Question 8.13: A slipper (slider) and plate (guide), both 0.5 m wide consti...
A slipper (slider) and plate (guide), both 0.5 m wide constitutes a bearing as shown in Fig. 8.19. Density of the fluid, ρ = 9.00 kg/m ^{2}‚ and viscosity, μ = 0.1 Ns/m ^{2}.
Find out the (a) load carrying capacity of the bearing, (b) drag, and (c) power lost in the bearing.
(a) Considering the width as b and using Eq. (8.70) for the load carrying capacity,
P=\frac{6 \mu U l^{2}}{\left(h_{1}-h_{2}\right)^{2}}\left[\ln \frac{h_{1}}{h_{2}}-2\left\{\frac{h_{1}-h_{2}}{h_{1}+h_{2}}\right\}\right] (8.70)
P=\frac{6 \pi U l^{2} b}{\left(h_{1}-h_{2}\right)^{2}}\left[\ln \frac{h_{1}}{h_{2}}-2 \frac{h_{1}-h_{2}}{h_{1}+h_{2}}\right]
=\left[\frac{6 \times 0.1 \times 1 \times 0.2 \times 0.2 \times 0.5}{(0.005-0.002)^{2}}\right] \times
\left[\ln \frac{0.005}{0.002}-2 \frac{0.005-0.002}{0.005+0.002}\right]
=\frac{0.012}{9.0 \times 10^{-6}}(0.9163-0.8571)=78.93 N
(b) Making use of Eq. (8.72) for width b, the drag force may be written as
D=\int_{0}^{l} \tau_{0} d x=\frac{\mu U l}{h_{1}-h_{2}}\left[4 \ln \frac{h_{1}}{h_{2}}-6 \frac{h_{1}-h_{2}}{h_{1}+h_{2}}\right] (8.72)
D=\frac{\mu U l b}{h_{1}-h_{2}}\left[4 \ln \frac{h_{1}}{h_{2}}-6 \frac{h_{1}-h_{2}}{h_{1}+h_{2}}\right]
=\frac{0.1 \times 1 \times 0.2 \times 0.5}{(0.005-0.002)}\left[4 \ln \frac{0.005}{0.002}-6 \frac{0.005-0.002}{0.005+0.002}\right]
=\frac{0.01}{0.003}(3.6651-2.5714)=3.645 N
(c) Power lost = drag × velocity
=3.645 \times 1=3.645 W