Question 21.17: In a journal bearing diameter of the shaft is 75 mm, L/D = 1......

Given

L = D = 15 mm

C_r=0.05 \mathrm{~mm}

h_0=0.02 \mathrm{~mm}

N^*=\frac{400}{60}=6.67 \mathrm{~rps}

W = 3.5 N

p=\frac{W}{L D}=\frac{3500}{75 \times 75}=0.622 \mathrm{~N} / \mathrm{mm}^2

\frac{h_0}{C_r}=\frac{0.02}{0.05}=0.4

Corresponding to \frac{h_0}{C_r}=0.4,

Sommerfeld number, S = 0.121

CFV = 3.22

FV = 4.33

S=\left(\frac{\mu N^*}{p}\right)\left(\frac{r}{C_r}\right)^2=\left(\frac{\mu \times 6.67}{0.622 \times 10^6}\right)\left(\frac{37.5}{0.05}\right)^2=0.121

\mu=\frac{0.121 \times 0.622 \times 10^6}{6.67} \times\left(\frac{0.05}{37.5}\right)^2=0.02 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^2

Temperature rise variable is

\frac{\rho C_p \Delta T}{p}=\frac{\mathrm{CFV}}{\mathrm{FV}} \times 4 \pi=\frac{3.22}{4.33} \times 4 \pi=9.345

Temperature rise is

\Delta T=\frac{9.345 \times p}{\rho C_p}=\frac{9.345 \times 0.622 \times 10^6}{900 \times 1750}=3.69^{\circ} \mathrm{C}=4.0^{\circ} \mathrm{C}

\frac{r}{C_r} f= coefficient of friction variable = CFV = 3.22

f=\frac{3.22 \times 0.05}{37.5}=0.00429

Power loss in friction

H_f=f W \times V=f W\left(\frac{\pi D N}{60,000}\right)=0.00429 \times 3500\left(\frac{\pi \times 75 \times 400}{60,000}\right)=23.59 \mathrm{~W}