Question 4.10: Simple Slider Bearing with a Given Volumetric Flow Rate...

Simple Slider Bearing with a Given Volumetric Flow Rate

Concepts Assumptions Sketch
• Reynolds lubrication equation • Steady unidirectional laminar flow in a very small gap
• Conservation of mass • Negligible inertia terms
{\textrm{}}|\mathrm{h}_{2}-\mathrm{h}_{1}\mid\ll \mathrm{L} • No end effects
• Constant fluid properties
Step-by-Step
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Again, the basic problem solution is a combination of Couette flow and Poiseuille flow (see Eq. (4.44)):

\rm{u}({x},{y})={U}\!\left\lgroup1-{\frac{{y}}{{h}({x})}}\right\rgroup -{\frac{\left[{h}({x})\right]^{2}}{2\mu}}\left\lgroup{\frac{\mathrm{d}{p}}{\mathrm{d}{x}}}\right\rgroup {\frac{{y}}{{h}({x})}}\left\lgroup1-{\frac{{y}}{{h}({x})}}\right\rgroup         (4.44)
\mathrm{u}(\mathrm{x},\mathrm{y})=\mathrm{U}\left\lgroup1-{\frac{\mathrm{y}}{\mathrm{h}}}\right\rgroup -{\frac{\mathrm{h}^{2}}{2\mathrm{{\mu}}}}{\left\lgroup{\frac{\mathrm{dp}}{\mathrm{dx}}}\right\rgroup \!\frac{\mathrm{y}}{\mathrm{h}}\left\lgroup1-{\frac{\mathrm{y}}{\mathrm{h}}}\right\rgroup \qquad(\mathrm{E}.4.10.1)}

where here

\rm\mathrm{h}\equiv\mathrm{h}(\mathrm{x})=\mathrm{h}_{1}+{\frac{\mathrm{h}_{2}-\mathrm{h}_{1}}{\mathrm{L}}}\,\mathrm{x}\qquad\qquad(E.4.10.2)

The flow rate Q is:

\rm Q=\begin{array}{l}\int^{\mathrm{h}({ x})} _0\end{array}\mathrm{u}({ x},{ y})\;\mathrm{d}{ y}=\frac{\mathrm{h}_{1}\mathrm{h}_{2}}{h_1+h_2} U=⊄\qquad\qquad(E.4.10.3)

And from Eq. (4.45)

\rm\mathrm{Q}={\int_0^{\mathrm{h}({x})}}{{\mathrm{u}({x},\,{y})\,\,\mathrm{d}{y}=-\,\frac{\left[\mathrm{h}({x})\right]^{3}}{12\mathrm{\mu}}\left\lgroup\frac{\mathrm{d}{p}}{\mathrm{d}{x}}\right\rgroup +\frac{\mathrm{h}\mathrm{U}}{2}}}\qquad\qquad(4.45)
\rm\frac{\mathrm{d}\mathrm{p}}{\mathrm{d}\mathrm{x}}=-\,\frac{12\mu}{\mathrm{h}^{3}}\left\lgroup\mathrm{Q}-{\frac{\mathrm{h}\mathrm{U}}{\mathrm{2}}}\,\right\rgroup\qquad\qquad(E.4.10.4)

The maximum pressure occurs at an x-location where dp/dx = 0, i.e.,

\rm{ h}_{\mathrm{opt}}={\frac{2{ Q}}{ U}}={\frac{2{ h}_{1}{h}_{2}}{{ h}_{1}+{h}_{2}}}={h}_{1}+{\frac{{h}_{2}-{h}_{1}}{L}x_{\mathrm{opt}}}\qquad\qquad(E.4.10.5a)

Thus,

\rm{x_{\mathrm{opt}}}={\frac{{h_{1}}\mathrm{L}}{{h_{1}}+{ h_{2}}}}\qquad\qquad({ E.4.l0.5b)}

The sustaining force per unit width is:

\rm\mathrm{F}=\int_{0}^{\mathrm{{L}}}(\mathrm{p}-\mathrm{p}_{0})\,\mathrm{{d}}\mathrm{x}\qquad\qquad(E.4.10.6)

where with Eq. (E.4.10.4) we obtain:

\rm\mathrm{p(x)}=\mathrm{p}_{0}+\mathrm{6\mu U}\int_{0}^{\mathrm{x}}{\frac{\mathrm{d}{x}}{\mathrm{h}^{2}}}-12\mathrm{\mu Q}\int_0^x{\frac{\mathrm{dx}}{\mathrm{h}^{3}}}\qquad\qquad(E.4.10.7)

Tasks:

• Solve (E.4.10.7) and (E.4.10.6) with \rm\mathrm{h}({x})=\mathrm{h}_{1}+\frac{\mathrm{h}_{2}-\mathrm{h}_{1}}{\mathrm{L}}{ x}

• Plot p(x) and indicate \rm x_{opt} and \rm p_{max} for a reasonable geometry (see Graph a)

• Plot Eq. (E.4.10.1) profiles (see Graph b)

Graphs:

Comments:

(a) When \mathrm{x}=\mathrm{x}_{\mathrm{opt}}={\frac{\mathrm{h}_{1}\mathrm{L}}{\mathrm{h}_{1}+\mathrm{h}_{2}}}, p(x) reaches its maximum \rm p_{max} .

(b) The u(x,y)-distributions vary greatly relative to the x position; as x increases, the backflow region decreases.

example 4.10
example 4.10 1

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