Midterm Exam 2004 solutions.doc

# Answer d is correct the continuity equation states

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Answer d is correct. The continuity equation states that the amount of matter entering a control volume minus the amount leaving the control volume (i.e. the net amount of matter entering the control volume) must cause an incrase in the total amount of matter within the volume. 13. Consider a flat disk of infinite radius rotating at constant angular velocity above a stationary surface, as shown to the right. The equation for momentum for the Newtonian fluid between the plates in the v direction is: a. non-linear b. dependent on c. 0 1 2 z v z r v r r r r d. all of the above e. none of the above While the equations of momentum are non-linear in general, the lack of changes in the direction and lack of velocity components in the r and z directions cause the non-linear terms on the left hand side of the momentum equations to disappear. Therefore the equations turn out to be linear in this case, so a is not a correct answer. The symmetry of the problem with respect to direction causes the equation to be independent of , so that answer b is incorrect. Since a and be are incorrect, d is obviously incorrect. It remains to be seen if c is a correct answer. The momentum equation for v in cylindrical coordinates is found in appendix B, table B.5 (equation B.5-5). This is:  g r z r r dr r p r r v v z v v v r v r v v t v r r z r r z r r 1 1 1 2 2 6

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Steven A. Jones BIEN 501 Physiological Modeling Midterm Exam April 7, 2004 In this case, gravity is in the z direction, so the gravity term is zero. The terms on the left hand side of the equation disappear because: term 1: flow is unsteady. term 2: no velocity in the r direction. term 3: no changes in the direction ( 0 ). term 4: no velocity in the z direction. term 5: no velocity in the r direction. The pressure term goes to zero because there are no changes in the direction. The second term in brackets (  term) is zero because there are no changes in the direction. The last term in brackets is zero because r r . Consequently, the only terms left are: z r r r z r 2 2 1 0 . From table B.1 for stresses in cylindrical coordinates, r r v r r v r r 1 , and z v v r z z 1 .
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