2019-11-13T22:35:28-05:00
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
2019-11-18T08:25:23-0500
F o u t = π ∗ D 2 / 4 = 3.1415 ∗ 500 / 4 = 196343.75 Fout=\pi*D^2/4=3.1415*500/4=196343.75 F o u t = π ∗ D 2 /4 = 3.1415 ∗ 500/4 = 196343.75
F i n = π ∗ d 2 / 4 = 3.1415 ∗ 40 0 2 / 4 = 125660 Fin=\pi*d^2/4=3.1415*400^2/4=125660 F in = π ∗ d 2 /4 = 3.1415 ∗ 40 0 2 /4 = 125660
Δ F = F o u t − F i n = 70683.75 \Delta F=Fout-Fin=70683.75 Δ F = F o u t − F in = 70683.75
F e f f = F i n + 0.5 ∗ Δ F = 125660 + 0.5 ∗ 70683.75 = 161002 Feff=Fin+0.5*\Delta F=125660+0.5*70683.75=161002 F e ff = F in + 0.5 ∗ Δ F = 125660 + 0.5 ∗ 70683.75 = 161002
PRESSURE
p = N / F e f f = 500000 N / 161002 m m 2 = 3.11 M P a p=N/Feff=500000N/161002mm^2=3.11MPa p = N / F e ff = 500000 N /161002 m m 2 = 3.11 MP a
FLOW
Q = ( p ∗ h 3 / ( 12 ∗ ρ ) ) ∗ ( ( D − d ) / 2 ) ∗ ( 1 / ( π ∗ D ) ) = 3.11 M P a ∗ ( 1 m m ) 3 ∗ ( 500 m m − 400 m m ) / ( 12 ∗ 500 ∗ 1 0 ( − 9 ) M P a ∗ s ∗ 2 ∗ 3.1415 ∗ 500 m m ) = 16500 m m 3 / s = 0.0165 l / s Q=(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 Q = ( p ∗ h 3 / ( 12 ∗ ρ )) ∗ (( D − d ) /2 ) ∗ ( 1/ ( π ∗ D )) = 3.11 MP a ∗ ( 1 mm ) 3 ∗ ( 500 mm − 400 mm ) / ( 12 ∗ 500 ∗ 1 0 ( − 9 ) MP a ∗ s ∗ 2 ∗ 3.1415 ∗ 500 mm ) = 16500 m m 3 / s = 0.0165 l / s
FRICTION TORQUE
M = ρ ∗ Δ F ∗ ω ∗ ( D − d ) 2 / ( h ∗ 2 ∗ 2 2 ) = 500 ∗ 1 0 ( − 9 ) M P a ∗ s ∗ 70684 m m 2 ∗ 3 r a d / s ∗ ( 500 m m − 400 m m ) 2 / ( 1 m m ∗ 2 ∗ 4 ) = 10800 N ∗ m m = 10.8 N m M=\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 M = ρ ∗ Δ F ∗ ω ∗ ( D − d ) 2 / ( h ∗ 2 ∗ 2 2 ) = 500 ∗ 1 0 ( − 9 ) MP a ∗ s ∗ 70684 m m 2 ∗ 3 r a d / s ∗ ( 500 mm − 400 mm ) 2 / ( 1 mm ∗ 2 ∗ 4 ) = 10800 N ∗ mm = 10.8 N m
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