Pipe Flow. Lecture 17


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1 Pipe Flow Lecture 7
2
3 Pipe Flow and the Energy Equation For pipe flow, the Bernoulli equation alone is not sufficient. Friction loss along the pipe, and momentum loss through diameter changes and corners take head (energy) out of a system that theoretically conserves energy. Therefore, to correctly calculate the flow and pressures in pipe systems, the Bernoulli Equation must be modified. P /γ + V /g + z = P /γ + V /g + z + H maj + H min Energy line with no losses Energy line with major losses H maj
4 Major Losses Major losses occur over the entire pipe, as the friction of the fluid over the pipe walls removes energy from the system. Each type of pipe has a friction factor, f, associated with it. H maj = f (L/D)(V /g) Energy line with no losses Energy line with major losses H maj
5 Minor Losses Unlike major losses, minor losses do not occur over the length of the pipe, but only at points of momentum loss. Since Minor losses occur at unique points along a pipe, to find the total minor loss throughout a pipe, sum all of the minor losses along the pipe. Each type of bend, or narrowing has a loss coefficient, K L to go with it. Minor Losses
6 Major and Minor Losses Major Losses: f = friction factor V = Velocity Minor Losses: H maj = f (L/D)(V /g) L = pipe length D = pipe diameter g = gravity H min = K L (V /g) K L = sum of loss coefficients V = Velocity g = gravity When solving problems, the loss terms are added to the system at the second analysis point P /γ + V /g + z = P /γ + V /g + z + H maj + H min
7 Minor Head Loss Minor loss at entrance (h ) Minor loss at exit (h ) Loss in submerged discharge Loss in moving water at outlet Minor loss at contraction or expansion (h 3 ) Loss at sudden contraction Loss at gradual contraction
8 Minor (secondary) losses in pipes (cont.)  Pipe with reservoir reservoir h L = losses due to entrance v h = l k g v entrance losses = 0.5 g v = velocity of entrance k k 0. 0 pipe h L = losses due to exit v h l = k g v = velocity of exit pipe reservoir k.0
9 Secondary losses (Minor or transient losses)  Sudden enlargement v v H. G. L T. E. L v g v g piezometric head line ( ) g v k g v D D g v A A g v v v g v v h L = = = = = For Gradually enlargement 0 0. k
10 Secondary losses (Minor or transient losses) (cont.) 3. Sudden contraction v v A A A c g v k g v D d g v A A h c L ) ( = = =.0 < = D D k φ
11 Minor Loss Minor loss at fittings Loss in valve Loss in flange Minor loss in bends and elbows
12 Secondary losses (Minor or transient losses) (cont.) 3. Losses due to bends or elbows v h L = k g v D For bend R ( θ, D R) k = φ, bend For elbow k = φ ( θ ) v elbow
13 Secondary losses (Minor or transient losses) (cont.) 4. Losses due to valves v h L = k g plug valve butter fly valve gate valve
14 Loss Coefficients
15 Bernoulli's Equation with Major and Minor Loss L B B A A A h g V D L f z P g V z P g V = + + γ γ B h h h h h h L = g V k h i i = Minor Loss Term
16 Pipe Flow Example r/d = 0 Z =? 60 m r/d = γ oil = 8.8 kn/m 3 f =.035 Z = 30 m 7 m K out = 30 m If oil flows from the upper to lower reservoir at a velocity of.58 m/s in the 5 cm diameter smooth pipe, what is the elevation of the oil surface in the upper reservoir? Include major losses along the pipe, and the minor losses associated with the entrance, the two bends, and the outlet.
17 Pipe Flow Example r/d = 0 Z =? 60 m 7 m 30 m Z = 30 m K out = r/d = If oil flows from the upper to lower reservoir at a velocity of.58 m/s in the 5 cm diameter smooth pipe, what is the elevation of the oil surface in the upper reservoir? Include major losses along the pipe, and the minor losses associated with the entrance, the two bends, and the outlet.
18 Pipe Flow Example r/d = 0 Z =? 60 m 7 m 30 m Z = 30 m K out = r/d = Apply Bernoulli s equation between points and : Assumptions: P = P = Atmospheric = 0 V = V = 0 (large tank) Z = m + H maj + H min H maj = (f L V ) /(D g)=(.035 x 97m x (.58m/s) )/(.5 x x 9.8m/s ) H maj = 5.85m
19 Pipe Flow Example r/d = 0 Z =? 60 m r/d = 7 m 30 m Z = 30 m K out = Z = m m + H min H min = K bend V /g + K ent V /g + K out V /g From Loss Coefficient table: K bend = 0.9 K ent = 0.5 K out = H min = (0.9x ) x (.58 /x9.8) H min = 0.4 m
20 Pipe Flow Example r/d = 0 Z =? 60 m 7 m 30 m Z = 30 m K out = r/d = Z = m + H maj + H min Z = m m + 0.4m Z = 36. m
21 Numerical Example: Minor Loss Convert the piping system to an equivalent length of 6 pipe.
22 Numerical Example: Minor Loss Solution This problem will be solved by using the Bernoulli equation, A to M, datum M, as follows: (0+0+h) = (0.05x50/+8.0+x )(V /(g) +(0.0x00/ x )(V 6 /(g) ) ( 6 6 g V k g V k g V D L f g V D L f z P g V z P g V i c i i c = + + γ γ 0 0 h 0 0 0
23 Numerical Example: Minor Loss Solution Then V A = V 6 A 6 4V = V 6 V = (/6)V 6 h =4.5V /(g) + 5.7V 6 /(g) = =(4.5x(/6)+5.7) [V 6 /(g)]=6.6v 6 /(g) For any available head h, the lost head is 6.6V 6 /(g). The lost head in L E ft of 6 pipe is f(l E /D)[V 6 /(g)] Equating the two values, 6.6V 6 /(g)=0.0(l E /0.5)[(V 6 /(g)] L E =45ft
24 Design of Single Pipe Flow Type I: Head loss problem Given flow rate and pipe combinations, determine the total head loss in a system Type II: Discharge problem Given allowable total head loss and pipe combinations, determine the flow rate Type III: Sizing problem Given allowable total head loss and flow rate, determine the pipe diameter
25 Type I: Head loss problems Given the pipe material, pipe geometry, and flow rate, find out how much friction loss we may encounter over the pipe. Application: pumping station design.
26 Type I: Head loss Problem A 0indiameter galvanized iron pipe miles long carries 4 cfs of water at 60 o F. Find the friction loss using Moody diagram
27 Unit Conversion of Length mile = 760x3 = 5,80 ft ft = in = 30.4 cm in =.54 cm mile =.6 Km
28 Type I: Head loss Problem e = ft, so e/d = ()/0 = L= mile=*(580 ft/mile)= 0,560 ft 4Q 4(4) V = = = fps πd π (0/) for water at 60 o F: ν =.7x05 ft /sec R e DV (0 /).8333 = = = υ Turbulent flow
29 Absolute Pipe Roughness
30 Type I: Head loss Problem Based on e/d and R e, we find our f = 0.07 in the Moody Diagram, then we may obtain: L V h f = f = 0.07( )( ) = D g 0 / (3.) ft
31 Type II: Discharge Problem Given the pipe material and friction loss gradient, find out how much fluid we may send over the pipe. Application: leak detection.
32 Type II: Discharge Problem Water at 0 o C flows in a 500mm diameter welded steel pipe. If the friction loss gradient is 0.006, determine the flow rate.
33 Type II: Discharge Problem For welded steel pipe: e = mm; e/d =0.046/500 = At 0 o C, υ =.003x06 m /sec h f /L = is given S = h f L = f fv = 0.5() V = / f V D g
34 Dimensional Analysis of Pipe Flow Moody Chart Laminar Marks Reynolds Number independence
35 Absolute Pipe Roughness
36 Type II: Discharge Problem For e/d = f min ~0.08 (Moody Diagram) Try f = 0.08 then V = 0.43/(0.08) 0.5 =.3 m/sec R e VD υ (0.5)(.3) = = = 6 With e/d = and R e =.4x0 6; we may get f = 0.03 Try one more time and show this results have converged Q = AV = 0.46 m 3 /sec
37 Type III: Sizing Problem Given the boundary condition of water head in the pipe system and the allowable head loss, determine the pipe diameter. Application: designing the pipe system between the reservoir and the end user.
38 Type III: Sizing Problem The allowable head loss is 5 m/km of pipe length. Estimate the size of a uniform, horizontal weldedsteel pipe installed to carry 500 l/sec of water at 0 o C.
39 Type III: Sizing Problem Solution Bernoulli s equation can be applied to pipe section km apart For a uniform, horizontal pipe with no localized head losses V =V z = z L B B B A A A h g V D L f z P g V z P g V = + + γ γ B
40 Type III: Sizing Problem The Bernoulli s equation reduces to P P A B = h 5 f = m γ γ (per each Km) h f = f L D V f L Q 5 D g( D / 4) g D g = π = 8 flq π 5 8 flq D = = gπ f
41 Type III: Sizing Problem Therefore L = 000m, and h f = 5m. At 0 o C, v =.3x06 m /sec. Assuming weldedsteel roughness to be in the lower range of riveted steel, e = 0.9 mm. Apply try and error procedure: Assume D = 0.8 m
42 Type III: Sizing Problem 3 Q 0.5m /sec V = = = 0.995m /sec A π (0.4m) R e VD (0.995)(0.8) = = = υ.3 0 e/d = (0.9mm)/(800mm) = 0.00 Entering these value to the Moody diagram, we get f = 0.0 5
43 Type III: Sizing Problem A better estimate of D can be obtained by substituting the latter values into: 5 8 f LQ D = = gπ f Which gives D = 0.6m A new iteration provides V =.7 m/sec, R e = 8.0x0 5, e/d = 0.005, f = 0.0, and D = 0.6m
44 Software Applications
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