Lecture 15.pptx - Lecture 15 LAMINAR FLOW THROUGH...

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LAMINAR FLOW THROUGH NON-CIRCULAR CHANNEL Q= (P 1 -P 2 ).( .D o 4 )/( x. .128) Circular tube Q= (P 1 -P 2 ).( lh 3 )/( x. .12) Parallel plates Q= (P 1 -P 2 ). .(D o 2 – Di 2 ) [(D o 2 +D i 2 ) - D o 2 -D i 2 /ln(D o /D i )]/( x. .128) Annulus Lecture 15
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TURBULENT FLOW THROUGH NON-CIRCULAR CHANNEL Pressure force = shear force P. (area perpendicular to flow)= (wall shear stress). (wetted perimeter). x P/ x = . (wall shear stress)/ (area perpendicular to flow) HR= area perpendicular to flow/ wetted perimeter= D/4 circular pipe  P/ x) non-circular /  P/ x) circular = (1/HR)/(4/D)=D/4HR f non-circular =f. x.V 2 /2.HR For Re and /D, replace D with 4HR
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EXAMPLE PROBLEM- Air at 1 atm and 68 o F flowing in a long rectangular duct whose cross section is 1 ft x 0.5 ft with a velocity of 40 ft/s. The roughness of the duct is 0.00006 in. What is the pressure drop per unit length? air = 0.075 lbm/ft 3 air =0.018 cP Conversion factor cP=6.72 .10 -4 lbm/ft.s
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FLOW AROUND SUBMERGED OBJECTS Drag force = F = .r 2  air .V 2 /2 F/A=C d .  .V 2 /2 (C d =drag coefficient) R p = particle diameter. velocity. fluid density/fluid viscosity
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