058 0160 Professor Fred Stern Fall 2009 Chapter 6 part4 1 Chapter 6 Viscous Flow in Ducts 6 4 Turbulent Flow in Pipes and Channels using meanvelocity correlations 1 Smooth circular pipe Recall laminar flow exact solution 8 uave d f 2w 64 Re d Re d 2000 uave A turbulent flow approximate solution can be obtained simply by computing uave based on log law u 1 yu ln B u v Where u u y 0 41 B 5 u w y R r V uave Or R 1 Q yu 1 2 u ln B 2 r dr A R 0 v 1 2 Ru 3 u ln 2 B 2 v 058 0160 Professor Fred Stern Chapter 6 part4 2 Fall 2009 V Ru 2 44 ln 1 34 u v f 1 2 1 99 log Re d f 1 2 1 02 2 log Re d f 1 2 0 8 EFD Adjusted constants f only drops by a factor of 5 over 104 Re 108 Since f equation is implicit it is not easy to see dependency on V and D 4000 ReD 105 Blasius 1911 power law curve fit to data f pipe 0 316 Re D1 4 L v2 f hf D 2g p 3 4 1 4 5 4 7 4 p 0 158 L D V Turbulent Flow Nearly linear Only slightly with 0 241L 3 4 1 4 D 4 75Q1 75 Laminar flow p 8 LQ R 4 Near quadratic as expected Drops weakly with pipe size 058 0160 Professor Fred Stern Chapter 6 part4 3 Fall 2009 p turbulent decreases more sharply than p laminar for same Q therefore increase D for smaller p 2D decreases p by 27 for same Q umax u r 0 1 Ru ln B u u Combine with 1 Ru 3 V ln B 2 u u max V u max 3 3u 3u V u max 1 2 2 V 2 V u u Also 2 u w u w u and f f 2 2 1 8 V 1 8 V V 2 umax 3u 3 1 1 V 2 V 2 Or f 8 1 1 3 f V For Turbulent Flow u 1 1 3 f max Recall laminar flow V u max 0 5 1 f 8 058 0160 Professor Fred Stern Chapter 6 part4 4 Fall 2009 2 Turbulent Flow in Rough circular pipe U f y k U 1 f f Re d k d ln y B B k Log law shifts downward which leads to three roughness regimes 1 k 4 2 4 k 60 3 k 60 f hydraulically smooth transitional roughness Re dependence full rough no Re dependence k d Moody diagram 2 51 2 log 1 2 3 7 Re d f 6 9 k d 1 11 Approximate explicit 1 8 log formula Re 3 7 d 1 2 There are basically four types of problems involved with uniform flow in a single pipe 1 Determine the head loss given the kind and size of pipe along with the flow rate Q A V 2 Determine the flow rate given the head kind and size of pipe 3 Determine the pipe diameter given the type of pipe head and flow rate 058 0160 Professor Fred Stern Chapter 6 part4 5 Fall 2009 4 Determine the pipe length given Q d hf ks g 1 Determine the head loss The first problem of head loss is solved readily by obtaining f from the Moody diagram using values of Re and ks D computed from the given data The head loss hf is then computed from the Darcy Weisbach equation f f ReD ks D p p h z1 z 2 1 2 L V2 hf f h D 2g p z ReD ReD V D 2 Determine the flow rate The second problem of flow rate is solved by trial using a successive approximation procedure This is because both Re and f Re depend on the unknown velocity V The solution is as follows 1 solve for V using an assumed value for f and the DarcyWeisbach equation 1 2 2gh f 1 2 V f L D known from given data note sign 058 0160 Professor Fred Stern Chapter 6 part4 6 Fall 2009 2 using V compute Re 3 obtain a new value for f f Re ks D and reapeat as above until convergence 1 2 Or can use Re f D 3 2 2 gh f L 1 2 scale on Moody Diagram 1 compute 2 read f Re f 1 2 and ks D L V2 3 solve V from h f f D 2g 4 Q VA 3 Determine the size of the pipe The third problem of pipe size is solved by trial using a successive approximation procedure This is because hf f and Q all depend on the unknown diameter D The solution procedure is as follows 1 solve for D using an assumed value for f and the DarcyWeisbach equation along with the definition of Q 8LQ 2 D 2 gh f 1 5 known from given data f 1 5 058 0160 Professor Fred Stern Chapter 6 part4 7 Fall 2009 2 using D compute Re and ks D 3 obtain a new value of f f Re ks D and repeat as above until convergence 4 Determine the pipe length The four problem of pipe length is solved by obtaining f from the Moody diagram using values of Re and ks D computed from the given data Then using given hf V D and calculated f to solve L from L 2 g Dh f V2 f 058 0160 Professor Fred Stern Fall 2009 Chapter 6 part4 8 058 0160 Professor Fred Stern Chapter 6 part4 9 Fall 2009 3 Concept of hydraulic diameter for noncircular ducts For noncircular ducts w f perimeter thus new 8 w 2 w f C definitions of and are required f 2 V 2 V Define average wall shear stress 1P w w ds P0 ds arc length P perimeter Momentum z pA w PL AL 0 L W h p z wL A P A P Rh Hydraulic radius R 2 for circular pipe and h wL R 2 058 0160 Professor Fred Stern Chapter 6 part4 10 Fall 2009 Energy h hL w wL A P A h A dh A d p z A d p P L P dx P dx P dx non circular duct Recall for circular pipe w R dp D dp 2 dx 4 dx In analogy to circular pipe Dh 4A d p A Dh A dp Dh w 4 P P P dx 4 dx Hydraulic diameter For multiple surfaces such as concentric annulus P and A based on wetted perimeter and area f 8 w f Re Dh Dh V 2 Re Dh VDh w L V 2 f L L V2 h hL f Rh 8 Rh Dh 2 g However accuracy not good for laminar flow 40 and marginal turbulent flow 15 058 0160 Professor Fred Stern Chapter 6 …
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