Question 21.13: Design a bearing and journal to support a load of 5.5 kN at ......
Design a bearing and journal to support a load of 5.5 kN at 650 rpm using hardened steel journal and bronze backed babbitt bearing. The bearing is lubricated by oil rings. Consider room temperature as 22°C and oil temperature as 85°C.
Assumption \frac{L}{D}=1
Oil SAE 20
The absolute viscosity of SAE 20 oil at temperature 85°C is 0.00879 Ns/m²
\mu=8.79 \mathrm{~mPa} \cdot \mathrm{s}
Assume \frac{C_d}{D}=0.001
N^*=\frac{650}{60}=10.833 \mathrm{~rps}
From the design data the minimum bearing modulus for given combination of bearing and shaft material is 2.8.
The operating value of bearing modulus should be
\frac{\mu N}{p}=3 \times 2.8=8.4
p=\frac{0.00879 \times 650}{8.4}=0.68 \mathrm{~N} / \mathrm{mm}^2
Now p=\frac{W}{L D}=\frac{5.5 \times 1000}{D^2}=0.68
D = 89.9 = 90 mm
L = 90 mm
The actual bearing pressure is
p=\frac{W}{L D}=\frac{5.5 \times 1000}{90 \times 90}=0.679 \mathrm{~MPa}
Sommerfeld number is
S=\left(\frac{\mu N^*}{p}\right)\left(\frac{r}{C_r}\right)^2=\left(\frac{0.00879 \times 10.833}{0.679 \times 10^6}\right)(1000)^2=0.14
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{0.00879 \times 10.833}{0.679 \times 10^6}\right)(1000)+0.002=0.00477
Heat generated is
H_g=f W \times V
where
V=\frac{\pi D N}{60 \times 1000}=\frac{\pi \times 90 \times 650}{60 \times 1000}=3.063 \mathrm{~m} / \mathrm{s}
H_g=0.00477 \times 5500 \times 3.063=80.36 \mathrm{~W}
Heat dissipated is given as
H_d=k A \Delta T
The value of k for still air is 390.00 W/m² °C
\Delta T=\frac{1}{2}(85-22)=31.5^{\circ} \mathrm{C}
H_d=390.00 \times 0.090 \times 0.09 \times 31.5=99.71 \mathrm{~W}
As it is more than the heat generation, the bearing is safe.
Note: Heat generation is calculated on the basis of heat dissipation coefficient k. One can use the method as used in Example 21.12
The general values of heat dissipation coefficient are
k=\left\{\begin{array}{ll}11.4 \mathrm{~W} / \mathrm{m}^2{ }^{\circ} \mathrm{C} & \text { for still air } \\15.3 \mathrm{~W} / \mathrm{m}^2{ }^{\circ} \mathrm{C} & \text { for shaft-stirred air } \\33.5 \mathrm{~W} / \mathrm{m}^2{ }^{\circ} \mathrm{C} & \text { for air moving at } 25.4 \mathrm{~m} / \mathrm{s}\end{array}\right.