Answer:
i) 154°F
ii) 0.0114 mm
iii) 0.075 btu/s
iv) 0.080 in^3/s
Explanation:
<u>i)Determine the average film Temperature </u>
( 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
<u>ii) Calculate the minimum film thickness</u>
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
<u>iii)Determine the heat loss rate </u>
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
<u>iv)Calculate lubricant side-flow rate for minimum clearance assembly </u>
Side flow rate = 0.315 * Total volume flow rate
= 0.315 * ( 3.8 * 1.5 * 0.0015 * 10 * 3 )
= 0.080 in^3/s.
