1. We are restricting the problem to the case where inertia forces are much weaker than viscous forces. Thus we expect that the interia terms of the Navier-Stokes equation should be much smaller than the viscous and pressure terms and that thus they can be neglected. For this case the Reynolds number is very small. If we make the equations dimensionless all terms are no larger in magnitude than about unity. Thus the parameters that appear in these equations, and which can have values much different from on, determine which terms are needed for the solution. In the nondimensional equations, the Reynolds number multiplies the intertia terms and these will consequently be neglected in the solution.
2. Since the Reynolds number is small, the fluid goes only where specifically pushed and it will stop if the forcing is stopped. Thus the geometry of the flow field is determined by the boundaries.
3. Because viscous forces dominate the flow field, the fluid can never accelerate above the free stream value even if an obstacle causes the fluid to be squeezed. Thus the velocity in the region of the sphere just slows down and then returns to the free stream value.
4. Both normal stresses and tangential stresses contribute to the drag on the sphere. These can be termed form drag and skin drag.
5. Consistent with the fluid not accelerating, the pressure never increases above the free stream value. The fluid has no inertia that would cause a pressure increase as the fluid slows down.
6. The velocity decays slowly (as ![[Graphics:../Images/creeping_sphere_gr_5.gif]](../Images/creeping_sphere_gr_5.gif)
) and thus the disturbance is felt very far away from the sphere. This makes it difficult to do a real experiment, in a reasonable size container, that allows that sphere to fall at a speed specified by the drag that is predicted from the analysis here. The very high Reynolds number case decays much faster.
1. Linear and nonlinear partial differential equations: In general nobody can make much progress solving nonlinear partial differential equations analytically. Thus, if we are to solve problems involving the Navier Stokes equations, there will have to be physically based simplifications that allow us to solve linear PDE's or we will be a mathematical technique (e.g., a perturbation method) that linearizes the equations. In the present problem the vanishing inertia, which allows us to neglect the left side of the equations results in a linear problem.
2. Solving PDE's: You may not have solved a set of coupled partial differential equations before. It is not as hard as it sounds.
i. The general procedure is to turn a PDE into an ordinary differential equation -- preferably an equation that you know how to solve. In this case (and in many situations) this is done by Separation of Variables. Separation of variables involves using a solution form, say F(x,y) that is a function which is product of the functions of each of the independent variables, say X(x) Y(y).
ii. This assumed form of the solution is substituted into the equations and if it is going to work, you will either have one of functions appearing as an identical factor in every term (the case for this problem) or you will be able to group all of the X(x) functions together and all of the Y(y) terms together. Since x and y can be varied independently, each of these groups of terms should be equation to the same constant.
3. You may not have solved a set of coupled partial ordinary equations before. It is not as hard as it sounds.
i. You solve these by eliminating dependent variables (just as you would for a set of algebraic equations) and derivatives of independent variables in terms of the variable that you have chosen.
ii. If a dependent variable, that you need to eliminate (in this case Pressure) appears only as a derivative in two similar complex equations, it is often possible to take an extra derivative in one or both of the equations to create an identical term in each equation. Then the equations can be subtracted to eliminate the variable.
4. Once you have gone through the process, you will have created an Euler differential equation.
5. An Euler equation can be solved by assuming that the solution is a polynomial in the independent variable.