56 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,µ the fluid viscosity, D the pipe diameter and ρ 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 θ. 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. L U A B θ57 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): ( ) ( ) ( )V(t) S(t),t dV ,t ,t dSt∂ρ= −ρ⎡⎤⎣⎦∂∫∫rnrvr⋅ (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 = 24ADUρπ− Face B: contribution to the right hand side = 24BDUρπ 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 ABUUU==. 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): ( ) ( )( ) ( ) ( )otherVfixed Sfixed Vfixed SfixedddV dS dV P dSdt== ==ρ + ρ⋅ = ρ + − ++∫∫ ∫∫vvvng ntF (57) Let us examine this term by term. ( )VfixedddVdt=ρ∫v = 0 as the flow is steady.58 ( )( )SfixeddS=ρ⋅∫vvn gets contributions from faces A and B across which flow occurs. But 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, ( )( )SfixeddS=ρ⋅∫vvn = 0. Inside this control volume that we consider there are no forces other than the body force due to gravity and surface forces. Thus, otherF= 0. Therefore, equation (57) simplifies to ( ) ( )Vfixed SfixeddV P dS==ρ + − +=∫∫gnt0 (74) Note that this is a vector equation. Let us consider the component of this equation along the axis of the pipe. First consider ( )VfixeddV=ρ∫g. The contribution from this term is 2DL g sin4−π ρ θ where we have chosen the direction along the flow (see figure) as the positive direction. Next consider ( )SfixedPdS=−∫n. Contribution to axial momentum from this term 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.)59 If ABP and P denote the average pressure on faces A and B, then the contribution is ( )2ABDPP4π−. Finally consider( )SfixeddS=∫t. Contribution to axial momentum from this term 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 wτ (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 wDLπτ−. Putting these together, we find the axial component of equation (74) to be ( )22AB wDDLgsin P P DL 044−π ρ θ+ π−−πτ= (75) Simplifying, ( )AB w4L gsin P P LD−ρ θ+ − = τ (76) In the absence of fluid flow, we expect the wall shear stress to be zero, so that No flow: ( )ABLgsin P P 0−ρ θ+ − =, 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 AB and PP such that ( )ABsinABLg P Pρθ− = − + −PP (77)60 Combining equations (76) and (77), we get AB w4LDτ− =PP or equivalently, ABw4DLτ−=PP (78) The frictional (dynamic) pressure drop per unit length, ABL L−Δ≡PPP, is thus clearly related to the wall shear stress. We have derived this relationship (equation 78) by applying the integral momentum balance. Let us now