CIE4491 Lecture. Hydraulic design

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1 CIE4491 Lecture. Hydraulic design Marie-claire ten Veldhuis Delft University of Technology Challenge the future

2 Hydraulic design of urban stormwater systems Focus on sewer pipes Pressurized and part-full pipe flow Saint-Venant equations and its simplifications Differences between part-full and pipe flow Part-full flow in pipes Local losses in sewer networks Orifice Gully pot, manhole Hydraulic jump Weir

3 Stormwater systems design steps Quantify stormwater design flow rate Run-off model, rational method Determine pipe capacity based on full pipe flow Computationally easier Max Capacity is reached at 85% filling rate Actual capacity is only 5% greater than for fully pipe, but not stable Determine required D h Select standard D > D h D h = 4A Ω = 4R h Account for full pipe flow condition at maximum capacity Account for part-full flow for all other conditions (esp. combined systems dry weather flow) 3

4 De Saint-Venant momentum equation Q Q yz t g A W t x A x r I III IV II Q, A, z, W, t, r are discharge, flow area, bottom elevation, wet perimeter, wall shear stress and density. y is the water depth (part-full flow) or pressure head (pipe flow) 0 Applicable for part-full flow and pressurized pipe flow! It comprises 4 parts: I. Advective acceleration (inertia) II. Convective acceleration (changes in flow cross-section) III. Gravity force IV. Friction force 4

5 De Saint-Venant continuity equation y B t Q x 0 B is the surface width, y is water depth (part-full flow) or pressure head (pressurised flow) Part full flow: Internal storage at the free surface. Each element behaves like a storage tank with surface area B*Dx. The net discharge DQ is stored in the element. Full pipe flow: B = 0, pipe elasticity and liquid compressibility are neglected Q 0 x 5

6 Typical applications of De Saint Venant terms momentum equations a. Rainfall event (sewer network filling and draining) b. Rapid filling event at maximum hydraulic grade line c. Steady varied flow (in part-full pipe) d. Steady full pipe flow e. Part-full flow at normal depth 6

7 Typical applications of De Saint Venant terms momentum equations a. Rainfall event (sewer network filling and draining) b. Rapid filling event at maximum hydraulic grade line Complete De Saint-Venant equations: Terms I, II, II, IV Terms I, III c. Steady varied flow (in partfull pipe) ga W 0 Terms II, III, IV Q yz t x A x r d. Steady full pipe flow y t Terms III, IV ga W 0, y D x r e. Part-full flow at normal depth Q yz ga 0 t x z t ga W x r 0 Terms III, IV 7

8 Typical applications of De Saint Venant terms Rainfall event (sewer network filling and draining) Rapid filling event at maximum hydraulic grade line Steady varied flow (in part-full pipe) Steady full pipe flow Independent of bottom slope! Part-full flow at normal depth Friction slope=bottom slope! Complete De Saint-Venant equations Q yz ga 0 t x Q yz t ga W 0 x A x r y t ga W 0, y D x r z t ga W 0 x r 8

9 Friction term relations Colebrook-White/Darcy-Weisbach Typical values l = Darcy/Fanning friction factor (UK) Typical values f = Chezy friction factor Typical values C = Manning Typical values n = (SI units) n is dependent of units used L v DH = l. D g L v D H = f. f 1. R g 4 l h D v C R H 8g h C L l v h n R D h = 4A Ω = 4R h h DH L 9

10 Wall shear stress correlations (turbulent conditions) Colebrook-White Darcy/Fanning Chezy l 1,51 k t rv ; = - log + 8 l Re l 3,71D f t rv v t rg C Manning vn t rg R 1/3 h 10

11 Moody diagram (Colebrook-White) 11

12 Hydraulic sewer design Given a sewer pipeline with the following dimensions: Q=A*I L 1- :00m, D 1- :600mm Connected surface area: ha (70% impervious) Design rainfall intensity: 90 l/s/ha 1 1

13 Hydraulic sewer design Question: what is the design discharge? Q=A*I L 1- :00m, D 1- :600mm Connected surface area: ha (70% impervious) Design rainfall intensity: 90 l/s/ha 13

14 Hydraulic sewer design Answer: Q=A*I*C runoff =*0.7*90=16 l/s or 0.16m 3 /s Q=A*I L 1- :00m, D 1- :600mm Connected surface area: ha (70% impervious) Design rainfall intensity: 90 l/s/ha 14

15 Hydraulic sewer design Question: Calculate the energy levels in nodes 1 and? Q=A*I L 1- :00m, D 1- :600mm 1 Connected surface area: ha (70% impervious) Design rainfall intensity: 90 l/s/ha 15

16 Hydraulic sewer design Question: energy levels in nodes 1 and? Q=A*I +mnap +0.4mNAP L 1- :00m, D 1- :600mm 1 +1mNAP Connected surface area: ha (70% impervious) Design rainfall intensity: 90 l/s/ha Additional data: Groundlevel: +mnap; Waterlevel at outflow: +1mNAP 16

17 Hydraulic sewer design Answer: energy levels in nodes 1 and Q=A*I +mnap +0.4mNAP L 1- :00m, D 1- :600mm 1 1mNAP Note: pipe outflow below surface water level Flow condition: full pipe, pressurised flow Apply Darcy Weisbach: dh = λ L D g (De Saint-Venant terms III and IV: gravity and friction) v ; l=0.0[-]; A=0.8m 17

18 Hydraulic sewer design Answer: energy levels in nodes 1 and Q=A*I +mnap +0.4mNAP L 1- :00m, D 1- :600mm 1 1mNAP Apply Darcy Weisbach: dh = λ L D dh=0,07m Energy levels:? v g ; l=0.0[-]; A=0.8m ; v=0,45m/s 18

19 Hydraulic sewer design Answer: energy levels in nodes 1 and Q=A*I +0.4mNAP +1.07mNAP L 1- :00m, D 1- :600mm 1 +mnap +1mNAP Energy loss according to Darcy Weisbach: dh=0,07m Energy levels: node 1: H 1 =+1mNAP; node : H =+1.07m 19

20 Hydraulic sewer design NB: we have not defined a bottom slope! Is this relevant? Q=A*I +1.07mNAP L 1- :00m, D 1- :600mm +mnap +0.4mNAP S b? 1 +1mNAP Energy loss according to Darcy Weisbach: dh=0,07m Energy levels: node 1: H 1 =+1mNAP; node : H =+1.07m 0

21 Hydraulic sewer design, part-full flow Question: What is the expected water depth in the sewer? Given following conditions : Flow rate = 640 m3/h Sewer diameter = 500 mm Slope = 1:00 Friction Factor =

22 Hydraulic sewer design, part-full flow Question: What is the expected water depth in the sewer? Given following conditions : Flow rate = 640 m3/h Sewer diameter = 500 mm Slope = 1:00 Friction Factor = Check: flow rate for full pipe flow conditions, assuming flow at normal depth

23 Hydraulic sewer design, part-full flow Answer: step 1, determine full-pipe flow Given following conditions : Flow rate = 640 m3/h Sewer diameter = 500 mm Slope = 1:00 Friction Factor = dh L = λ v gd From Darcy-Weisbach, assuming flow at normal depth (hydraulic gradient equals bottom slope 1:00): v full = 1.81 m/s; Q full = 180 m3/h Note: part-full flow conditions for given flow rate (640 m3/h) 3

24 Part-full geometric formulae Water depth y, angle a Width water surface, B Wet area, A Wetted perimeter, W Hydraulic diameter, D h cosa 1 y R sin y y B R a R R R B A ar R y W a R 4A Dh W A y a 4

25 Part-full vs full pipe geometric relations Wet area ratio Wetted perimeter ratio A 1 cos 1 y B y 1 1 Af R D R W 1 cos 1 y 1 W R f Hydraulic diameter ratio R R h h, f AA WW f f A y a 5

26 y/r Part-full vs full pipe flow W/W f A/A f R h /R h,f B/D

27 Part-full vs full pipe flow properties Assuming: Constant dh/dx equal to bottom slope Normal flow in part-full situation Constant pipe friction factor Velocity ratio Discharge ratio v v f R R h h, f Q A R Q A R h f f h, f A y a Reynolds number ratio Re Re R R h R R h f h, f h, f 7

28 y/r Part-full vs pipe flow properties Re/Re f v/v f Q/Q f

29 Hydraulic sewer design, part-full flow Answer: step, determine filling rate Q/Q f Given following conditions : Flow rate = 640 m3/h Sewer diameter = 500 mm Slope = 1:00 Friction Factor = dh L = λ v gd From Darcy-Weisbach, assuming flow at normal depth (hydraulic gradient equals bottom slope 1:00): v full = 1.81 m/s; Q full = 180 m3/h Q / Q f = 0.5. Pipe is half full - use: Q/Q f in previous slides 9

30 y/r Hydraulic sewer design, part-full flow Answer: step 3, determine flow depth in sewer Pipe is half full: Flow depth = 0.5D = 0.5 m Q/Q f v/v f Re/Re f Flow velocity: v full = 1.81 m/s; Q full = 180 m3/h Q / Q f = 0.5. v / v f = 1; v = 1.81 m/s

31 Hydraulic sewer design, special structures (weirs, valves etc.) Specific energy E: water depth + velocity head Specific energy E defined for free-surface flows Specific energy is energy relative to the bottom/bed E v g y 31

32 Relevance of specific energy E drives steady, varied flow equation Bottom slope -S b v y z t W 0 x g x x r ga x v g y Sb S f Friction slope S f At a given discharge, E is minimum at the critical depth At a given E, the discharge is maximum at the critical depth E y Q y y c y c 0 0 3

33 E in rectangular and circular conduit E y y c 0 de Q da dy ga dy A c = y c B c Rectangular conduit v c gy c Fr 1 c Circular conduit v c A g B c c E min 3 y c E min A c B c y c 33

34 Shooting Flow, Tranquil Flow Critical depth marks transition from tranquil to shooting flow At critical depth Fr = 1 If y >y c then: tranquil flow or subcritical flow Information can be propagated upstream and downstream If y < y c then: shooting flow or supercritical flow Information can only be propagated downstream 34

35 Varied flow profiles (mild slope) Mild slope: y n > y c M1 backwater curve (during filling of pump pit and sewers) M drawdown curve (during dry-weather-flow towards pit) normal depth M1 M3 M critical depth 35

36 Varied flow (steep slopes) Steep slope y n < y c S1 approaches the horizontal S drawdown curve in steep slope (transition mild steep) S3 may occur downstream of a gate valve in a steep slope critical depth S S1 S3 normal depth 36

37 Weir h 1 Water levels h 1 always relative to weir crest short crested weir Broad-crested weir (critical) Broad-crested weir (subcritical) 3/ Q b g h1 3 Q 1.7 bc h, h h 3 Q bc h g h h, h h 3 3/

38 Weir Depth at control point determines discharge (critical flow) control point M S 38

39 Manhole Entrance point Multiple pipes connected Vertical transport Typical head losses (for smooth bottom dz = 0) Inflow from manhole Outflow from pipe into manhole h L V 0.5 g h L V g 39

40 Local losses in sewer networks Local losses are NOT minor losses in sewer networks General concept of deceleration losses No energy loss in accelerating flow Bernoulli The energy is lost in decelerating flow Momentum balance 40

41 Example: contribution of local losses Question: What is the relative contribution from the local manhole losses to the total head loss? If: 1 manhole every 50 metres; local loss factor k = 1 Sewer diameter 500 mm, flow velocity 1 m/s friction factor

42 Olofsbuurt Design assignment, this week: - Define design requirements: Return period T for Westerkwartier flooding and for critical rainfall - Choose applicable IDF-curve - Determine catchment areas per pipe segment - Determine runoff coefficients for catchment areas - Quantify wastewater flow based on population and industry - Quantify stormwater flow applying rational method - Determine Poptahof dimensions of sewer pipes (and weirs etc.) - Start calculation hydraulic gradients for stormwater systems 47

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