Need to solve ode equation which are dependent

Need to solve following equations, where the output of eq. 7 used as the input of eq. 8. All values are on right side in equations are known. you can use random value for solving those. Initial conditions are given in the image.
but here are the values
n = 99000
do_fiber = 175 * 10^-6
di_fiber = 85 * 10^-6
Jw = 0.026
Js = 0.0026

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Alan Stevens
Alan Stevens el 5 de Oct. de 2020

1 voto

Your equations can be manipulated as follows
Equations (1) and (2) can be solved directly as functions of z.
Equations (3) and (4) actually have analytical solutions also, though it might be simpler to use ode45 to solve them within Matlab.

7 comentarios

Dhaval Patel
Dhaval Patel el 5 de Oct. de 2020
And How can I use result of ODE45 into another equation? (beacuse ther result is like "41 x 1 double")?
Alan Stevens
Alan Stevens el 5 de Oct. de 2020
Editada: Alan Stevens el 5 de Oct. de 2020
You should have values of Q (and C) for every value of z (presumably there are 41 values). What are the other equations in which you want to use the Q's (and C's)?
I could probaby make more useful comments if you were to upload your coding.
Dhaval Patel
Dhaval Patel el 5 de Oct. de 2020
I want outlet value of Cds and Cfs at perticular z value, the array gives the value, but during programing how can i use?, because for different input, the value of Cds and Cfs at outlet is diffrent(manual input is not an option i think).
Alan Stevens
Alan Stevens el 5 de Oct. de 2020
Have you coded anything at all? If so upload it, otherwise it is difficult to help further.
Dhaval Patel
Dhaval Patel el 5 de Oct. de 2020
Code is based on following algorithm,
Hmm. This is so much more than is in the original question, so I've taken what seems to be the relevant data, solved all of QDS, QFS, CDS and CFS analytically, and produced the following. I can't tell if the results are sensible in the context of the larger problem. I'll leave it to you to decide if the fllowing is of any use ata ll:
L_M = 0.74;
n = 99000;
di_fiber = 85 * 10^-6;
k = n*pi*di_fiber;
Jw = 2.5;
Js = 0.5;
QDSinlet = 1.7 *(10^-3/60);
QFSinlet = 2.7 * (10^-3/60);
CDSinlet = 0.005818;
CFSinlet = 0.770021;
dz = 10^-3;
z = 0:dz:L_M;
QDS = k*Jw*z + QDSinlet; QDSoutlet = QDS(end);
QFS = -k*Jw*z + QFSinlet; QFSoutlet = QFS(end);
CDS = (CDSinlet + Js/Jw)*QDSinlet./(k*Jw*z+QDSinlet) - Js/Jw;
CFS = (CFSinlet + Js/Jw)*QFSinlet./(-k*Jw*z + QFSinlet) - Js/Jw;
disp('QDSoutlet, QFSoutlet')
disp([QDS(end) QFS(end)])
disp('CDSoutlet, CFSoutlet')
disp([CDS(end) CFS(end)])
This produces
QDSoutlet, QFSoutlet
48.9076 -48.9075
CDSoutlet, CFSoutlet
-0.2000 -0.2000
Dhaval Patel
Dhaval Patel el 5 de Oct. de 2020
As you told it is quite diffrent from the larger context. But, Thanks for the help, I have to go thorough through flowchart and Code.

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