48 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): [ ]SV(t) S(t)d dV ( ) dSdtr = -×r-òònvv (42) The corresponding momentum balance is given by (see p. 47): ( ) ( ) ( )( )( ) ( )sotherV(t ) S(t ) V(t ) S(t )ddV dS dV P dSdtr + r-×= r + - ++òò òòvvvvng ntF (56) Let us now solve a few problems by using these equations. vnvsS(t)V(t)49 Example: Force on a U-bend: An incompressible liquid (density, r) is flowing through a stationary U-shaped pipe (with a 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, ( )2124DAA Asayp== =. (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 1xv=ve so that 1v 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 2xv= -veso that 2vis still a positive quantity. Patm50 Let us now construct the macroscopic mass balance. First examine the terms in equation (42), one at a time. V(t)d dV 0dtr =ò as the fluid density is constant everywhere and the control volume does not change with time. [ ] [ ] [ ] [ ]SSSS12S(t) S S S()dS ()dS ()dS ()dS¢-×r- = -×r- - ×r- -×r-òòòònvv nvv nvv nvv Here, as S,=v0 [ ]( )[ ]S11 1x1x 1 11SS S()dS vdS vdS vA×r - = -×r = -r = -ròò ònvv e e. Here, 111S1vvdS1A=ò (59) is the average velocity on face S1. Similarly, [ ]( )[ ]S22x2x 22SS()dS vdS vA×r - = -×-r = ròònvv e e. (60) [ ]SS()dS0¢×r - =ònvv, 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 11 2 20v v0AA= r-r + Combining this with equation (95), we conclude that 12vvU== (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.51 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 ( )( ) ( ) ( )SVSdS dV P dSr× = r + - +òòòvvn g nt (62) Here I have also recognized that there is no otherF 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. ( )( ) ( )( ) ( )( ) ( )( )12SSS SdS dS dS dS¢r× = r× + r× + r×òòòòvvn vvn vvn vvn ( )( ) ( ) ( ) ( )( )( )11 1221x 1x x x 1 1 xSS SdS v v dS v dS v1Ar× = r×-= -r = -ròò òvvn e e e e e Here, 21v is defined according to ( )12211SvdS v1A=ò. (63) Similarly, ( )( ) ( ) ( ) ( )( )( )222x 2x xSS22x2 2xS2dS v v dS v dS v2Ar× = -r - × -= -r = -ròòòvvn e e eee (64) As the fluid does not penetrate the tube wall, ( )( )SdS 0¢r× =òvvn Thus, ( )( )2212xSdS v v12AAéùr× = -r + rëûòvvn e (65) The horizontal component of the body force term, ( )VdVròg, is clearly zero. Now, ( ) ( ) ( ) ( )12SSS SPdS PdS PdSPdS¢- +=- ++- ++- +òòòònt nt nt nt52 where ( )1SPdS- +ònt are the traction forces acting on surface S1, ( )2SP dS- +ònt are the traction forces acting on surface S2, and ( )S'PdS(say)- +=ònt f are the traction forces 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.] ( ) ( ) ( )111xx1SSSPdS PdSPdS1AP- + »- ==òòònt n e e where 11SPdS1AP=ò. (66) Similarly,