76 CBE341 Lecture 10 (10/4) – Differential momentum balances Viscous stress Let us now return to the integral linear momentum balance equation (equation 56 in page 47) and study the viscous traction term more closely. ( ) ( ) ( )( )( ) ( )sotherV(t ) S(t) V(t ) S(t)ddV dS dV P dSdtρ + ρ−⋅= ρ + − ++∫∫ ∫∫vvvvng ntF (56) The control volume over which we apply this momentum balance is sketched below. Let us suppose that the control volume contains only the fluid whose flow is being analyzed and that there are no other forces, otherF. Then, equation (56) becomes: ( ) ( ) ( )( )( ) ( )sV(t ) S(t) V(t ) S(t)ddV dS dV P dSdtρ + ρ−⋅= ρ + − +∫∫ ∫∫vvvvng nt (116) v n vs S(t) V(t) s = -Pn + t77 Here, s is the total traction acting on the control surface (as a result of the action of the material immediately outside the control volume) and this is partitioned into a component due to the pressure and the remainder arising from viscous action. t is the viscous traction acting on the surface, and ( )S(t)dS∫t = Total force due to the action of the viscous traction on the control surface. At this point, I will present some results without going into the details of the proof. For the purpose of this course, you can take the following as results presented without proof. Statement #1: It can be shown that at every point in the fluid one can define a tensorial quantity referred to as the deviatoric stress tensor, τ. The viscous traction at any point on the surface S(t) can be related to the deviatoric stress tensor as = ⋅tnτ. (117) Recall that any tensor can be written as ij i ji1,3j1,3==τ∑eeτ = (118) and ijτ denotes the components of this stress tensor. Clearly, there are 9 stress tensor components. Note that at each point in space, there is a single stress tensor. We can consider an infinite set of surface elements dS containing this point, and each of them will have a different orientation of the unit normal n. Each of them will also experience, in general, a different traction t. However, all of them can be found from the same stress tensor at this location through the equation (117). This is conceptually similar to what we discussed previously in the context of the pressure. Recall that at each point there is a single pressure. However, different surface78 elements containing this point will experience different pressure traction and we wrote this pressure traction as (–Pn). That is, the orientation of the unit normal coupled with the magnitude of the unique pressure at a point determines the pressure traction. In the same way, the orientation of the unit normal coupled with the magnitude of the unique set of stress tensor components at a point determines the viscous traction. Statement #2: For virtually every practical system, the deviatoric stress tensor is symmetric. t=ττ. That is, ij jiτ = τ. (119) Thus, there are only six independent components (and not 9) in the stress tensor. These are 11 22 33 12 21 13 31 23 32,,, , , and .ττττ= ττ= ττ= τ This statement comes from the application of Newton’s law of angular momentum balance. We will now examine the implication of equation (117) through an example. Consider a Cartesian coordinate system and a differential volume (see figure on next page). The coordinates of the circled point are (x, y, z) The sides of the box are ( ),,dx dy dz. The coordinates of other points readily follow. Consider the shaded face. The unit outward normal on this face is ze. Then the viscous traction acting on this face is given by ( ),, ,,,,shaded z z ij i j zj jixyz jxyzjxyzττ===⎛⎞⎜⎟⋅ = ⋅ =⎜⎟⎜⎟⎝⎠∑∑t=e e ee eτ (120) Note that the first subscript denotes the surface on which the stress acts. For example, zyτ acts on a surface of constant z (an xy plane). The second subscript denotes the direction of the stress and is considered positive when exerted by the fluid in the region of greater79 first subscript. For instance, zyτexerts stress in the y direction and is positive at |zy z dzτ+, and negative at |zy zτ. Thus, only three of the six independent components of the stress contribute towards the traction on this face. Note that, in general, the traction shadedt is NOT oriented perpendicular to the surface, unlike the pressure traction which is always oriented normal to the surface. and zy zxττ are oriented in the tangential plane and they represent shear stresses. zzτ is oriented normal to the surface and this is a viscous normal stress. Thus, viscous action can produce both normal and shear tractions. If we combine the pressure and viscous traction on this surface, we would get Total traction = ( ),,zshaded z zjj zzzzxxzyyjxyzPP Pττττ=− = − = − +++∑e+t e+ e e e e. (121) It should also be noted that the term ( )zzPτ− + is also written as ( )zzσ. ( ) ( )( ) ( )( ) ( ); ; . xx xx yy yy zz zzPPPτσ τσ τσ− + ≡−+ ≡−+ ≡ (122) dydxdzxyzτzz|z+dzτzy|z+dzτzx|z+dz80 Now consider the opposite face. The unit outward normal on this face is z−e. Then the viscous traction acting on this face is given by ( )( ),, ,, ,,,,opposite z z ij i j zj j zj jixyz jxyz jxyzjxyzτττ====⎛⎞⎜⎟⋅ = −⋅ = − = −⎜⎟⎜⎟⎝⎠∑∑∑t=-e e ee e eτ (123) Again, and zy zxττ are oriented in the tangential plane and they represent shear stresses. zzτ is oriented normal to the surface and this is a viscous normal stress. Thus, viscous action can produce both normal and shear tractions. If we combine the pressure and viscous traction on this surface, we would get: Total traction ( ) ( )( ),,z opposite z zj jjxyzPPτ== −− = −− −∑e+t e+ e( )( ) ( )( )zz z zx x zy yPτττ= − + − + − + −eee (124) Once again, we write ( )zzPτ− + as ( )zzσ. Note that the vector directions in the traction are now ( )x−e, ( )y−e and ( )z−e. dydxdzxyzτzz|zτzy|zτzx|z81 We can proceed in this manner for the other four faces. Always keep the following rule in mind: The pressure traction on a surface point is ( )P− n where n is the unit outward normal and the viscous traction is ( )⋅nτ. We now return to