Given:
μ=1.2poise=0.12N⋅s/m2
Diameter of shaft D=10cm=0.1m
Distance between shaft and journal bearing dy=1.0mm=10−3m
Speed of shaft N=200 r.p.m.
Tangential speed of shaft is given by
u=60πDN=60π⋅0.1m⋅200 r.p.m.=3πm/s Change of velocity between shaft and bearing
du=u−0=3πm/s Mathematically
τ=μdydu=0.12N⋅s/m2×10−3m3πm/s=
=40πN/m2≈125.6637N/m2
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