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 V 1 2 / 2 + p 1 /  + gz 1 =  V 2 2 /2 + p 2 /  + gz 2 + h lT

 V 1 2 / 2 + p 1 /  + gz 1 =  V 2 2 /2 + p 2 /  + gz 2 + h lT. H E A D L O S S. h lT = h l + h m. Convenient to break up energy losses, h lT , in fully developed pipe flow to major loses, h l , due to frictional effects along the pipe and minor losses, h lm ,

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 V 1 2 / 2 + p 1 /  + gz 1 =  V 2 2 /2 + p 2 /  + gz 2 + h lT

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  1. V12/ 2 + p1/ + gz1= V22/2 + p2/ + gz2 + hlT H E A D L O S S hlT = hl + hm

  2. Convenient to break up energy losses, hlT, in fully developed pipe flow to major loses, hl, due to frictional effects along the pipe and minor losses, hlm, associated with entrances, fittings, changes in area,… Minor losses not necessarily < Major loss , hl, due to pipe friction.

  3. Minor losses traditionally calculated as: hlm = KV2/2 (K for inlets, exits, enlargements and contractions) where K is the loss coefficient or hlm = (Cpi – Cp)V2/2 (Cpi & Cp for diffusers) where Cp is the pressure recovery coefficient or hlm = f(Le/D)V2/2 (Le for valves, fittings, pipe bends) where Le is the equivalent length of pipe. Both K and Le must be experimentally determined and will depend on geometry and Re, uavgD/. At high flow rates weak dependence on Re.

  4. V1avg2/ 2 + p1/ + gz1= V2avg2/2 + p2/ + gz2 + hlT hlT = hl + hm;hlm = KV2/2 inlets, sudden enlargements & contractions; gradual contractions and exits

  5. Minor losses due to inlets: hlm= p/ = K(V2/2); V2 = mean velocity in pipe If K=1, p =  V2/2

  6. V1avg2/ 2 + p1/ + gz1= V2avg2/2 + p2/ + gz2 + hlT hlT = hl + hm

  7. Head is lost because of viscous dissipation when flow is slowed down (2-3) and in violent mixing in the separated zones For a sharp entrance ½ of the velocity head is lost at the entrance!

  8. vena contracta separation K = 0.78 unconfined mixing as flow decelerates r K = 0.04 r/D > 0.15 D

  9. V12/ 2 + p1/ + gz1= V22/2 + p2/ + gz2 + hlT hlT = hl + hlm;hlm = KV2/2 inlets,sudden enlargements & contractions; gradual contractions and exits

  10. Minor losses due to sudden area change: hlm= p/ = K(V2/2); V2 = faster mean velocity pipe • hlm head losses are primarily due to separation • Energy is dissipated deceleration after separation leading to violent mixing in the separated zones

  11. NOTE SOME BOOKS (Munson at al.): hlm = K V2/(2g) our Hlm!!!

  12. AR < 1 AR < 1 hlm = ½ KV2fastest V1 V2

  13. AR < 1 Why is Kcontraction and Kexpansion = 0 at AR =1?

  14. V1avg2/ 2 + p1/ + gz1= V2avg2/2 + p2/ + gz2 + hlT hlT = hl + hlm;hlm = KV2/2 inlets,sudden enlargements & contractions; gradual contractions and exits

  15. Entire K.E. of exiting fluid is dissipated through viscous effects, V of exiting fluid eventually = 0 so K = 1, regardless of the exit geometry. hlm = KV2/2 hydrogen bubbles hydrogen bubbles Only diffuser can help by reducing V. Water, velocity = 14 cm/s, width of opening = 30 mm, Re = 4300

  16. V2 ~ 0 Which exit has smallest Kexpansion?

  17. K =1.0 K =1.0 K =1.0 K =1.0 MYO

  18. V1avg2/ 2 + p1/ + gz1= V2avg2/2 + p2/ + gz2 + hlT hlT = hl + hlm;hlm = KV2/2 inlets,sudden enlargements & contractions; gradual contractions and exits

  19. GRADUAL CONTRACTION AR < 1 Where average velocity is fastest

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  21. V1avg2/ 2 + p1/ + z1=V2avg2/2 + p2/ + gz2 + hlT hlT = hl + hlm;hlm = p/ = (Cpi – Cp) 1/2V12 gentle expansions ~ diffusers

  22. ugly DIFFUSERS 20 cm/sec 3 cm/sec bad good

  23. assume fully developed ….. ? > < = P1 P2 Fully developed laminar flow, is: P1 greater, less or equal to P2? What if fully developed turbulent flow? What if developing flow?

  24. P1, V1 P3, V3 P1, V1 P2, V2 Is P2 greater, less than or equal to P1? Is P2 greater, less than or equal to P1? Is P likely to be greater, less than or equal to P?

  25. DIFFUSERS Diffuser data usually presented as a pressure recovery coefficient, Cp, Cp = (p2 – p1) / (1/2 V12 ) Cp indicates the fraction of inlet K.E. that appears as pressure rise [ hlm = p/ = (Cpi – Cp) 1/2V12] The greatest that Cp can be is Cpi, the case of zero friction.

  26. DIFFUSERS Diffuser data usually presented as a pressure recovery coefficient, Cp, Cp = (p2 – p1) / (1/2 V12 ) Cp indicates the fraction of inlet K.E. that appears as pressure rise [ hlm = p/ = (Cpi – Cp) 1/2V12] Cp will get from empirical data charts. It is not difficult to show that the ideal (frictionless) pressure recovery coefficient is: Cpi = 1 – 1/AR2, where AR = area ratio

  27. Cp = (p2 – p1) / (1/2 V12 ) Cpideal = 1 – 1/AR2 AR = A2/A1 > 1 p1 + ½ V12 = p2 + ½ V22 (BE - ideal) p2/ – p1/ = ½ V12 - ½ V22 A1V1 = A2V2 (Continuity) V2 = V1 (A1/A2) p2/ – p1/ = ½ V12 - ½ [V1(A1/A2)]2 p2/ – p1/ = ½ V12 - ½ V12(1/AR)2 (p2 – p1)/( ½  V12) = 1 – 1/AR2 Cpi= 1 – 1/AR2

  28. Relating Cp to Cpi and hlm p1 / + ½ V12 = p2/ + ½ V22 + hlm (z1 = z2 = 0) hlm = V12/2 - V22/2 – (p2 – p1)/ hlm = V12/2 {1 + V22/V12 – (p2 – p1)/( 1/2V12)} A1V1 = A2V2 Cp = (p2 – p1)/( 1/2V12) (Cp is positive & < Cpi) hlm = V12/2 {1 - A12/A22 – Cp} Cpi = 1 – 1/AR2 hlm = V12/2 {Cpi – Cp} Q.E.D. (see Ex. 8.10)

  29. hlm = (Cpi – Cp)V2/2; Cpi= 1 – 1/AR2 Cp

  30. N/R1 = 0.45/(.15/2) = 6 * AR ~ 2.7 Cp 0.62 Pressure drop fixed, want to max Cp to get max V2; minimum hlm

  31. If flow too fast or angle too big may get flow separation. Cp for Re > 7.5 x 104, “essentially” independent of Re

  32. V1avg2/ 2 + p1/ + z1=V2avg2/2 + p2/ + gz2 + hlT hlT = hl + hlm; hlm = f(Le/D)V2/2 valves and fittings

  33. hlm = f(Le/D)V2/2

  34. Head loss of a bend is greater than if pipe was straight (again due to separation).

  35. Nozzle Problem

  36. A Neglecting friction, is flow faster at A or B or same?

  37. = 0 V12/ 2 + p1/ + z1=V22/2 + p2/ + gz2 + hlT VA2/ 2 + patm/ + d = VB2/ 2 + patm/ + d A If flow at B did not equal flow at A then could connect and make perpetual motion machine. A

  38. C d

  39. C d 0 V12/ 2 + p1/ + z1=V22/2 + p2/ + gz2 + hlT = 0 = 0 VT2/ 2 + patm/ + d = VC2/ 2 + patm/ + d

  40. ? neglect friction V12/ 2 + p1/ + z1=V22/2 + p2/ + gz2 + hlT

  41. C d Nozzle 0 V12/ 2 + p1/ + z1=V22/2 + p2/ + gz2 + hlT = 0 = 0 VT2/ 2 + patm/ + d = VC2/ 2 + patm/ + d

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  43. Pipe Flow Examples ~ Solving for pressure drop in horizontal pipe

  44. V1avg2/2 + p1/ + gz1 – (V2avg2/2 + p2/ + gz2) = hlT = hl + hlm = f [L/D][V2/2] +  f [Le/D][V2/2] + K[V2/2] Laminar flow ~ f = 64/ReD Turbulent flow ~ 1/f0.5 = -2.0 log{(e/D)/3.7 + 2.51/(ReD f0.5) (f = 0.316/ReD0.25 for Re < 105) p2- p1 = ?; Know hlT , L, D, Q, e,  , , z2, z1

  45. p2- p1 = ?; Know L, D, Q, e,  , , z2, z1 Compute the pressure drop in 200 ft of horizontal 6-in-diameter asphalted cast-iron pipe carrying water with a mean velocity of 6 ft/s.* V1avg2/2 + p1/ + gz1 - V2avg2/2 - p2/ - gz2 = f [L/D][V2/2] +  f [Le/D][V2/2] + K[V2/2] p1/ - p2/ = f [L/D][V2/2] = hlm

  46. p2- p1 = ?; Know L, D, Q, e,  , , z2, z1 p1/ - p2/ = f [L/D][V2/2] = hl f(Re, e/D); ReD = 270,000 & e/D = 0.0008 1/f0.5 = -2.0 log{(e/D)/3.7 + 2.51/(ReD f0.5); f = 0.0197 f ~ 0.02 p2 – p1 = hl = f(Re, e/D)[L/D][V2/2] = 280 lbf/ft2

  47. Pipe Flow Examples ~ Solving for pressure drop in non-horizontal pipe

  48. p2- p1 = ?; Know L, D, Q, e,  , , z2, z1 Oil with  = 900 kg/m3 and = 0.00001 m2/s flows at 0.2 m3/s through 500m of 200 mm-diameter cast iron pipe. Determine pressure drop if pipe slopes down at 10o in flow direction. V1avg2/2 + p1/ + gz1 - V2avg2/2 - p2/ - gz2 = f [L/D][V2/2] +  f [Le/D][V2/2] + K[V2/2] p1/ + gz1 - p2/ - gz2= f [L/D][V2/2] = hlm

  49. p2- p1 = ?; Know L, D, Q, e,  , , z2, z1 p1/ + gz1 - p2/ - gz1= f [L/D][V2/2] f(Re, e/D); ReD = 128,000 & e/D = 0.0013 1/f0.5 = -2.0 log{(e/D)/3.7 + 2.51/(ReD f0.5) f = 0.0227 f ~ 0.023 p2 – p1 = hl - g500(sin 10o) = 265,000 kg/m-s2

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