Design of bearings and flywheel

DESIGN OF BEARINGS AND
FLYWHEEL UNIT V

PROBLEMS IN BALL BEARINGS
A single row deep groove ball bearing is subjected to a radial load of 8 kN
and thrust load of 3 kN. The service imposes light shock and the bearing will
be in use for 40 hours/week for 5 years. The speed of the shaft is 1200 rpm.
Select suitable ball bearing for this application. The minimum acceptable
diameter of the shaft is 75 mm.
Given:
Radial Load, Fr = 8 kN = 8000 N
Axial Thrust Load, Fa = 3 kN = 3000 N
Working hour = 40 hours / week for 5 years
Number of years of service = 5 years
Speed of shaft = 1200 rpm
Minimum diameter of shaft, d = 75 mm

Step I: Select a deep grove ball bearing & Find the static & dynamic capacity
from data book
Select by trial & check for meeting the required dynamic capacity. Select a
bearing for first trial at the middle series to reduce number of trials. For
example, deep groove ball bearing has 4 series such as 61, 62, 63 & 64, in
which 63 or 62 series may be selected for first trial.
Selected bearing Number for Trail 1 : 6315
Static capacity, Co = 7200 kgf = 7200 x9.81 = 70632 N
[PSG 4.14]
Dynamic capacity, C = 9000 kgf = 9000 x9.81 = 88290 N
[PSG 4.14]

PROBLEMS IN BALL BEARINGSStep II: Calculate the equivalent load [PSG Data book 4.2]
Equivalent Load, P = (XFr + YFa) S
X- Radial factor
Y – Thrust Factor
S – Service Factor
S= 1.2 for rotary machine with no impact [PSG 4.2]
For selecting, X and Y using table in PSG 4.4, Calculate Fa/Co & Fa/Fr & “e”
Fa/Co = 0.042
Calculated (by interpolation) “e” = 0.242 Fa/Fr = 0.375
In this problem, Fa/Fr > e . Hence X=0.56 and Y = 1.787 (Calculated by
Interpolation)
Equivalent Load, P = {(0.56 x 8000) + (1.787 x 3000)} x 1.2 = 11809.2 N

Step III: Calculate the required dynamic capacity for required working
hours
Since, the rated dynamic capacity (88290 N) is less than the required dynamic
capacity (107236.4 N), bearing 6315 is not suitable for this application. Hence a
higher series bearing is selected for next trial.

Step IV: Selected bearing Number for Trail 2 : 6415
Static capacity, Co = 10160 kgf = 10160 x9.81 = 99669.6 N [PSG 4.15]
Dynamic capacity, C = 12000 kgf = 12000 x9.81 = 117720.0 N [PSG
4.15] Fa/Co = 0.030
Calculated (by interpolation) “e” = 0.227
Fa/Fr = 0.375
Interpolation)

Since, the rated dynamic capacity of bearing (117720 N) is above than
the required dynamic capacity (112009 N), the bearing 6415 is
suitable for this application.
Dimensions of selected bearing number : 6415
ID = 75 mm OD = 190mm Width = 45 mm

A bearing for an axial flow compressor is to carry a radial load of 2500 N
and thrust of 1500 N. The service imposes light shock and the bearing will
be in use for 40 hours/week for 5 years. The speed of the shaft is 1000
rpm. Select suitable ball bearing for the purpose and give the required
tolerances on the shaft and the housing. Diameter of the shaft is 50 mm.
Given:
Radial Load, Fr = 2500 N
Axial Thrust Load, Fa = 1500 N
Working hour = 40 hours / week for 5 years
Number of years of service = 5 years
Speed of shaft = 1000 rpm
Minimum diameter of shaft, d = 50 mm

Step I: Select a deep grove ball bearing & Find the static & dynamic
capacity from data book
Select by trial & check for meeting the required dynamic capacity.
Select a bearing for first trial at the middle series to reduce number of
trials. For example, deep groove ball bearing has 4 series such as 61,
62, 63 & 64, in which 63 or 62 series may be selected for first trial.
Selected bearing Number for Trail 1 : 6310
Static capacity, Co = 3550 kgf = 3550 x9.81 = 34825.5 N
[PSG 4.14]
Dynamic capacity, C = 4800 kgf = 4800 x9.81 = 47088.0 N
[PSG 4.14]

Step II: Calculate the equivalent load [PSG Data book 4.2]
X- Radial factor
Y – Thrust Factor
For selecting, X and Y using table in PSG 4.4, Calculate Fa/Co & Fa/Fr &
“e” Fa/Co = 0.04
Calculated (by interpolation) “e” = 0.24 Fa/Fr = 0.060
In this problem, Fa/Fr > e . Hence X=0.56 and Y = 1.8
Equivalent Load, P = {(0.56 x 2500) + (1.8 x 1500)} x 1.2 = 4920 N

Step III: Calculate the required dynamic capacity for required working hours

A single row deep groove ball bearing No 6002 is subjected to an
axial thrust load of 1000 N and a radial load of 2200 N. Find the
expected life that 50% of the bearings will complete under this
condition.
Given:
Bearing No: Deep Groove Ball Bearing No 6002
Radial Load, Fr = 2200 N
Axial Thrust Load, Fa = 1000 N
Required Reliability, p = 50% =0.5

Step I: Find the static & dynamic capacity of given bearing from data
book
For 6002 bearing,
Static capacity, Co = 225 kgf = 225 x9.81 = 2207.3 N [PSG 4.12]
Dynamic capacity, C = 440 kgf = 440 x9.81 = 4316.4 N [PSG 4.12]
Step II: Calculate the equivalent load [PSG Data book 4.2]
X- Radial factor
Y – Thrust Factor

For selecting, X and Y using table in PSG 4.4, Calculate Fa/Co & Fa/Fr & “e”
Fa/Co = 0.453
Calculated (by interpolation) “e” = 0.426
Fa/Fr = 0.454
Interpolation)
Equivalent Load, P = {(0.56 x 2200) + (1.038 x 1000)} x 1.3
= 2951 N

Step III: Calculate the life of the bearing for given loads [PSG Data
book 4.2]

Step IV: Calculate the life of the bearing for given loads for required
reliability of 50%
[PSG Data book 4.2]

PROBLEMS IN JOURNAL BEARINGS
Design a journal bearing for a centrifugal pump from the following data. Load on the
journal = 20000 N; Speed of the journal = 900 rpm; Type of oil is SAE 10, for which
the absolute viscosity at 550C = 0.017 kg/m-s; Ambient temperature of oil = 15.50C.
Maximum bearing pressure for the pump = 1.5 N/mm2. Calculate the mass of the
lubricating oil required for artificial cooling, if rise of temperature of oil be limited to
100C. Heat dissipation coefficient = 1232 W/m2/0C.
Given:
Load on Journal, W = 20000 N
Speed of Journal, N = 900 rpm
Viscosity of oil at 550C, Z = 0.017 kg/m-s
Oil temperature, To = 550C
Ambient temperature of oil, Ta = 15.50C
Maximum bearing pressure for the pump, P = 1.5 N/mm2
Allowable Rise of oil temperature = 100C
Heat dissipation coefficient, C = 1232 W/m2/0C Selected:
Specific heat of oil, S = 1900 J/kg/0C

Step I: Select L/D ratio & Calculate the diameter and length of
Journal

Step II: Calcu
late the coefficient of friction using McKEES Equation [PSG 7.34]

Step III: Calculate Heat Generated

Step IV: Calculate Heat Dissipated

Step V: Calcula
te the Mass of oil Required for Artificial Cooling

A full journal bearing of 50 mm diameter and 100 mm long has a bearing
pressure of 1.4 N/mm2. The speed of the journal is 900 rpm and the
ratio of journal diameter to the diametral clearance is 1000. The bearing
lubricated with oil whose absolute viscosity at the operating temperature
of 750C may be taken as 0.011 kg/m-s. The room temperature is 350C.
Find i. The amount of artificial cooling required ii. The mass of the
lubricating oil required, if the difference between the outlet and inlet
temperature of oil is 100C. Take specific heat of oil as 1850 J / kg / 0C.
Given:
Speed of Journal, N = 900 rpm Diameter of Journal, D = 50
mm Length of Journal, L = 100 mm Viscosity of
oil at 750C, Z = 0.011 kg/m-s Oil temperature, To = 750C
Ambient temperature of oil, Ta = 350C Bearing pressure, P =
1.4 N/mm2 Allowable Rise of oil temperature = 100C
Specific heat of oil, S = 1850 J/kg/0C Diameter/Clearance
(D/C) = 1000

Step II: Calculate the coefficient of friction using McKEES Equation
[PSG 7.34]

Step III: Calculate Heat Generated
Heat generated, Hg = µ W v
Hg – Heat generated in W(Watts)
W – Load on Journal in N
v – Surface speed of Journal in m/s

Step V: Calculate the Mass of oil Required for Artificial Cooling
Mass of oil required for artificial cooling = 0.18
kg/min

Design a journal bearing for a centrifugal pump to following
specification Diameter of the Journal = 75 mm
Speed of Journal = 1140 rpm
Load on each journal = 11500 N
Given:
Load on Journal, W = 11500 N
Diameter of Journal, D = 75 mm Selected:
Diametral clearance, C = 150 micron = 0.150 mm [PSG 7.32]
Oil temperature, To = 600C
Ambient temperature of oil, Ta = 250C
Bearing pressure for centrifugal pump, P = 1.5 N/mm2 [PSG 7.31]
Allowable Rise of oil temperature = 100C
Specific heat of oil, S = 1900 J/kg/0C

Step I: Select L/D ratio & Calculate the diameter and length of
Journal

Step V: Calculate the Mass of oil Required for Artificial Cooling

A journal bearing, 100 mm in diameter and 150 mm long carries a
radial load of 7 kN at 1200 rpm. The diametral clearance is 0.075
mm. Find the viscosity of the oil being used at the operating
temperature, if 1.2 kW power is wasted in friction.
Given:
Load on Journal, W = 7 kN = 7000 N
Diameter of Journal, D = 100 mm
Length of Journal, L = 150 mm
Diametral Clearance, C = 0.075 mm
Power wasted in friction = 1.2 kw = 1200 W

Step I: Calculate the bearing pressure

Step III: Calculate the Viscosity using Heat Generated

Following data is given for 3600 hydrodynamic bearing
Journal diameter = 100 mm
Radial load = 50 kN
Journal speed = 1440 rpm
Radial clearance = 0.12 mm
Bearing length = 100 mm
Viscosity of lubricant = 16 cp Calculate (i) Minimum film thickness
(ii) Coefficient of friction (iii) Power lost due to friction

Given:
Load on Journal, W = 50 kN = 50000 N
Journal speed, n = 1440 rpm
Diameter of Journal, D = 100 mm
Length of Journal, L = 100 mm
Viscosity of oil, Z = 16 cp
Radial Clearance = 0.12 mm

PROBLEMS IN FLYWHEEL
A single cylinder four stroke oil engine develops 20 kW at 300 RPM.
The work done by the gases during expansion stroke is 2.3 times the
work done on gases during the compression and the work done
during the suction and exhaust strokes is negligible. The speed is to
be maintained within ± 1 %. Determine the mass moment of inertia of
the flywheel.
Given:
Power, P = 20 kW = 20000 W
Speed, N = 300 rpm
Workdone during expansion = 2.3 x Workdone during compression
Fluctuation of speed = ±1%

Step IV: Calculate the Mass Moment of Inertia (I)
ΔE = I ω2 Cs
Angular speed, ω = 2π N / 60

Design and draw suitable flywheel for four stroke four cylinder 133
kW engine running at 375 rpm. Due to space restriction the flywheel
diameter should not exceed 1.2 m.

Design a CI flywheel for a 4 stroke engine developing 150 kW at 200 rpm.
Calculate the mean diameter of the flywheel, if hoop stress is not to exceed
4 MPa. Total fluctuation of speed is to be 4% of mean speed. Work done
during power stroke is 1.5 times of average work done during the cycle.
Density of CI is 7200 kg/m3.
Given:
Power, P = 150 kW = 150000 W
Speed, N = 200 rpm
Allowable hoop stress, σ = 4 MPa = 4 N/mm2
Workdone during power stroke = 1.5 x Average Workdone
Fluctuation of speed = 4% of Mean speed
Density of CI = 7200 kg/m3

Design of bearings and flywheel

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