Oil SG089 enters at section 1 in Fig 1 at a weight flow of 2

Oil (SG_0.89) enters at section 1 in Fig. 1 at a weight flow of 250 N/h to lubricate a thrust bearing. The steady oil flow exits radially through the narrow clearance between thrust plates. Compute the outlet volume flux in mL/s and the average outlet velocity in cm/s

Solution

The specific weight of the oil is (0.89)(9790) = 8713.1 N/m3. Then
Q2 =Q1 = (250/3600)/8713.1 = 7.97 *10^-6 m^3/s = 7.97 mL/s
But Q2=V₂ (0.1 m)(0.002 m)= 7.97*10^-6

solve for V2 = 1.268 cm/s