Question 4.11: Long, Eccentric Journal Bearing Consider a rotating shaft of......

Long, Eccentric Journal Bearing

Consider a rotating shaft of radius \rm R_1 inside a stationary housing of inner radius \rm R_2 and both of length L. The difference \rm R_2 − R_1 ≡ c is the clearance and the ratio e/c ≡ ε is the eccentricity, where e is the off-center distance (see Sketch). Clearly, it is the “confined wedge-flow effect” (as with the slider bearing in Example 4.10) that generates the high, load-bearing pressure distribution. Again, the gap h << 1, where h ≈ c(1 + ε cos θ) which implies h(θ = 0) = \rm h_{max} = c(1 + ε) = c + e and h(θ = π) = \rm h_{min} = c-e.

Derive an expression for the internal pressure in the form \hat p(\theta), where \rm\hat{p}=\frac{{ p-p}_{0}}{{\mu}\omega_{0}({ R_{1}}/{ c})^{2}}

\rm\frac{1}{\mu} [\frac{\partial}{\partial x} (h^3\frac{\partial p}{\partial x} )+\frac{\partial}{\partial z} (h^3\frac{\partial p}{\partial z} )]=6U\frac{\partial h}{\partial x} +12\frac{\partial h}{\partial t} \qquad\qquad\qquad(4.54)
Concepts Assumptions Sketch
• Reynolds lubrication equation (4.54) • Steady laminar flow
• Slow shaft rotation
• Reynolds pressure conditions at θ = \rm θ_{cavity}: p = 0 and \rm\frac{dp}{d\theta}=0 \rm h/R_1 << 1; R_1/L << 1
• h = h(e, θ) as given
• Coordinate x = \rm_1R θ
• Constant properties
• No end effects
Step-by-Step
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Nondimensionalizations

\rm\theta=\frac{{ x}}{{ R}_{1}},\ \ {\hat{ h}}=\frac{{ h}}{{ c}},\ \ {\hat{ z}}=\frac{{ z}}{{ L}/2},\,\,and\,\,\, {\hat{ p}}=({ p}-{ p}_{0})/[\mu\omega_{0}({ R}_{1}/{c})^{2}]

Employing Eq. (4.54) with \rm (R_1 / L)^2 ≈ 0 we have:

\rm\frac{\partial}{\partial \theta } \left\lgroup \hat h^3\frac{\partial\hat p}{\partial \theta } \right\rgroup =6\frac{\partial\hat h}{\partial \theta } \qquad\qquad\qquad(E.4.11.1)

Integrating twice yields:

\rm\hat{\mathrm{p}}(\theta)=6\int_{0}^\theta \frac{\hat{\mathrm{h}}-\mathrm{C}}{\hat h^3} \mathrm{d}\theta\qquad\qquad\qquad(E.4.11.2a)

where the integration constant depends on the pressure boundary conditions assumed. For example, setting p = 0 and dp/dθ = 0 when a vapor cavity starts to form at θ = θcav, we get (Szeri 1980):

\mathrm{C}=\mathrm{\hat{h}}(\theta_{\mathrm{cav}})\qquad\qquad\qquad{(\mathrm{E}.4.11.2b)}

Typical θcav-values are 249.2° when ε = 0.1 and 219.7 when ε = 0.5. For the latter case, \rm\hat p(θ) is graphed (see also Panton 2005).

Graph:

Comments:

• The pressure distribution in the journal bearing follows qualitatively that of planar bearings (see Example 4.10).

• Again, the magnitude of h and the pressure boundary conditions greatly determine \rm\hat p(θ) .

example 4.11

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