Question 21.14: A 100 mm diameter shaft operating at 2000 rpm is supported b......
A 100 mm diameter shaft operating at 2000 rpm is supported by means of a 150 mm long full journal bearing which is supported to a radial load of 43 kN. Assume \frac{\mu N}{p}=30 \times 10^{-6}. Determine
(a) the coefficient of friction using McKee equation which is given by
f=2 \pi^2\left(\frac{\mu N^*}{p}\right)\left(\frac{r}{C_r}\right)+0.002, \text { take } D / C_d=1000
(b) bearing pressure in MN/m²
(c) heat generated.
Given that D = 100 mm, L = 150 mm, N = 200 rpm and W = 43 kN
(a) f=2 \pi^2\left(\frac{\mu N^*}{p}\right)\left(\frac{r}{C_r}\right)+0.002
=\frac{2 \pi^2}{60}\left(30 \times 10^{-6}\right) \times 1000+0.002=0.01186
(60 is divided in the equation as bearing modulus is given in terms of speed in rpm.)
(b) Bearing pressure
p=\frac{W}{L D}=\frac{43000}{100 \times 150}=2.87 \mathrm{~MPa}
(c) Heat generated
H_g=f W \times V
where
V=\frac{\pi D N}{60 \times 1000}=\frac{\pi \times 100 \times 2000}{60 \times 1000}=10.472 \mathrm{~m} / \mathrm{s}
H_g=0.01186 \times 43000 \times 10.472=5341 \mathrm{~W}=5.3 \mathrm{~kW}