The volume flux per unit width in the z direction of fluid along the channel is

# The volume flux per unit width in the z direction of

This preview shows page 178 - 180 out of 282 pages.

The volume flux per unit width (in the z -direction) of fluid along the channel is thus Q = integraldisplay d 0 v x d y = G ( x ) d 3 ( x ) 12 μ + 1 2 U d ( x ) . (9.68) Of course, in a steady state, this flux must be independent of x . Hence, dp dx = G ( x ) = 6 μ bracketleftBigg U d 2 ( x ) 2 Q d 3 ( x ) bracketrightBigg , (9.69) where d ( x ) = d 1 α x . Integration of the above equation yields p ( x ) p 0 = 6 μ α U parenleftBigg 1 d 1 d 1 parenrightBigg Q 1 d 2 1 d 2 1 , (9.70) where p 0 = p (0). Now, assuming that the sliding block is completely immersed in fluid of uniform ambient pressure p 0 , we would expect the pressures at the two ends of the lubricating layer to both equal p 0 , which implies that p ( l ) = p 0 . It follows from the above equation that Q = U parenleftBigg d 1 d 2 d 1 + d 2 parenrightBigg , (9.71) Incompressible Viscous Flow 179 and p ( x ) p 0 = 6 μ U α [ d 1 d ( x )] [ d ( x ) d 2 ] d 2 ( x ) ( d 1 + d 2 ) . (9.72) Note that if d 1 > d 2 then the pressure increment p ( x ) p 0 is positive throughout the layer, and vice versa. In other words, a lubricating layer sandwiched between two solid bodies in relative motion only generates a positive pressure, that is capable of supporting a normal load, when the motion is such as to drag (by means of viscous stresses) fluid from the wider to the narrower end of the layer. The pressure increment has a single maximum in the layer, and its value at this point is of order μ l U / d 2 , assuming that ( d 1 d 2 ) / d 1 is of order unity. This suggests that very large pressures can be set up inside a thin lubricating layer. The net normal force (per unit width in the z -direction) acting on the lower plane is f y = integraldisplay l 0 [ p ( x ) p 1 ] dx = 6 μ U α 2 bracketleftBigg ln parenleftBigg d 1 d 2 parenrightBigg 2 parenleftBigg d 1 d 2 d 1 + d 2 parenrightBiggbracketrightBigg . (9.73) Moreover, the net tangential force (per unit width) acting on the lower plane is f x = integraldisplay l 0 μ parenleftBigg ∂v x ∂y parenrightBigg y = 0 dx = 2 μ U α bracketleftBigg 2 ln parenleftBigg d 1 d 2 parenrightBigg 3 parenleftBigg d 1 d 2 d 1 + d 2 parenrightBiggbracketrightBigg . (9.74) Of course, equal and opposite forces, f x = f x = μ U l 2 d 2 0 3 2 k 2 bracketleftBigg ln parenleftBigg 1 + k 1 k parenrightBigg 2 k bracketrightBigg , (9.75) f y = f y = μ U l d 0 1 k bracketleftBigg 2 ln parenleftBigg 1 + k 1 k parenrightBigg 3 k bracketrightBigg , (9.76) act on the upper plane. Here, d 0 = ( d 1 + d 2 ) / 2 is the mean channel width, and k = ( d 1 d 2 ) / ( d 1 + d 2 ). Note that if 0 < d 2 < d 1 then 0 < k < 1. The e ff ective coe ffi cient of friction, C f , between the two sliding bodies is conventionally defined as the ratio of the tangential to the normal force that they exert on one another. Hence, C f = f x f y = f x f y = 4 3 d 0 l H ( k ) , (9.77) where H ( k ) = k bracketleftBigg ln parenleftBigg 1 + k 1 k parenrightBigg 3 k 2 bracketrightBiggslashBiggbracketleftBigg ln parenleftBigg 1 + k 1 k parenrightBigg 2 k bracketrightBigg . (9.78) The function H ( k  #### You've reached the end of your free preview.

Want to read all 282 pages?

• • •  