Question

01 ) During a study of a certain flow system the following equation was developed: P2=C+4.6'f LV P. P.Dg where pi an p2 are pressure between points 1 and 2, C is dimensionless constant, V is velocity, L is the distance between the points, D is the diameter, g is acceleration of gravity and f is drag coefficient. Determine the primary dimensions of the coefficient f so that this equation is homogenous. b) A 25-mm-diameter shaft is pulled through a cylindrical bearing of 25.5- mm inner diameter as shown. The lubricant that fills the 0.3-mm gap between the shaft and bearing is an D=25,5 mm oil having a kinematic viscosity of 8.0x10m²/s and a specific gravity of 0.91. Determine the force P required to pull the shaft at a velocity of 3 m/s. Assume the velocity distribution in the gap is linear. 13 III 11

Answer 1

Q.! P c + 4.7FLV Pi og is dimensionless :: P C there for 4.6°FLV . Pog V also should be dimensionless to satisfy dimensional homogenity of eguation. let Dimension of f be I LOMYTJ . 406 [L*M°T2] ['mºt°][L'M° T4) FILM TOJ [1'M' 7*2] [Lim?7•J [ L'Mo Tre] ... 4.6° [**? MY 72"] - [L'Mºre] [['m' try :. 4.? [Let'm yy 72+] = [L"Mºtº] equating powers of L.M and T on both sides, x + 1 = 0 = 1 =- у= =6 = 9 =) 2+3 =O → Z=-3 of f = [[X MY T] [f] = [ L'M'T3] : Dimension

Given: n. 1) diameter of shaft, d= 25 mm (2) Diameter of bearing iD=25.5 mm (3) length of beaning ,l=0.5 m = 500 mm (4) Oil film thickness, h = 0.3 mm (5) kinematic viscosity L = 8X10 m2 (6) specific gravity, so G=0.9, (7) velocity , V = 3 m = 3000 mm Solution Beaning A 16 S shaft e ce e velocity profile Clynamic Viscosity, U=vxs = Le x 0.91 X1000 i u = 8810-40.91 X 1000 = 0.728 NS By Newton's law of viscosity, shear stress, t= P = UX V . for linear A ħ relocity profile mz :: PE UX AX = UX TTd Ox W : p= 0.728 x TTX 0.025 x 0.5 X 3 0.38 103 P= 285.88 N