Answer Happy

I have attached the memo to the question, please clarify how the
calculations of the Ave Lubricant Temperature, Viscocity, and
others in the table were done. thanks
60 um A full journal bearing is required for a shaft with the following specifications: Nominal diameter 40 mm Diametric clearance Rotational speed (rpm) 1450 Load on the bearing 2100 N Inlet temperature of the lubricant 60°C Bearing length to diameter 1 Assume the lubricant is SAE 10 and an initial temperature change is 15°C. 1.1 Determine the overall temperature rise in the bearing 1.2 Determine the flow rate of the lubricant required. Show all your calculations. Note: If no calculations are shown your work will not be marked! [25]
D=40 mm (0) N=1450 rpm, Ns=24.17 rps W=2100 N T1=60°C Propose L/D=1 L=40 mm ✓ Hence c=30 um. , (2) this !! w 2100 P = 1.313x10Pa (2) LD 0.04 0.04 Reasonable loading. From Figure 4.10 for N = 1450 rpm and D=40 mm, 2C=60 um. *Most students did not get Try SAE 10 Assume AT=15°C Scrie students igueel this into!! ( 7) (3) Temperature rise, AT (°C) 15 10.5 10.8 Average lubricant temperature Tax (°C) 67.5 7 students 65.3 65.4 Average lubricant viscosity (Pa.s) 0.011 0.012 did not show Converged Sommerfield number, S 0.09 0.0982 that the calu Coefficient of friction variable (r/c)f 2.7 7 2.8 Flow variable Q/rcNsL 4.41 4.4 Side flow to total flow ratio Q,JQ 0.735 0.72 From Figure 4.11 with S=0.098 and L/D=1, the design is within the optimum zone for ( 2 ) did not minimum f and maximum W. answ Q=rcNsLx4.4 = 4.4x0.02 0.03x1024.17%0.04 mºs= 2552 mmºs! V/ (2) MOST had onverged! N mast Stucuts th