Chapter 3 Unidirectional transport - Chemical Engineering ...

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32 CHAPTER 3. UNIDIRECTIONAL TRANSPORTNote that the force at (x + ∆x) is negative, since momentum leaves thedifferential volume at this surface. We can now add up all the contributionsto get−2πrLτ rz | r+2πrLτ rz | r+∆r+2πr∆rρvz2 ∣ −2πr∆rρv 2 ∣z=0z∣ z=L+2πr∆rLρg+2πr∆r(p 0 −p L ) = 0(3.123)Since the fluid is considered to be incompressible, v z is equal at z = 0 andz = L. Therefore, the momentum flux due to fluid motion cancels out, andwe can divide by 2πL∆r and take the limit ∆r → 0 to get1 d(rτ rz )+r dr( ) P0 − P L= 0 (3.124)Lwhere P = p − ρgz. This equation can be integrated once to giveτ rz = −( ) P0 − P Lr + C 12L r(3.125)If the momentum flux is not infinite at r = 0, the constant C 1 is zero and sowe get( ) P0 − P Lτ rz = −r (3.126)2LThe velocity profile is calculated from the relationThis is used to obtainv z = −τ rz = µ dv zdr(3.127)( ) P0 − P Lr 2 + C 2 (3.128)4LThe constant C 2 is determined from the condition that the fluid velocity iszero at the wall, v z = 0 at r = R.v z =( (P0 − P L )R 2 ) ( )1 − r24µLR 2(3.129)This is the familiar parabolic velocity profile for the flow in a tube. Thevelocity is a maximum at the center, and decreases to zero at the walls. Theshear stress ia a maximum at the walls and decreases to zero at the center.
3.3. EFFECT OF PRESSURE ON MOMENTUM TRANSPORT 33The volumetric flow rate is determined by integrating the velocity profileover the radius of the tube.Q = π(P 0 − P L )R 48µL(3.130)This shell balance analysis is also valid only for steady flows, where thestreamlines are straight. This occurs in a fully developed flow, away fromthe entrance or exit of the pipe, and also in a the Reynolds number is lessthan 2100, where the Reynolds number is defined as Re = ρv max R/µ, wherev max is the maximum velocity and R is the radius of the pipe.Oscillatory flow in a pipeAn oscillatory pressure gradient (∆p/L) = k cos (ωt) is applied across a pipeof length L. Determine the velocity profile in the pipe.The differential equation for the velocity profile is, The momentum equationin the flow x direction is∂u x∂t − 1 Re( ∂ 2 u x∂x 2)+∂2 u x= − ∂p∂r 2 ∂x(3.131)Setting u r = 0 for a unidirectional flow, and neglecting variations in thestreamwise direction, the equation for the velocity profile is∂u x∂t − 1 ( 1 dRe r dr r du )x= − cos (t) (3.132)drwhere Re = (ρωR 2 /µ). Use the solution for an oscillatory flow,The equation for ũ x becomesıũ x − 1 Reu x = ũ x exp (ıt) (3.133)( 1rddr r dũ )x= 1 (3.134)drThe solution for ũ x is divided into a homogeneous and a particular solution.The homogeneous solution is given byũ xh = CJ 0 ( √ −ıRer) (3.135)
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