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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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**V12/ 2 + p1/ + gz1= V22/2 + p2/ + gz2 + hlT**H E A D L O S S hlT = hl + hm**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.**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.**V1avg2/ 2 + p1/ + gz1= V2avg2/2 + p2/ + gz2 +**hlT hlT = hl + hm;hlm = KV2/2 inlets, sudden enlargements & contractions; gradual contractions and exits**Minor losses due to inlets:**hlm= p/ = K(V2/2); V2 = mean velocity in pipe If K=1, p = V2/2**V1avg2/ 2 + p1/ + gz1= V2avg2/2 + p2/ + gz2 +**hlT hlT = hl + hm**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!**vena contracta**separation K = 0.78 unconfined mixing as flow decelerates r K = 0.04 r/D > 0.15 D**V12/ 2 + p1/ + gz1= V22/2 + p2/ + gz2 + hlT**hlT = hl + hlm;hlm = KV2/2 inlets,sudden enlargements & contractions; gradual contractions and exits**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**NOTE SOME BOOKS (Munson at al.):**hlm = K V2/(2g) our Hlm!!!**AR < 1**AR < 1 hlm = ½ KV2fastest V1 V2**AR < 1**Why is Kcontraction and Kexpansion = 0 at AR =1?**V1avg2/ 2 + p1/ + gz1= V2avg2/2 + p2/ + gz2 +**hlT hlT = hl + hlm;hlm = KV2/2 inlets,sudden enlargements & contractions; gradual contractions and exits**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**V2 ~ 0**Which exit has smallest Kexpansion?**K =1.0**K =1.0 K =1.0 K =1.0 MYO**V1avg2/ 2 + p1/ + gz1= V2avg2/2 + p2/ + gz2 +**hlT hlT = hl + hlm;hlm = KV2/2 inlets,sudden enlargements & contractions; gradual contractions and exits**GRADUAL CONTRACTION**AR < 1 Where average velocity is fastest**V1avg2/ 2 + p1/ + z1=V2avg2/2 + p2/ + gz2 + hlT**hlT = hl + hlm;hlm = p/ = (Cpi – Cp) 1/2V12 gentle expansions ~ diffusers**ugly**DIFFUSERS 20 cm/sec 3 cm/sec bad good**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?**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?**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.**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**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**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)/( 1/2V12)} A1V1 = A2V2 Cp = (p2 – p1)/( 1/2V12) (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)**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**If flow too fast or angle too big may get flow separation.**Cp for Re > 7.5 x 104, “essentially” independent of Re**V1avg2/ 2 + p1/ + z1=V2avg2/2 + p2/ + gz2 + hlT**hlT = hl + hlm; hlm = f(Le/D)V2/2 valves and fittings**Head loss of a bend is greater**than if pipe was straight (again due to separation).**A**Neglecting friction, is flow faster at A or B or same?**= 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**C**d**C**d 0 V12/ 2 + p1/ + z1=V22/2 + p2/ + gz2 + hlT = 0 = 0 VT2/ 2 + patm/ + d = VC2/ 2 + patm/ + d**?**neglect friction V12/ 2 + p1/ + z1=V22/2 + p2/ + gz2 + hlT**C**d Nozzle 0 V12/ 2 + p1/ + z1=V22/2 + p2/ + gz2 + hlT = 0 = 0 VT2/ 2 + patm/ + d = VC2/ 2 + patm/ + d**Pipe Flow Examples ~**Solving for pressure drop in horizontal pipe**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**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**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**Pipe Flow Examples ~**Solving for pressure drop in non-horizontal pipe**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**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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