Tubular Flow with Laminar Flow (CHE 512) M.P. Dudukovic Chemical Reaction Engineering Laboratory (CREL), Washington University, St
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Tubular Flow with Laminar Flow (CHE 512) M.P. Dudukovic Chemical Reaction Engineering Laboratory (CREL), Washington University, St. Louis, MO 4. TUBULAR REACTORS WITH LAMINAR FLOW Tubular reactors in which homogeneous reactions are conducted can be empty or packed conduits of various cross-sectional shape. Pipes, i.e tubular vessels of cylindrical shape, dominate. The flow can be turbulent or laminar. The questions arise as to how to interpret the performance of tubular reactors and how to measure their departure from plug flow behavior. We will start by considering a cylindrical pipe with fully developed laminar flow. For a Newtonian fluid the velocity profile is given by r 2 u = 2u 1 (1) R u where u = max is the mean velocity. By making a balance on species A, which may be a 2 reactant or a tracer, we arrive at the following equation: 2C C D C C D A u A + r A r = A (2) z2 z r r r A t For an exercise this equation should be derived by making a balance on an annular cylindrical region of length z , inner radius r and outer radius r. We should render this equation dimensionless by defining: z r t C = ; = ; = ;c = A L R t C Ao where L is the pipe length of interest, R is the pipe radius, t = L/u is the mean residence time, C is some reference concentration. Let us assume an n-th order rate form. Ao The above equation (2) now reads: 2 D c c D L 1 c n1 c 212 + kC t c n = (2a) 2 () A0 u L u R R We define: u L L2 / D characteristic diffusion time() axial Pe = = = (3) a D L / u process time() characteristic convection time = axial Peclet number. 1 u R Pe = = radial Peclet number (4) r D u R2 R2 / D characteristic radial diffusion time = = = DL L / u characteristic convection time R Per x = ()radial Peclet number x() pipe aspect ratio L n 1 L/u process time Dan = kC A t = = (5) o 1 characteristic reaction time kCn1 Ao In terms of the above dimensionless groups which involve: characteristic reaction time, characteristic diffusion time, characteristic convection or process time and the aspect ratio, we can rewrite the above equations as: u L u dt L L (3a) Pea = = = Pe D D dt dt u R u dt R Pe Per = = = (4a) D D dt 2 1 L 1 Pe Per R 647 a4 8 64 7 4 8 2 1 d c 2 c 1 4 L 1 c n c t 21 + Da c = 2 () n Pe L Pe dt 1 L 1 Pe Per R 647 a4 8 64 7 48 2 1 d c 2 c 1 4 L 1 c n c t 21 + Da c = (2b) 2 () n Pe L Pe dt u dt where Pe = = Re Sc is the Peclet number for the flow. In terms of Pea and Per we have: D 2 1 c 2 c 1 L 1 c n c 21 Da c (2c) 2 () + n = Pea Pe r R The above equation can be simplified if we deal with a steady state reactor problem for which c = 0, or if we deal with a nonreactive tracer dynamic response, for which Da = 0. In either n 2 event we need to solve a cumbersome partial differential equation and need two boundary conditions in axial coordinate and two in radial coordinate . 4.1 Segregated Flow Model The question arises whether we really need to solve the above equation (2c) numerically at all times or whether we can find simple solutions which are valid under certain conditions. Since reactant or tracer dimensionless concentration, c, is a smooth function, based on physical arguments, the value of the function and its derivatives is of similar order of magnitude except perhaps at a finite number of points. It can be shown then that when Pea >> 1 and R Pe >> 1 the first and third term of eq (2c) can be neglected. For a steady state reactor r L problem this results in the following equation: c 21 2 Da cn = 0 (6) () n = 0 ;c=1 (7) The exit concentration c () =1, is a function of radial position, i.e of the stream line on which the reactant travels. The overall average exit concentration, or mixing cup concentration, is obtained as R R 2 r 2 ruC z = L,r dr 4 ru 1 C ()L,r dr C A () A A exit o o R c = = R = ex C R2u C Ao 2 ru C dr Ao Ao o 2 r where u = 2 u 1 . Using dimensionless variables we get: R 1 2 cex = 4 ()1 c()1, d (8) o For a first order reaction (n = 1) we readily find from eqs. (6) and (7) that Da 21() 2 c () = 1, = e (9) Using eqn (9) in eqn (8) we obtain the exit mixing cup concentration: 3 Da 1 2 21()2 cex = 4 ()1 e d (10) o 1 2 d Change variables to 2 = u; 2 = du to get 1 ()1 2 Da 2 3 Da u ()1 u e 2 Da c = 4 e 2 du = 2 du = 2 E (10a) ex 3 3 1 2 1 u 2 e xu where E ()x = du is the n-th exponential integral. n un 1 In contrast, the cross-sectional average concentration is: Da 1 1 2 ˜ 21() cex = 2 c ()1, d = 2 e d o o Da (11) u e 2 Da = du = E u2 2 2 1 Da Da Da Da c˜ = E = e 2 E (11a) ex 2 2 2 1 2 Thus, if we measure by an instrument the cross-sectional average concentration, and try to infer reactant conversion from it, our results may be in error since conversion is only obtainable from mixing cup (flow averaged) concentration and clearly there is a discrepancy between equation (11a) and eq (10a).. You should examine the deviation of eq (11a) compared to eq (10a) and plot it as a function of Da . The needed exponential integral are tabulated by Abramowitz and Stegun (Handbook of Mathematical Functions, Dover Publ. 1964). Using the following relationship among exponential integrals 1 E ()z = ez zE()z (12) n+1 n []n we get the following expression for conversion from eqn (10a) Da Da Da2 Da 1 x = c = 1 e 2 + E (10b) A ex 2 4 1 2 4 e xu e t where E ()x = du = dt 1 u t 1 x We should realize immediately, upon reflection, that by eliminating the diffusion terms in eq (2c) and in arriving at eq (6-7) we deal with the segregated flow system. Indeed, in absence of diffusional effects there is no mixing among various stream lines. The reactant that enters on a particular stream line exits on the same stream line, i.e at the same radial position , and hence is surrounded by elements of the same age as its own at all times during its journey through the reactor. Every stream line has a different residence time. The shortest residence time is experienced by the elements on the center line ()t / 2 , the mean residence time ()t is experienced by the fluid traveling on the stream line that has the mean velocity, i.e, at = 1/ 2 = 0.707, while infinite residence time is experienced by the elements on the stream line at the wall ( = 1). However, since the stream line at the wall receives infinitesimal amount of new fluid the mean residence time for the system exists and is t . We recall that for the segregated flow condition and first order reaction the exit concentration can be written as: Da (13) cex = e E () d = E ()s s = Da o This means that we have found the Laplace transform of the exit age density function for fully developed laminar flow of Newtonian fluid in a pipe s s s s2 s LE() = E ()s = 2 E = 1 e 2 + E (14) {} 3 2 2 4 1 2 However, even the extensive transform pairs of Bateman (Tables of Integral Transform Vol. 1) do not show this transform. We can, however, derive the RTD or the F function for laminar flow readily based on physical arguments. Let us imagine that we have switched from white to red fluid at the inlet at t = 0. Red fluid will appear at the outlet at z = L only starting at time t / 2 . The fraction of the outflow that is younger than t is given by the fraction of the fluid which is red. This is obtained by integrating the flow from the center stream line, where the red fluid is present at the outlet from time t /2, to the stream line at position r at which red fluid just at the outlet at time t and by dividing this flow rate by the total flow rate. 5 Recall that L L t = = u r 2 2u 1 R By definition the F curve is given by: r r 2 r' 2 r' udr' 4 r' u 1 dr' R Ft()= o = o (15) Q R2u t Ft()= 4 ' 1 '2 d' , for t (15a) () 2 o t t 1/ 2 where t = or = 1 . Hence, 21() 2 2t 2 4 2 t Ft()= 4 = 22 1 for t (16) 2 4 2 2 t 1 2 t t t t Ft()= 21 1 2t = 1 1 + = 1 2t 2 2t 2t 4t2 t 0 t < Ft t 2 2 (17) ()= 1 t 4t2 t 2 or t 2 t Ft()= 1 Ht (17a) 4t2 2 and 1 1 F() = 1 H (17b) 4 2 2 The exit age density function is: 6 dF t 2 t E(t) = = Ht (18) dt 2t3 2 The dimensionless exit age density function is: 1 1 E () = t E()t = H (18a) 2 3 2 For any reaction order we can then write for segregated flow laminar flow 1 1 c () c = c () H d = batch d (19) ex batch 23 2 2 3 o 1 2 R Previous studies have shown that this model is valid when Pe > 1000 and Pe > 85.