TAMU CHEN 304 - L10 (11 pages)

Previewing pages 1, 2, 3, 4 of 11 page document View the full content.
View Full Document

L10



Previewing pages 1, 2, 3, 4 of actual document.

View the full content.
View Full Document
View Full Document

L10

58 views


Pages:
11
School:
Texas A&M University
Course:
Chen 304 - Chem Engr Fluid Ops
Chem Engr Fluid Ops Documents
Unformatted text preview:

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 d v dV v v vs n dS g dV Pn t dS Fother dt V t S t V t S t 56 The control volume over which we apply this momentum balance is sketched below v n s Pn t vs V t S t Let us suppose that the control volume contains only the fluid whose flow is being analyzed and that there are no other forces Fother Then equation 56 becomes d v dV S t v v vs n dS V t g dV S t Pn t dS dt V t 76 116 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 t dS Total force due to the action of the viscous traction on the S t 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 The viscous traction at any point on the surface S t can be related to the deviatoric stress tensor as t n 117 Recall that any tensor can be written as ee ij i j 118 i 1 3 j 1 3 and ij 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 surface 77 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 That is ij ji 119 Thus there are only six independent components and not 9 in the stress tensor These are 11 22 33 12 21 13 31 and 23 32 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 e z Then the viscous traction acting on this face is given by t shaded e z e z ij ei e j i x y z j x y z zj e j 120 j x y z Note that the first subscript denotes the surface on which the stress acts For example zy 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 greater 78 first subscript For instance zy exerts stress in the y direction and is positive at zy z dz and negative at zy z zz z dz zx z dz zy z dz dz z y dx x dy Thus only three of the six independent components of the stress contribute towards the traction on this face Note that in general the traction t shaded is NOT oriented perpendicular to the surface unlike the pressure traction which is always oriented normal to the surface zy and zx are oriented in the tangential plane and they represent shear stresses zz 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 Pe z t shaded Pe z zj j x y z e j P zz e z zxe x zy e y 121 It should also be noted that the term P zz is also written as zz P xx xx P yy yy P zz zz 79 122 dz z zx z y zy z x dy dx zz z Now consider the opposite face The unit outward normal on this face is e z Then the viscous traction acting on this face is given by t opposite e z e z ij ei e j zj e j i x y z j x y z j x y z j x y z zj e j 123 Again zy and zx are oriented in the tangential plane and they represent shear stresses zz 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 P e z t opposite P e z P zz ez zx ex zy e y j x y z zj e j 124 Once again we write P zz as zz Note that the vector directions in the traction are now e x e y and e z 80 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 Pn where n is the unit outward normal and the viscous traction is n We now return to equation 116 and rewrite it as d v dV v v vs n dS g dV Pn n dS dt V t S t V t S t 125 Cauchy s equation of motion With a modest algebraic effort we can deduce the differential momentum balance equation from equation 125 First we apply the Leibniz rule see equation 35 in p 34 and find that d v dV v dV v vs n dS dt V t t V t S t When this is inserted into equation 125 we get t v …


View Full Document

Access the best Study Guides, Lecture Notes and Practice Exams

Loading Unlocking...
Login

Join to view L10 and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view L10 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?