TOPIC T2: FLOW IN PIPES AND CHANNELS AUTUMN 2013

Transcription

1 TOPIC T: FLOW IN PIPES AND CHANNELS AUTUMN 03 Objectives () Calculate te friction factor for a pipe using te Colebrook-Wite equation. () Undertake ead loss, discarge and sizing calculations for single pipelines. (3) Use ead-loss vs discarge relationsips to calculate flow in pipe networks. (4) Relate normal dept to discarge for uniform flow in open cannels.. Pipe flow. Introduction. Governing equations for circular pipes.3 Laminar pipe flow.4 Turbulent pipe flow.5 Expressions for te Darcy friction factor, λ.6 Oter losses.7 Pipeline calculations.8 Energy and ydraulic grade lines.9 Simple pipe networks.0 Complex pipe networks (optional). Open-cannel flow. Normal flow. Hydraulic radius and te drag law.3 Friction laws Cézy and Manning s formulae.4 Open-cannel flow calculations.5 Conveyance.6 Optimal sape of cross-section Appendix References Cadwick and Morfett (03) Capters 4, 5 Hamill (0) Capters 6, 8 Wite (0) Capters 6, 0 (note: uses f = 4c f = λ for friction factor ) Massey (0) Capters 6, 7 (note: uses f = c f = λ/4 for friction factor ) Hydraulics T- David Apsley

2 . PIPE FLOW. Introduction Te flow of water, oil, air and gas in pipes is of great importance to engineers. In particular, te design of distribution systems depends on te relationsip between discarge (Q), diameter (D) and available ead (). Flow Regimes: Laminar or Turbulent In 883, Osborne Reynolds demonstrated te occurrence of two regimes of flow laminar or turbulent according to te size of a dimensionless parameter later named te Reynolds number. Te conventional definition for round pipes is VD Re () ν were V = average velocity (= Q/A) D = diameter ν = kinematic viscosity (= μ/ρ) laminar turbulent For smoot-walled pipes te critical Reynolds number at wic transition between laminar and turbulent regimes occurs is usually taken as Re crit 300 () In practice, transition from intermittent to fully-turbulent flow occurs over 000 < Re < Development Lengt At inflow, te velocity profile is often nearly uniform. A boundary layer develops on te pipe wall because of friction. Tis grows wit distance until it fills te cross-section. Beyond tis distance te velocity profile becomes fully-developed (i.e., doesn t cange any furter wit downstream distance). Typical correlations for tis development lengt are (from Wite, 0): L dev 0.06 Re (laminar ) /6 (3) D 4.4 Re (turbulent) Te kinematic viscosity of air and water is suc tat most pipe flows in civil engineering ave ig Reynolds numbers, are fully turbulent, and ave a negligible development lengt. Example. ν water =.00 6 m s. Calculate te Reynolds numbers for average velocity 0.5 m s in pipes of inside diameter mm and 0.3 m. Estimate te development lengt in eac case. Answer: Re = 6000 and.50 5 ; L develop = 0.3 m and 9.6 m. Hydraulics T- David Apsley

3 . Governing Equations For Circular Pipes Fully-developed pipe flow is determined by a balance between tree forces: pressure; weigt (component along te pipe axis); friction. For a circular pipe of radius R, consider te forces wit components along te pipe axis for an internal cylindrical fluid element of radius r < R and lengt Δl. l p r z direction of flow mg p+p Note: () p is te average pressure over a cross-section; for circular pipes tis is equal to te centreline pressure, wit equal and opposite ydrostatic variations above and below. () Te arrow drawn for stress indicates its conventional positive direction, corresponding to te stress exerted by te outer on te inner fluid. In tis instance te inner fluid moves faster so tat, if V is positive, τ will actually be negative. Balancing forces along te pipe axis: p (πr ) ( p Δp)(πr ) mg sin θ net pressure force weigt τ(πr Δ l) 0 friction From te geometry, Δz m ρπr Δl, sin θ Δl Hence: Δp (πr ) ρπr g Δz τ(πr Δ l) 0 Dividing by te volume, πr Δl, Δ( p ρgz) τ 0 Δl r Writing p * = p + ρgz for te piezometric pressure and rearranging for te sear stress, Δp* τ r (4) Δ l Since te flow is fully-developed te sear stress and te gradient of te piezometric pressure are independent of distance. For convenience write G for te streamwise pressure gradient: Δp * dp * G ( constant ) (5) Δl dl (Te negative sign is included because we expect p* to drop along te pipe.) Hence, from (4), Hydraulics T-3 David Apsley

4 were τ Gr (6) d p pressure drop ρg f G * (7) dl lengt L G is te piezometric pressure gradient and f is te ead lost (by friction) over lengt L. (6) applies to any fully-developed pipe flow, irrespective of weter it is laminar or turbulent. For laminar flow it can be used to establis te velocity profile, because τ can be related to te velocity gradient du/dr (Section.3). For turbulent flow an analytical velocity profile is not available, but gross parameters suc as quantity of flow and ead loss may be obtained if te wall sear stress τ w can be related empirically to te dynamic pressure ½ρV (Section.4)..3 Laminar Pipe Flow Laminar flow troug a circular pipe is called Poiseuille flow or Hagen -Poiseuille flow. In laminar flow te sear stress is related to te velocity gradient: du τ μ dr r (8) R Hence, from (6) and (8), du G r dr μ Integrating and applying te no-slip condition at te wall (u = 0 on r = R), Laminar pipe-flow velocity profile G u ( R r ) (9) 4μ Example. Find, from te velocity distribution given above, (a) te centreline velocity u 0 ; (b) te average velocity V; (c) te volumetric flow rate Q, in terms of ead loss and pipe diameter; L V (d) te friction factor λ, defined by f λ ( ), as a function of Reynolds number, Re. D g Answer: (a) u 0 GR u(0) 4μ ; (b) V u GR 0 8μ ; (c) Q π 8 ρg f μl D 4 ; (d) λ 64 Re Part (d) of tis exercise demonstrates tat te friction factor λ is not constant. J.L.M Poiseuille ( ); Frenc pysician wo was interested in flow in blood vessels. G.L.H. Hagen; German engineer wo, in 839, measured water flow in long brass pipes and reported tat tere appeared to be two regimes of flow. Hydraulics T-4 David Apsley

5 .4 Turbulent Pipe Flow In turbulent flow one is usually interested in time-averaged quantities. Velocity usually implies time-averaged velocity and te sear stress τ is te time-averaged rate of transport of momentum per unit area; it is dominated by turbulent mixing rater tan viscous stresses. In turbulent flow tere is no longer an explicit relationsip between mean stress τ and mean velocity gradient du/dr because a far greater transfer of momentum arises from te net effect of turbulent eddies tan te relatively small viscous forces. Hence, to relate quantity of flow to ead loss we require an empirical relation connecting te wall sear stress and te average velocity in te pipe. As a first step define a skin friction coefficient c f by wall sear stress τ w c f (0) dynamic pressure ρv (Later, c f will be absorbed into a friction factor λ to simplify te expression for ead loss.) For te lengt of pipe sown, te balance of forces along te axis in fully-developed flow is: πd Δp mg sin θ τ w πdl 0 4 net pressure force weigt wall friction From te geometry, πd m ρ L 4 Δz sin θ L p L w direction of flow Substituting tese gives: πd πd Δp ρgδz τw πdl 4 4 πd Δ( p ρgz) τ w πdl 4 z mg p+p Dividing by te cross-sectional area (πd /4), L Δp* 4 τ w D Write: τ w c f ( ρv ) (definition of skin-friction coefficient) Substituting and rearranging gives for te drop in piezometric pressure: L Δp * 4c f ( ρv ) D Te quantity 4c f is known as te Darcy friction factor and is denoted by λ. Hydraulics T-5 David Apsley

6 Darcy 3 -Weisbac 4 Equation L Δp * λ ( ρv ) () D L pressure loss due to friction λ dynamic pressure D Dividing by ρg tis can equally well be written in terms of ead rater tan pressure: L V f λ ( ) () D g L ead loss due to friction λ dynamic ead D *** Very important *** Tere is considerable disagreement about wat is meant by friction factor and wat symbol sould be used to denote it. Wat is represented ere by λ is also denoted f by some autors and 4f by oters (including ourselves in past years and exam papers). Be very wary of te definition. You can usually distinguis it by te expression for friction factor in laminar flow: 64/Re wit te notation ere; 6/Re wit te next-most-common alternative. It remains to specify λ for a turbulent pipe flow. Metods for doing so are discussed in Section.5 and lead to te Colebrook-Wite equation. Since λ depends on bot te relative rougness of te pipe (k s /D) and te flow velocity itself (troug te Reynolds number Re VD/ν) eiter an iterative solution or a cart-based solution is usually required. Altoug te bulk velocity V appears in te ead-loss equation te more important quantity is te quantity of flow, Q. Tese two variables are related, for circular pipes, by πd Q VA V 4 were D is te pipe diameter. At ig Reynolds numbers λ tends to a constant (determined by surface rougness) for any particular pipe. In tis regime compare: Q f (turbulent) 5 D Q f (laminar) 4 D 3 Henri Darcy ( ); Frenc engineer; conducted experiments on pipe flow. 4 Julius Weisbac; German professor wo, in 850, publised te first modern textbook on ydrodynamics. Hydraulics T-6 David Apsley

7 .5 Expressions for te Darcy Friction Factor λ Laminar Flow (teory) 64 λ Re Turbulent Flow (smoot or roug pipes) Nikuradse 5 (933) used sand grains to rougen pipe surfaces. He defined a relative rougness k s /D, were k s is te sand-grain size and D te diameter of te pipe. His experimental curves for friction factor (see, e.g., Wite s textbook) sowed 5 regions:. laminar flow (Re < Re crit 000; rougness irrelevant). laminar-to-turbulent transition (approximately 000 < Re < 4000) 3. smoot-wall turbulent flow (λ is a function of Reynolds number only) 4. fully-roug-wall turbulent flow (λ a function of relative rougness only) 5. intermediate rougness (λ is a function of bot Re and k s /D) In te smoot- or roug-wall limits, Prandtl 6 and Von Kármán 7 gave, respectively: Smoot-wall turbulence: Roug-wall turbulence:.0 log λ.0 log λ 0 0 Re λ.5 3.7D Unfortunately, in practice, many commercial pipes lie in te region were bot rougness and Reynolds number are important, so tat te friction factor is not constant for any particular pipe but depends on te flow rate. Colebrook and Wite (937) combined smootand roug-wall turbulence laws into a single formula, te Colebrook-Wite equation. Colebrook-Wite Equation k s λ.0log 0 k s.5 (3) 3.7D Re λ Tis is te main formula for te friction factor in turbulent flow. Te main difficulty is tat it is implicit (λ appears on bot sides of te equation) and so must be solved iteratively. Tere are several explicit approximations to (3), accurate to witin a few percent for realistic ranges of Reynolds number see te references in Massey and Wite s textbooks. Equivalent Sand Rougness For commercial pipes te pattern of surface rougness may be very different to tat in te artificially-rougened surfaces of Nikuradse. Colebrook (939) and Moody (944) gatered 5 Joann Nikuradse ( ); PD student of Prandtl. 6 Ludwig Prandtl ( ); German engineer; introduced boundary-layer teory. 7 Teodore von Kármán (88-963); Hungarian matematician and aeronautical engineer; gave is name to te double row of vortices sed from a -d bluff body and now known as a Kármán vortex street. Hydraulics T-7 David Apsley

8 l Transition data to establis effective rougness for typical pipe materials. Typical values of k s are given in te Appendix. Moody Cart Grapical solutions of (3) exist. Te most well known is te Moody cart (λ versus Re for various values of relative rougness k s /D). Te curves are just solutions of te Colebrook- Wite equation. A ome-produced version is sown below k s /D Laminar l = 64/Re smoot-walled limit E0.0E03.0E04.0E05.0E06.0E07 Re = VD/n Hydraulics T-8 David Apsley

9 .6 Oter Losses Pipeline systems are subject to two sorts of losses: wall-friction, contributing a continuous fall in ead over a large distance; minor losses due to abrupt canges in geometry; e.g. pipe junctions, valves, etc. Eac type of loss can be quantified using a loss coefficient K, te ratio of pressure loss to dynamic pressure (or ead loss to dynamic ead): V pressure loss K( ρv ) or ead loss K( ) (4) g Typical values of K are given below. Commercial pipe fittings (approximate) Fitting K Globe valve 0 Gate valve wide open 0. Gate valve ½ open elbow 0.9 Side outlet of T-junction.8 Entry/exit losses Configuration K Bell-mouted entry 0 Abrupt entry 0.5 Protruding entry.0 Bell-mouted exit 0. Abrupt enlargement.0 Minor losses are a one-off loss, occurring at a single point. Frictional losses are proportional to te lengt of pipe L and, in te grand sceme of tings, usually dominate. For long pipelines minor losses are often ignored. Hydraulics T-9 David Apsley

10 .7 Pipeline Calculations Te objective is to establis te relationsip between available ead and quantity of flow. Available ead, H H = sum of ead losses along te pipe (5) Available ead is te overall drop in ead from start to end of te pipe, usually determined by still-water levels, sometimes supplemented by additional pumping ead. Head losses are proportional to te dynamic ead V /g. Fluid ten flows troug te pipe at precisely te rigt velocity V (or discarge Q) tat (5) is satisfied. Pipe parameters are illustrated below. Altoug a reservoir is indicated at eac end of te pipe, tis is simply a diagrammatic way of saying a point at wic te total ead is known. D Q L Typical pipeline problems are: given two of te following parameters, find te tird. Head loss: Quantity of flow: Q Diameter: D Oter parameters: lengt L, rougness k s, kinematic viscosity ν and minor loss coefficient K. Calculations involve: () Head losses e.g. wit friction factor λ and minor-loss coefficient K: L V (λ K)( ) (6) D g () Expressions for loss coefficients e.g. te Colebrook-Wite equation for friction factor: k.5.0log s 0 (7) λ 3.7D Re λ In most problems (6) and/or (7) must be solved iteratively. Te exception is te calculation of Q wen and D are known (and minor losses neglected) because in tis special case (Type in te examples wic follow) te Reynolds number can be expanded to give: k.5ν.0log s 0 (8) λ 3.7D D λv If minor-loss coefficient K = 0 ten te combination λv can be found from (6) and ence λ can be found. Knowledge of bot λv and λ gives V and tence Q. Hydraulics T-0 David Apsley

11 Overall Head Loss If tere is a free surface in still water ten bot gauge pressure and velocity tere are zero and so te total ead equals te surface elevation: H = z. If, owever, te pipe discarges to atmospere as a free jet ten te total ead includes te dynamic ead, V /g. If te discarge is to anoter reservoir (a) ten (wit a well-rounded exit): H z, H z and te loss in ead is just te difference in still-water levels: z z z z Alternatively, if te discarge is a free jet to atmospere, ten H z, H z V / g and te loss in ead is z z V / g (In terms of piezometric ead tis is equivalent to aving an exit loss coefficient.0.) z z V Te second case also applies if tere is an abrupt exit into a tank, since flow separation means tat te pressure in te jet leaving te pipe is essentially te ydrostatic pressure in te tank (piezometric ead z ) but tere is still a dynamic ead V / g wose energy is ultimately dissipated in te receiving tank. Again, tis is equivalent to an exit loss coefficient.0. For long pipelines, owever, tis is usually neglected. Hydraulics T- David Apsley

12 Type : Diameter D and ead difference known; find te quantity of flow Q. Example. A pipeline 0 km long, 300 mm diameter and wit rougness 0.03 mm, conveys water from a reservoir (top water level 850 m AOD) to a water treatment plant (700 m AOD). Assuming tat te reservoir remains full, and neglecting minor losses, estimate te quantity of flow. Take ν =.00 6 m s. Solution. List known parameters: L = 0000 m D = 0.3 m = 50 m k s = 30 5 m ν =.00 6 m s Since D and are known, te ead-loss equation enables us to find λv : L V gd λ ( ) λv m s D g L 0000 Rewriting te Colebrook-Wite equation, k s.5.0 log0 λ 3.7D Re λ k.5ν.0 log s D D λv log Hence, λ Knowledge of bot λv and λ gives V λv λ m s Finally, te quantity of flow may be computed as velocity area: πd π Q VA V ( ) m s Answer: quantity of flow = 0.8 m 3 s. Hydraulics T- David Apsley

13 Type : Diameter D and quantity of flow Q known; find te ead difference. Example. Te outflow from a pipeline is 30 L s. Te pipe diameter is 50 mm, lengt 500 m and rougness estimated at 0.06 mm. Find te ead loss along te pipe. Solution. List known parameters: Q = 0.03 m 3 s L = 500 m D = 0.5 m k s = 60 5 m ν =.00 6 m s Inspect te ead-loss equation: L V λ ( ) D g We can get V from Q and D, but to find we will require te friction factor. First V: V Q A Q πd / π m s /4 Inspect te Colebrook-Wite equation: k.5.0log s 0 λ 3.7D Re λ To use tis we require te Reynolds number: VD Re ν.0 0 Substituting values for k s, D and Re in te Colebrook-Wite equation and rearranging for λ: λ log 0 (.080 ) λ Iterating from an initial guess, wit successive values substituted into te RHS: Initial guess: λ = 0.0 First iteration λ = Second iteration λ = Tird iteration λ = Fourt iteration λ = λ can ten be substituted in te ead-loss equation to derive : L V λ ( ) m D g Answer: ead loss = 8.8 m. Hydraulics T-3 David Apsley

14 Type 3 (Sizing problem): Quantity of flow Q and available ead known; find te required diameter D. Example. A flow of 0.4 m 3 s is to be conveyed from a eadworks at 050 m AOD to a treatment plant at 000 m AOD. Te lengt of te pipeline is 5 km. Estimate te required diameter, assuming tat k s = 0.03 mm. Solution. List known parameters: Q = 0.4 m 3 s = 50 m L = 5000 m k s = 30 5 m ν =.00 6 m s Before iterating, try to write D in terms of λ. From te ead-loss equation: λ L D V ( ) g λl Q gd A / 5 λl Q gd πd /4 8LQ λ 5 π gd 8LQ D λ π g Substituting values of Q, L and gives a working expression (wit D in metres): / 5 D (.3 λ) (*) Te Colebrook-Wite equation for λ is: k.5.0log s 0 λ 3.7D Re λ Te Reynolds number can be written in terms of te diameter D: 5 VD Q D 4Q Re ν πd / 4 ν πνd D Substituting tis expression for Re we obtain an iterative formula for λ: λ D.0 log 0 ( ) D λ (**) Iterate (*) and (**) in turn, until convergence. Guess: λ = 0.0 D = 0.40 m Iteration : λ = D = m Iteration : λ = D = m Iteration 3: λ = D = 0.44 m Iteration 4: λ = D = 0.44 m Answer: required diameter = 0.44 m. In practice, commercial pipes are only made wit certain standard diameters and te next available larger diameter sould be cosen. Hydraulics T-4 David Apsley

15 Bot Pipe Friction and Minor Losses In some circumstances minor losses (including exit losses) actually contribute a significant proportion of te total ead loss and must be included in te ead-loss equation L V (λ K)( ) D g An iterative solution in conjunction wit te Colebrook-Wite equation is ten inevitable, irrespective of weter te problem is of type, or 3. Example. A reservoir is to be used to supply water to a factory 5 km away. Te water level in te reservoir is 60 m above te factory. Te pipe lining as rougness 0.5 mm. Minor losses due to valves and pipe fittings can be accommodated by a loss coefficient K = 80. Calculate te minimum diameter of pipe required to convey a discarge of 0.3 m 3 s. Answer: 0.44 m Hydraulics T-5 David Apsley

16 .8 Energy and Hydraulic Grade Lines Energy grade lines and ydraulic grade lines are grapical means of portraying te energy canges along a pipeline. Tree elevations may be drawn: pipe centreline: z geometric eigt ydraulic grade line (HGL): energy grade line (EGL): p ρg p g ρ z piezometric ead V z total ead g p is te gauge pressure (i.e. pressure difference from atmosperic pressure). Illustrations energy grade line Pipe friction only reservoir ydraulic grade line pipeline V /g p/g reservoir Pipe friction wit minor losses (exaggerated), including cange in pipe diameter. entry loss pipeline EGL HGL exit loss Pumped system EGL HGL pipeline pump Hydraulics T-6 David Apsley

17 Energy Grade Line (EGL) Sows te cange in total ead along te pipeline. Starts and ends at still-water levels. Steady downward slope reflects pipe friction (slope cange if pipe radius canges); small discontinuities correspond to minor losses; large discontinuities correspond to turbines (loss of ead) or pumps (gain of ead). Te EGL represents te maximum eigt to wic water may be delivered. Hydraulic Grade Line (HGL) Sows te cange in piezometric ead along te pipeline. For pipe flow te HGL lies a distance p/ρg above te pipe centreline. Tus, te difference between pipe elevation and ydraulic grade line gives te static pressure p. If te HGL drops below pipe elevation tis means negative gauge pressures (i.e. less tan atmosperic). Tis is generally undesirable since: extraneous matter may be sucked into te pipe troug any leaks; for very negative gauge pressures, dissolved gases may come out of solution and cause cavitation damage. An HGL more tan p atm /ρg ( 0 m of water) below te pipeline is impossible. Te HGL is te eigt to wic te liquid would rise in a piezometer tube. For open-cannel flows, pressure is atmosperic (i.e. p = 0) at te surface; te HGL is ten te eigt of te free surface. Te EGL is always iger tan te HGL by an amount equal to te dynamic ead For uniform pipes (constant V), te two grade lines are parallel. V / g. Hydraulics T-7 David Apsley

18 .9 Simple Pipe Networks For all pipe networks te following basic principles apply: () continuity at junctions (total flow in = total flow out); () te ead is uniquely defined at any point; (3) eac pipe satisfies its individual resistance law (i.e. ead-loss vs discarge relation): = αq Te last of tese comes from te proportionality between ead loss and dynamic ead, i.e. L V 4Q (λ K) were V D g πd λ is te friction factor and K is te sum of minor loss coefficients. For and calculations, α is often taken as a constant for eac pipe (altoug, in reality, it will vary sligtly wit flow rate). Tere is a close analogy wit electrical networks: ead H potential V discarge Q current I However, te ydraulic equivalent of Om s law is usually non-linear: ead loss H Q potential difference V I.9. Pipes in Series Q = Q H = H + H same flow add te ead canges.9. Pipes in Parallel H = H Q = Q + Q same ead cange add te flows Hydraulics T-8 David Apsley

19 .9.3 Branced Pipes Single Junction Te simplest case is tree pipes meeting at a single junction. If te flows are known ten te eads can be determined (relative to te ead at one point) by calculating te ead losses along eac pipe. If, owever, te eads H A, H B and H C are known (for example, from te water levels in reservoirs) ten we ave a classic problem known as te treereservoir problem. Te ead at J is adjusted (iteratively) to satisfy: (a) te loss equation ( αq ) for eac pipe; i.e: H H H J J J H H H A B C α α α JA JB JC Q Q Q JA JB JC (b) continuity at te junction J: net flow out of junction Q JA Q JB Q JC 0 A C J? B Note te sign convention: Q JA is te flow from J to A; it will be negative if te flow actually goes from A to J. Te direction of flow in any pipe is always from ig ead to low ead. Altoug we consider only 3 reservoirs, te problem and its solution metod clearly generalise to any number of reservoirs. Solution Procedure (0) Establis te ead-loss vs discarge equations for eac pipe; () Guess an initial ead at te junction, H J; () Calculate flow rates in all pipes (from te ead differences); (3) Calculate net flow out of J; (4) As necessary, adjust H J to reduce any flow imbalance and repeat from (). If te direction of flow in a pipe, say JB, is not obvious ten a good initial guess is to set H J = H B so tat tere is initially no flow in tis pipe. Te first flow-rate calculation will ten establis weter H J sould be lowered or raised and ence te direction of flow in tis pipe. Example. Reservoirs A, B and C ave constant water levels of 50, 0 and 90 m respectively above datum and are connected by pipes to a single junction J at elevation 5 m. Te lengt (L), diameter (D), friction factor (λ) and minor-loss coefficient (K) of eac pipe are given below. Pipe L (m) D (m) λ K JA JB JC (a) Calculate te flow in eac pipe. (b) Calculate te reading of a Bourdon pressure gauge attaced to te junction J. Hydraulics T-9 David Apsley

20 Solution. First, prepare ead-loss vs discarge relations for eac pipe: L V Q (λ K) were V D g A L 8 ( λ K) Q 4 D π gd Q πd /4 Substituting L, D, λ and K for eac pipe we obtain te ead-loss vs discarge relationsips: Pipe AJ: Pipe JB: Pipe JC: H H H J J J H or A B 4Q JA H or C 7488Q JB H or 634Q JB Q Q Q JA JB JC H H H J 50 4 J J Te value of H J is varied until te net flow out of J is 0. If tere is net flow into te junction ten H J needs to be raised. If tere is net flow out of te junction ten H J needs to be lowered. After te first two guesses at H J, subsequent iterations are guided by interpolation. Te working is conveniently set out in a table. H J (m) Q JA (m 3 s ) Tis is sufficient accuracy (0.0003/0.9 or about 0.5%). Te quantity of flow in eac pipe is given in te bottom row of te table, wit te direction implied by te sign. (b) A Bourdon gauge measures absolute pressure. From te piezometric ead at te junction: p H J z ρ g were p is te gauge pressure. Hence, ρg( H z) (3.8 5) p J H J Pa 50 4 Q JB (m 3 s ) Taking atmosperic pressure as 0000 Pa, te absolute pressure is ten = Pa H J Q JC (m 3 s ) H J Net flow out of J (m 3 s ) = Q JA + Q JB + Q JC Answer:.68 bar. Hydraulics T-0 David Apsley

21 .0 Complex Pipe Networks (Optional).0. Loop Metod (Hardy-Cross, 936) Used for networks made up of a series of closed loops, were te external flows are known. loop loop Basic Idea Start wit any flow satisfying continuity. Apply iterative flow corrections δq until te net ead cange round eac loop is 0. Adopt a suitable sign convention (e.g. Q positive if clockwise) in eac loop. Te signed ead loss for any particular pipe is ten sαq (9) were s = + if Q is positive and if Q is negative. Initially, te net ead loss round a closed loop probably won t be 0. To try to acieve tis after perturbing te flow in all pipes of a loop by δq we require sα( Q δq) 0 loop were δq is te same for every pipe in te loop. Expanding: s αq sαq δq sα δq 0 Neglecting te δq term and noting tat sαq δq α Q Tis is applied to every pipe in te loop. sq Q leads to a flow correction for tis loop of (0) Algoritm Divide te network into closed loops. Start wit any flow satisfying continuity. For eac loop in turn: sαq calculate δq ; α Q update all pipes in tis loop by δq; Repeat until te net ead cange around all loops is sufficiently small. An example wit two loops is given on te Example Seet. Hydraulics T- David Apsley

22 .0. Nodal Metod (Cornis, 939) Used for loops or brances were te external eads are known. H H Basic Idea Start wit guessed eads H i at eac internal junction and calculate te resulting flow in eac pipe. Apply iterative ead corrections δh i so as to satisfy continuity at eac junction. (Te 3-reservoir problem is te simplest case, wit a single junction). H 3 H 4 As in te previous subsection, ead canges at junctions i and j cause a cange in te flow between tem: H i H j δh i δh j sijαijqijδqij δqij () Q were Q ij is te flow rate from te it node to te jt node, wit appropriate sign. ij Initially, te net outflow at te i t junction won t be 0; to try to acieve tis we aim to perturb te flow so tat ( Q δ ) 0 or j j ij Q ij Qij Qij (δh i δh j ) 0, i,, 3, () ( H H ) j i j Taken over all junctions i tis gives a set of simultaneous equations for te δh i. A -junction example is given on te Example Seet. Hydraulics T- David Apsley

23 . OPEN-CHANNEL FLOW Flow in open cannels (e.g. rivers, canals, guttering,...) and partially-full conduits (e.g. sewers) is caracterised by te presence of a free surface were te pressure is atmosperic. Unlike pipe flow, open-cannel flow is always driven by gravity, not pressure. PIPE FLOW OPEN-CHANNEL FLOW Fluid: LIQUIDS or GASES LIQUIDS (free surface) Driven by: PRESSURE, GRAVITY or BOTH GRAVITY (down slope) Size: DIAMETER HYDRAULIC RADIUS Volume: FILLS pipe Depends on DEPTH Equations: DARCY-WEISBACH (ead loss) COLEBROOK-WHITE (friction factor) MANNING S FORMULA. Normal Flow Te flow is uniform if te velocity profile does not cange along te cannel. (Tis is at best an approximation for natural cannels like rivers were te cannel cross-section canges.) Te flow is steady if it does not cange wit time. Steady uniform flow is called normal flow and te dept of water is called te normal dept. Te normal dept depends on te discarge Q. V /g EGL HGL (free surface): p = 0 f L In normal flow equal ydrostatic pressure forces at any cross-section mean no net pressure force. Hence, te downslope component of weigt balances bed friction; Note. Te following assumes te slope to be sufficiently small for tere to be negligible difference between te dept measured vertically (wic determines te energy level) and tat perpendicular to te bed of te cannel (wic determines te flow rate). Hydraulics T-3 David Apsley

24 . Hydraulic Radius and te Drag Law A P In bot open cannels and partially-full pipes, wall friction occurs only along te wetted perimeter. Let A be te cross-sectional area occupied by fluid and P te wetted perimeter. For steady, uniform flow, te component of weigt down te slope balances bed friction: ( ρal) g sinθ τbpl were τ b is te average wall friction. Hence, τ A b ρg( )sin θ P Define: Hydraulic radius b L mg A cross - sectional area R (3) P wetted perimeter Note tat, in general, te ydraulic radius depends on dept. Hence, for normal flow, Normal flow relationsip τ b ρgr S (4) were S (= drop lengt) is te slope. (We ave assumed tan θ sin θ for small angles.) Examples. () For a circular pipe running full, A πr R R P πr (5) i.e. for a full circular pipe, te ydraulic radius is alf te geometric radius. (Sorry folks, tis is just one of tose tings!). As a result, it is common to define a ydraulic diameter D by D 4 R () For a rectangular cannel of widt b wit water dept, R A b P b / b For a very wide cannel /b <<, and ence R i.e. R is equal to te dept of flow. To progress we need an expression for te average bed stress τ b. Hydraulics T-4 David Apsley

25 .3 Friction Laws Cézy and Manning s Formulae From te balance of forces above: τ b ρgr S In principle, a (skin-)friction coefficient can be used to relate te (average) bed sear stress to te dynamic pressure. Hence, c ( ρv ) ρgr S (6) f Friction factors λ = 4c f based on te Colebrook-Wite equation (using 4R as te ydraulic diameter) are unsatisfactory for open conduits because te sear stress is not constant on te wetted perimeter. Instead, engineers tend to use simpler empirical formulae due to Cézy 8 and Manning 9. Rearranging equation (6) gives: g V RS c wence Cézy s Formula C ( f V C R S (7) g/c f ) is Cézy s coefficient. Tis gives te variation wit slope for a particular cannel, but it is not a elpful design equation because C varies wit cannel rougness and ydraulic radius. Te most popular correlation for C is tat of Manning wo proposed, on te basis of a review of experimental data, tat C R / 6 function of rougness wic e cose to write as / 6 R C n Combined wit Cézy s formula (7), tis yields: Manning s Formula V n / 3 / R S (8) Very important. Bot Cézy s C and Manning s n are dimensional and depend on te units used. Typical values of n in metre-second units are given in te Appendix. Typical figures for artificiallylined cannels and natural water courses are 0.05 m /3 s and m /3 s respectively. 8 Antoine Cézy (78-798); Frenc engineer wo carried out experiments on te Seine and on te Courpalet Canal. 9 Robert Manning (86-897); Iris engineer. Actually, if you live on te wrong side of te Englis Cannel ten wat we call Manning s equation is variously ascribed to Gauckler and/or Strickler. Hydraulics T-5 David Apsley

26 .4 Uniform-Flow Calculations Assuming tat te cannel slope, sape and lining material are known, tere are two main classes of problem: (Type A - easy) Given te dept () determine te quantity of flow (Q) Calculate: () cross-sectional area A and wetted perimeter P from geometry of cannel; () A ydraulic radius R. P (3) / 3 / average velocity from Manning s formula: V R S n (4) quantity of flow as velocity area: Q = VA. (Type B - arder) Given te quantity of flow (Q) determine te dept () () Follow te steps for Type A above to write algebraic expressions for, successively, A and P, R, V, Q in terms of dept. () Invert te Q vs relationsip grapically or numerically. Example. A smoot concrete-lined cannel as trapezoidal cross-section wit base widt 6 m and sides of slope V:H. If te bed slope is in 500 and te normal dept is m calculate te quantity of flow. Solution. We are given slope S = From te Appendix, Manning s n is 0.0 m /3 s. m 6 m 4 m Break te trapezoidal section into rectangular and triangular elements to obtain, successively: Area: A m Wetted perimeter: P m A 0 Hydraulic radius: R.339 m P 4.94 Average velocity: V R n Quantity of flow: Q VA / 3 / / m s S R m 3 s Answer: 9 m 3 s. Hydraulics T-6 David Apsley

27 Example. For te cannel above, if te quantity of flow is 40 m 3 s, wat is te normal dept? Solution. Tis time we need to leave all quantities as functions of eigt. In metre-second units we ave te following. Area: A 6 Wetted perimeter: P 6 5 Hydraulic radius: A R P / 3 Average velocity: V 3.77 R Quantity of flow: Q VA Now try a few values of. (m) A (m s ) P (m) R (m) V (m s ) Q (m 3 s ) After te first two guesses, subsequent coices of ome in on te solution by interpolating/extrapolating from previous results. Answer: =.30 m If desired, te sequence of calculations could be assembled into a single formula for Q(): / 3 5 / 3 A / S A Q VA S A / 3 n P n P Wit te given values of S and n, and te formulae for A and P: 5 / 3 (6 ) Q 3.77 / 3 (6 5) A sequence of trial values of is ten used as above. (Small numerical differences are due to rounding). (m) Q (m 3 s ) Note. Microsoft Excel is good for inverting Q() (use te Goal-Seek or Solver tools). Hydraulics T-7 David Apsley

28 .5 Conveyance Combining Manning s formula for average velocity ( V / 3 / ( / n) R S ydraulic radius ( R A/ P ) and discarge ( Q VA ) we obtain: were ) wit expressions for / Q KS (9) 5 / 3 A K (30) / 3 n P K is a function of te cannel geometry and te rougness of its lining. It is called te conveyance of te cannel and is a measure of te cannel s discarge-carrying capacity. Te primary use of K is in determining te discarge capacity of compound cannels for example river and flood plain. By adding te contribution to total discarge from individual components wit different rougness: / / Q Q K S K S i i te total conveyance is simply te sum of te separate conveyances: K K eff i eff 3 flood plain river flood plain.6 Optimal Sape of Cross-Section Expressions for A, P and R for important cannel sapes are given below. rectangle trapezoid circle b b R cross-sectional area A wetted perimeter P ydraulic radius R b b / b b tanα b sin α b / tanα b / sin α R ( θ sin θ) R θ R sin θ θ Te most ydraulically-efficient sape of cannel is te one wic can pass te greatest quantity of flow for any given area. Tis occurs for te minimum ydraulic radius or, equivalently, for te minimum wetted perimeter corresponding to te given area. A semi-circle is te most ydraulically-efficient of all cannel cross-sections. However, ydraulic efficiency is not te only consideration and one must also consider, for example, Hydraulics T-8 David Apsley

29 fabrication costs, excavation and, for loose granular linings, te maximum slope of te sides. Many applications favour trapezoidal cannels. Trapezoidal Cannels For a trapezoidal cannel: cross-sectional area: wetted perimeter: A b tanα P b sin α b Wat dept of flow and wat angle of side give maximum ydraulic efficiency? To minimise te wetted perimeter for maximum ydraulic efficiency, we substitute for b in terms of te fixed area A: A A P b ( ) ( ) (3) sin α tanα sin α sin α tanα To minimise P wit respect to water dept we set P A ( ) 0 sin α tanα and, on substituting te bracketed term into te expression (3) for P, we obtain P A Te ydraulic radius is ten A R P In oter words, for maximum ydraulic efficiency, a trapezoidal cannel sould be so proportioned tat its ydraulic radius is alf te dept of flow. Similarly, to minimise P wit respect to te angle of slope of te sides, α, we set P ( cos α sec α) ( cos α) 0 α sin α tan α sin α Tis occurs wen cos α = ½. Te most efficient side angle for a trapezoidal cannel is 60. Substituting tese results for and α into te general expression for R one obtains / b 3/ ; i.e. te most ydraulically-efficient trapezoidal cannel sape is alf a regular exagon. Circular Ducts In similar fasion it can be sown tat te maximum quantity of flow for a circular duct actually occurs wen te duct is not full in fact for a dept about 94% of te diameter (Exercise. Prove it; ten try to explain in words wy you migt expect tis). Hydraulics T-9 David Apsley

30 Appendix Material k s (mm) Riveted steel Concrete Wood stave Cast iron 0.6 Galvanised iron 0.5 Aspalted cast iron 0. Commercial steel or wrougt iron Drawn tubing Glass 0 (smoot) Table. Typical rougness for commercial pipes (from Wite, 0). n (m /3 s) Artificial lined cannels: Glass 0.0 Brass 0.0 Steel, smoot 0.0 painted 0.04 riveted 0.05 Cast iron 0.03 Concrete, finised 0.0 unfinised 0.04 Planed wood 0.0 Clay tile 0.04 Brickwork 0.05 Aspalt 0.06 Corrugated metal 0.0 Rubble masonry 0.05 Excavated eart cannels: Clean 0.0 Gravelly 0.05 Weedy 0.03 Stony, cobbles Natural cannels: Clean and straigt 0.03 Sluggis, deep pools 0.04 Major rivers Floodplains: Pasture, farmland Ligt brus 0.05 Heavy brus Trees 0.5 Table. Typical values of Manning s n (from Wite, 0). Hydraulics T-30 David Apsley