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1 TO: Joseph M. Powers FROM: Christian D. Hanson DATE: 18 April 1997 RE: ME 334 Project (Part II) 1 Problem Description AFortune 500 hundred company, Gooey Foods, Inc., wants to develop a piping system which transports its carbohydrate-laden product from a large vat in its warehouse to its packaging plant, L, 100 m away. In the vat, which has a `very large' diameter (with respect to the height), a constant height, h, of 25 m of uid is maintained. The uid has the following properties: = 1100 kg m 3, ; where = 4:2Pa s, is the density of the uid, rz is the shear stress, is the absolute viscosity, u is the velocity of the uid, and r is the radial distance from the center of the pipe. Top management at Gooey Foods, Inc. insists that this project, unlike other recent endevors, stay within budget. Thus, the proposed design must: deliver at peak capacity 1000 kg s of product, minimize the number of pipes required, and respond to uctuating demand with a 2 s response time. 2 Recommended Design After careful review, it is recommended that Gooey Foods, Inc. use 65 pipes, each 17.2 cm in diameter, to transport its product to its packaging plant. These pipes should be low on the vat in order to gain advantage of pressure eects in the vat. This design will meet all design specications, and, also, require the minimum number of pipes. 1

2 3 Technical Analysis The uid ow in the pipe follows the unsteady z momentum equation for uni-directional ow of an incompressible non-newtonian uid under the inuence of a constant pressure =,P L + (r rz); (1) where P is the pressure change which the uid experiences in the pipe. The following initial conditions exist: u z (r; 0)=0;u z (R; j r=0 =0(R the radius of the pipe, m). Time is set to innity and the analytic solution is achieved (as described in the preliminary report, Appendix B) to be: u z = P 4L (r2, R 2 ): (2) Using the recommended radius of m, Eq.(2) yields a maximum uid velocity in the pipe of m. This results closely compares with the one the numerical simulation s generates, m, after 2 s. Figure 1 shows this response. (Please reference Appendix s A for the program used to perform this simulation, and Appendix B which describes the development of the program.) Because these two results are within 0.2% of one another, one may assume with condence that the 2 s response time criteria is satised. The steady-state average velocity, u z,isnow calculated with use of the analytical solution in the following equation: u z A = Z A u z da; (3) where u z is given in Eq. (2), and A represents the cross sectional area of the pipe. After integrating and simplifying Eq. (3), the following result is obtained for the average steadystate velocity: u z = PLR2 8 Applying mass continuity, the mass ow rate is: (4) _m = u z AK: (5) From Eq. (5), it is determined that K, the number of pipes required, is 65 for the previously determined pipe diameter, 17.2 cm. Figure 2 shows the radial distance from the center of the pipe vs. the velocity of the uid at four dierent times, 0.5 s, 1.0s, 1.5 s, and 2.0 s. The analytical solution for Eq. (1) is now found without the assumption that t!1. For simplication, Eq. (1) along with the bounday and initial conditions are non-dimensionalized and simplied to give: 2

3 1.2 1 Maximum Velocity (m/s) Time (s) Figure 1: Maximum Velocity vs. Time 3

4 0.09 Radial Distance from Center of Pipe (m) t=0.5 s t=1.0 s t=1.5 s t=2.0 s Velocity of Fluid in Pipe (m/s) Figure 2: Radial Distance from the Center of the Pipe vs. the Velocity of the Fluid at Four Dierent Times 4

5 @u (6) where the following scales are used: u (r ; 0) = 0;u r=0 =0 (7, 9) t = t ;u = u ;r = r t c u c R ;t c = R2 ;u c =,PR 2 : (10, 14) L The velocity, u(r;t) is now divided into its transient, u T (r;t), and steady state, u S (r), components; note that the former is dependent on distance and time and the latter is simply dependent on distance. (The 0 s indicating non-dimensional numbers have been omitted for simplicity and will continue to be omitted until otherwise noted.) The transient and steady state components are summed to give the total velocity: u(r;t)=u T (r;t)+u S (r) (15) The total velocity is dierentiated once with respect to t and, then, twice with respect to r. After non-dimensionalizing and simplifying the results, they are substituted into Eq. (6) which T + T 2 where the following initial and boundary conditions exist: u T (1;t)=0;u T (r; 0) =,u S (r) =r 2, This equation is then solved by separation of variables using u T creates two equations, one in A and one in B: da (16) r=0 =0 (17, 19): = A(t)B(r) which dt + 2 A =0 (20) d 2 B dr + 1 db 2 r dr + B2 =0 (21) where is an arbitrary constant 1. Eq. (20) may be solved as follows: A(t) =C 1 e,2 t The relation s = r is introduced into Eq. (22) to make it a standard Bessel equation whose solution is: (22) B(r) =C 2 J 0 (s) =C 2 J 0 (r): (23) The transient velocity is thus the product of A(t) and B(r): 5

6 Bessel Zero, m Corresponding Coecient,C m Table 1: Bessel Zeroes and their Corresponding Coecients u T (r;t)= 1X m=1 and the total non-dimensionalized equation is: u(r;t)= " 1 X m=1 C m e,2 m t J 0 ( m r) (24) C m e,2 m t J 0 ( m r) # +(1, r 2 ) (25) This equation, Eq. (25), may be solved by using the initial condition, Eq. (18), where t = 0,which yields: r 2, 1= 1X m=1 C m J 0 ( m r): (26) Both sides of this equation are then multiplied by another Bessel function, J o, and a weighting function, r, and integrated over the interval 0! 1:, Z 1 0 J o ( n r)(1, r 2 )rdr = 1X m=1 Z 1 0 J o ( m r)j o ( n r)rdr; (27) where ma nd n are counters beginning at 1. Values of C m are then founds for dierent Bessel zeros as listed in Table 1. The nal solution is, thus, Eq. (24) with the values of Table 1. 6

7 endnotes 1 The following was used as a guide to Bessel equations and functions: Kreyszig, Erwin. Advanced Engineering Mathematics, 6th ed., John Wiley & Sons, New York

8 APPENDIX A PROGRAM SLUDGE IMPLICIT NONE REAL dr, dt, rho, L, Rpipe, r, mu, dp, A, B, C, u(31,3000) INTEGER i, n, istep, imax, tmax, f OPEN(UNIT=69, FILE="rrrs", STATUS="replace") OPEN(UNIT=12, FILE="velocity", STATUS="replace") OPEN(UNIT=13, FILE="maxvelocity", STATUS="replace") OPEN(UNIT=14, FILE="time", STATUS="replace") imax=31; rho = ; L = 100.0; Rpipe = ; mu = 4.2; dp = dr = Rpipe/(imax-1) dt = dr*dr*rho/(2*mu) tmax=3000 A = dp*dt/(rho*l) B =-mu*dt/(rho*2*dr) C = mu*dt/(rho*dr*dr) DO i=1,imax u(i,1)=0.0 END DO DO n = 1,tmax-1 r=rpipe u(1,n) = 0.0 DO i = 2,imax-1 r=r-dr u(i,n+1) = -A + (B/r)*(u(i+1,n)-u(i-1,n)) + C*(u(i+1,n) - & 2*u(i,n) + u(i-1,n)) + u(i,n) END DO u(imax,n+1) = u(imax-1,n+1) END DO DO i=1,imax WRITE(12,*) u(i,tmax) WRITE(69,*) (Rpipe - i*dr) END DO DO f=1,tmax WRITE(13,*) u(imax,f) WRITE(14,*) f*dt END DO END 8

9 TO: Joseph M. Powers FROM: Christian D. Hanson DATE: 26 March 1997 RE: ME 334 Project (Part I) Appendix B A Fortune 500 hundred company, Gooey Foods, Inc., wants to develop a piping system which transports its carbohydrate-laden product from a large vat in its warehouse to its packaging plant,l, 100 m away. In the vat, which has a `very large' diameter (with respect to the height), a constant height, h, of25m of uid is maintained. The uid has the following properties: = 1100 kg m 3, rz where = 4:2Pa where is the density of the uid, rz is the shear stress, is the absolute viscosity, u is the velocity of the uid, and r is the radial distance from the center of the pipe. The pipe used to transport the uid has a radius of.05 m. Also, the ow of the uid in the pipe follows the unsteady z momentum equation for uni-directional ow of an incompressible non-newtonian uid under the inuence of a constant pressure =,P L + (r rz); (1) where P is the change in pressure that the uid experiences in the pipe, and the following initial conditions exist: u z (r; 0) = 0;u z (R; j r=0 =0(R the radius of the pipe, 0.05 m). When time is set to innity, the analytic solution is achieved through the following steps: Set the left hand side of the equation Eq.(1) equal to zero (because time tends to innity), Substitute rz into the right hand side of the equation and solve the resulting second order dierential equation. The initial conditions yield the solution: u z =,160:575r 2 + :4014 When t is not taken to be innity, Eq.(1) can be solved by discretizing the space and time elements of the equation. The following formula is obtained: " # u n+1 i, u n i =, P t L + u n " # i+1, u n i,1 u n + i+1, 2u n i + u n i,1 r 2r (r) 2 9

10 Numerical Solution Analytic Solution 0.04 Distance from Center of Pipe (m) Velocity of Fluid in Pipe (m/s) Figure 3: Radial Position from the Center of the Pipe vs. Velocity of the Fluid for Analytic and Numerical Methods where n is the time component and i is the space component. The parameters u(i,1)=0 and u(1,n)=0, which mean that the ow along the time and space boundaries ( n=0 and i=0 respectively) is zero, are used when solving the equation. Also, the ow exactly centered in the middle of the pipe is assumed to be the same as the ow a dierential amount radially outward. This constraint allows for denition of the middle boundary condition. Both methods of analysis, the analytic (t! 1) and the numerical solution (t=1 s) yield very similar results, as seen in Figure 3. This gure describes the uid's velocity as a function its radial position in the pipe. Thus, it may be concluded that for approximate results, either method may be used. One may also conclude that velocity denitely does change as its location varies in a pipe. 10

P 1 P * 1 T P * 1 T 1 T * 1. s 1 P 1

P 1 P * 1 T P * 1 T 1 T * 1. s 1 P 1 ME 131B Fluid Mechanics Solutions to Week Three Problem Session: Isentropic Flow II (1/26/98) 1. From an energy view point, (a) a nozzle is a device that converts static enthalpy into kinetic energy. (b)