Friction Factor
2
This sheet gives the value of friction factor f from the Colbrook-W Enter the parameters in the coloured boxes Colebrook white Moody Barr Enter the parameters here ks (mm) 0.03 d (mm) 300 Re 664000 Solution f λ Moody 0.003018 0.012072 Barr 0.003515 0.01406 Colebrook-White 0.003495 0.01398 Colebrook-White iteration f(n) a b LHS RHS Error 1 0.003018 2.69542E-05 3.45E-05 18.20297 16.84461 1.358352 2 0.003524 2.69542E-05 3.2E-05 16.84461 16.919 -0.07438 3 0.003493 2.69542E-05 3.21E-05 16.919 16.91484 0.004157 4 0.003495 2.69542E-05 3.21E-05 16.91484 16.91507 -0.00023 5 0.003495 2.69542E-05 3.21E-05 16.91507 16.91506 1.3E-05 6 0.003495 2.69542E-05 3.21E-05 16.91506 16.91506 -7.2E-07 7 0.003495 2.69542E-05 3.21E-05 16.91506 16.91506 4.04E-08 8 0.003495 2.69542E-05 3.21E-05 16.91506 16.91506 -2.3E-09 9 0.003495 2.69542E-05 3.21E-05 16.91506 16.91506 1.26E-10 10 0.003495 2.69542E-05 3.21E-05 16.91506 16.91506 -7E-12 11 0.003495 2.69542E-05 3.21E-05 16.91506 16.91506 3.94E-13 12 0.003495 2.69542E-05 3.21E-05 16.91506 16.91506 0 13 0.003495 2.69542E-05 3.21E-05 16.91506 16.91506 0 1 √ f =−4 log 10 ( k s 3.71 d + 1.26 Re √ f ) f n+1 =1 / [ −4log 10 ( k s 3.71 d + 1.26 Re √ f n ) ] 2 f n+1 = 1 [ − 4log 10 ( a +b) ] 2 f=0 . 001375 [ 1+ ( 200 k s d + 10 6 Re ) 1/ 3 1 √ f =−4 log 10 [ k s 3.71 d + 5.1286 Re 0.89 ] f=1 / [ −4log 10 ( k s 3.71 d + 5.1286 Re 0.89 ) ] 2
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Friction Factor
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This sheet gives the value of friction factor f from the Colbrook-White, Moody and Barr formuleaEnter the parameters in the coloured boxesColebrook white MoodyBarrEnter the parameters hereks (mm) 0.03d (mm) 300Re 664000olution f !Moody 0.003018 0.012072Barr 0.0031 0.01406Colebrook-White 0.0034! 0.013!8Colebrook"#hite iteratio$%($) a b &'( R'( )rror *1 0.003018 2.6!42)"00 3.)"00 18.202!7 16.84461 1.3832 0.0120722 0.00324 2.6!42)"00 3.2)"00 16.84461 16.!1! "0.074384 0.0140!73 0.0034!3 2.6!42)"00 3.2)"00 16.!1! 16.!1484 0.00417 0.013!744 0.0034! 2.6!42)"00 3.2)"00 16.!1484 16.!107 "0.000232 0.013!81 0.0034! 2.6!42)"00 3.2)"00 16.!107 16.!106 1.3)"00 0.013!86 0.0034! 2.6!42)"00 3.2)"00 16.!106 16.!106 "7)"007 0.013!87 0.0034! 2.6!42)"00 3.2)"00 16.!106 16.!106 4.0)"008 0.013!88 0.0034! 2.6!42)"00 3.2)"00 16.!106 16.!106 "2)"00! 0.013!8! 0.0034! 2.6!42)"00 3.2)"00 16.!106 16.!106 1.3)"010 0.013!810 0.0034! 2.6!42)"00 3.2)"00 16.!106 16.!106 "7)"012 0.013!811 0.0034! 2.6!42)"00 3.2)"00 16.!106 16.!106 3.!)"013 0.013!812 0.0034! 2.6!42)"00 3.2)"00 16.!106 16.!106 0 0.013!813 0.0034! 2.6!42)"00 3.2)"00 16.!106 16.!106 0 0.013!81f =4 log10(ks3. 71d + 1. 26Re f )f n+1=1/[4 log10(ks3. 71d+ 1. 26Ref n )]2f n+1=1[4 log10 ( a+b)]2f =0. 001375[1+(200ksd+106Re )1/ 3]1f =4 log10[ks3. 71d+5. 1286Re0. 89 ]f =1/[4 log10(ks3. 71d +5. 1286Re0 . 89 )]2This sheet gives the value of friction factor f from the Colbrook-White, Moody and Barr formuleaf =0. 001375[1+(200ksd+106Re )1/ 3]1f =4 log10[ks3. 71d+5. 1286Re0. 89 ]f =1/[4 log10(ks3. 71d +5. 1286Re0 . 89 )]2
