### Cu06997 lecture 8_exercise

1. Combined sewer / gemengd rioolstelsel 50 m Ø300 PVC Ø500 concrete Ø250 PVC pump P4 P3 P2 GL (ground level) +6.00 m IL +4,00 m IL +3,90 m IL +3,73 m Rain water Waste water Rain Waste +5,5 m IL (Invert level) +3,53 m P1 1
2. Length Chezy formula Chezy formula describes the mean velocity of uniform, turbulent flow ΔH 𝑉 = 𝐶 ∙ 𝑅 ∙ 𝑆𝑓 𝑉 = Mean Fluid Velocity [m/s] R = Hydraulic Radius [m] 𝑆𝑓 = Hydraulic gradient [1] 𝐶 = 8𝑔 𝜆 Chezy coefficient [m1/2/s] 𝑆𝑓 = ΔH 𝐿 3 Total Head Pressure Head
3. Head loss sewer pipe 𝑉 = 𝐶 ∙ 𝑅 ∙ 𝑆𝑓 𝑄 = 𝑉 ∙ 𝐴 𝑆𝑓 = 𝑖 = ∆𝐻 𝐿 Combine ∆𝐻 = 𝐿 𝑄2 𝐶2 ∙ 𝑅ℎ ∙ 𝐴 𝑠 2 ∆𝐻 = Head Loss, energy loss [m] Q = discharge pipe [m3/s] L = length of the pipe [m] C = Chezy coefficient [m1/2/s] R = Hydraulic Radius [m] A = Wetted Area, flow surface [m2] Sf ,i = slope of hydraulic gradient [-] 3
4. Question 1 50 m Ø300 PVC Ø500 concrete Ø250 PVC Pump P4 P3 P2 GL +6.00 m IL +4,00 m IL +3,90 m IL +3,73 m Rain Waste Rain Waste +5,5 m IL +3,53 m P1 5
5. Question 2 50 m Ø300 PVC Ø500 concrete Ø250 PVC Pump P4 P3 P2 GL +6.00 m IL +4,00 m IL +3,90 m IL +3,73 m Rain=0 Waste=10l/s Rain=0 Waste=10l/s +5,5 m IL +3,53 m Q=20 l/s Q=10 l/s P1 5
6. Partially filled pipe 𝐼𝑛𝑝𝑢𝑡: 𝑄 𝑝𝑎𝑟𝑡 𝑄 𝑓𝑢𝑙𝑙 = 0,17 𝑂𝑢𝑡𝑝𝑢𝑡: ℎ 𝐷 = 0,27 𝑂𝑢𝑡𝑝𝑢𝑡: 𝑢 𝑝𝑎𝑟𝑡 𝑢 𝑓𝑢𝑙𝑙 = 0,75 5
7. Table 5
8. Question 3 50 m Ø300 PVC Ø500 concrete Ø250 PVC Pump P4 P3 P2 GL +6.00 m IL +4,00 m IL +3,90 m IL +3,73 m Rain=1,1 ha Waste=10 l/s Rain=3,75 ha Waste=10 l/s +5,5 m IL +3,53 m P1 5
9. Question 3c 50 m Ø300 PVC Ø500 beton Ø250 PVC Pump P4 P3 P2 GL +6.00 m Rain=66 l/s Waste=10 l/s Rain=225 l/s Waste=10 l/s +5,5 m Q=66 l/s Q=291 l/s P1 In example m = 1,85 Q=0 l/s
10. s/m1/3
11. Strategy [situation with overflow] Preparation Information available for each pipe - Diameter, R, L, k, C - Discharge and Velocity Information Overflow / weir - Width, m - Discharge - Level crest in m N.A.P. 𝐶 = 18 ∙ 𝑙𝑜𝑔 12𝑅 𝑘 5
12. 1. Calculate H at weir 2. Calculate ∆H each pipe 3. Water level at weir (P1) = level crest weir + H at weir 4. Water level at P2 = Water level at weir + ∆Hweir(p1) – p2 5. Water level at P3 = Water level at P2 + ∆H p2– p3 6. Water level at P4 = Water level at P3 + ∆H p3– p4 Strategy [situation with overflow] Steps All levels in m N.A.P. 𝑄 = 𝑚 ∙ 𝐵 ∙ 𝐻 3 2 ∆𝐻 = 𝐿 𝑄2 𝐶2 ∙ 𝑅ℎ ∙ 𝐴 𝑠 2 5
13. Little error or inaccuracy: Distance between H en y is velocity head. At manhole (bigger than sewer pipe) we assume velocity head is = 0