Report

Inlet and Outlet Manifolds and Plant Hydraulics In which Kinetic Energy BECOMES SIGNIFICANT (Thanks to A.A. Milne) Nomenclature: a start Symbol Description Sub Q Flow P Port A Area M Manifold H Piezometric head D Diffuser hL Total Head Loss HGL Hydraulic Grade Line EGL Energy Grade Line Cp Pressure Coefficient (includes shear and expansion effects) Pvc Area of the vena contracta divided by the orifice area = 0.62 D Diameter P Dimensionless ratio n Number of ports The Problem • How can we deliver water uniformly • between sed tanks • into the bottom of the sedimentation tank and • between StaRS layers • Within layers of the StaRS? • Extract water uniformly • from above the plate settlers and • How can we make it so that the water doesn’t preferentially take the easy path? 1 2 n-1 n • Draw manifold in lake picture • Define Pi.Q= Qp1/Qpn • Define H average (=vjet^2/2g) and deltaH • • • • • (vpipe^2/2g) showing manometers Note that shape of inlet (pitot tube) matters H proportional to 1/A^2 (from orifice eq) How do you get Pi.Q = 1? Hbar> dH Ajet<Apipe How can we make water choose equally between several paths? • Draw a manifold with ports that you think would give unequal flow • Draw a manifold with ports that you think would give equal flow • What do you think is important? 1 2 n-1 n Will the flow be the same? NO! An example to illustrate the concepts Dh Long K=1 p1 V12 p2 V22 z1 z2 hL g 2g g 2g K=0.5 K=0.2 Short K=1 Head loss for long route = head loss for short route if KE is ignored Q for long route< Q for short route Flow Division Analysis V2 hl C p 2g hl C p C pShort Q L V2 hf f D 2g L Cp K f D 2 8Q g 2 D 4 hl C p Q 2 Short V2 he K 2g Short path 2 hl C pShort QShort 2 C pLong Q 2 Long PQ Long path 2 hl C pLong QLong QLong QShort C pShort C pLong How did the flow divide? Long PQ K=1 Improve this? Short K=0.2 PQ QLong QShort C pShort C pLong K Control C pShort K Control C pLong K Control 2 PQ C pLong C pShort Set PQ to 0.95 2 1 PQ 2 0.2 0.26 3 K.Control P.Q C.P Long C.P Short 2 1 P.Q 25.718 Plant Flow Distribution • Equal flow between sed tank bays? • Equal flow through diffusers into sed tank? • Equal flow between plate settlers? • Equal flow through ports into sludge drain Where can we use flow restrictions? ___________________ After flocs are removed Terminology • Flow into tank (out of manifold) – Inlet Manifold • Flow out of tank (into manifold) – Outlet launder • Overflow Weir • Submerged pipe with orifices (head loss through orifices is set to be large relative to construction error in level of weir Ease of construction, avoid floating flocs Manifold: Flow Calculations • We will derive equations in terms of Hydraulic • • • p z g Grade Line (HGL) because piezometric head controls the port flow 2 a p g V Port flow • based on orifice _______ equation Piezometric head change (DH) across port • flow expansion In manifold Piezometric head change (DH) between ports • Darcy-Weisbach and Swamee-Jain Head Loss due to Sudden Expansion 2 pin pout Vin2 Vout Energy hex g 2g 2 2 Ain Vout Vin Aout Momentum pin pout g g hex 2 2 Vout V V 2 out 2 in 2 Vin Vin2 Vout 2g 2g hex Ain Vout Mass A V out in 2 Vout 2VinVout Vin2 2g Ain A in Vin Vout K ex 1 hex 1 hex A Aout out 2g Discharge into a reservoir?_________ Kex=1 2 2 in V 2g 2 2 Inlet Manifold Vin Vout Vout pout pin g g pin pout g DH expansion EGL 2 out V V 2 in Ain Aout g Vin Vout Vout g Pressure recovery HGL 1 Major head loss 2 n-1 n What is total SDHexpansion as a function of n? DH expansion n 1 DH i 1 Vin Vout Vout g QM n i QM 1 nAM g i 1 nAM n1 expansion n 1 DH expansion i 1 n 1 DH i 1 expansion 2 n 1 M V g n i i 1 VM2 n 1 g 2n n2 Vin Vout QP AM QP Vouti QM n n i QM nAM VM2 Approaches for large n 2g _______________ All kinetic energy is recovered for very gradual expansion. Outlet Manifold (Launder) Flow contractions, thus no significant minor loss! EGL HGL VM2 All of the changes at the ports sum to 2g 1 2 n-1 n Head Loss in a Manifold (same for inlet or outlet) between first and last ports L V2 hf f D 2g Define manifold length as 2 LP VM i hfi f i DM 2 g VM i QM nAM n i LM LP n 1 EGL HGL 1 LM 1 QM h f fi i DM 2 g AM i 1 n 1 n 1 n i i 1 n 1 hfi fi i 1 2 2 1 n 1 n 2 n n 1 2n 1 n i 2 2n 1 n-1 n 2 i 1 n1 1 2 ( n 1) n i 1 6 LM 1 QM DM 2 g AM n 1 2 2 ( n i) simplify 1 3 1 6 n EGL HGL 1 2 n-1 6n Head loss in a manifold is 1/3 __ of the head loss with constant Q. n EGL Change in Piezometric Head in an Outlet Manifold LM 1 QM hfi f i DM 2 g AM i 1 n 2 DH total 2 6n 2n 1 1 6n 1 2 n-1 VM2 2g 2n 1 1 QM LM fi 2 g AM DM HGL Total change in piezometric head C plong Note: We have factored out the friction factor knowing 0.25 that f and thus f is not constant 2 5.74 log 3.7 D Re0.9 n EGL HGL Change in Piezometric Head in an Inlet Manifold LM 1 QM hfi f i DM 2 g AM i 1 n 2 DH total 2 1 n 1 DH 2n 1 i 1 6n LM 1 QM n 1 f i 2 g AM n DM 2 expansion n-1 VM2 n 1 2g n 2n 1 6n C plong This equation gives the difference in piezometric head between the first port and the last port. Since the two terms have opposite signs the maximum difference could be at an intermediate port. We need to determine if one of these terms dominates to see if the maximum difference really is between the first and last ports. n Calculating the Control (Orifice) Pressure Coefficients 0 K Control 2 PQ C plong C pshort 2 1 PQ 2 PQ C plong K Control 2 1 PQ EGL EGL HGL HGL 1 C plong For a manifold the short path head loss is zero (not including the flow control head loss) 2 n-1 n LM 2n 1 1 fi DM 6n 1 2 n-1 n n 1 LM 2n 1 f n D 6 n M Minor Loss Coefficient for an Orifice Port (in or out) heP K eP Ke has a value of 1 for an exit and is close to 1 for an entrance nQP QM Vvc2 2g But this V is the vena contracta velocity. The control coefficient analysis normalizes everything to the maximum velocity in the manifold. So let’s get the velocity ratio nP vc DP2 VM QM AP P vc 2 Vvc AM QP DM heP K eP 2 D VM2 2 n P D vc P 2g 2 M K Control Vvc VM KControl 2 DM nP vc DP2 D K eP 2 n P D vc P 2 M 2 Solution Path • The length of the manifold will be determined by the plant geometry • The spacing of the ports will be set by other constraints • We need to determine the diameter of the manifold and the diameter of the ports EGL Launder: Traditional Design Guidelines • Recommended port velocity is 0.46 to 0.76 HGL 1 2 n-1 m/s (Water Treatment Plant Design 4th edition page 7.28) • The corresponding head loss is 3 to 8 cm • • through the orifices How do you design the diameter of the launder? (coming up…) Would this work if head loss through the manifold were an additional 10 cm? _____ NO! Q P vc AOrifice 2 g Dh 1 VPort Dh 2 g P vc 2 n Design Constraints • For sed tank Inlet Manifold the port velocities • and the manifold diameter are set by the _____________________________________ energy dissipation rate in the flocculator For the launder that takes clear water from the top of the sed tank bays the goal will be to keep head loss low and greater than construction errors in level of weir (we aim for about 5 cm) • For Outlet Manifold that takes sludge from the bottom of the sed tank bays the goal is to be able to drain the tanks in a reasonable length of time (perhaps 30 minutes) (this means that the initial flow rate would be able to drain the tank in 15 minutes: remember the hole in a bucket analysis) EGL Design for Outlet Launder HGL 1 2 n-1 n • Given target head loss between sed tank and clear water channel (5 cm for AguaClara) 8QM 2 CPTotal hl g 2 DM 4 DM 8QM 2 CPTotal 2 g h l CPTotal CPLong K Control Minor loss equation 1 4 Solve the minor loss 0 equation for the manifold diameter 2 PQ C pLong C pShort K Control 2 1 PQ 2 PQ C pLong K Control 2 1 PQ Outlet Launder Diameter: Iterative EGL HGL 1 solution for DM CPTotal CPLong K Control CPTotal C pLong f C pLong DM P Q2 1 2 1 P Q C pLong 2 1 P Q K Control 2 n-1 2 PQ C pLong 2 1 PQ 0.25 5.74 log 3.7 D Re0.9 2 LM 2n 1 f 1 DM 6n 8QM 2 C pLong g 2 h 1 P 2 l Q 1 4 The iterative solution will converge quickly because f varies slowly with Re. n Example Code for Iteration y0 First guess at solution Error ← 1 MaxError ← _____ While Error > MaxError Set error to be large to ensure that loop executes once a f y0 y1 f a Error y0 y1 Return y1 y0 y1 y0 y1 Improved guess Dimensionless error Launder Diameter (Approximate Solution) 8QM 2 C pLong DM 2 g h 1 P 2 l Q 8QM 2 1 DM 2 g h 1 P 2 l Q 1 4 1 4 In this equation the head loss is the total head loss for both the orifices and the pipe flow C pLong LM 2n 1 f 1 DM 6n Here we are omitting the major (wall shear) head loss contribution Example: Launder What is the minimum launder diameter for a plant flow rate of 50 L/s divided between 8 bays if we use 5 cm of head loss? For an approximate solution you can omit the effect of the major losses. Use a value of 0.8 for the minimum flow ratio between the last and first orifice PQ 0 .8 8QM 2 1 DM g 2 h 1 P 2 l Q 1 4 50 Q 8 h l 5cm L s 6 .25 L s 1 4 8 Q2 1 1 1.6cm DM g 2 h 1 P 2 l Q Example: Launder C pLong LM 2n 1 f 1 DM 6n • What is the effect of the shear force? • How can we estimate the length of the launder? We will assume that the sed bay has a width of 1 m. • What is the length of the sedimentation tank? V↑ = 1 mm/s 3 0.05 m s LSed 8 m 1m 0 .00 1 s 6 .25m Example: Launder C pLong LM 2n 1 f 1 DM 6n • n is the number of orifices (ports). If the port spacing is 10 cm how many are there? 62 2n 1 6n 1 3 LM 2n 1 f DM 6n C pLong For large n 6 .25m 1 0 .02 0 .35 9 0 .11 6m 3 LM 2n 1 f 1 1.36 DM 6n f 0.25 5.74 log 3.7 D Re0.9 2 More exact solution… 8QM 2 C pLong DM 2 g h 1 P 2 l Q 0 .02 8QM C pLong DM 2 g h 1 P 2 l Q 2 6 .94m 1 0 .42 1 0 .11m 3 1 4 1 What diameter launder do you recommend? 6 inches 4 8 Q2 1 .36 DM 1 1.8cm g 2 h 1 P 2 l Q 1 4 Why is the launder diameter so large? • (50L/s /9) launder of 6 inches • The head loss in the launder is small and it would be • • • tempting to use a smaller pipe Why is such a large pipe necessary?______________ Equal orifice flow Why do we even need a launder pipe? For uniform flow distribution between (and within) ___________________________________________ plate settlers ___________ What is the max velocity above the plate settlers given a 1 m wide tank, 25 cm of water above the plates, a single launder? __________ 2 mm/s What is the horizontal velocity above the plate settlers without a launder? HSup er 2 5cm mm VU p 1 s LSed 6m LSed HSup er LSed VU p 24 mm VSup er 24 HSup er s 2 VSuper 2g 29m This velocity is very large compared with the head loss through the plate settlers (about 1 m) and thus elimination of the launder would result in preferential flow through the plate settlers closest to the exit Approach to Find Port Diameter EGL HGL 1 • Calculate the head loss in L 1 QM hf f M DM 2 g AM the manifold • Subtract 50% of that head loss from the target head loss (5 cm) to estimate the port head loss • Calculate the port diameter directly using the orifice equation 2 2 n-1 2n 1 6n QP AP P vc 2 g Dh Dorificio 4Q P vc 2 g Dh n What about Inlet Manifold Design? EGL HGL 1 2 n-1 • Total head loss is not a constraint (it will be VERY small) • Energy dissipation rate at the inlet of the manifold determines the manifold diameter • Energy dissipation rate at the inlet to the diffuser pipes will set the diffuser diameter • Available pipe sizes for inlet manifold and for the diffusers is a constraint n Schulz and Okun guidelines: Note these cause floc breakup! • VPort = 0.2 to 0.3 m/s (assumes no diffusers) • “The velocity through the ports should be 4x higher than any approaching velocities.” (but to prevent sedimentation approach velocities need to be 0.15 m/s which would give velocities of 0.6 m/s!) These guidelines result in extremely high energy dissipation rates! Max VPort Jet P vc DPort P vc 3 Schulz and Okun famous quote… “In practice, one can rarely meet all four basic requirements because they conflict with one another; thus a reasonable compromise must be attained.” Conclusion of inlet design for sedimentation tanks. Page 135 in Surface Water Treatment for Communities in Developing Countries Flow Distribution Equation for Inlet Manifold C pLong EGL HGL 1 n 1 LM 2n 1 f n D 6 n M K Control 0 PQ n-1 D K eP 2 n P D vc D 2 M n 2 Control resistance by orifice C pShort K Control C pLong K Control 2 PQ 2 2 DM K eP 2 nP vc DD 2 2 n 1 DM LM 2n 1 f K eP 2 n D 6 n n P D M vc D DM2 VD 2 nP vc DD VM n 1 n ? f LM DM What can we play with to get a better flow distribution? 2n 1 6n Area ratio if the DM and DD cause the same Max 6 7 AM n AD 1 7 1 Area ratio to achieve equal ε.Max 4 Jet QM 1 7 3 6 Max P vc AM 6 AP 7 4 Jet QM n 1 7 3 6 n Max P vc 0.9 0.8 0.7 0.6 0.5 0 20 40 60 80 100 Number of ports per manifold But apparently energy dissipation rate doesn’t matter! EGL Importance of Area A A Ratio HGL 1 M 2 n-1 n Ratio of actual port flow to average port flow D 0 .55 0 .6 0 .65 0 .7 1.2 Area Area Area Area 1.1 Ratio Ratio Ratio Ratio of of of of 0.55 0.6 0.65 0.7 AM AD 7 6 6 36 2 0 1 2 ports Effect of pressure recovery 1 0.9 Max 0.8 0.7 0 0.2 0.4 0.6 0.8 Normalized distance along man ifold 1 3 JetVJet DJet 1 .2 1 .72 8 3 One more Issue: Vena Contracta with High Velocity Manifold • The vena contracta at each port must be much more pronounced (small Pvc) when the velocity inside the manifold is high. • If the vena contracta, Pvc, is smaller, then the velocities are higher and the energy dissipation rate is higher. • This requires further investigation Manifold Conclusions • Outlet manifolds (launder) require an iterative • design to get the manifold diameter Inlet manifold design has complex constraints… • Avoid breaking flocs • Don’t let flocs settle (ignore if ports are on bottom) • Distribute flow uniformly • Eliminate horizontal velocity in the sed tank • Produce jets to resuspend flocs to form floc blanket Pérdida de Carga Acu 20 Head loss in an AguaClara Plant 10 50 Pérdida de Cargahead Acumulada (cm)(cm) loss Cumulative Pérdida de Carga Acumulada (cm) 0 0.01 40 40 30 30 20 20 0.495 1 Orificio de laOrifice Mezcla Rápida Rapid Mix Tubo la Mezcla RapiddeMix Pipe Rápida Floculador Flocculator Launder Tubo de Recolección Settled water weir Decantada Vertedero de Agua • Why isn’t there much head loss between 10 10 • 0 0 0.01 0.01 the flocculator and the launder pipe? How do we ensure that the flow divides equally between sedimentation tanks? 0.495 1 0.495 1 10 50 Orificio de la Mezcla Rápida Orificio de la Mezcla Rápida L/sTubo de la Mezcla Rápida Settled Water Weir: Controls the Plant Level H is water level measured from the top of the weir 2 P vcW 2 g H 3/2 With a maximum H of 5 3 cm the sedimentation tank water level can change a 3 Q W total of 10 cm! Launders 3/2 2 P vc 2 g H have 5 cm of head loss also. Q 3 3 W 2 0.62 m 0.05 s m 3/2 2 9.8 2 0.05m s 3 Q 0 .05 m s H 5cm W 3 Q 2 3 Kv c 2g H 2 2 .44 3m Hydraulic Conclusions • The water level in the plant is set by the settled • • • • water weir The most significant head loss in the sedimentation tank is the orifices in the launder The water level increases through the flocculator. The entrance tank water level is significantly higher than the flocculator due to head loss in the rapid mix orifice The stock tanks have to be even higher to be able to flow by gravity thru the chemical doser and into the entrance tank.