Page 1 ME-662 CONVECTIVE HEAT AND MASS TRANSFER A. W. Date Mechanical Engineering Department Indian Institute of Technology, Bombay Mumbai - 400076 India LECTURE-15 FULL Y -DEVELOPED LAMINAR FLOWS-1 () February 20, 2011 1 / 17 Page 2 ME-662 CONVECTIVE HEAT AND MASS TRANSFER A. W. Date Mechanical Engineering Department Indian Institute of Technology, Bombay Mumbai - 400076 India LECTURE-15 FULL Y -DEVELOPED LAMINAR FLOWS-1 () February 20, 2011 1 / 17 LECTURE-15 FULLY-DEVELOPED LAMINAR FLOWS-1 1 De?nition 2 Friction Factor - Circular Cross-section 3 Friction Factor - Annular Cross-section 4 Friction Factor - Rectangular and Annular Sectors () February 20, 2011 2 / 17 Page 3 ME-662 CONVECTIVE HEAT AND MASS TRANSFER A. W. Date Mechanical Engineering Department Indian Institute of Technology, Bombay Mumbai - 400076 India LECTURE-15 FULL Y -DEVELOPED LAMINAR FLOWS-1 () February 20, 2011 1 / 17 LECTURE-15 FULLY-DEVELOPED LAMINAR FLOWS-1 1 De?nition 2 Friction Factor - Circular Cross-section 3 Friction Factor - Annular Cross-section 4 Friction Factor - Rectangular and Annular Sectors () February 20, 2011 2 / 17 De?nition - L15( 1 15 ) 1 Fully-developed ?ow region occupies greater part of the tube length in ducts of large L / ( D * Re ). 2 Fully-developed ?ow friction factors f fd provide the lower bounds to the apparent f app and local f l friction factors. 3 In laminar ?ows, f fd × Re = const for the given duct 4 f fd is evaluated from force balance ? p× A c = t w × P× ? x where t w is average wall shear stress. Thus, f fd = t w ? u 2 /2 = 1 2 | dp dx | D h ? u 2 D h = 4× A c P 5 This is called the Fanning’s Friction Factor () February 20, 2011 3 / 17 Page 4 ME-662 CONVECTIVE HEAT AND MASS TRANSFER A. W. Date Mechanical Engineering Department Indian Institute of Technology, Bombay Mumbai - 400076 India LECTURE-15 FULL Y -DEVELOPED LAMINAR FLOWS-1 () February 20, 2011 1 / 17 LECTURE-15 FULLY-DEVELOPED LAMINAR FLOWS-1 1 De?nition 2 Friction Factor - Circular Cross-section 3 Friction Factor - Annular Cross-section 4 Friction Factor - Rectangular and Annular Sectors () February 20, 2011 2 / 17 De?nition - L15( 1 15 ) 1 Fully-developed ?ow region occupies greater part of the tube length in ducts of large L / ( D * Re ). 2 Fully-developed ?ow friction factors f fd provide the lower bounds to the apparent f app and local f l friction factors. 3 In laminar ?ows, f fd × Re = const for the given duct 4 f fd is evaluated from force balance ? p× A c = t w × P× ? x where t w is average wall shear stress. Thus, f fd = t w ? u 2 /2 = 1 2 | dp dx | D h ? u 2 D h = 4× A c P 5 This is called the Fanning’s Friction Factor () February 20, 2011 3 / 17 Circular Tube - 1 - L15( 2 15 ) 1 When ?ow is fully-developed, v r = v ? = ?u/?x =0 and dp / dx = const ( negative ) 2 Hence, the axial momentum equation reduces to 1 r ? ?r (r ?u ?r )= 1 µ d p d x = Constant (1) with boundary conditions, u = 0 at r = R ( tube wall ) and ?u/?r = 0 at r = 0 ( symmetry ). 3 Integrating equation 1 twice with respect to r and using bcs, u =- R 2 4µ d p d x (1- r 2 R 2 ) (2) 4 Hence, u = R R 0 u r d r R R 0 r d r =- R 2 8µ d p d x or u u = 2(1- r 2 R 2 ) (3) () February 20, 2011 4 / 17 Page 5 ME-662 CONVECTIVE HEAT AND MASS TRANSFER A. W. Date Mechanical Engineering Department Indian Institute of Technology, Bombay Mumbai - 400076 India LECTURE-15 FULL Y -DEVELOPED LAMINAR FLOWS-1 () February 20, 2011 1 / 17 LECTURE-15 FULLY-DEVELOPED LAMINAR FLOWS-1 1 De?nition 2 Friction Factor - Circular Cross-section 3 Friction Factor - Annular Cross-section 4 Friction Factor - Rectangular and Annular Sectors () February 20, 2011 2 / 17 De?nition - L15( 1 15 ) 1 Fully-developed ?ow region occupies greater part of the tube length in ducts of large L / ( D * Re ). 2 Fully-developed ?ow friction factors f fd provide the lower bounds to the apparent f app and local f l friction factors. 3 In laminar ?ows, f fd × Re = const for the given duct 4 f fd is evaluated from force balance ? p× A c = t w × P× ? x where t w is average wall shear stress. Thus, f fd = t w ? u 2 /2 = 1 2 | dp dx | D h ? u 2 D h = 4× A c P 5 This is called the Fanning’s Friction Factor () February 20, 2011 3 / 17 Circular Tube - 1 - L15( 2 15 ) 1 When ?ow is fully-developed, v r = v ? = ?u/?x =0 and dp / dx = const ( negative ) 2 Hence, the axial momentum equation reduces to 1 r ? ?r (r ?u ?r )= 1 µ d p d x = Constant (1) with boundary conditions, u = 0 at r = R ( tube wall ) and ?u/?r = 0 at r = 0 ( symmetry ). 3 Integrating equation 1 twice with respect to r and using bcs, u =- R 2 4µ d p d x (1- r 2 R 2 ) (2) 4 Hence, u = R R 0 u r d r R R 0 r d r =- R 2 8µ d p d x or u u = 2(1- r 2 R 2 ) (3) () February 20, 2011 4 / 17 Circular Tube - 2 - L15( 3 15 ) Further, wall shear stress is evaluated as t w =- µ ( ?u ?r ) r=R =- R 2 d p d x = 4µ u R (4) Hence, f fd = t w ? u 2 /2 = 1 2 | d p d x | D ? u 2 = 16 Re (5) Note that f fd × Re = 16 = const. Also, for a circular tube, D h = D and t w is circumferentially uniform. () February 20, 2011 5 / 17Read More

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