A full journal bearing has a shaft diameter of 3.000 in with a unilateral tolerance of -0.0004 in. The l/d ratio is unity. The bushing has a bore diameter of 3.003 in with a unilateral tolerance of 0.0012 in. The SAE 40 oil supply is in an axial-groove sump with a steady-state temperature of 140o F. The radial load is 675 lbf. Estimate the average film temperature, the minimum film thickness, the heat loss rate, and the lubricant side-flow rate for the minimum clearance assembly, if the journal speed is 10 rev/s.

i) 154°F

ii) 0.0114 mm

iii) 0.075 btu/s

iv) 0.080 in^3/s

Explanation:

i)Determine the average film Temperature

( from Viscosity-temperature chart in US customary units for SAE10 )

at Temp =  154°F

absolute viscosity = 4.25 rev

and ΔT = 2 ( 154 - operating temp ) = 28°F

where : operating temp = 140°F as given in question

also from the chart applying Raimondi and Boyd boundary conditions

ΔT = 29°F  hence we can pick 154°F as the average film temperature

ii) Calculate the minimum film thickness

Cmin = Bore diameter - Journal shaft diameter / 2

         = 3.003 - 3 / 2 = 0.0015 in

Given that : h₀ / Cmin = 0.76

there h₀ = 0.0015 * 0.76 = 0.0114 mm

iii)Determine the heat loss rate

Also known as power loss ratio ( H ) = ( 2π*w*f*r*N ) / ( 778 *12 )

( 2π * 0.0132 * 675 * 1.5 * 10 ) / ( 9336 )

Heat loss rate = 0.075 btu/s

iv)Calculate lubricant side-flow rate for minimum clearance assembly

Side flow rate = 0.315 * Total volume flow rate

                       = 0.315 * ( 3.8 * 1.5 * 0.0015 * 10 * 3 )

                      = 0.080 in^3/s.

answer: i have no clue tbh

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