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8. fm 9 flow in pipes major loses co 3 copy

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8. fm 9 flow in pipes major loses co 3 copy

  1. 1. FLUID MECHANICS – 1 Semester 1 2011 - 2012 Week – 8, 9 and 10 FLOW IN PIPES- CO3 Compiled and modified by Sharma, Adam
  2. 2. Review• Conservation of mass (mass balance)• Continuity Equation• Bernoulli Equation• Momentum Equation
  3. 3. Objectives• Understand Laminar and Turbulent flow in pipes• Identify types of flow using Reynolds number• Explain minor losses and major losses for flow in pipes• Determine friction factor and major losses using moody chart and simplified Colebrook equation• Calculate minor losses, major losses, pressure losses and head losses• Understand equivalent length• Solve piping system using Bernoulli equation considering all losses. 3
  4. 4. 1. Introduction• There are two kind of flowa. Internal flow ( flow in pipes)b. External flow ( flow over bodies, drag, lift) – will be covered in fluids mechanics 2 (BMM2543)• Examples of internal flow1. Water flow in pipes2. Blood flow3. Oil and Gas industry4. Cooling system of a car (Radiator)5. Air Conditioning (Chilled water system) 4
  5. 5. examples & pictures 5
  6. 6. 6
  7. 7. 2. Types of flow• Flow in pipes has three types, Laminar, Transition, Turbulent • Laminar flow ― Smooth streamlinesTransition flow (flow) ― Highly ordered motion ― Short in length ― normally appeared in high viscosity flow and small pipe/passage i.e. oil in small pipe 7
  8. 8. 2. Types of flow (cont’)• Turbulent flow― Rough streamlines (flow)― Highly disordered motion― Most flow in reality is turbulent― High momentum, thus high friction• Transition flow― flow from laminar does not suddenly change to turbulent, it will enter transition flow first― Normally ignored in calculation― fluctuation of laminar and turbulent randomly 8
  9. 9. Is there other method to distinguishthese flow? • Yes, use Reynolds number, Re ν = kinematic viscosity, (m 2 s ) µ = dynamic viscosity, (kg m.s ) ρ = density of flowing fluid (kg/m 3 ) Vavg = Average pipe velocity D = Internal diameter of pipe • Re < 2300, Laminar flow • 2300 < Re < 4000, Transition flow • Re > 4000, Turbulent flow 9
  10. 10. Is there other method to distinguishthese flow?(cont)• For non-circular pipes, hydraulic diameter, Dh is used for calculating the Reynolds number. 4 × Crosssectional Area (internal) 4 Ac Dh = = Internal Perimeter p• however, in practical, type of flow depend on smoothness of pipe, vibrations and fluctuation in the flow. 10
  11. 11. Reynolds Number (example)• Q1: Water at 20°C flow with average velocity of 2cm/s inside a circular pipe. Determine flow type if the pipe diameter, a) 2 cm, b) 15 cm, and c) 30 cm From table (Cengel book, 2010) At 20°C µ = 1.002 × 10-3 kg m.s , ρ = 998kg/m 3 ρVave D 998(0.02)(0.02)a) Re = = = 398 • (a) Laminar flow µ 0.001002 ρVave D 998(0.02)(0.15)b) Re = = = 2988 • (b) Transition flow µ 0.001002 ρVave D 998(0.02)(0.3)c) Re = = = 5970 • (c) Turbulent flow µ 0.001002 11
  12. 12. Reynolds Number (example) cont’ • Q2: Water at 20°C flow in a circular pipe of 3.5 cm diameter. Determine the range for the average velocity so the flow is always transition flow From table (Cengel book, 2010) At 20°C µ = 1.002 × 10-3 kg m.s , ρ = 998kg/m 3 ρVave D 998(Vave ,min )(0.035)Re min imum = 2300 = = µ 0.001002 2300 × 0.001002 Vave ,min = = 0.066 m/s 998(0.035) ρVave D 998(Vave ,max )(0.035)Re maximum = 4000 = = µ 0.001002 4000 × 0.001002 Vave ,max = = 0.115 m/s 998(0.035) Range ∴ 0.066 m/s < Vave < 0.115 m/s 12
  13. 13. Reynolds Number (example) cont’ • Q3: Air at 35°C flow inside rectangular pipe of 2cmx5cm. Determine the maximum flow velocity for the pipe before the flow enter transition region From table (Cengel book, 2010), Air At 35°C, ν = 1.655 ×10-5 m 2 s 4 Ac 4(0.02)(0.05)Dh = = = 0.029 m p 2(0.02) + 2(0.05) Vave D Vave ,max (0.029)Re max imum = 2300 = = ν 1.655 × 10-5 2300 ×1.655 ×10 −5 Vave ,max = = 1.313m/s 0.029 13
  14. 14. 4. Entrance Region• Between the entrance and fully developed flow• Uniform velocity profile at entrance• because of no slip boundary condition, friction at the wall reduce the velocity of flow near the wall• to conserve mass, velocity at the center increase (compensate velocity decreased near the wall) 14
  15. 15. 4. Entrance Region (cont’)• As the fluid move deeper in the pipe, the velocity near the wall decreased further and velocity at center increase (developing velocity profile)• Both (up & down @ left & right) velocity profile increase till it merge with the other side• fully developed flow is when the velocity profile stop to develop as it flows deeper inside the pipe 15
  16. 16. 4. Entry Region (cont’)• hydrodynamic entry length, Lh is evaluated from pipe entrance to where wall shear stress achieve 2% of fully developed value or approximately Lh ,la min ar ≅ 0.05ReD Lh ,turbulent ≅ 10 D shorter length for turbulent pipe flow 16
  17. 17. 4. Entry Region (example)• Q4: Determine the hydrodynamic entry region for Q1 (a & c). Lh ,la min ar ≅ 0.05ReD Lh ,turbulent ≅ 10 D ρVave D 998(0.02)(0.02)a) Re = = = 398 • (a) Laminar flow µ 0.001002Lh ,la min ar ≅ 0.05ReD = 0.05 × 398 × 0.02 = 0.398 m ρVave D 998(0.02)(0.3)c) Re = = = 5970 • (c) Turbulent flow µ 0.001002Lh ,turbulent ≅ 10 D = 10 × 0.02 = 0.2 m 17
  18. 18. 4.1. Turbulence velocity profile VS Laminar velocity profile (fully developed)• Velocity profile based on analysis • Velocity profile is based on analysis and empirical• Consist of 1 layer • Consist of 4 layer• Small velocity gradient • High velocity gradient• The average velocity in fully developed laminar pipe flow is ½ of the maximum velocity u max = 2Vave 18
  19. 19. 5. Losses in piping system• There are two type of losses which is major losses and minor losses. Losses are mainly due to friction and obstruction• Total losses, hL is major losses + minor losses 5.1 Major Losses (pg345)• Major losses, hL, major , is also known pressure losses or head losses• The major losses is solely depend on the pipe, nothing else 19
  20. 20. 5.1 Major Losses (cont’)• Pressure drop across pipe can be formulated as dP P − P2 = 1 dx L 8µLVave 32 µLVave∆P = P − P2 = 1 2 = R D2 L ρVave 2 Pressure loss (Smooth or Rough) ∆PL = f D 2 ∆PL 2 L Vave major head loss, hL ,major = = f ρg D 2g• f, Darcy friction factor. There is another friction factor, Fanning friction factor, but will not be covered• for fully developed laminar flow, friction factor is obtained from combining Darcy equation with Pressure drop. 20
  21. 21. 5.1 Major Losses (cont’) 32 µLVave L ρVave 2 ∆P = ∆PL ⇒ 2 = f D D 2 32µ ρVave ρVave 32µ = f f = D 2 2 D ρVave D f = 32 × 2 µ f Re = 64 64 f = (friction factor for fully developed laminar flow) Re• from the eq. ,friction factor in Laminar flow is independent of roughness• The head loss represents the additional height that the fluid needs to be raised by a pump in order to overcome the frictional losses in the pipe 21
  22. 22. 5.1 Major Losses (cont’)• For fully developed turbulent flow pressure losses or head losses, the equation is more complex• Unlike laminar flow, friction factor, f , for turbulent flow is depend on internal pipe roughness and Reynolds number base on Colebrook equation. 1 ε D 2.51  = -2.0 log  +  (turbulent flow) f  3.7 Re f   • Whereas, ε is roughness of pipe and D is pipe diameter.• However this equation is implicit and cannot be solved directly. Using iteration (numerical method) is possible but the method is tedious 22
  23. 23. 5.1 Major Losses (cont’)• Hence, there are two option, (a), Moody chart, (b) Modified Colebrook equation f =0.025 ε/D=0.02 Re=100,000• In Moody Chart, friction factor can be obtain by knowing the roughness ratio, ε/D and Reynolds number 23
  24. 24. 5.1 Major Losses (cont’)• Haaland modified Colebrook Equation (1983). 1  6.9  ε D 1.11  = -1.8 log  +   (Modified for turbulent flow) f  Re  3.7     -2   1.11  f =  -1.8 log  6.9 +  ε D     (simplified)   Re  3.7     • Easier to calculate• The formula has about 2% error compare to original equation• Formula for major head loss for turbulent is the same 2 L Vave hL ,major = f D 2g 24
  25. 25. 5.1 Major Losses (cont’)• In real pipe application (turbulent flow), friction is unwanted because the rougher the surface, the higher the friction• Old piping system such as Cast Iron, GI, the performance deteriorate through time as corrosion reduce smoothness and size of internal pipe.• New piping (HDPE), Anti-corrosion metal pipes, smoothness maintain, thus performance is maintained 25
  26. 26. 5.1 Major Losses (example)• Q5: Water at 40°C (ρ = 992.1 kg/m3 and µ = 0.653×10-3 kg/m.s) is flowing steadily in a 5cm-diameter horizontal pipe made of stainless steel at a rate of 300 l/min. Determine the pressure drop, the head loss, and the required pumping power input for flow over a 50m-long section of the pipe. Q 300 / (1000)/60 Q = AcVavg → Vavg = = = 2.55m/s Ac π (0.05) / 4 2 ρVavg D 992.1(2.55)(0.05) Re = = = 193710, Turbulent µ 0.000653 Roughness ratio, ε D = 0.002mm / 50mm = 0.00004 from Moody Chart, ε D = 0.00004 & Re 1.94 ×105 , f ≈ 0.016 -2   6.9  0.00004    1.11 f =  −1.8log  +   = 0.016  193710  3.7       L ρVave 2 50 992.1(2.55) 2 ∆PL = f = 0.016 × × = 51609 Pa D 2 0.05 2 ∆PL 51609 hL = = = 5.3 m ρg 992.1(9.81) W = Q∆PL = 300 / 1000 / 60 × 51609 = 258 Watt 26
  27. 27. 5.1 Major Losses (example cont’)• Q6: Air at 40°C (ρ = 1.127 kg/m3, ν = 1.702×10-6 m2/s) is flowing steadily in a 50cm-diameter horizontal pipe made of plastic at a rate of 30 l/min. Determine the head loss, and the required pumping power input for flow over a 150m-long section of the pipe. Q 30 / (1000)/60 Q = AcVavg → Vavg = = = 2.55 ×10 −3 m/s Ac π (0.5) / 4 2 Vavg D (2.55 × 10 −3 )(0.5) Re = = = 748, Laminar ν 0.000001702 64 64 f = = = 0.0856 Re 748 2 L Vave 150(2.55 ×10 −3 ) 2 hL = f = 0.0856 = 8.51×10 −6 m D 2g 0.5(2)(9.81) ∆PL = ∆hL × ρg = 8.51×10 −6 (1.127)(9.81) = 9.41×10 −5 Pa W = Q∆PL = 30 / 1000 / 60 × 9.41×10 −5 = 4.71×10 −8 Watt 27
  28. 28. 5.2 Minor Losses• In piping system there are various fittings, valves, bends, elbows, tees, inlets, exits, enlargements, and contractions in addition to the pipes.• These component Interrupt the smoothness of flow and cause additional losses• Usually, these additional losses is called minor losses and smaller than pipe losses .• Rarely, minor losses will be greater than major losses especially when the component installed frequently along the pipe system (between short distance) 28
  29. 29. 5.2 Minor Losses (cont’) 29
  30. 30. 5.2 Minor Losses (cont’) 30
  31. 31. 5.2 Minor Losses (cont’) • usually expressed in terms of the loss coefficient, KL • KL , is provided by manufacturer and the value varies for different components. ρVavg 2 ∆PL = K L 2 • Consider a component, valve, the pressure losses is losses due to valve minus losses by imaginary pipe section without valves• Independent of Re, Reynolds number 31
  32. 32. 5.2 Minor Losses (cont’)• Usually expressed in term of head losses, hL 2 Vavg hL = K L = minor loss due to components 2g• Also, In industrial application, the manufacturing data is expressed in terms of equivalent length, Lequiv.• for this method, simply add Lequiv to the total length of pipe for the total head losses calculation.• However, the former expression will be used thoroughly 2 2 Vavg Lequiv Vavg D hL = K L = f → Lequiv = KL 2g D 2g f 32
  33. 33. 5.2 Minor Losses (cont’) • Component with sharp edge such as sharp edge exit has higher loss coefficient compare to well rounded • Sharp edge introduce recirculating flow due to fluid unable to make sharp 90° turn especially at high speed • Same for sudden expansion/contraction 33
  34. 34. 5.2 Minor Losses (cont’) 34
  35. 35. 5.2 Minor Losses (cont’)• These loss coefficient depends on the manufacturer data 35
  36. 36. 5.2 Minor Losses (cont’) 36
  37. 37. 5.2 Minor Losses (cont’) 37
  38. 38. Ball valve Globe valve Swing checkAngle valve valve 38
  39. 39. 5.2 Minor Losses (cont’)• Total head losses, hL, total 2  L  Vavg hL ,total = hL ,Major + hL ,Minor = f + ∑KL   D  2g• Calculation of Minor losses is straight forward. If the piping system consist same components such as bends, 2  the loss coefficient with the number of the simply multiply Li  Vavg,i hL ,total = ∑ f i  D + ∑K L i  2g  same bends.  i  (for pipe system with more than one pipe diameter) Vavg is different for different pipe diameter 39
  40. 40. 6 Piping system• Normally, as an engineer and consultant, piping system for example water storage, sprinkler system, hose reel system is the main concern.• To solve piping system, extended Bernoulli equation is required which the total losses is placed at the right hand side (at point 2) P (V1 ) 2 P2 (V2 ) 2 1 + + z1 = + + z 2 + hL ,total ρg 2g ρg 2g• Piping system usually constructed to deliver fluid at higher level or to create a pressurized system 40
  41. 41. 6 Piping system (cont’)• Two principles in analyzing piping system which area) Conservation of mass throughout the system must be satisfiedb) Pressure drop (and thus head loss) between two junctions must be the same for all paths between the two junctions 2 h 1 point 2, Just above water level,point 1, Just above water level, P2=?, V2 = ?, Z2 = ?P1=?, V1 = ?, Z1 = ? 41
  42. 42. 6 Piping system (example)• Q7:A piping system delivering water at 25°C from tank 1 to tank 2. The system consist two 45º, a sharp entrance and a sharp exit. The diameter of the stainless steel pipe is 2cm and length of 55 m. Determine h so that the flowrate is 83.3 L/min. 1 h=?m Tank 1 45° 2 Tank 2 45°point 2, Just above water level, point 1, Just above water level,P2=0, V2 = 0, Z2 = h P1=0, V1 = 0, Z1 = 0 42
  43. 43. 6 Piping system (example) Q 83.3 / (60000)Q = AcVavg → Vavg = = = 4.42m/s Ac π (0.01) 2 ρVavg D 997(4.42)(0.02)Re = = = 98916, Turbulent µ 0.891×10 -3Stainless steel, Roughness , ε = 0.002mm = 2 × 10 −6 mRoughness ratio , ε / D = 2 × 10 − 6 ÷ 0.02 = 0.0001 -2   6.9  0.0001  1.11  f =  −1.8log  +   = 0.018  98916  3.7      2  L  Vavg  55  (4.42) 2 hL ,major = f  =  0.018  = 49.29 m H 2O  D  2g  0.02  2(9.81) 43
  44. 44. 6 Piping system (example) 2 Vavg (4.42) 2hL ,minor = ( ∑K L ) = ( 0.5 + 2 × 0.4 +1) = 2.29 m H 2O 2g 2(9.81)hL ,total = hL ,Major + hL ,Minor = 49.29 + 2.29 = 51.58m 0 0 2 0 02 Reference point P (V1 ) P2 (V2 ) 1 + + z1 = + + z 2 + hL ,total ρg 2g ρg 2g0 + 0 + h = 0 + 0 + 0 + hL ,total h = 51.58 m 44
  45. 45. 6 Piping system (example)• Q8:A piping system delivering water at 25°C from pressurised tank 1 to tank 2. Initial plastic pipe diameter is 2 cm has a sharp entrance(inlet), fully open globe valve and two 90º smooth flanged bend, and a sudden expansion at the 1/3 of the pipe length. After expansion, Galvanized Iron (GI) pipe diameter is installed with 8 cm diameter and along the GI pipe, there are two 90º miter bend, a fully open angle valve and a sharp exit. The total pipe length is 75 m. Determine gauge pressure at tank 1 required to deliver water at 226 liter/hour. 2 GI pipe Tank 2 h=25m 1 Tank 1 Plastic point 2, Just above water level, pipe P2=0, V2 = 0, Z2 = hpoint 1, Just above water level, 45P1≠0, V1 = 0, Z1 = 0
  46. 46. 6 Piping system (example) Q 226 / (3600 × 1000)Q = Ac1Vavg1 → Vavg1 = = = 0.2m/s Ac1 π (0.01) 2 Q 226 / (3600000)Q = Ac 2Vavg2 → Vavg2 = = = 0.0125m/s Ac 2 π (0.04) 2 ρVavg1D 997(0.2)(0.02)Re1 = = = 4475, Turbulent µ 0.891× 10 -3 ρVavg2 D 997(0.0125)(0.08)Re 2 = = = 1119, Laminar µ 0.891×10-3Plastic pipe, Roughness , ε = 0, smooth pipes -2   1.11 f1 =  − 1.8log  6.9 +  0     = 0.039 (turbulent friction factor)   4475  3.7      64 64f2 = = = 0.057 (laminar friction factor) Re 1119 46
  47. 47. 6 Piping system (example) 2 2  D12   0.02 2  K L ,sudden expansion = 1 − 2  = 1 −    0.08 2  = 0.88   D2    2  L1  Vavg1hL ,total plastic =  f1  D + K L ,inlet + K L ,g.valve + 2 K L ,smooth flange bend + K L ,sudden expansion  2 g   1  2  75 / 3  0.2hL ,total plastic =  0.039 + 0.5 + 10 + 2(0.3) + 0.88  = 0.124m  0.02  2(9.81) 2  L2  Vavg2hL ,total GI  f2 = + 2 K L ,miter bend + K L ,angle valve + K L ,sharp exit   2g  D2   75 − 75 / 3  0.0125 2hL ,total GI =  0.057 + 2(1.1) + 5 + 1 = 0.00035m  0.08  2(9.81)hL ,total = hL ,total plastic + hL ,total GI = 0.124 + 0.00035 = 0.12435m 47
  48. 48. 6 Piping system (example) Reference point 0 2 0 02P (V1 ) P2 (V2 ) 1 + + z1 = + + z 2 + hL ,totalρg 2g ρg 2gP1 + 0 + 0 = 0 + 0 + 25 + 0.12435ρgP = 25.12435( 997 )( 9.81) = 245730Pa 1Most of the pressure at tank 1 is to overcomethe elevation difference as losses is very small 48
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