The following data is given for $360^0$ hydrodynamic bearing:

Journal diameter $\hspace{2cm}$: 110 mm

Bearing length $\hspace{2.3cm}$ : 55 mm

Journal speed $\hspace{2.5cm}$: 1400 r.p.m

Minimum oil-film thickness $\hspace{0.6cm}$: 15 microns

Viscosity of lubricant $\hspace{1.6cm}$: 30cP

Specific gravity of lubricant $\hspace{0.6cm}$: 0.86

Specific heat of lubricant $\hspace{0.9cm}: 2 KJ/kg^0C$

Calculate:

i) The load carrying capacity of bearing; ii) the coefficient of friction; iii) the power lost in friction iv) the side leakage; and v) the temperature rise

Mumbai University > Mechanical Engineering > Sem 7 > Machine Design 2

Marks: 10M

Year: Dec 2016

For fit $H_8C_8$ $C_{avg} = 0.126mm$

$\frac{L}{D} = 0.5\\ \frac{2h_0}{c} = \frac{2 \times 15 \ times 10^{-3}}{0.126}= 0.23809$

By interpolation

s = 0.1347

$\mu \frac{P}{C} = 4.1817\\ \frac{4 q}{DCn'L} = 5.3033\\ \frac{q_s}{q} = 0.8465\\ \frac{3c' \triangle to}{P} = 17.1328$

Sammer field no $s = \frac{z'n'}{P}(\frac{P}{C})^2\\ Z' = \frac{Z}{9.81 \times 10^7}\\ Z\ = 3.0581 \times 10^{-7}kgf sec/cm^2\\ n' = 23.33\\ P = 40.368 kgf/cm^2\\ P = 4.03 N/cm^2\\ P = \frac{W}{LD}\\ \therefore w = 24.42 KN$

$\mu \frac{P}{C} = 4.1817 \mu = 4.789 \times 10^{-3}\\ Hg = W \mu V\\ Hg = 24422 \times 4.789 \times 10^{-3} \times \frac{\pi \times 0.4 \times 1400}{60}\\ Hg = 943.073 W$

$\frac{3 c \triangle to}{P} = 17.1328\\ \triangle to = 40^0 C\\ \frac{4q}{DCn'L} = 5.3033\\ q = 2.3579 \times 10^{-5} kg/m^3\\ \frac{q_s}{q} = 0.8465\\ q_s = 1.99 \times 10^{-5} kg/m^3$