Question 21.12: Design a self-contained journal bearing to support a load of......

Assume

\frac{L}{D}=2

Assume L = 100 mm D = 50 mm

Assume SAE 20 oil

The absolute viscosity of SAE 20 oil at temperature 80°C is 0.01 kg/ms (Ns/m²)

Assume \frac{C_d}{D}=0.001

p=\frac{W}{L D}=\frac{4500}{100 \times 50}=0.9 \mathrm{~N} / \mathrm{mm}^2=0.9 \times 10^6 \mathrm{~N} / \mathrm{m}^2

N^*=\frac{600}{60}=10 \mathrm{~rev} / \mathrm{s}

From the McKee equation the coefficient of friction is given as

f=2 \pi^2\left(\frac{\mu N^*}{p}\right)\left(\frac{r}{C_r}\right)+0.002=2 \pi^2\left(\frac{0.01 \times 10}{0.9 \times 10^6}\right)(1000)+0.002=0.00419

S=\left(\frac{\mu N^*}{p}\right)\left(\frac{r}{C_r}\right)^2=\left(\frac{0.01 \times 10}{0.9 \times 10^6}\right)(1000)^2=0.111

Corresponding to S = 0.111, from graph, we get

\left(\frac{Q}{r C_r N^* L}\right)=2.8

Hence the total volumetric flow rate is

Q=2.8 r C_r N^* L=2.8 \times 25 \times(0.001 \times 25) \times 10 \times 100=1750 \mathrm{~mm}^3 / \mathrm{s}=1.75 \times 10^{-6} \mathrm{~m}^3 / \mathrm{s}

Heat generated is given by

H_g=\frac{(f W)(2 \pi N r)}{60}=\frac{(f W)(\pi N D)}{60}=\frac{(0.00419) \times 4500\left(\pi \times 600 \times \frac{50}{1000}\right)}{60}=29.61 \mathrm{~W}

Heat dissipated is given by

H_d=\rho Q C_p \Delta T=860 \times 1.75 \times 10^{-6} \times 1764 \times 29.5=78.31 \mathrm{~J} \cdot \mathrm{s}=79 \mathrm{~W}

where \rho=860 \mathrm{~kg} / \mathrm{m}^3 \text { and } C_p=1764 \mathrm{~J} /\left(\mathrm{kg}^{\circ} \mathrm{C}\right)

\Delta T=\frac{1}{2}\left(T_0-T_a\right)=\frac{1}{2}(80-21)=29.5^{\circ} \mathrm{C}

Since heat dissipation is more than heat generation, the design is safe.