Question 21.6: A journal bearing is to be designed for gear reducer shaft w......

The bearing is to be used in a gear reducer unit. The unit pressure range for reducing gears is from 0.8 to 1.5 MPa for steady load application.

Assume p = 1.5 MPa

The LID ratio for gear reducer unit ranges between 2 and 4.

Because of the non-availability of data for L/D = 2, it is assumed that L = D and further calculations are made.

p=\frac{W}{L D}=\frac{5000}{D^2}=1.5

D=\sqrt{\frac{5000}{1.5}}=57.7 \mathrm{~mm}

Take D = 60 mm

The bearing pressure will be

p=\frac{5000}{60 \times 60}=1.39 \mathrm{~Mpa}

From Raimondi and Boyd charts, for L/D ratio of 1.0 the limiting values of S for optimum conditions of load and friction are obtained as follows.

Corresponding to \frac{h_0}{C_r}=0.3 \text { and } 0.53, Sommerfeld numbers are 0.083 and 0.193 respectively. The other parameters corresponding to these two extreme values are obtained from charts and given below.

The temperature rise corresponding to these two extreme values of S are obtained as follows and given below.

For S = 0.083

\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{2.5}{4.47} \times 4 \pi=7.03

\Delta T=\frac{7.03 \times p}{\rho C_p}=\frac{7.03 \times 1.39 \times 10^6}{860 \times 1760}=6.45^{\circ} \mathrm{C}

The rough estimate of heat dissipation can be obtained as

H_d=k A \Delta T

where k is the overall heat transfer coefficient

k = 140 to 420 W/m²/°C without ventilation

= 490 to 1400 W/m²/°C with ventilation.

Taking 500 W/m²/°C

H_d=k A \Delta T=500 \times 0.06 \times 0.06 \times 6.45=11.61 \mathrm{~W} (21.71)

Heat generated

H_g=f W V=f \times 5000 \times \frac{\pi \times 60 \times 300}{1000 \times 60}=4712.4 f \mathrm{~W} (21.72)

As the bearing is under controlled condition and under reasonable temperature, equating (21.71) and (21.72), we get

4712.4 f=11.61

f=\frac{11.61}{4712.4}=2.46 \times 10^{-3}

\left(\frac{r}{C_r}\right) f=2.5

C_r=\frac{30 \times 2.46 \times 10^{-3}}{2.5}=0.02952 \mathrm{~mm}

\frac{h_0}{C_r}=0.3

h_0=0.3 \times 0.02952=8.856 \times 10^{-3}

Checking with the limiting value of minimum film thickness as given in Eq. (21.67),

h_0 \geq 0.005+0.00004 D, \quad \mathrm{D} \text { and } h_0 \text { in } \mathrm{mm} (21.67)

\geq 0.005+0.00004 \times 60=7.4 \times 10^{-3}

which is less than the calculated value, i.e., 8.856 \times 10^{-3}. Hence, the higher clearance should be considered.

Q=4.47 r C_r N^* L=4.47 \times 30 \times 0.02952 \times 5 \times 60=1188 \mathrm{~mm}^3 / \mathrm{s}

\frac{Q_s}{Q}=0.76, \text { side flow is } Q_s=0.76 \times 1188=903 \mathrm{~mm}^3 / \mathrm{s}

Power loss =4712.4 f=4712.4 \times 2.46 \times 10^{-3}=11.59 \mathrm{~W}

Considering the second limiting value of S = 0.193

\frac{\rho C_p \Delta T}{p}=\frac{\mathrm{CFV}}{\mathrm{FV}} \times 4 \pi=\frac{4.51}{4.16} \times 4 \pi=13.623

\Delta T=\frac{13.623 \times p}{\rho C_p}=\frac{13.623 \times 1.39 \times 10^6}{860 \times 1760}=12.51^{\circ} \mathrm{C}

The rough estimate of heat dissipation can be obtained as

H_d=k A \Delta T

H_d=k A \Delta T=500 \times 0.06 \times 0.06 \times 12.51=22.518 \mathrm{~W} (21.73)

Equating (21.73) and (21.72), we get

4712.4 f=22.518

f=\frac{22.518}{4712.4}=4.779 \times 10^{-3}

\left(\frac{r}{C_r}\right) f=4.51

C_r=\frac{30 \times 4.779 \times 10^{-3}}{4.51}=0.0318 \mathrm{~mm}

\frac{h_0}{C_r}=0.53

h_0=0.53 \times 0.0318=0.0169 \mathrm{~mm} which is more than the critical value obtained above.

Q=4.16 r C_r N^* L=4.16 \times 30 \times 0.0318 \times 5 \times 60=1191 \mathrm{~mm}^3 / \mathrm{s}

\frac{Q_s}{Q}=0.59, \text { side flow is } Q_s=0.59 \times 1191=703 \mathrm{~mm}^3 / \mathrm{s}

Hence for L = D = 60 mm, radial clearance is C_r=0.03 \mathrm{~mm}

(b) To plot the variation with clearance, an intermediate value of S = 0.121 corresponding to \frac{h_0}{C_r}=0.4 is considered.

\frac{\rho C_p \Delta T}{p}=\frac{\mathrm{CFV}}{\mathrm{FV}} \times 4 \pi=\frac{3.22}{4.33} \times 4 \pi=9.35

\Delta T=\frac{9.35 \times p}{\rho C_p}=\frac{9.35 \times 1.39 \times 10^6}{860 \times 1760}=8.6^{\circ} \mathrm{C}

The rough estimate of heat dissipation can be obtained as

H_d=k A \Delta T

H_d=k A \Delta T=500 \times 0.06 \times 0.06 \times 8.6=15.48 \mathrm{~W} (21.74)

Equating Eqs. (21.72) and (21.74), we get

4712.4 f=15.48

f=\frac{15.48}{4712.4}=3.285 \times 10^{-3}

\left(\frac{r}{C_r}\right) f=3.22

C_r=\frac{30 \times 3.285 \times 10^{-3}}{3.22}=0.0306 \mathrm{~mm}

\frac{h_0}{C_r}=0.4

h_0=0.4 \times 0.0306=0.01224 \mathrm{~mm}

Q=4.33 r C_r N^* L=4.33 \times 30 \times 0.0306 \times 5 \times 60=1193 \mathrm{~mm}^3 / \mathrm{s}

\frac{Q_s}{Q}=0.68 \text {, side flow is } Q_s=0.68 \times 1193=811.24 \mathrm{~mm}^3 / \mathrm{s}

Note: Other points may be taken for smooth plotting of the results.

Results are given in the following table.

The variation of the bearing parameters with clearance is shown in Figure 21.24.

Generally the clearances are specified as unilateral or bilateral clearance. In unilateral clearance, the shaft and the bearing are specified as follows.

The diameter of bearing is D_{b-b}^{+0.0} and diameter of journal is D_{j_{-0.00}}^{+c} where b and c are the tolerance limits of bearing and journal respectively.