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}}
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 |
Nondimensionalizations
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) .