http://faculty.washington.edu/markbenj/CEE345/ CEE 345, Spring 2012 Part 2: Hydraulics and Open Channel Flow Instructor: Mark Benjamin, 335 More Hall; 543-7645, [email protected] Text: Munson et al.., Fundamentals of Fluid Mechanics, 6th ed. • Pipe networks, pump selection, hydraulic transients • Flow dependence on depth & slope in rivers, streams, culverts, storm and sanitary sewers • • • • Weekly HW, due beginning of class; no late homework One lab in HHL; no write-up, but data used in HW Four Thursday sessions – computer lab, HW, practice problems Exam on hydraulics on Friday, May 18; open channel questions on final exam Our Plan – Weeks 6 and 7 • Review energy relationships in single pipes • Extend analysis to progressively more complex systems – Pipes in parallel or series – Pipe networks with known flow direction in each pipe – Interconnected pipe loops and reservoirs where flow direction is not obvious • Consider key factors in selection of pumps to add energy to fluid in a system • Consider some special cases of transients in pipe systems – cavitation and water hammer Overview of Pipe Networks • ‘Pipe flow’ generally refers to fluid in pipes and appurtenances flowing full and under pressure • Examples: Water distribution in homes, industry, cities; irrigation • System components – – – – – Pipes Valves Bends Pumps and turbines Storage (often unpressurized, in reservoirs, tanks, etc.) Energy Relationships in Pipe Systems • Energy equation between any two points: E 2 E 1 h pum p hturb z2 p2 2 V2 2g z1 p1 V1 h L, f 2 2g h pum p hturb h L, f • Analysis involves writing expressions for hL in each pipe and for each link between pipes (valves, expansions, contractions), relating velocities based on continuity equation, and solving subject to system constraints (Q, p, or V at specific points). Energy Losses in Piping Systems • Darcy-Weisbach equation for headlosses in pipes (major headlosses): hL f l V 2 D 2g Estimating f Graphically Trends in f • f declines with increasing Re, e.g., increasing V at fixed D. • In laminar region, f = 64/Re • In turbulent region, for given e/D, f declines more slowly than in laminar region; eventually, the decline stops altogether. Mathematical Expressions for f • Colebrook and Haaland eqns yield good estimates of f in turbulent flow e D 2.71 2 log f Re f 3.7 1 1.11 1 6.9 e D 1.8 log R e f 3.7 • Useful for calculations in spreadsheets or special software for pipe flow analysis Understanding Headloss in Pipes • Darcy-Weisbach equation: hL f l V 2 D 2g • For travel distance of one pipe diameter (l D ): E nergy lost due to friction w hen hL f V 2 2g fluid travels a distance l equal to D K E of the fluid • f is the ratio of energy lost via friction (i.e., shear) to the kinetic energy of the water when the water travels a distance of one pipe diameter Example • Compare the velocity and pressure heads for typical conditions in a street main: V = 1.5 m/s; D = 0.5 m; p = 500 kPa V 2 2g p 500 1.5 m /s 2 9.8 m /s 2 2 0.115 m kP a 1000 N /m 3 9800 N /m 2 kP a 51.0 m • If f = 0.02, hL for each 0.5 m of pipe is 2% of the velocity head, or 0.0023 m, corresponding to 0.0045% of the pressure head. Typical Pipe Flow Problems • Type I: Pipe properties (e, D, l) and V known, find hL. • Determine f from Moody diagram or an equivalent equation, and hL from the DW eqn hL f l V 2 D 2g Example A 20-in-diameter galvanized pipe (e = 0.0005 ft) 2 miles long carries 4 cfs at 60oF. Find hL using (a) the Moody diagram and (b) the Colebrook eqn. a) V Q 3 A Re DV 4 ft /s 1.67 ft 1.67 ft 1.83 1.22 x10 e D 2 5 0.0005 ft 1.83 ft/s 4 ft/s 2 2.51x10 ft /s 0.00030 1.67 ft f 0.017 5 hL f l V 2 D 2g 0.017 b) Colebrook eqn: 9 10 11 12 13 14 2 5280 ft F e/D Re f LHS RHS LHS - RHS 1.67 ft 1.83 ft/s 2 32.2 ft/s 2 2 e D 2.71 2 log 3.7 f Re f 1 G 0.0003 251000 0.03 =1/SQRT(G11) =-2*LOG(G9/3.7 + 2.71/G10*G12) =G12-G13 5.59 ft H 0.0003 2.51E+05 0.03 5.774 7.687 -1.913 e/D Re f LHS RHS LHS - RHS hL f l V 2 D 2g 0.0174 0.0003 2.51E+05 0.017422 7.576 7.576 2.55E-07 2 5280 ft 1.67 ft 1.83 ft/s 2 32.2 ft/s 2 2 5.72 ft Typical Pipe Flow Problems • Type II: Pipe properties (e, D, l) and hL known, find V. • Guess V, determine f and hL as in Type I, iterate until hL equals known value, or • Solve Colebrook and DW eqns simultaneously to eliminate V, yielding: Solving Type II Pipe Problems: Iterative Approach hL f l V 2 D 2g f 0 .2 5 lo g 3 gD 0 .3 1 7 2 h L l 1 2 2 h L gD fl e / D 3 .7 2 1/ 2 Rearranged D-W eqn: V V 2 2 gD h L l e D 2.51 log 3.7 D 2 gD h L l Example For the pipe analyzed in the preceding example, what is the largest flow rate allowable if the total frictional headloss must remain <8 ft? Example For the pipe analyzed in the preceding example, what is the largest flow rate allowable if the total frictional headloss must remain <8 ft? V 2 2 gD h L l e D 2.51 log 3.7 D 2 gD h L l Substituting known values, V 2.19 ft s Q V A 2.19 ft s 1.67 ft 4 2 4.80 ft s 3 Typical Pipe Flow Problems • Type III: e, l, V, and hL known, find D. • Several approaches, all iterative; e.g., Guess D, determine V as in Type II, iterate until V equals known value Example What diameter galvanized pipe would be required in the preceding examples if a flow rate of 10 cfs was needed, while keeping the total frictional headloss at <8 ft? Solving Type III Pipe Problems: Iterative Graphical Approach hL f l V 2 D 2g Solving Type III Pipe Problems: Iterative Analytical Approach What diameter galvanized pipe would be required in the preceding examples if a flow rate of 10 cfs was needed, while keeping the total frictional headloss at <8 ft? V 2 2 gD h L l Q 2 e D 2.51 log 3.7 D e D 2.51 2 gD h L log l D 3.7 2 gD h L l D 2 2 gD h L 4 l Q 2 e D 2.51 2 gD h L log l D 3.7 g hL l eps nu D_guess LHS = Q RHS LHS RHS 32.2 8 10560 0.0005 1.22E-05 2 10 7.72E+00 2.28E+00 D 2 2 gD h L 4 l g hL l eps nu D_guess LHS = Q RHS LHS RHS 32.2 8 10560 0.0005 1.22E-05 2.206594 10 1.00E+01 -8.22E-07 Dependence of hL on D and V hL f l V 2 D 2g Dependence of hL on D and V • In laminar region: For a given pipe 64 l V 2 32 l ' hL V k Q lam 2 gD DV D 2g • In turbulent region, when f becomes constant: h L f full turb l V 2 D 2g For a given pipe k full Q 2 turb • Under typical water distribution conditions, hL in a given pipe can be expressed as kQn with n slightly <2. Example For the systems analyzed in the first two examples, what value of n causes the data to fit the equation hL = kQn? hL ,2 h L ,1 log kQ kQ hL ,2 h L ,1 n log h L , 2 h L ,1 log Q 2 Q1 n 2 n 1 Q2 Q1 n log n Q2 Q1 log 5.72 ft / 8 ft log 4 cfs 4.8 cfs 1.84 Alternative Equations for Flow - Headloss Relationships in Turbulent Pipe Flow • Hazen-Williams equation – widely used for hL as function of flow parameters for turbulent flow at typical velocities in water pipes: V 0.849 C H W R h L 10.7 l Q 1.85 D 4.87 0.63 h hL l 1 1.85 C HW 0.54 A flow D 2 4 D R Rh Pw etted D 4 2 Coefficients shown are for SI units; for BG units, replace 0.849 by 1.318 and 10.7 by 4.73. Comparison of Equations for Transitional and Turbulent Curves on the Moody Diagram D-W 2 gD Q hf 1 l V D Manning* 0.63 0.849 C H W R h hL (=S*l) g 2 0.54 0 .5 0 2.50 Q 2 1 n S S 0 .5 0 0.50 f f 0 .5 0 0.50 0.354 D 0.63 S l D 5 f 0.54 0.67 Rh S 0.50 C H W 0.397 D 0.67 S 0.50 1 n 0.278 D 2.63 S 0.54 4 8 S f 2g D 2g H-W* 10.7 Q 1.85 D C HW l 1 4.87 1.85 HW C 0.312 D 2.67 S 0.50 1 n 10.3 Q l 2 D 5.33 1 n 2 *Coefficients shown are for SI units (V in m/s, and D and R in m); for BG units h (ft/s and ft), replace 0.849 by 1.318; 0.354 by 0.550; 0.278 by 0.432; 10.7 by 4.73; 1/n by 1.49/n; 0.397 by 0.592; 0.312 by 0.465; and 10.3 by 4.66. Energy Losses in Bends, Valves, and Other Transitions (‘Minor Losses’) • Minor headlosses generally significant when pipe sections are short (e.g., household, not pipeline) • Caused by turbulence associated with flow transition; therefore, mitigated by modifications that ‘smooth’ flow patterns • Generally much greater for expansions than for contractions • Often expressed as multiple of velocity head: h L K m in o r • K is the ratio of energy lost via friction in the device of interest to the kinetic energy of the water (upstream or downstream, depending on geometric details) V 2 2g Energy Losses in Contractions 2 hc k c V2 2g All images from Finemore & Franzini (10e, 2002) Energy Losses in Expansions hx h x,discharge V 2 2g Vc V Vc 2 2g 2 2g h x,discharge V 2 2g Vc 2 2g All images from Finemore & Franzini (10e, 2002) Energy Losses in Expansions Conical diffuser h cone k cone V1 V 2 2 2g k’, rough k’, smooth All images from Finemore & Franzini (10e, 2002) Energy Losses in Pipe Fittings and Bends hb k b V 2 2g All images from Finemore & Franzini (10e, 2002) Example A 5-in-diameter pipe with an estimated f of 0.033 is 110 feet long and connects two reservoirs whose surface elevations differ by 12 feet. The pipe entrance is flushed, and the discharge is submerged. (a) Compute the flow rate. (b) How much would the flow rate change if the last 10 ft of the pipe were replaced with a smooth conical diffuser with a cone angle of 10o? 5 " 0.417 ft a h L , tot h L , pipe h L , m inor V 2 gh L , tot fl 1.5 2 2 2 V f 0.5 1 f 1.5 D 2g 2g D 2g l V V 2 32.2 ft/s 2 12 ft 0.033 110 ft D Q V A 8.70 ft/s l 8.70 ft/s 1.5 0.417 ft 0.417 ft 4 2 1.19 ft /s 3 b h L , tot h L , pipe h L , entrance h L , cone h L , exit l1 V1 f 2 D1 2 g k entrance V1 2 2g k cone V1 V 2 2g 2 2 k exit V2 2g D 2 D1 2 L cone tan 5 0.417 ft 2 10 ft 0.0875 2.17 ft o 2 2 D1 0.417 ft 0.0370 V1 D 2 2.17 ft V2 From graph, for a smooth, 10o cone, kcone = 0.175 h L , tot f l1 V1 2 D1 2 g 0.033 k entrance V1 2 2g 100 ft V1 2 0.5 0.417 ft 2 g 0.175 k cone V1 V1 V 2 2g 2 2 k exit V2 2g 2 2g V1 0.037V1 2 1.0 0.037V1 2g 2g V1 9.49 ft/s Q V1 A 9.49 ft/s 0.417 ft 4 2 1.29 ft /s 3 2

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