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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | fp Dimension-based coefficient, lin ed ith actual applied load | | | | | | | | | | ■P 1 0,9 0,8 0,7 0,6 0,5 0,4 0,3 0,2 0,1 0 | | | | | | Admissible operating thermal speed The admissible operating speed nadm is the rotating speed at which the mean bearing temperature reaches the admissible limit value, in actual operating conditions. | | | | | | | | | | 0 | | 50 | | 100 | | 150 | | 200 | | 250 | | 300 | | 350 | | 400 | | 450 | | 500 dm (mm) | | | | | | | | | | | | | | | | | | fv Load-based coefficient, lin ed ith actual lubricant viscosity | | | | | | | | | | 1,2 | | | | | | | | | | To obtain an approximate calculation of the operating thermal speed, multiply the reference speed n9r by the coefficients fp and fv that originate from the table opposite. nadm = n9r* fpX fv To perform a more accurate analysis based on standards, use the formulae listed below. | | | | | | 1,1 | | | | | | | | | | 0,9 | | | | | | | | | | 0,8 | | | | | | | | | | 0,7 | | | | | | | | | | 0,6 | | | | | | | | | | 0,5 | | | | | | | | | | | | | | 0 0,05 0,1 0,15 0,2 0,25 0,3 0,35 0,4 0,45 0,5 PC | | | | | | | | | | athematic calculation based on Standards The standard permits a calculation based on all parameters which can be indicated, in lieu of the reference values. In particular, the following parameters can be taken into account: - actual load P, - deviation between ambient temperature and internal bearing temperature AG, - actual viscosity v. | | | | | | | | | | The equation to determine nadm is ^ X nadm X 10-7forX (V X nadm)2 " X dm3 + fir X Px dm = 103 X qrX Ar 30 dm = mean diameter (D+d)/2 Ar = bearing journal area Ar = 71 (D+d)B qr = 0.016 (AG/50) (Ar/50000)-034w/mm2 if Ar 50000 mm2 f1r: friction coefficient linked to load, drawn from ISO 15312 standard, AppendiX A (values for information) f0r: friction coefficient linked with speed, drawn from ISO 15312 standard, AppendiX A (values for information) An iterative process is necessary to solve this equation. tarting from a ero or low initial value for nadm the nadm value is incremented until equality is obtained. | | | | | | | | | | 17 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
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