Question 21.7: A 360° journal bearing of length 45 mm and L/D = 1 has been ......

Given L = 45 mm, D = 45 mm, W = 900 N, N* = 3000/60 = 50 rps, C_r=0.02 \mathrm{~mm}

Let the temperature rise be ΔT = 20° C

Average temperature is

T_{\text {average }}=T_{\mathrm{in}}+\frac{\Delta T}{2}=60+10=70^{\circ} \mathrm{C}

For SAE 10 oil, from graph at temperature 70°C, μ = 9.2 mPa·s = 0.0092 Ns/m²

p=\frac{W}{L D}=\frac{900}{45 \times 45}=0.44 \mathrm{~MPa}

Sommerfeld number is

S=\left(\frac{\mu N^*}{p}\right)\left(\frac{r}{C_r}\right)^2=\left(\frac{0.0092 \times 50}{0.44 \times 10^6}\right)\left(\frac{22.5}{0.02}\right)^2=1.33

Corresponding to S = 1.3, the temperature rise variable is determined as follows.

From Table 21.7 corresponding to S = 1.33

CFV = 26.4 and FV = 3.37

\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{26.4}{3.37} \times 4 \pi=98.44

\Delta T=\frac{98.44 \times p}{\rho C_p}=\frac{98.82 \times 0.44 \times 10^6}{860 \times 1760}=28.62^{\circ} \mathrm{C}

In the second iteration, the average temperature is

T_{\text {average }}=T_{\text {in }}+\frac{\Delta T}{2}=60+\frac{28.62}{2}=74.31^{\circ} \mathrm{C}=75^{\circ} \mathrm{C}

Viscosity corresponding to temperature 75°C is μ = 1.5 mPa·s = 0.0075 Ns/m²

Sommerfeld number is

S=\left(\frac{\mu N^*}{p}\right)\left(\frac{r}{C_r}\right)^2=\left(\frac{0.0075 \times 50}{0.44 \times 10^6}\right)\left(\frac{22.5}{0.02}\right)^2=1.078

Corresponding to S = 1.078 from graph,

FV = 3.4

CFV = 20

\frac{\rho C_p \Delta T}{p}=\frac{\mathrm{CFV}}{\mathrm{FV}} \times 4 \pi=\frac{20}{3.4} \times 4 \pi=73.92

\Delta T=\frac{73.92 \times p}{\rho C_p}=\frac{73.92 \times 0.44 \times 10^6}{860 \times 1760}=21.5^{\circ} \mathrm{C}

In the second iteration, the average temperature is

T_{\text {average }}=T_{\mathrm{in}}+\frac{\Delta T}{2}=60+\frac{21.5}{2}=70.75^{\circ} \mathrm{C}=71^{\circ} \mathrm{C}

which is very close to the first iteration value.

Viscosity corresponding to temperature 71°C is μ= 9.0 mPa·s = 0.009 Ns/m²

Sommerfeld number is 1.3

Hence μ = 9.0 mPa·s = 0.009 Ns/m²

f=2 \pi^2\left(\frac{\mu N^*}{p}\right)\left(\frac{r}{C_r}\right)+0.002=2 \pi^2\left(\frac{0.009 \times 50}{0.44 \times 10^6}\right)\left(\frac{22.5}{0.02}\right)+0.002=0.0247

Power loss

H=f W V=0.0247 \times 900 \times \frac{\pi \times 45 \times 3000}{60,000}=157.135 \mathrm{~W}

If viscosity of the oil increases by 10%,

\mu^{\prime}=1.1 \mu=1.1 \times 9=9.9 \mathrm{~mPa} \cdot \mathrm{s}=0.0099 \mathrm{~Ns} / \mathrm{m}^2

f^{\prime}=2 \pi^2\left(\frac{\mu^{\prime} N^*}{p}\right)\left(\frac{r}{C_r}\right)+0.002=2 \pi^2\left(\frac{0.0099 \times 50}{0.44 \times 10^6}\right)\left(\frac{22.5}{0.02}\right)+0.002=0.02698

The power loss due to increase in viscosity will be

H^{\prime}=f^{\prime} W V=0.02698 \times 900 \times \frac{\pi \times 45 \times 3000}{60,000}=171.64 \mathrm{~W}

\text { Percentage change in power loss }=\frac{171.64-157.135}{171.64} \times 100=8.45 \%

which is greater than the specified percentage, i.e. 5%. So the bearing must be changed.