Once again Thesis Tuesday is upon us, so we'll be continuing our discussion of the Conservation of Momentum equation. Today we'll change gears a bit, moving toward examination of the differential form of the momentum equation. When I taught this introductory fluid mechanics class, this form of the equations was the most loved and most hated at the same time by my students- loved because it was much simpler to work with and hated because I always made them start from the full form of the (inviscid or viscous) equation and tell me why they ended up with the final result.
Recall that the linear momentum equation is actually a vector equation that is composed of 3 scalar equations (for a 3D problem). The differential (scalar) form of the linear momentum equation is:
d/dt(rho*u) + u*d/dx(rho*u) + v*d/dy(rho*u) + w*d/dz(rho*u) = rho*gx + d/dx(Sxx) + d/dy(Txy) + d/dz(Txz) + sum(Fx)
d/dt(rho*v) + u*d/dx(rho*v) + v*d/dy(rho*v) + w*d/dz(rho*v) = rho*gy + d/dx(Tyx) + d/dy(Syy) + d/dz(Tyz) + sum(Fy)
d/dt(rho*w) + u*d/dx(rho*w) + v*d/dy(rho*w) + w*d/dz(rho*w) = rho*gz + d/dx(Tzx) + d/dy(Tzy) + d/dz(Szz) + sum(Fz)
We've discussed each of the components in these equations before except for the S and T terms. These terms are ones from the stress tensor/matrix- S represents the normal stresses on a fluid due primarily to pressure forces; T represents the shear stresses on a fluid due primarily to the viscosity of the fluid. These stress terms reduce to different forms depending on whether we are dealing with a viscous or an inviscid fluid. Today we'll examine the viscous fluid equations.
The viscous form of the momentum equations is achieved by applying Stokes' postulates:
1) Each stress component is a linear function of strain rate (Newtonian)
2) The properties of a fluid are isotropic (not varying with direction)
3) when all strain rates are equal to zero, hydrostatic pressure is the only stress present
and Stokes' assumption:
lambda = -(2/3)*mu, where lambda is the bulk viscosity (related to the volumetric dilatation of the fluid) and mu is the viscosity
These allow the stress terms in the equations above to be rewritten as:
d/dx(Sxx) + d/dy(Txy) + d/dz(Txz) = -d/dx(p) + mu*(d/dx(d/dx(u))+d/dy(d/dy(u)) + d/dz(d/dz(u)))
d/dx(Tyx) + d/dy(Syy) + d/dz(Tyz) = -d/dy(p) + mu*(d/dx(d/dx(v))+d/dy(d/dy(v)) + d/dz(d/dz(v)))
d/dx(Tzx) + d/dy(Tzy) + d/dz(Szz) = -d/dz(p) + mu*(d/dx(d/dx(w))+d/dy(d/dy(w)) + d/dz(d/dz(w)))
Placing these expressions in the initial differential form of the momentum equation gives us the Navier Stokes Equations. What these scalar equations physically represents is:
Left side- "mass times acceleration" of a fluid particle (per unit volume)
Right side-
First term- force of gravity of a fluid particle per unit volume; if positive, this terms acts to accelerate the particle in the +x-direction
Second term- pressure difference across the fluid particle; if negative, than this term acts to accelerate the particle in the +x-direction
Third-fifth terms- viscous stresses acting on the fluid particle; if sum is positive, than this group acts to accelerate the particle in the +x-direction
Sixth term- any other forces (support, etc.) per unit volume acting on the fluid; this term is often neglected
By combining these equations with the continuity equation, we can provide a complete mathematical description of the flow of an incompressible, Newtonian fluid: 4 equations allows us to solve for 4 possible unknowns (p, u, v, and w). If we have exactly four unknowns, our problem is well-defined. Fewer and our problem is over-defined; more (unknown external forces, etc.) and our problem is ill-posed and we cannot solve it unless more information is obtained.
These equations may seem a bit intimidating, but remember, we can always employ simplifying assumptions in order to obtain a more manageable set. In fact, that's what we'll look at next week- application of the Navier Stokes Equations. So until next time, happy studying!
