TAMU CHEN 304 - L6 (9 pages)

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L6



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L6

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9
School:
Texas A&M University
Course:
Chen 304 - Chem Engr Fluid Ops
Chem Engr Fluid Ops Documents
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CBE341 Lecture 6 9 25 Dimensional analysis of fully developed flow in a pipe Consider steady incompressible viscous flow of a fluid through a long straight pipe Let U be the average velocity the fluid viscosity D the pipe diameter and the fluid density We consider the possibility that the inner wall of the pipe is rough and characterize the wall roughness by an average roughness height k Let the axis of the pipe be inclined to the horizontal at an angle B A L U Intuitively one would expect that the flow patterns will change as a function of axial position along the length of the pipe both near the inlet where the flow changes from that in the upstream region to what is typical for the pipe under consideration and near the outlet where the flow field begins to adjust to conditions prevailing in the downstream region If the pipe is long then one can reasonably expect that there will be a region of pipe in the middle where the flow pattern is essentially fully developed where the flow behavior is for all practical purposes independent of the axial position Our goal in this example is to analyze this pressure drop per unit length in the fully developed region Consider the body of fluid inside a section of length L in the fully developed region as shown in the figure below Let us consider conservation of mass and linear momentum over the shaded region in the figure 56 Let A and B denote the two faces of the shaded region see figure The shaded region is thus bounded by the flow regions A B and the pipe wall Conservation of mass equation 38 r t dV n r t v r t dS t V t S t 38 Here the control volume is stationary and the fluid is specified to be incompressible so the left hand side is equal to zero The right hand side gets contributions from faces A and B across which flow occurs Face A contribution to the right hand side Face B contribution to the right hand side D2 UA 4 D2 UB 4 Here UA and UB denote the average velocities at faces A and B respectively These two contributions to the mass conservation equation clearly add up to zero if and only if U A U B U In other words according to mass conservation principle the average velocity U is independent of axial location Of course this is intuitively obvious Conservation of linear momentum equation 57 d v dV v v n dS g dV Pn t dS Fother dt V fixed S fixed V fixed S fixed Let us examine this term by term d v dV 0 as the flow is steady dt V fixed 57 57 v v n dS gets contributions from faces A and B across which flow occurs But S fixed if the flow is fully developed then the velocity profile over the tube cross section is the same at the two faces A and B Thus for every point on face A there is a corresponding point on face B at the same radial position where the density and velocity are identical however the outward normal at these two points have opposite signs Thus the contribution to this integral from face B will be exactly opposite of that from face A Physically if the velocity profile is not changing with axial location the convective inflow of momentum through face A is exactly the same as the convective outflow through face B Thus v v n dS 0 S fixed Inside this control volume that we consider there are no forces other than the body force due to gravity and surface forces Thus Fother 0 Therefore equation 57 simplifies to g dV Pn t dS 0 V fixed 74 S fixed Note that this is a vector equation Let us consider the component of this equation along the axis of the pipe First consider g dV The contribution from this term is V fixed D2 L g sin 4 where we have chosen the direction along the flow see figure as the positive direction Next consider Pn dS Contribution to axial momentum from this term S fixed comes via both faces A and B The curved wall of the pipe does not contribute as the unit outward normal n on this wall does not have any component in the axial direction 58 If PA and PB denote the average pressure on faces A and B then the contribution is D2 PA PB 4 Finally consider t dS Contribution to axial momentum from this term S fixed comes via faces A and B and the curved wall of the pipe As the flow is fully developed the contributions from faces A and B will cancel each other leaving only the contribution from the curves inner wall of the pipe Also as the flow is fully developed one expects all points on the pipe wall to contribute the same traction to the fluid Let w referred to as the wall shear stress denote the magnitude of the stress force per unit area in the axial direction that the wall exerts on the fluid Physically one would expect the wall to resist the motion of the fluid and try to bring the fluid to rest that is the wall shear stress is acting in direction opposite to that of the flow The total contribution from the wall is then DL w Putting these together we find the axial component of equation 74 to be D2 D2 L g sin PA PB DL w 0 4 4 75 Simplifying L g sin PA PB 4 L w D 76 In the absence of fluid flow we expect the wall shear stress to be zero so that No flow L gsin PA PB 0 and this equation says that the pressure variation is simply hydrostatic In the presence of fluid flow the left hand side of equation 76 denotes the pressure difference over and above the hydrostatic contribution and this pressure drop is commonly referred to as frictional pressure drop or dynamic pressure drop We will define the dynamic pressures PA and PB such that PA PB L g sin PA PB 59 77 Combining equations 76 and 77 we get PA PB 4 L w D or equivalently PA PB 4 w L D 78 The frictional dynamic pressure drop per unit length P PA PB is thus clearly L L related to the wall shear stress We have derived this relationship equation 78 by applying the integral momentum balance Let us now consider a dimensional analysis problem where we wish to correlate the frictional pressure drop per unit length with parameters affecting the flow P f U D k L 79 A set of dimensionless parameters which can be used to correlate the data is deduced as follows It is clear from equation 79 that there are six dimensional parameters where we have treated the frictional pressure drop P as a single L dimensional parameter The dimensional matrix is as follows m L t P L 1 …


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