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Question

# A lightly loaded full journal bearing has journal diameter of 50 mm, bush bore of 50.05 mm and bush length of 20 mm. If rotational speed of journal is 1200 rpm and average viscosity of liquid lubricant is 0.03 Pa s, the power loss (in W) will be

A
118
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B
74
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C
237
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D
37
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Solution

## The correct option is D 37Tangential velocity of shaft, u=πDN60=π×50×10−3×120060 =3.14 m/s Clearance, y=50.05−502=0.025 mm By τ=μdudy Shear stress on shaft τ=0.03×3.140.025×10−3=3768 N/m2 Shear force on shaft, F=τ×Area=3768×πD×L =3768×π×50×10−3×20×10−3 =11.83 N Torque : T=F×D2=11.83×50×10−32 =0.2953 Nm Power loss =2πNT60=2π×1200×0.295360 =37.1 W

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