1、06,Viscous Flow in Ducts,06 - 2,Objectives,Have a deeper understanding of laminar and turbulent flow in pipes and the analysis of fully developed flow.,Calculate the major and minor losses associated with pipe flow in piping networks and determine the pumping power requirements.,06 - 3,Recall - beca
2、use of the no-slip condition, the velocity is not uniform, and at the walls of a duct flow is zero.,Incompressible Flow: Velocity and Mass Flow Rate,We are often interested only in Vavg, which we usually call just V (drop the subscript for convenience).,Keep in mind that the no-slip condition causes
3、 shear stress and friction along the pipe walls.,06 - 4,Laminar and Turbulent Flows,Can be steady or unsteady.,Always unsteady.,There are always random, swirling motions in a turbulent flow.,Can be one-, two, three-dimensional.,Always three-dimensional.,However, it can be 1-D or 2-D in the mean.,Has
4、 regular, predictable behavior.,Has irregular or chaotic behavior.,Cannot predict exactly there is some randomness associated with it.,Analytical solution possible.,No analytical solution exits so far!,Its too complicated.,Occurs at low Reynolds numbers.,Occurs at high Reynolds numbers.,Laminar Flow
5、,Turbulent Flow Flow,06 - 5,Laminar and Turbulent Flows: Reynolds Number,Definition,Physical Significance,These values are approximate, different books may give slightly different values.,For a given application, these values depends upon Pipe roughness Vibrations Upstream fluctuations, disturbances
6、 (valves, elbows, etc. that may disturb the flow), .,Critical Reynolds Number,06 - 6,The Entrance Region,Entrance Length,A boundary layer is a layer of fluid in the immediate vicinity of a bounding surface, over which the velocity gradient is not equal to zero.,Boundary Layer,The profile develops do
7、wnstream over several diameters called the entry length Le. Le/d is a function of Re.,06 - 7,Entrance Length of Circular Pipes,06 - 8,Head Loss,Fx_visc = twAsurf = tw(2pRL),SFx = DppR2 + prgR2Dz 2ptwRL,Fx_other = 0,= prgR2Dz,= rg(pR2L) sinf, Divided by prgR2,(1),(2),06 - 9,Head Loss Friction Factor,
8、Head Loss,To predict head loss hf, we need to be able to calculate w.,Darcy-Weisbach Equation,(e = wall roughness height),(3),(4),Friction Factor f,e is significant for turbulent flows, but not for laminar flows.,Pressure Drop,06 - 10,Friction Factor: Fully Developed Lamina Flow,Velocity Profile,Wal
9、l Shear Stress,Friction Factor f,Head Loss hf,06 - 11,Friction Factor : Turbulent Flow,Background,Need EES, Excel, Matlab or other software to solve for f.,Haaland Formula,The Colebrook formula is valid for both smooth and rough walls.,The Haaland formula varies less than 2% from Colebrook.,The Cole
10、brook formula is accurate to 15% due to roughness size, experimental error, curve fitting of data, etc.,e has no effect on laminar flow,(e = wall roughness height),Since analytical solutions for turbulent flows do not exist, all available formulas are all from interpolation of experimental data.,Col
11、ebrook Formula,06 - 12,Turbulent Flow: Moody Chart,Moody Chart plot of the Colebrook formula and the lamina flow equation.,Probably the most famous and useful figure in fluid mechanics with 15% accuracy.,06 - 13,Recommended Roughness, e, for Commercial Ducts*,* From White 6th ed. Table 6.1,06 - 14,E
12、xample 6.1 (White 6th ed. Ex. 6.7),Oil, with = 900 kg/m3 and = 0.00001 m2/s, flows at 0.2 m3/s through 500 m of 200 mm-diameter cast iron pipe. Determine (a) the head loss and (b) the pressure drop if the pipe slopes down at 10 in the flow direction.,Problem,V,Red,e/d,Solution,Use the Colebrook form
13、ula (or Moody Chart).,Turbulent flow. Use Colebrook formula.,06 - 15,Example 6.1: f, hf and Dp,hf,hf = 117 m,Dp = 266 kPa,Dp,f,f = 0.0227 (use Excel),06 - 16,Four Types of Pipe Flow Problems,Given d, L, and V or Q, , , and g, computer the head loss or Dp (head loss problem, Red is known).,Given d, L
14、, hf, , , and g, compute the velocity V or flow rate Q (flow rate problem, Red is unknown).,Given Q, L, hf, , , and g, compute the diameter d of the pipe (sizing problem, Red is unknown).,Given Q, d, L, hf, , , and g, computer the pipe length L (Red is known).,In design and analysis of piping system
15、s, 4 problem types are encountered.,Types 1 and 4 are well suited to the Moody chart.,Types 2 and 3 are common engineering design problems, i.e., selection of pipe diameters to minimize construction and pumping costs. However, iterative approach required since either V or D is unknown, and both are
16、in the Reynolds number.,With the help of suitable computer software (EES, Matlab, Mathcad, Mathematica, etc.), the iteration may not be needed.,06 - 17,Type 2: Given d, L, hf, r, m & g, Compute V or Q,Iteration,Explicit Formula,Guess an initial f, based on e/d,where,06 - 18,Example 6.2 (White 6th ed
17、. Ex. 6.9): Direct Solution,Oil, with = 950 kg/m3 and = 0.00002 m2/s, flows through a 30-cm-diameter pipe 100 m long with a head loss of 8 m. The roughness ratio e/d = 0.0002. Find the average velocity and flow rate.,Problem,z,Red,V,V = 4.84 m/s,Q,Q = 0.342 m3/s,06 - 19,Example 6.2: Iterative Soluti
18、on,Simplify Equation,Initial Guess,2nd Iteration,f2 = 0.0195 (not close enough to f1),(Turbulent),06 - 20,Example 6.2: Iterative Solution (Contd),Q = 0.342 m3/s,3rd Iteration,f3 = 0.0201 (not close enough to f2),4th Iteration,f4 = 0.0201 (close enough to f3 !),V,V = 4.84 m/s,Q,This is the final solu
19、tion: f = 0.201,(Turbulent),06 - 21,Type 3: Given Q, L, hf, r, m & g, Compute d,Iteration,Explicit Formula,Guess an initial f,06 - 22,Example 6.3 (White 6th ed. Ex. 6.10): Direct Solution,Work example 6.2 backward, assuming that Q = 0.342 m3/s and = 0.06 mm are known but that d (30 cm) is unknown. R
20、ecall L = 100 m, =950 kg/m3, and = 2E5 m2/s, hf = 8 m.,Problem,Swamee and Jain,d = 0.303 m = 30.3 mm,Check e/d,Between 106 and 102, OK!,Check Red,OK!,06 - 23,Example 6.3: Iterative Solution,Simplify Equation,Initial Guess,Obtain the relation between Red and d, and e/d and d,Procedure,Guess f, comput
21、e d from (1), Red from (2) and e/d from (3). Then compute a new f from Colebrook. Repeat until convergence.,(1),(2),(3),06 - 24,Example 6.3: Iterative Solution (contd),2nd Iteration,f2 = 0.0204 (not close enough to f1),3rd Iteration,f3 = 0.0201 (close enough to f2 !),Final d,d= 30 cm,This is the fin
22、al solution: f = 0.201,(Turbulent),06 - 25,Flow in Non-Circular Ducts,The wetted perimeter is the perimeter that is actually in contact with the flow.,A : cross-sectional area P : wetted perimeter tw: average wall stress,The calculation of tw is more complicated for non-circular ducts.,The next ques
23、tion is: f = ?,06 - 26,The calculation of f is very complicated for non-circular ducts.,Friction Factor for Non-Circular Ducts,This is an approximation, bringing additional error. Get a better solution whenever possible.,In engineering applications, we approximately use the hydraulic diameter Dh in
24、the Moody Chart to calculate f:,06 - 27,Flow Between Parallel Plates,Approximate Solution,Better Solution,Instead of using Dh, an effective diameter Deff is used in calculating the friction factor f:,06 - 28,Example 6.4 (White 6th ed. Ex. 6.13),Fluid flows at an average velocity of 6 ft/s between ho
25、rizontal parallel plates a distance of 2.4 in apart. Find the head loss and pressure drop for each 100 ft of length for = 1.9 slugs/ft3 and (a) = 0.00002 ft2/s and (b) = 0.002 ft2/s. Assume smooth walls.,Problem,Analysis,Will try both the approximate and better solutions to solve the problem.,06 - 2
26、9,Example 6.4: Part (a), n = 0.00002 ft2/s,ReDh,f,From Colebrook, f = 0.0173,hf,Dp ,hf = 2.42 ft ,Dp = 148 psf ,ReDeff,f,From Colebrook, f = 0.0289,hf,Dp,hf = 2.64 ft ,Dp = 161 psf ,Approximate Solution,Better Solution,06 - 30,Example 6.4: Part (b), n = 0.002 ft2/s,ReDh,f,hf,Dp,hf = 7.45 ft ,Dp = 45
27、5 psf ,ReDeff,f,hf,Dp,hf = 11.2 ft ,Dp = 685 psf ,Approximate Solution,Better Solution,06 - 31,Flow through a Concentric Annulus,Approximate Solution,Better Solution,06 - 32,Minor Losses,Minor Loss,Loss Coefficient,In addition to the major loss hf, the following components in the system bring additi
28、onal losses called minor losses, denoted as hm.,Pipe entrance or exit. Sudden expansion or contraction. Bends, elbows, tees, and other fitting. Valves, open or partially closed. Graduate expansion or contraction.,(Recall the major loss ),06 - 33,Total Loss and Pump/Turbine Selection,Total Loss,Pump
29、Selection,Pump/ Turbine Power,06 - 34,Typical Commercial Valves,White 6th ed. Fig. 6.17 (a) Gate valve; (b) globe valve; (c) angle valve; (d) swing-check valve; (e) disk-type gate valve,06 - 35,Open Valves, Elbows & Tees (White 6th ed Table 6.5),06 - 36,Partially Open Valve: Loss Coefficient,(White
30、6th ed. Fig. 6.18b),06 - 37,Butterfly Valve: Loss Coefficient,White 6th ed. Fig. 6.19 (a) geometry; (b) loss coefficients for 3 different manufactures,06 - 38,Smooth Bend: Loss Coefficient,White 6th ed. Fig. 6.20: Resistance (loss) coefficients for smooth-walled 45o, 90o, and 180o bends at Red = 200
31、,000.,06 - 39,Entrance and Exit: Loss Coefficient,Reentrant inlets,Rounded and beveled inlets,Exit losses are K 1.0 for all shapes of exit (reentrant, sharp, beveled or rounded),White 6th ed. Fig. 6.21,06 - 40,Sudden Expansion & Contraction: Loss Coefficient,White 6th ed. Fig. 6.22,06 - 41,Gradual C
32、onical Expansion & Contraction,Expansion,Contraction,White 6th ed. Fig. 6.23,06 - 42,Example 6.5 (White 6th ed. Ex. 6.16),Water = 1.94 slugs/ft3 and = 0.000011 ft2/s, is pumped between two reservoirs at 0.2 ft3/s through 400 ft of 2-in-diameter pipe and several minor losses. The roughness ratio is /
33、d = 0.001. Compute the pump horsepower required if the pump efficiency is 80%.,Problem,Analysis,The minor losses for this problem include Sharp Entrance: K = 0.5 Open global valve : K = 6.9 12-in bend: K = 0.25 Regular 90 elbow: K = 0.95 Half-closed gate valve: K = 3.7 Sharp exit : K = 1.0,06 - 43,Example 6.5: Solution,Energy Equation,V in the pipe,Minor Losses,06 - 44,Example 6.5: Solution (contd),Major Loss,Pump Head,f = 0.02156,Pump Power,Darcy friction factor from Colebrook,(Turbulent flow),
