Question #98651
A hydrostatic bearing is used to rotate a load. The load 500kN, fluid viscosity 500 mPa-s. The bearing has an outside diameter of 500mm and an inside diameter of 400mm.

a) What flow Q is required if the film thickness is 1mm?
b) What Pressure is required?
c) what is the friction torque if it is rotated at 3 rad/s?
1
Expert's answer
2019-11-18T08:25:23-0500
Fout=πD2/4=3.1415500/4=196343.75Fout=\pi*D^2/4=3.1415*500/4=196343.75

Fin=πd2/4=3.14154002/4=125660Fin=\pi*d^2/4=3.1415*400^2/4=125660

ΔF=FoutFin=70683.75\Delta F=Fout-Fin=70683.75


Feff=Fin+0.5ΔF=125660+0.570683.75=161002Feff=Fin+0.5*\Delta F=125660+0.5*70683.75=161002

PRESSURE

p=N/Feff=500000N/161002mm2=3.11MPap=N/Feff=500000N/161002mm^2=3.11MPa

FLOW

Q=(ph3/(12ρ))((Dd)/2)(1/(πD))=3.11MPa(1mm)3(500mm400mm)/(1250010(9)MPas23.1415500mm)=16500mm3/s=0.0165l/sQ=(p*h^3/(12*\rho))*((D-d)/2)*(1/(\pi*D))=3.11MPa*(1mm)^3*(500mm-400mm)/(12*500*10^(-9) MPa*s*2*3.1415*500mm)=16500mm^3/s=0.0165l/s

FRICTION TORQUE

M=ρΔFω(Dd)2/(h222)=50010(9)MPas70684mm23rad/s(500mm400mm)2/(1mm24)=10800Nmm=10.8NmM=\rho*\Delta F*\omega*(D-d)^2/(h*2*2^2)=500*10^(-9)MPa*s*70684mm^2*3rad/s*(500mm-400mm)^2/(1mm*2*4)=10800N*mm=10.8Nm


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