## L8

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## L8

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- Pages:
- 9
- School:
- Texas A&M University
- Course:
- Chen 304 - Chem Engr Fluid Ops

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CBE341 Lecture 8 9 28 Application of the macroscopic balances We have already discussed the integral forms of the mass conservation equation and the linear momentum balance These equations are also known as Macroscopic balance equations We will now go through some examples to illustrate the application of these equations The macroscopic mass balance applied to the control volume below is given by equation 42 see p 37 v n vs V t S t d dt V t r dV n r v v dS S 42 S t The corresponding momentum balance is given by see p 47 d rv dV S t rv v vs n dS V t rg dV S t Pn t dS Fother dt V t Let us now solve a few problems by using these equations 48 56 Example Force on a U bend An incompressible liquid density r is flowing through a stationary U shaped pipe with a Patm uniform ID D at an average velocity of U Pressures at the inlet and the outlet are P1 and P2 respectively Both arms of the bend are at the same elevation Assume that the flow is in the turbulent regime and that the velocity profile at the inlet and outlet can be approximated as plug flow i e the velocity is approximately independent of the radial location Neglect the viscous stresses over the inlet and outlet flow surface areas a What is the horizontal force exerted by the water on the U bend To do this problem consider a control volume enclosing just the liquid inside the tube b If the U tube is held in place with a clamp what is the horizontal force exerted by the clamp on the tube while the liquid is flowing through the pipe To do this problem consider a control volume that encloses the tube and the liquid inside it This control volume will cut the pipe wall at the liquid inlet and outlet neglect the traction exerted by pipe wall outside the control volume on the pipe wall inside the control volume at these inlet and outlet locations Solution a Since we seek the force exerted by the fluid on the tube wall it makes sense to consider a stationary control volume V around the fluid inside the U tube The surface S enclosing this control volume can be partitioned into three parts S1 tube cross sectional area at the inlet Let the area of this face be A1 S2 tube cross sectional area at the outlet Let this area be A2 S Curved area representing the pipe wall Clearly A1 A2 p D2 A say 4 58 We will take the inflow direction as the positive x direction Then at the inlet the velocity at various radial locations can be written as v v1e x so that v1 is positive everywhere and can in general be a function of the radial position At the exit where we know that the flow is pointing in the negative direction we will write that v v2ex so that v 2 is still a positive quantity 49 Let us now construct the macroscopic mass balance First examine the terms in equation 42 one at a time d dt r dV 0 as the fluid density is constant everywhere and the control volume V t does not change with time n r v vS dS n r v vS dS n r v v S dS n r v v S dS S t S1 S S2 Here as v S 0 n r v v dS e rv e dS r v dS rA x S S1 1 x 1 S1 1 v1 S1 Here v1 1 v1dS A1 S 1 59 is the average velocity on face S1 Similarly n r v v dS e rv e dS rA x S S2 2 x 2 v2 60 S2 n r v v dS 0 S S as there is no flow in the direction of the normal since the fluid does not penetrate through the tube wall Introducing all these into equation 42 we get 0 rA1 v1 rA2 v2 0 Combining this with equation 95 we conclude that v1 v2 U 61 Of course we could have written this down intuitively and indeed it is already implied in the problem statement In more complex problems we may not be able to see the answer intuitively and the systematic application of the mass conservation equation will lead to the right answer 50 Let us now consider the momentum balance As the flow is steady in a macroscopic sense and as the control surface velocity is zero equation 56 simplifies to rv v n dS rg dV Pn t dS S V 62 S Here I have also recognized that there is no Fother acting on the body of fluid Equation 62 is a vector equation and we can look at its component in any direction we want In the problem we are asked to look at the components lying on the horizontal plane Again let us take one term at a time rv v n dS rv v n dS rv v n dS rv v n dS S S1 S S2 rv v n dS rv e v e e dS e rv dS rA 1 x S1 1 x x 2 1 x S1 1 v12 ex S1 Here v12 is defined according to v dS A 2 1 1 v12 63 S1 Similarly rv v n dS rv e v e e dS 2 x S2 S2 2 x x e x rv dS rA2 v e x 2 2 2 2 64 S2 rv v n dS 0 As the fluid does not penetrate the tube wall S Thus rv v n dS rA 1 S v12 rA2 v22 e x The horizontal component of the body force term 65 rg dV is clearly zero V Now Pn t dS Pn t dS Pn t dS Pn t dS S S1 S2 51 S where Pn t dS are the traction forces acting on surface S1 Pn t dS are the S1 S2 traction forces acting on surface S2 and Pn t dS f say are the traction forces S acting on S We note that the traction forces acting on S are exerted by the tube wall on the fluid and we are seeking the horizontal force of the fluid on the U bend Thus we seek fh where fh is the horizontal component of f We are also given that on S1 and S2 the viscous terms are unimportant As a newcomer to this field you have no basis for anticipating this result In general for most flow problems involving Newtonian fluids this turns out to be a good assumption In any case to estimate the viscous terms we need to either know the velocity profiles or have empirical friction factor correlations Friction factor correlations could be used to estimate the contribution of the viscous traction term on the tube wall but not on the inlet and outlet faces Pn t dS Pn dS Pe dS A x S1 S1 …

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