Question 21.9: A shaft rotates at 1750 rpm and applies a radial load of 3.5......

Assuming that the load is relatively steady, the unit bearing pressure in most of the applications (gear unit, turbine, electric motors, pumps, etc.) is in the range of 0.6 to 2 MPa. Now for square bearing, assuming the unit bearing pressure is 1.5 MPa, we have

p=\frac{W}{L D}=\frac{3500}{D^2}=1.5

D = 48.3 mm

Taking L = D = 50 mm, the calculations are made as follows. The modified bearing pressure is

p=\frac{W}{L D}=\frac{3500}{50 \times 50}=1.4 \mathrm{~MPa}

For L/D = 1.0, the optimum values are \frac{h_0}{C_r}=0.30 for the minimum friction and 0.53 for maximum load capacity.

The Sommerfeld number corresponding to these two values are as follows.

When \frac{h_0}{C_r}=0.30

S=0.121-\frac{0.121-0.0446}{0.4-0.2}(0.4-0.3)=0.0828

When \frac{h_0}{C_r}=0.53

S=0.264-\frac{0.264-0.121}{0.6-0.4}(0.6-0.53)=0.214

For SAE 10 oil and operating temperature as 85°C, μ = 0.01 kg/ms and the clearance is obtained from Sommerfeld number as follows.

S=\left(\frac{\mu N^*}{p}\right)\left(\frac{r}{C_r}\right)^2=\left(\frac{0.01 \times 29.17}{1.4 \times 10^6}\right)\left(\frac{25}{C_r}\right)^2=0.0828

Solving, we get

C_r=0.0397

The other limiting clearance is C_r=0.0247 corresponding to S = 0.214. The results are

S = 0.214    0.0828

\frac{h_0}{C_r}=0.53    0.3

h_0=0.53 \times 0.0247=0.0131 \mathrm{~mm} \quad h_0=0.3 \times 0.0397=0.0119 \mathrm{~mm}

The limiting value of the oil film thickness is

h_0 \geq 0.005+0.00004 D, D \text { and } h_0 \text { are in } \mathrm{mm}

\geq 0.005+0.00004 \times 50=0.003 \mathrm{~mm}

Hence if the lowest value, i.e. 0.0119 mm is considered, it is safe as the value is more than the limiting value of 0.003 mm. Before accepting this as design value, it must be checked for temperature rise and power lost.

Corresponding to S = 0.0828, the coefficient of friction variable is

\left(\frac{r}{C_r}\right) f=3.22-\frac{3.22-1.70}{0.121-0.0446}(0.121-0.0828)=2.46

f=2.46 \times \frac{0.0397}{25}=0.00391

V=\frac{\pi D N}{60 \times 1000}=\frac{\pi \times 50 \times 1750}{60 \times 1000}=4.58 \mathrm{~m} / \mathrm{s}

Power lost =f W V=0.00391 \times 3500 \times 4.58=62.68 \mathrm{~W}

This much heat must be taken away by the oil. Hence the temperature rise will be obtained from the flow variable and the coefficient of friction variable.

Flow variable =\mathrm{FV}=\frac{Q}{r C_r N^* L}=4.4 \text { corresponding to } S=0.0828

\text { Temperature rise variable }=\frac{\rho C_p \Delta T}{p}=\frac{860 \times 1760 \Delta T}{p} \text { and } p \text { must be in } \mathrm{N} / \mathrm{m}^2

From Eq. (21.62)

\frac{\rho C_p \Delta T}{p}=\frac{f\left(\frac{r}{C_r}\right)}{\left(\frac{Q}{r C_r N^* L}\right)}(4 \pi) (21.62)

\frac{\rho C_p \Delta T}{p}=\frac{f\left(\frac{r}{C_r}\right)}{\left(\frac{Q}{r C_r N^* L}\right)}(4 \pi)=\frac{\mathrm{CFV}}{\mathrm{FV}} \times 4 \pi=\frac{2.46}{4.4} \times 4 \pi=7.03

\Delta T=\frac{7.03 \times p}{\rho C_p}=\frac{7.03 \times 1.4 \times 10^6}{860 \times 1760}=6.5^{\circ} \mathrm{C}

Hence the oil leaves at temperature of 85+\frac{6.5}{2}=88.25^{\circ} \mathrm{C} to have an average temperature of 80°C. This is safe as per the requirement of oil.

Hence the recommended values are

D = L = 50 mm

C_r=0.04 \mathrm{~mm}