1. Flow rate across a fluid element with top and bottom faces moving Fluid element h dx dy Consider the fluid element with dimensions dx, dy, and h (film thickness) Consider bottom face moving up with velocity w 2 and top face moving up with velocity w 1 Rate of flow out in x direction Rate of flow in, in x direction Rate of flow out in y direction Rate of flow in, in y direction Volume change in z direction due to w 2 Volume change in z direction due to w 1

2. Top and bottom faces moving vertically Net rate inflow in x direction = Net rate inflow in y direction = Rate volume change in z direction = Therefore equating the net rate inflow in x and y directions to the rate of volume change in z direction we get

3. Top and bottom faces moving vertically We have seen earlier that Substituting into eqn. (7) we get:

4. Top and bottom faces moving vertically Assuming viscosity is constant, U 1 and U 2 do not vary with x, and V 1 and V 2 do not vary with y Which on rearranging and simplification gives:

5. Net vertical velocity of top surface whwh U1U1  dx dh Downward velocity due to U 1 is U 1 tan  (new position of surface is lower than original position) Vertical externally applied velocity = w h Therefore net vertical velocity of upper surface = w h – U 1 tan  = w h – U 1 dh/dx If V1 and are both non zero, then the net vertical velocity of the top surface will be:

6. Both surfaces inclined and translating whwh U1U1  dx dh wowo U2U2  Bottom surface: Horizontal velocity U 2, angle of inclination  o, vertically applied velocity w o Subscript “o” refers to the bottom surface Therefore, similar to the top surface, net velocity of the bottom surface in the vertical direction:

7. Both surfaces inclined and translating  Substituting for w 1 and w 2 in Reynold’s equation which is  We get

8. 8 P1. Velocity at a given co-ordinate Bottom surface moves with velocity U Upper surface is stationary h Calculate the velocity of flow at a height 0.5 mm above the bottom surface when distance of separation is 0.8 mm. The variation of pressure with x at distance of separation 0.8 mm was found experimentally as 10 Pa/cm. U = 5 cm/s,  = 50 cp. NOTE: 1 cp = 0.001 Pa.s 1 Pa = 1N/m 2 z h z=0 z=h U1 U2

9. 9 Solution P1 Therefore velocity u = 1.875-0.015 = 1.86 cm/s

10. 10 P2. Finding volume flow rate Top surface moves h The top surface is part of a rotating journal rotating at 120 RPM. The diameter of the journal is 20 cm. The angle of inclination of the wedge formed is  = 15 o The minimum height of separation is 10  m. Using Reynold’s equation in 2 dimensions, the pressure variation in the x direction at a distance x = 20  m is measured as 10 Pa/cm. Calculate the volume flow rate.  = 50 cp. (consider velocity of top surface = speed of journal circumference) NOTE: 1 cp = 0.001 Pa.s 1 Pa = 1N/m 2 z h z=0 z=h U1 U2 x 

11. 11 Solution problem 2 Velocity of top surface U = 2  r. RPM/60 = 2  r.120/60 = 1.248 m/s x = 20.10 -6 m, h = xtan15 o + 10.10 -6 = 15.36.10 -6 m Therefore