# Solving Descrete Differential Equations

Solving Descrete Differential Equations 150 150 Affordable Capstone Projects Written from Scratch

3.2  BOUNDARY CONDITIONS

Using hot water the flow parameters at the inlet pipe are while P0 =110,000 Pa and T0=370K. While Reynolds number: Re=1500, pressure coefficient: PC =6.5 and Prandtl number: Pr = 1.75, Figure 3.1 show the position of a conical ring insert in the cylindrical tube, with the axis indicated. The DCR insert radius is 18mm and the outlet radius is 25mm, with the length along z axis as 70mm.

 j
 k
 r
 z

Figure 3.1 Flow past a diverging conical ring insert with slits.

1. DISCRETIZATION OF THE GOVERNING EQUATIONS

Given that and representing a plane along the cylindrical tube such that (i,k) =(0,0) is at the centre of the inlet on a  DCR insert

 i-1
 i+1
 k-1
 k+1

r

 i

z

 k

figure 3.2. Discrete grid points.

SOLVING THE DISCRETIZED EQUATIONS

Factoring in the boundary conditions and non dimension numbers for the flow of hot water in a cylindrical tube within the region of the conical ring insert if Δz=0.01, Δr =0.001, axial velocity at the DCR  inlet it is uz=4m/s then it starts reducing as radial velocity increases from zero, is the radial distance from the centre of the tube with a singularity existing at . Parameters are predetermined at the centre line with boundary conditions Po=1.1×105Pa,  To =370K, Re=1500, Pr =1.75 and PC = 6.5 then the governing equations are solved to obtain the various flow profiles.

Equation (3.4) is expressed as

(3.11)

At the DCR inlet boundary values are estimated from a flow simulation by solid works to reduce the equation and determine radial velocities with change in radial distance. Equation (3.11) is expressed as a tri diagonal matrix to solve the radial velocities.

 Step size Δ z radial  distance, with step size  Δi = 0.001m 0.0001 0.001 0.002 0.008 0.009 0.01 0.016 0.017 0.018 = 0.01m i=0 i=1 i=2 i=8 i=9 i=10 i=16 i=17 i=18 k=0 6.4534 6.4245 6.4100 6.2367 6.2078 6.1934 5.9280 5.8761 5.7722 k=1 6.4385 6.4109 6.3971 6.2317 6.2041 6.1903 5.9276 5.8757 5.7719 k=2 6.4527 6.4239 6.4095 6.2368 6.2080 6.1936 5.9492 5.9026 5.8092 k=3 6.4534 6.4216 6.4058 6.2154 6.1837 6.1678 5.8567 5.8101 5.7170 k=4 6.4536 6.4522 6.4259 6.2418 6.2155 6.1892 5.8496 5.7779 5.3962 k=5 6.4538 6.4276 6.4013 6.2436 6.2173 6.1910 5.8519 5.7802 5.7086 k=6 6.4409 6.4177 6.3945 6.2551 6.2319 6.2086 5.9060 5.8283 5.7507 k=7 6.4544 6.4281 6.4018 6.2441 6.2178 6.1915 5.8522 5.7805 5.7088 k=8 6.4540 6.4288 6.4036 6.2523 6.2270 6.2018 5.8878 5.8160 5.7443 k=9 7.1526 7.1531 7.1536 7.0219 6.9325 6.8432 6.1246 5.9744 5.8242 k=10 8.3323 8.3246 8.3169 8.2910 8.3050 8.3190 8.0736 7.9558 7.8380 k=11 9.1584 9.1592 9.1600 9.1821 9.1942 9.2063 8.6922 8.5367 8.3811 k=12 9.4957 9.5476 9.5996 9.7026 9.6772 9.6518 8.5441 8.2799 8.0156 k=13 9.2044 9.3436 9.4828 9.5092 9.4336 9.3581 7.5506 7.2042 6.8729 k=14 8.9132 9.1054 9.2977 9.0598 8.9428 8.8259 6.5478 6.2688 5.9898 k=15 8.6219 9.1542 9.3414 8.3254 8.0552 7.6841 5.6096 5.3906 5.1716 k=16 8.3307 8.7228 9.1149 7.4597 7.1072 6.7547 4.9913 4.8145 4.6377 k=17 8.0394 7.8275 7.6156 6.3042 6.0392 5.7742 4.4270 4.2429 4.0588 k=18 7.3542 7.2549 7.0565 5.8660 5.6675 5.4691 4.2786 4.0801 3.8817 k=19 6.1844 6.0913 6.0292 5.6567 5.5947 5.5326 5.0610 4.9659 4.8708 k=20 6.0760 6.0400 6.0040 5.8058 5.7698 5.7338 5.4266 5.3602 5.2939 k=21 6.0978 6.0677 6.0526 5.8720 5.8419 5.8268 5.6108 5.5718 5.4938

Table 3.1: Velocities UZ along the plains on a DCR insert.

Values of uz’s  provide for differences on the column in the right hand side  of figure 3.4 to enable us solve the tri-diagonal matrix given by equation (3.11) to determine the value of ur’s. At the DCR insert inlet k=0 for i=1 to i=18; then at K=1 the rows are tabulated starting from i=1 to i=18 in the diverging conical ring insert where i=0 is on the centre line of our pipe.

The value of the RHS as been determined early and provided in the attached excel mark sheet.

Figure 3.5: A tri diagonal matrix to determine radial velocities in the DCR insert.

The matrix is solved using a Matlab algorithm to obtain values of ur’s

That are save in an array to be used when solving equation 1 and 2.

which together with uz’s  provided in the table 3.1 above are used to solve equation (1) to obtain the pressure profile and equation (2) to obtain temperature profiles within the DCR insert.  PRESURE PROFILE

(1)

To determine the pressure profile, velocities in axial direction and radial distance are used in equation (1) with the step size in axial and radial direction provided. TEMPERATURE PROFILE

(2)

Equation (2) is used to determine temperature variations by substituting non dimensional numbers for the flow and boundary conditions estimated by simulating the flow.

SKIN FRICTION.

From values generated in equation (1) and the input velocities in the table 3.1, we should be able to plot a graph of skin friction along the pipe from k=0 to k=21 given density of the fluid is ρ = 957.9kg/m3 and Re=1500 to obtain .        (3)

For different values of Re ( Reynolds number) we should be able to plot different line graphs.

NOTE:

1. At the DCR insert inlet the temperature is constant and also let for the first time, while the values of  are obtain form equation 3.11.
2. From the provided excel sheet we are able to plot the line graphs at i=1, i=9 and i=17.

For pressure against distance (k=1 to k=20) and temperature against distance (k=1 to k=20).

1. The equations in green are the final equations to be solved using matlab to plot graphs of equation 3.12 and 3.13

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