Bearing Design: Power Loss Reduction and Ceramic Bearing Recommendation

Student number:
Section B: Bearing design
Solution : (a) Load (P) = 500 KN
Starup load = 0
Shaft diameter = 500 mm
Inside dia (di) = 550 mm
N = 3000 rpm
Cdmin = 0.0015
Considering ISO Grade 46
Viscosity index = 95
Bearing inlet temperature = 600c
Ra = 0.5 μm
Actual mean pressure < 2.5 MPa
Total power loss (KW) = 2 π Tr n
= 2 π x 50 x 3000 = 942.477 kw
Minimum film thickness allowed (h0) = R – (r + ε)
h0 = 275 – (250 + 0.338) = 24.662 μm
ε =0.338
Diametric clearance (Cd) = d x Cdmin
= 500 x 0.0015 = 750 μm
Cr = 0.5 Cd = 375 μm
hmin = Cr (1 -ε ¿ = 375 (1-0.338) = 248.25 μm
Actual mean pressure = P/π(R2-r2) = 500/ π (5502 -5002) = 30.315 MPa
Actual
mean
pressur
e (MPa)
Pad
insid
e dia.
(mm)
Pad
outer
dia.
(mm)
No
of
pads
Min.
film
thicknes
s (μm)
Min film
thicknes
s
allowed
(μm)
Maximu
m
allowable
roughnes
s (μm)
Max.
pad
temp
(C)
Total
power
loss
(kW)
1

[Type here]
30.315 550 500 2 248.25 24.662 0.828 40 942.47
7
(a) Indicate below why this design is acceptable.
Answer : This design acceptable because this is within the limit and minimum film thickness allowed
is 24.662 μm .
Answer (b) : Due to relative motion between shaft and bearing there is always a power loss occur in
overcoming the frictional resistance and also wear occurs due to metal to metal contact.
Total power loss (KW) = 2 π Tr n
Where ,
Tr = Torque in N-m
n= rpm
Eccentricity (e )= c – h0
Minimum film thickness (h0) = R – (r + e)
Radial clearance should be small to provide necessary velocity gradient and ideally c = r
1000
2