situated,  = 0,6 and  = 0,025, fresh watera. Calculate the value of totalb. Calculate the water level upstream of the culvert.c. Make a sketch of the energy and pressure line. Include the valuesChannel u=0,3 m/sUpstreamWL ??? DownstreamWL +2,0 m NAPL= 45 m3 m wide, 2 m high, u = 1,4 m/s Channel u=0,5 m/sOnderdeel aR=A/O=(3 x 2)/(3+3+2+2)=0,60 m,211i =0,44Rlw4 =0,471uksi-totaal=1,91
6. 6. Question 3In a channel the culvert above is situated,  = 0,6 and  = 0,025, fresh watera. Calculate the value of totalb. Calculate the water level upstream of the culvert.c. Make a sketch of the energy and pressure line. Include the valuesChannel u=0,3 m/sUpstreamWL ??? DownstreamWL +2,0 m NAPL= 45 m3 m wide, 2 m high, u = 1,4 m/s Channel u=0,5 m/s2guξΔΗ22totaala  =0,187 m Q = u x A = 1,4 x 6 = 8,4 m3/sSnelheidshoogte bovenstrooms = 0,32/20 = 0,005 mSnelheidshoogte benedenstrooms = 0,52/20 = 0,013 mWaterstand bovenstrooms = 2,0 + 0,013 + 0,187 – 0,005 = 2,195 m𝑦1 + 𝑧1 +𝑢122𝑔= 𝑦2 + 𝑧2 +𝑢222𝑔+ ∆𝐻1−2
7. 7. Question 3In a channel the culvert above is situated,  = 0,6 and  = 0,025, fresh watera. Calculate the value of totalb. Calculate the water level upstream of the culvert.c. Make a sketch of the energy and pressure line. Include the valuesChannel u=0,3 m/sUpstreamWL ??? DownstreamWL +2,0 m NAPL= 45 m3 m wide, 2 m high, u = 1,4 m/s Channel u=0,5 m/sOnderdeel cSnelheidshoogte duiker 1,42/20= 0,098 mEnergieverlies instroom = 0,44 x 0,098 = 0,043 mEnergieverlies wrijving = 0,47 x 0,098 = 0,046 mEnergieverlies uitstroom = 0,098 mDip in druklijn tgv contractie.Acontractie = A x 0,6 Q=v.a Ucontractie = 1,4/0,6 = 2,33 m/sSnelheidshoogte 0,27 m
8. 8. Question 4In the picture above you see a sewer pipe designed for the discharge of rainfall directlyto surface water. The water level surface water is +2 m, the velocity is low. The waterlevel in the manhole is +2,95 m. The equivalent roughness k is 1 mma. Calculate the discharge in the sewer pipe. Assume the flow is turbulent.b. Check if the flow is turbulent.c. Suppose the intensity of the rainfall is 60 l/s.ha. Calculate the paved area which isconnected to the sewer. As discharge use the result from question a.d. Explain why the design of this sewer is actually bad.Water level + 2 mSurface level +3 mRainfall, discharge???60 mD=500 mmManhole(inspectieput)Onderdeel aWaterstandsverschil is 2,95 – 2 = 0,95 mI=dh/L=0,95/60=0,0158 1:63 R=D/4=0,500/4=0,125 m A=0,196 m2kRC12log18 =57,17𝑉 = 𝐶 ∙ 𝑅 ∙ 𝑆𝑓=2,55 m/sQ=u x a =0,5 m3/s=500 l/s
9. 9. Question 4In the picture above you see a sewer pipe designed for the discharge of rainfall directlyto surface water. The water level surface water is +2 m, the velocity is low. The waterlevel in the manhole is +2,95 m. The equivalent roughness k is 1 mma. Calculate the discharge in the sewer pipe. Assume the flow is turbulent.b. Check if the flow is turbulent.c. Suppose the intensity of the rainfall is 60 l/s.ha. Calculate the paved area which isconnected to the sewer. As discharge use the result from question a.d. Explain why the design of this sewer is actually bad.Water level + 2 mSurface level +3 mRainfall, discharge???60 mD=500 mmManhole(inspectieput)b turbulent]1[4ReRvw =1275000 > 4000 ja𝜐 = 1,00 ∙ 10−6
10. 10. Question 4In the picture above you see a sewer pipe designed for the discharge of rainfall directlyto surface water. The water level surface water is +2 m, the velocity is low. The waterlevel in the manhole is +2,95 m. The equivalent roughness k is 1 mma. Calculate the discharge in the sewer pipe. Assume the flow is turbulent.b. Check if the flow is turbulent.c. Suppose the intensity of the rainfall is 60 l/s.ha. Calculate the paved area which isconnected to the sewer. As discharge use the result from question a.d. Explain why the design of this sewer is actually bad.Water level + 2 mSurface level +3 mRainfall, discharge???60 mD=500 mmManhole(inspectieput)Onderdeel cQ=500 l/s 500/60 = 8,33 haOnderdeel dHet riool staat altijd vol water.