Lubrication theory applied to a step (Rayleigh) bearing: 22-Mec-B6 Advanced Fluid Mechanics, May 2015, Q4

Question 4: Lubrication theory applied to a step (Rayleigh) bearing (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Symbol	Value
Slider velocity	$U$	0.5 m/s
Oil dynamic viscosity	$\mu$	3.85 N·s/m²
Oil density	$\rho$	900 kg/m³
First-section gap and length	$h_1,\ L_1$	1 mm, 30 mm
Second-section gap and length	$h_2,\ L_2$	2 mm, 100 mm
Boundary conditions in each section	—	$u=0$ at $y=0$; $u=-U$ at $y=h$
Gauge pressure at inlet and outlet	$P$	0

The gaps are thin ($h\ll L$), the bearing is wide ($\partial/\partial z=0$), and the Reynolds number is negligible.

Find. (a) the reduced momentum equations, (b) the volume flow rate per unit width and the pressure change over a section, (c) the pressure distribution over the whole bearing, and (d) the load capacity per unit width for the stated data.

Figure 4.1 — The step bearing in the slider frame, with the local coordinate pairs $(x_1,y_1)$ and $(x_2,y_2)$ used in the analysis, and the resulting triangular gauge-pressure distribution. The peak occurs at the step.

Approach. Reduce the Navier–Stokes equations under the lubrication approximation, integrate twice for the Couette–Poiseuille profile in each section, impose continuity of volume flow across the step to obtain the step pressure, and integrate the resulting triangular pressure distribution for the load.

Part (a) — scale the Navier–Stokes equations. With $h\ll L$ the transverse velocity is smaller than the axial one by $O(h/L)$, and with $\rho Uh/\mu\ll1$ every inertia term is smaller than the viscous term by that same Reynolds number. Of the two viscous derivatives, $\partial^{2}u/\partial y^{2}$ exceeds $\partial^{2}u/\partial x^{2}$ by $(L/h)^{2}$. Gravity acts across a film only microns to millimetres thick and is absorbed into a modified pressure. What survives is $$\boxed{\;\begin{aligned} \frac{\mathrm{d}P}{\mathrm{d}x}&=\mu\frac{\mathrm{d}^{2}u}{\mathrm{d}y^{2}}\cr \frac{\partial P}{\partial y}&=0 \end{aligned}\;}$$ The second statement is the crucial one: the pressure is uniform across the film, so $P=P(x)$ only, and the first equation may be integrated with $\mathrm{d}P/\mathrm{d}x$ treated as a constant within each constant-gap section. These are the Reynolds lubrication equations.
Integrate for the velocity profile. Integrating twice and applying $u(0)=0$ and $u(h)=-U$ gives the superposition of a Poiseuille parabola and a Couette shear, $$u(y)=\frac{1}{2\mu}\frac{\mathrm{d}P}{\mathrm{d}x}\left(y^{2}-hy\right)-\frac{Uy}{h}.$$ Both conditions are satisfied by inspection, and the two contributions may be scaled independently.
Part (b) — integrate the profile for the volume flow rate. Per unit width in $z$, $$q=\int_{0}^{h}u\,\mathrm{d}y =-\frac{h^{3}}{12\mu}\frac{\mathrm{d}P}{\mathrm{d}x}-\frac{Uh}{2},$$ the first term being the pressure-driven flow and the second the drag flow carried by the moving wall. Rearranging for the pressure gradient and integrating over a section of constant gap between any two stations $x_a$ and $x_b$, $$\boxed{\;P_b-P_a=-\frac{12\mu}{h^{3}}\left(q+\frac{Uh}{2}\right)(x_b-x_a)\;}$$ The pressure therefore varies linearly within each section, which is why Figure 3 shows straight lines rather than curves.
Part (c) — impose continuity across the step. The same $q$ must pass both sections. With zero gauge pressure at both ends and the step pressure written $P_s$, the gradients are $\mathrm{d}P/\mathrm{d}x=+P_s/L_1$ in section 1 and $-P_s/L_2$ in section 2, so $$q=-\frac{h_1^{3}}{12\mu}\frac{P_s}{L_1}-\frac{Uh_1}{2} =+\frac{h_2^{3}}{12\mu}\frac{P_s}{L_2}-\frac{Uh_2}{2}.$$ Collecting the $P_s$ terms, $$\boxed{\;P_s=\frac{6\mu U\,(h_2-h_1)} {\dfrac{h_1^{3}}{L_1}+\dfrac{h_2^{3}}{L_2}}\;}$$ and the distribution itself is the triangle $$\begin{aligned} P(x_1)&=P_s\frac{x_1}{L_1}, &\quad 0&\le x_1\le L_1\cr P(x_2)&=P_s\left(1-\frac{x_2}{L_2}\right), &\quad 0&\le x_2\le L_2 \end{aligned}$$ Note that $P_s>0$ requires $h_2>h_1$: in the slider frame the bearing surface moves in $-x$, so the lubricant is dragged from the wide gap towards the narrow one, and it is that converging passage that generates the pressure. A step in the other sense would produce suction and no load.
Part (d) — substitute the data. Working in SI, $$\begin{aligned} \frac{h_1^{3}}{L_1}&=\frac{(1\times10^{-3})^{3}}{0.030}=3.333\times10^{-8}\ \text{m}^{2}\cr \frac{h_2^{3}}{L_2}&=\frac{(2\times10^{-3})^{3}}{0.100}=8.000\times10^{-8}\ \text{m}^{2} \end{aligned}$$ whose sum is $1.1333\times10^{-7}\ \text{m}^{2}$, while the numerator is $6(3.85)(0.5)(1\times10^{-3})=1.155\times10^{-2}\ \text{N/m}$. Hence $$P_s=\frac{1.155\times10^{-2}}{1.1333\times10^{-7}}=1.019\times10^{5}\ \text{Pa} =102\ \text{kPa (gauge)},$$ a little over one atmosphere of lift generated by a film one to two millimetres thick.
Integrate the triangle for the load capacity. The load per unit width is the area under the pressure distribution, and for a triangle of peak $P_s$ and base $L_1+L_2$ that is simply $$W'=\int_0^{L_1+L_2}\!\!P\,\mathrm{d}x=\frac{P_s\,(L_1+L_2)}{2} =\frac{(1.019\times10^{5})(0.130)}{2}$$ $$\boxed{\;W'=6.62\times10^{3}\ \text{N/m}=6.62\ \text{kN per metre of bearing width}\;}$$ As a check on the arithmetic the flow rate is the same from either section, $q=-3.24\times10^{-4}\ \text{m}^{2}/\text{s}$, and the Reynolds number is $\rho Uh_2/\mu=(900)(0.5)(0.002)/3.85=0.234$, comfortably inside the creeping-flow regime the question assumes.
Comment on the proportions chosen. For fixed $h_1$, $h_2$ and total length, $P_s$ is largest when $L_1/(L_1+L_2)=1/[1+(h_2/h_1)^{3/2}]=0.261$. The paper’s geometry uses $30/130=0.231$, which yields 101.9 kPa against a best possible 102.4 kPa — within half a per cent of optimal. The step bearing is therefore quite forgiving of its length split, which is one reason the geometry survives in thrust washers and in gas-lubricated spindle bearings.

Quantity	Result
(a) Reduced momentum equations	$\mathrm{d}P/\mathrm{d}x=\mu\,\mathrm{d}^{2}u/\mathrm{d}y^{2}$; $\partial P/\partial y=0$
Velocity profile in a section	$u=\dfrac{1}{2\mu}\dfrac{\mathrm{d}P}{\mathrm{d}x}(y^{2}-hy)-\dfrac{Uy}{h}$
(b) Flow rate per unit width	$q=-\dfrac{h^{3}}{12\mu}\dfrac{\mathrm{d}P}{\mathrm{d}x}-\dfrac{Uh}{2}$
(b) Pressure change over a section	$\Delta P=-\dfrac{12\mu}{h^{3}}\left(q+\dfrac{Uh}{2}\right)\Delta x$ (linear)
(c) Step pressure	$P_s=\dfrac{6\mu U(h_2-h_1)}{h_1^{3}/L_1+h_2^{3}/L_2}$
(c) Distribution	$P=P_sx_1/L_1$ then $P=P_s(1-x_2/L_2)$
(d) Step pressure, numerical	102 kPa gauge
(d) Load capacity per unit width	6.62 kN/m
Volume flow rate per unit width	$-3.24\times10^{-4}$ m²/s (towards the narrow gap)
Film Reynolds number	0.234 (creeping flow confirmed)

Topic. Reynolds lubrication theory and the Rayleigh step bearing — the simplest geometry that generates hydrodynamic load capacity.

Key relations. Reduced momentum $\mathrm{d}P/\mathrm{d}x=\mu\,\mathrm{d}^{2}u/\mathrm{d}y^{2}$ with $\partial P/\partial y=0$; Couette–Poiseuille profile; flow rate $q=-\tfrac{h^{3}}{12\mu}\tfrac{\mathrm{d}P}{\mathrm{d}x} -\tfrac{Uh}{2}$; continuity $q_1=q_2$ across the step; load $W'=\int P\,\mathrm{d}x$.

Why this works. Hydrodynamic lubrication rests on one observation: a moving wall drags fluid into the gap at a rate proportional to $Uh$, whereas a pressure gradient drives flow at a rate proportional to $h^{3}$. If the gap were uniform the two sections would demand the same drag flow and no pressure would build, but a change in $h$ makes the drag flows incompatible, and the only way continuity can be satisfied is for a pressure field to arise and push the difference back. The step bearing makes this argument visible because $h$ is piecewise constant, so the pressure is piecewise linear and a single unknown — the step pressure — carries the whole solution. The film is extraordinarily stiff: the load scales as $\mu U L^{2}/h^{2}$, so halving the clearance quadruples the capacity, and this is why a journal bearing rides on a film far thinner than the surface finish of the shaft. The same equations, generalised to a smoothly varying $h(x)$, give the Reynolds equation that underlies all thrust-pad and journal-bearing design.

Common pitfall. The single largest error is applying the ordinary Poiseuille boundary conditions ($u=U$ at the moving wall, $u=0$ at the stationary one) instead of the inverted set the question defines; the resulting sign of $P_s$ then comes out negative and the bearing appears to have no load capacity. A second is assuming $\mathrm{d}P/\mathrm{d}x$ is constant over the whole bearing rather than within each section — it changes sign at the step. A third is forgetting that $W'$ is a load per unit width, so multiplying by a bearing width in metres is still required before it can be compared with an applied force. Finally, confirm the Reynolds number: lubrication theory is an approximation, and at $\rho Uh/\mu$ approaching unity the neglected inertia terms begin to matter.

Source. F. M. White, Viscous Fluid Flow, 3rd ed., §3.9 (lubrication theory, slider and step bearings); F. M. White, Fluid Mechanics, 8th ed., §4.11 and Prob. 4.9x series; B. R. Munson et al., Fundamentals of Fluid Mechanics, 8th ed., §8.x (viscous film flows).

Question 4 of 6: Lubrication theory applied to a step (Rayleigh) bearing

Question 4: Lubrication theory applied to a step (Rayleigh) bearing (20 marks)