1. Mercury Marine Problem Basically what we are doing here is we are gluing a rubber seal on a painted aluminum part. It is sometimes difficult to keep the seal in place during the curing process, so our solution was to glue the seal on the part and then put a weight on it, which would hold the seal in place. By using the weight, we are also looking to control the amount of squeeze out and the adhesive thickness between the seal and the part. Basically, we want to do a calculation that says if we apply a known volume of adhesive with a known viscosity to the aluminum part, how heavy would our weight have to be to get x.xx inches of glue thickness and x amount of squeeze out before x amount of time. The time factor is there because eventually the glue is going to cure and no longer move. The best analogy that I can come up with would be two parallel surfaces: the bottom surface being the aluminum part, the top surface being the seal, and adhesive between the surfaces. We would then apply a downward force to the top surface, or the rubber seal, and because of that force it would cause the glue thickness to decrease and squeeze out. Thank you very much for your time and any help is greatly appreciated. Christopher Wackerle Propulsion/Integration Group Mercury Marine Ph: (920) 929-5373

2. Viscosity-fluid property that influences the rate of fluid flow under stress. The viscosity is the ratio between the shear stress and the velocity gradient between the plates, or y How can we attack this problem? (1) Define viscosity Newton’s Law of Viscosity

3. Units of viscosity

4. Layers of fluid particles with top layer moving faster than bottom layer energy Position Activation energy

5. Momentum transfer The top plate drags the fluid particles in the top fluid layer, which then drag the particles in the adjacent lower layer, which then drag the particles in the next lowest layer, and so on—thus giving rise to momentum transfer

6. Units on shear stress and pressure Force/area=mass*length/time 2 * 1/area =mass * length/time * 1/time * 1/area = mass*velocity * 1/(time * area) = momentum/(time*area) = momentum flux Forces balance at steady state (equilibrium) Rate of momentum in =rate of momentum out at steady state. A momentum flux (or stress) multiplied by a cross sectional area is a FORCE!

7. ‘control volume’ y x z 2. Determine velocity profile. If flow is fully developed, the fluid velocity only depends on y. P applied P atm V(x) y 0 - High PressureLow Pressure Hydrostatic pressure varies with x while shear stress varies with y! We only have to consider the shear stress acting normal to the xz plane and the hydrostatic pressure component acting normal to the yz plane.

8. Velocity can only depend on y Function of x unless P=constant or P linear with x. - - C is the integration constant from indefinite integration

9. C=0 If the velocity is a local maximum at y=0 (center in between plates), then

10. To obtain the volume of glue being ‘squeezed out’ per unit time, it is necessary to integrate over the cross sectional area. For one side of a four sided ‘sandwich’ with plate width, W, the volume flow rate is Check the units m/s m 2 /s m 3 /s

11. Flow Between Parallel Plates of spacing 2 delta and width W y x L 2 delta

12. Plate P 1 =Load/Area P1P1 P atm Fluid Approximation 1: Linear pressure gradient between the fluid under the center of the plate(=applied pressure) and pressure in the fluid flowing out of the plate. Approximation 2: fluid is incompressible so that the volume of fluid left between the plates is equal to initial volume - volume removed from between the plates. Approximation 3: flow is laminar L

13. Given these assumptions, a first order (approximate) solution follows This approximate approach is developed as a public service. Anything more precise will require finite element analysis. The solution to the above ODE is shown on the next slide. We can’t just integrate Q because we don’t know how δ changes with time!!!

14. Reasonable values for the parameters were substituted into these solutions and it was found that one solution was negative (not realistic) and the other was positive (pick this one). Two solutions 1. 2. The solution to the ODE by integration is

15. Plotting the solution in SI units Initial half thickness of glue is.05 m, with a load of 0.1 MPa above atm, width=1 m, half length=0.5 m and a linear dependence of viscosity on time.