Viscous Flow in Ducts


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1 Dr. M. Siavashi Iran University of Science and Technology Spring 2014
2 Objectives 1. Have a deeper understanding of laminar and turbulent flow in pipes and the analysis of fully developed flow 2. Calculate the major and minor losses associated with pipe flow in piping networks and determine the pumping power requirements Fluid Mechanics I 2
3 Introduction Average velocity in a pipe Recall  because of the noslip condition, the velocity at the walls of a pipe or duct flow is zero We are often interested only in V avg, which we usually call just V (drop the subscript for convenience) Keep in mind that the noslip condition causes shear stress and friction along the pipe walls Friction force of wall on fluid For a circular tube Fluid Mechanics I 3
4 Introduction V avg V avg For pipes of constant diameter and incompressible flow V avg stays the same down the pipe, even if the velocity profile changes Why? Conservation of Mass Fluid Mechanics I 4 same same same
5 Introduction For pipes with variable diameter, m is still the same due to conservation of mass, but V 1 V 2 D 1 D 2 V 1 m V 2 m 2 1 Fluid Mechanics I 5
6 Laminar and Turbulent Flows Fluid Mechanics I 6
7 Laminar and Turbulent Flows Definition of Reynolds number Critical Reynolds number (Re cr ) for flow in a round pipe Re < 2300 laminar 2300 Re 4000 transitional Re > 4000 turbulent Note that these values are approximate. For a given application, Re cr depends upon Pipe roughness Vibrations Upstream fluctuations, disturbances (valves, elbows, etc. that may disturb the flow) Fluid Mechanics I 7
8 Laminar and Turbulent Flows For nonround pipes, define the hydraulic diameter D h = 4A c /P A c = crosssection area P = wetted perimeter Example: open channel A c = 0.15 * 0.4 = 0.06m 2 P = = 0.7m Don t count free surface, since it does not contribute to friction along pipe walls! D h = 4A c /P = 4*0.06/0.7 = 0.34m What does it mean? This channel flow is equivalent to a round pipe of diameter 0.34m (approximately). Fluid Mechanics I 8
9 The Entrance Region Consider a round pipe of diameter D. The flow can be laminar or turbulent. In either case, the profile develops downstream over several diameters called the entry length L e. L e /D is a function of Re. L e Fluid Mechanics I 9
10 The Entrance Region Dimensional analysis shows that the Reynolds number is the only parameter affecting entry length In turbulent flow the boundary layers grow faster, and L e is relatively shorter Fluid Mechanics I 10
11 Fully Developed Pipe Flow Comparison of laminar and turbulent flow There are some major differences between laminar and turbulent fully developed pipe flows Laminar Can solve exactly Flow is steady Velocity profile is parabolic Pipe roughness not important It turns out that V avg = 1/2U max and u(r)= 2V avg (1  r 2 /R 2 ) Fluid Mechanics I 11
12 Fully Developed Pipe Flow Turbulent Cannot solve exactly (too complex) Flow is unsteady (3D swirling eddies), but it is steady in the mean Mean velocity profile is fuller (shape more like a tophat profile, with very sharp slope at the wall) Pipe roughness is very important Instantaneous profiles V avg 85% of U max (depends on Re a bit) No analytical solution, but there are some good semiempirical expressions that approximate the velocity profile shape. See text Logarithmic law (Eq. 846) Power law (Eq. 849) Fluid Mechanics I 12
13 Fully Developed Pipe Flow Wallshear stress Recall, for simple shear flows u=u(y), we had = du/dy In fully developed pipe flow, it turns out that = du/dr Laminar Turbulent w w = shear stress at the wall, acting on the fluid w w,turb > w,lam Fluid Mechanics I 13
14 Fully Developed Pipe Flow Pressure drop There is a direct connection between the pressure drop in a pipe and the shear stress at the wall Consider fully developed, and incompressible flow in a pipe Let s apply conservation of mass, momentum, and energy to this CV (good review problem!) Fluid Mechanics I 14
15 Fully Developed Pipe Flow Pressure drop Conservation of Mass Conservation of xmomentum Fluid Mechanics I 15
16 Fully Developed Pipe Flow Pressure drop Conservation of Energy there are no shaftwork or heattransfer effects since V 1 = V 2 and 1 = 2 (shape not changing) now reduces to a simple expression for the frictionhead loss h f Fluid Mechanics I 16
17 Fully Developed Pipe Flow Friction Factor From momentum CV analysis From energy CV analysis Equating the two gives h f 4 w L g D To predict head loss, we need to be able to calculate w. How? Laminar flow: solve exactly Turbulent flow: rely on empirical data (experiments) In either case, we can benefit from dimensional analysis! Fluid Mechanics I 17
18 Fully Developed Pipe Flow Friction Factor w = func( V,, D, ) analysis gives = average roughness of the inside wall of the pipe Fluid Mechanics I 18
19 Fully Developed Pipe Flow Friction Factor Now go back to equation for h L and substitute f for w h f 4 w L gd 2 LV hf f D 2 g Our problem is now reduced to solving for Darcy friction factor f Recall Therefore Laminar flow: f = 64/Re (exact) But for laminar flow, roughness does not affect the flow unless it is huge Turbulent flow: Use charts or empirical equations (Moody Chart, a famous plot of f vs. Re and /D) Fluid Mechanics I 19
20 Fully Developed Pipe Flow Laminar Flow We would like to find the velocity profile in laminar flow & dp/dx = const. and u=0 at r=r Fluid Mechanics I 20
21 Fully Developed Pipe Flow Laminar Flow (I) (II) and (I) & (II) Fluid Mechanics I 21
22 Fully Developed Pipe Flow Laminar Flow Inclined Pipe Q The results already obtained for horizontal pipes can also be used for inclined pipes provided that P is replaced by: PgL sin Fluid Mechanics I 22
23 Example 1 Oil at 20 C (=888 kg/m 3 and = kg/m s) is flowing steadily through a D= 5 cm diameter L=40 m long pipe. P in = 745 kpa, P out =97 kpa. Determine the Q assuming the pipe is (a) horizontal, (b) inclined 15 upward, (c) inclined 15 downward. Also verify that the flow through the pipe is laminar. Solution: 2 2 P1 V1 P2 V2 z1 z2 h g 2g g 2g V V 1 2 sin P1 P2 L V 64 L V 64 L sin 2 L z z L f V g D 2g VD D 2g D 2g Q 4Q 128LQ V P glsin 2 4 A D D Q 4 D P glsin 128L Fluid Mechanics I 23 f
24 Example 1 a) Horizontal 0 sin D P Q 128L b) Uphill =15 o 15 o Q 4 D P glsin 128L 4 o sin c) Downhill =15 o 15 o m /s 4 o m /s sin15 10 Q max m /s 4Q Vavg 1.8m/s 2 D Vavg D ReD m /s Q The flow is always Laminar Fluid Mechanics I 24
25 The Moody Diagram Fluid Mechanics I 25
26 Fully Developed Pipe Flow Friction Factor Moody chart was developed for circular pipes, but can be used for noncircular pipes using hydraulic diameter Colebrook equation is a curvefit of the data which is convenient for computations Implicit equation for f which can be solved using the rootfinding algorithm in EES Both Moody chart and Colebrook equation are accurate to ±15% due to roughness size, experimental error, curve fitting of data, etc. Fluid Mechanics I 26
27 Example 2 Solution For cast iron =0.26 mm from the Moody diagram Fluid Mechanics I 27
28 Types of Fluid Flow Problems In design and analysis of piping systems, 3 problem types are encountered 1. Determine p (or h f ) given L, D, V (or flow rate) Can be solved directly using Moody chart and Colebrook equation 2. Determine V (or Q), given L, D, p 3. Determine D, given L, p, V (or flow rate) 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 both V and D are in the Reynolds number. Fluid Mechanics I 28
29 Example 3 Known: kg/m, 2 10 m /s, d 0.3m L 100m, h 8m, / d Q?, V? Iterative Solution: f we only need to guess f, compute V, then get Re d, compute a better f from the Moody chart, and repeat. Fluid Mechanics I 29
30 Example 3 We have converged to three significant figures. Thus our iterative solution is Fluid Mechanics I 30
31 Example 4 Work example 3 backward, assuming Q=0.342 m 3 /s and =0.06mm are known but that d is unknown. Recall L=100m, =950 kg/m 3, =2E5 m 2 /s and h f =8m. Known: kg/m, 2 10 m /s, Q 0.342m /s L 100m, h 8m, 0.06 mm d? Iterative Solution: f Also write the Re and /d in terms of d: Fluid Mechanics I 31
32 Example 4 We guess f, compute d then compute Re d and /d and compute a better f from the Moody chart: 1 st try: 2 nd try: Last try: Fluid Mechanics I 32
33 Minor Losses For any pipe system, in addition to the Moodytype friction loss computed for the length of pipe, there are additional socalled minor losses due to 1. Pipe entrance or exit 2. Sudden expansion or contraction 3. Bends, elbows, tees, and other fittings 4. Valves, open or partially closed 5. Gradual expansions or contractions 2 V h K m 2 g h m is minor losses. K is the loss coefficient which: is different for each component. is assumed to be independent of Re. typically provided by manufacturer or generic table. Fluid Mechanics I 33
34 Minor Losses Total head loss in a system is comprised of major losses h f (in the pipe sections) and the minor losses h m (in the components) h h h tot f m 2 LV Vj h f K d 2g 2g 2 i i tot i j i i j i th pipe section j th componet If the piping system has constant diameter Fluid Mechanics I 34
35 Minor Losses Typical commercial valve geometries: (a) gate valve (b) globe valve (c) angle valve (d) swingcheck valve (e) disktype gate valve Fluid Mechanics I 35
36 Minor Losses Resistance coefficients K for open valves, elbows, and tees: This table represents losses averaged among various manufacturers, so there is an uncertainty as high as ±50%. loss factors are highly dependent upon actual design and manufacturing factors. Fluid Mechanics I 36
37 Minor Losses Averageloss coefficients for partially open valves: Fluid Mechanics I 37
38 Minor Losses Sudden Expansion (SE) and Sudden Contraction (SC): up to the value d/d = 0.76, above which it merges into the suddenexpansion prediction Note that K is based on the velocity in the small pipe. Fluid Mechanics I 38
39 Minor Losses Entrance and Exit loss coefficients: Exit losses are K 1.0 for all shapes of exit to large reservoir Fluid Mechanics I 39
40 Minor Losses Note that K is based on the velocity in the small pipe. Gradual Expansion Gradual Contraction Fluid Mechanics I 40
41 Minor Losses Fluid Mechanics I 41
42 Example 5 A 6cmdiameter horizontal water pipe expands gradually to a 9cmdiameter pipe. The walls of the expansion section are angled 30 from the horizontal. The average velocity and pressure of water before the expansion section are 7 m/s and 150 kpa, respectively. Determine the head loss in the expansion section and the pressure in the largerdiameter pipe. Assumptions: 1 Steady & incompressible flow. 2 Fully developed and turbulent flow with α Properties: water = 1000 kg/m 3 The loss coefficient for gradual expansion of =60 total included angle is K L = 0.07 Fluid Mechanics I 42
43 Example 5 Solution: V2 A2 D A D 0.06 m m V A V A V V V 3.11m/s h m P 2 2 V1 7 K m 2 g V 2g 169kPa z 2 P2 V2 2 z2 hturbine 2g 2 2 P1 V1 P2 V2 1 2 h 2g 2g m hpump h V V 2 P2 P1 ghm P tot Fluid Mechanics I 43
44 Multiple Pipe Systems Pipes in Series Volume flow rate is constant Head loss is the summation of parts Since V 2 and V 3 are proportional to V 1 based on the mass conservation equation: Fluid Mechanics I 44
45 Multiple Pipe Systems Pipes in Series If Q is given, the righthand side and hence the total head loss can be evaluated. If the head loss is given, a little iteration is needed, since f 1, f 2, and f 3 all depend upon V 1 through the Reynolds number. Begin by calculating f 1, f 2, and f 3, assuming fully rough flow, and the solution for V 1 will converge with one or two iterations. Fluid Mechanics I 45
46 Example 6 Given is a threepipe series system. The total pressure drop is P A P B =150000Pa, and the elevation drop is z A z B =5m. The pipe data are as the table. The fluid is water, =1000 kg/m 3 and = m 2 /s. Calculate the Q in m 3 /h through the system. Fluid Mechanics I 46
47 Example 6 Solution: The total head loss across the system is: From the continuity relation: and: Neglecting minor losses and substituting into the head loss Eq., we obtain: (1) Begin by estimating f 1, f 2, & f 3 from the Moodychart fully rough regime Fluid Mechanics I 47
48 The Moody Diagram Fluid Mechanics I 48
49 Example 6 Solution (cont.): (1) Substitute in Eq. (1) to find: Thus the first estimate is V 1 =0.58 m/s from which: Hence, using the Re number and roughness ratio and from the Moody chart, Substitution into Eq. (1) gives the better estimate A second iteration gives Q=10.22 m 3 /h, a negligible change Fluid Mechanics I 49
50 Multiple Pipe Systems Pipes in Paralel Volume flow rate is the sum of the components Pressure loss across all branches is the same If the h is known, it is straightforward to solve for Q i in each pipe and sum them. The problem of determining Q i when h f is known, requires iteration. Each pipe is related to h f by the Moody relation h, where 5 f f L d V 2g f Q C C gd 8L Thus head loss is related to total flow rate by: Fluid Mechanics I 50
51 Multiple Pipe Systems Pipes in Paralel Q i Ch i f i f Since the f i vary with Reynolds number and roughness ratio, one begins the above Eq. by guessing values of f i (fully rough values are recommended) and calculating a first estimate of h f. Then each pipe yields a flowrate estimate Q i and hence a new Reynolds number and a better estimate of f i. Then repeat this Eq. to convergence. Fluid Mechanics I 51
52 Example 7 Assume that the same three pipes in Example 6 are now in parallel with the same total head loss of 20.3 m. Compute the total flow rate Q, neglecting minor losses. Fluid Mechanics I 52
53 Example 7 Solution: From the above equation we have: Guess fully rough flow in pipe 1: f 1 =0.0262, V 1 =3.49 m/s; hence Re 1 =273,000. From the Moody chart read f 1 =0.0267; recompute V 1 =3.46 m/s, Q 1 =62.5 m 3 /h. Next guess for pipe 2: f 2 =0.0234, V 2 =2.61m/s; then Re 2 =153,000, and hence f 2 =0.0246, V 2 =2.55 m/s, Q 2 =25.9 m 3 /h. Finally guess for pipe 3: f 3 =0.0304, V 3 =2.56 m/s; then Re 3 =100,000, and hence f 3 =0.0313, V 3 =2.52 m/s, Q 3 =11.4 m 3 /h. Fluid Mechanics I 53
54 Multiple Pipe Systems Pipes in Junction If all flows are considered positive toward the junction, then The pressure must change through each pipe so as to give the same static pressure p J at the junction. Let the HGL at the junction have the elevation Assuming p 1 =p 2 =p 3 =0 (gage) Fluid Mechanics I 54
55 Multiple Pipe Systems Pipes in Junction We guess the position h J and solve the following Eq. for V 1,V 2 & V 3 and hence Q 1,Q 2 & Q 3 : iterating until the flow rates balance at the junction according to Eq. If we guess h J too high, the sum flow sum will be negative and the remedy is to reduce h J,and vice versa. Fluid Mechanics I 55
56 Example 8 Take the same three pipes as in Example 7, and assume that they connect three reservoirs at these surface elevations. Find the resulting flow rates in each pipe, neglecting minor losses. Fluid Mechanics I 56
57 Example 8 Solution: As a first guess, take h J equal to the middle reservoir height, z 3 = h J = 40 m. This saves one calculation (Q 3 = 0) Fluid Mechanics I 57
58 Example 8 Solution (cont.): Since the sum of the flow rates toward the junction is negative, we guessed h J too high. Reduce h J to 30 m and repeat: Fluid Mechanics I 58
59 Example 8 Solution (cont.): This is positive Q=12.9, and so we can linearly interpolate to get an accurate guess: h J 34.3 m. Make one final list: we calculate that the flow rate is 52.4 m 3 /h toward reservoir 3, balanced by 47.1 m 3 /h away from reservoir 1 and 6.0 m 3 /h away from reservoir 3. Fluid Mechanics I 59
60 Multiple Pipe Systems Piping network Fluid Mechanics I 60
61 Multiple Pipe Systems Piping network This network is quite complex algebraically but follows the same basic rules: 1. The net flow into any junction must be zero. 2. The net head loss around any closed loop must be zero. In other words, the HGL at each junction must have one and only one elevation. 3. All head losses must satisfy the Moody and minorloss friction correlations. By supplying these rules to each junction and independent loop in the network, one obtains a set of simultaneous equations for the flow rates in each pipe leg and the HGL (or pressure) at each junction. Fluid Mechanics I 61
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