numerical analysis term project_yash_ravi_final report

8/4/2019 Numerical Analysis Term Project_yash_ravi_final Report

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NUMERICAL ANALYSIS TERM PROJECT:

YASH SHAH 9003001

RAVI AGARWAL 9003017

PROJECT: SOLVING DIFFERENTIAL EQUATIONS FOR HEAT RECOVERY

SYSTEM GENERATOR USING MATLAB


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DESCRIPTION OF HEAT RECOVERY STEAM GENERATOR:

Heat Recovery Steam Generator (HRSG) is used to produce steam by using the waste heatproduced by a gas turbine or some processes where heat is a by-product.

The steam produced in the HRSG can be used to generate power (electrical or mechanical orboth) or it can be directly fed into some process where steam is required as input.

Because of better utilization of energy, the efficiency of the process increases with the useof HRSG.

Lumped Element Analysis:

We have taken Lumped parameters while dealing with the differentia equations of the system.

Discrete entities that approximate the behaviour of the distributed system under certainassumptions. Average values are taken over the system.

Mathematical models are developed for different HRSG components, using lumpedparameter, time dependent, mass and energy balances, with suitable simplifying

assumptions.

The system is linearised about a steady-state operating point. An optimal linear-quadraticregulator (Belanger, 1993) for this system is designed by choosing a suitable performance

index.


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CIRCUIT DIAGRAM OF HRSG:

IMPORTANT COMPONENTS OF HRSG:

1. EVAPORATOR2. ECONOMISER3. SUPERHEATER 14. SUPERHEATER 2

The system of equations has been linearised to 10 linear equations with 10 variables.

These are equations are linearised and there are 10 dependent variables

VARIABLES:

1. Evaporator (T1, h1)2. Economiser (T2, h2)3. Superheater 1 (T3, h3)4. Superheater 2 (T4, h4)5. Density of the fluid g6. Volume of the liquid

And one independent variable t.


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CONSTANTS:

1. Cd is the coefficient of discharge of the turbine inlet valve.2. Wd is the flow rate of cold fluid passing through the system3. Watt is the flow rate of the fluid in attemperator

The 10 differential equations are:

All the variables in this system of equations indicate the change in parameter(and not the value of

the parameter) at different time periods.

BOUNDARY CONDITION: At the steady state all these variables are zero.

1 = -0.157*T1+0.142*h1;

1 = -0.00114-0.048h1+0.051*T1;

2 = -0.090*T2+0.017*g;

2= -0.00312*h2+1.48*T2+0.00337*T1-0.279*g;

=0.00411;

g =0.0038-0.395*g+0.0374;

3 =-0.037*T3+0.069*h3+0.00672*g;

3 =-0.28*h3 +0.024*h2-0.0134*h1+0.128*T3-0.01*g+0.000642;

4 =-0.037*T4+0.094*h4+0.0036*g;

4 =0.032*h3-0.022*h2+0.012*h1-0.57*h4+0.212*T4-

0.00638*g+0.000382+0.000255


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EXPERIMENTAL PROCEDURE:

At t=0 the system is at the steady state, i.e. all the variables are zero We perturbed the system by changing parameters of any one chamber from the steady state

and then observe the system at different time intervals.

The deviations in each of these variables are obtained by solving the system of these 10equations by Runge-Kutta method (we have propose)

Graphs are plotted between various dependent and independent-dependent variables andthe behaviour of the whole system is studied.

ALGORITHM:

Runge-Kutta Method:

4th

order RK Method has been used for the system of 10 equations to get the required results

10 variables:

All these variables refer to the above mentioned Variables respectively.

1. a(1,i)2. a(2i)3. a(3,i)4. a(4,i)5.

a(5,i)

6. a(6,i)7. a(7,i)8. a(8,i)9. a(9,i)10.a(10,i)Here irefer to time instant.

Boundary Condition: All the variables are zero at t=0


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One of the chambers parameters (T1 and h1 to start with are given values instead of

zero to analyze its effect on the other variables)

Logic of the Matlab Program:

STEP 1:

A file is made which bears the system of linear equations and is saved as sole.m. This

file is called every time we solve the equations in another file using various numerical

analysis methods.

STEP 2

In a new file we type the syntax to solve the equations after calling the file sole.m

Program Body:

The system is perturbed and the initial values ofTiand hi are changed from zero.

For (j=0; j


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For a[i, j], y=i and t=j

General Runge-Kutta Method:

a (i, j+1) = a(i, j) + h/6(k1 + 2k2 + 2k3 + k4) /* herej refers to n in the following formula*/

Example:

In this when we are dealing with a[2,j] only a(2) will be changed and we will take the latest value

(values calculated in the preceding step) of the remaining variables present in the particular

equation.

All the values of p, q, r, s will be found for each variable a[i] at different time intervals depending on

the step size and finally the values of each variable at different time intervals will be found out.

We repeat this method for 4 step sizes: 0.1, 1.0

STEP 3

Step 2 is repeated 3 times by perturbing values of (T2, h2), (T3, h3), (T4, h4) sequentially.

STEP 4

We solve the same set of differential equations using the inbuilt Runge-Kutta function ode45.


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STEP 5

We plot the graphs of all the variables versus time for each perturbation

1. (T1, h1)2. (T2, h2),3. (T3, h3),4. (T4, h4)

Syntax:

Plot (0: h: 5, a(: ; : ))

STEP 6

To check the validity of our results we plot values of one variable calculated for different

step sizes and the values obtained using inbuilt function ode45 versus time.

Syntax:

For (i=1; i


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Figure 1

Scale:

X-axis: 1 unit = 0.5 seconds

Y-axis: 1 unit = 2 Degree Celsius/ 2 Joules


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Case 2:

Change in the temperature (T2) and enthalpy (h2) of the Economiser

Conclusions:

a. Temperature of the Economiser starts decreasing once it is disturbed and it alsoaffects the temperature (T1) of the Evaporator (decreases)

b. Enthalpy of the Economic increases once it is disturbed and it affects enthalpy of theSuperheaters one 1 (increases)&2 (decreases)

c. There is no change in the volume of the drum and density of the fluid


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Figure 2

Scale:




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Case 3:

Change in the Temperature (T3) and Enthalpy (h3) of the Superheater 1

Conclusions:

a. The temperature of the Superheater 1 keeps on increasing once it is disturbed fromthe equilibrium and the temperature of the other chambers is not affected

b. Enthalpy of the Superheater 1 keeps on decreasing once it is disturbed and it willalso affect the enthalpy of the Superheater 2 (increases) and enthalpy in chambers is

unaltered



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Figure 3

Scale:




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Case 4:

Change in the Temperature (T4) and Enthalpy (h4) of the Superheater 2

Conclusions:

a. Temperature in the Superheater 2 increases once it is disturbed and thetemperatures in the other chambers are unaffected

b. Enthalpy in the Superheater 2 starts decreasing once it is disturbed and the enthalpyin other chambers is not affected.



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Figure 4

Scale:




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DISCUSSIONS:

We sincerely thank Professor Dhiren Modi for introducing the concept of Engineering

Projects in this course which motivated us to learn Matlab, helped us to come across

various real life engineering problems and introduced us to the concept of report writing.

Why heat recovery steam generation?

Heat Recovery Steam Generator

We have taken this project from Professor Manmohan Pandey. Two students from IIT

Guwahati have written research paper as their B.Tech Project on HRSG under Prof. Pandeys

guidance. The paper focuses on converting the non-linear equations into dimensionless

linear equations and solving them. After studying the equations an open and closed loop

control systems are generated for the Heat Recovery System Generation.

We wanted to know about the types of B.Tech projects taken by students in other Deemed

Engineering Institutes which would help us in deciding the level of our B.Tech projects twoyears hence. Prof. Pandey introduced us to this project and we were more than happy to go

through it and solve the differential equations and make a beginning on doing projects and

report writing.

Also today Energy Conservation has become very important and almost any power plant,

manufacturing unit, vehicles have systems installed for energy efficiency. Heat Recovery

system generation is one such important system installed in power plants for generating

steam from the waste heat and using it to rotate turbines and generate power. Thus the

installation of the Heat Recovery System Generator (HRSG) is very important in present

power plants to increase the efficiency and from the economic point of view.


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Why have we used a multi-step method such as Runge-Kutta method?

General Point of View:

A single-step numerical method has a short memory. The only information passed

From one step to the next is an estimate of the proper step size and, perhaps, the

Value of f (tn; yn) at the point the two steps have in common.

A multistep method has a longer memory. After an initial start-up phase, a pth

-order

multistep method saves up to perhaps a dozen values of the solution, yn-p+1; yn-p+2 upto yn-1;

yn, and uses them all to compute yn+1. In fact, these methods can vary the order p, and the

step size, h.

From our System point of view:

Solving the system of equations we can conclude that our system is in non-equilibrium state.

Thus perturbing one variable leads to either continuous increment or decrement in its value and

other dependent variables values.

During analysis if we get even a slight unwanted change in one of the dependent variables,

the analysis may not be upto the mark.

As we are dealing with huge infrastructure oriented systems like Heat Recovery system

Generator within a power plot it is of utmost importance that we have the most accurate analysis.

As Runge-Kutta method is the most accurate we have studied till date, we have decided to

solve the equations using this method.

Accurate changes in all components of the systems can be calculated using Matlab (for

solving differential equations) upto infinite number of cycles in order to check the behaviour of the

system and take appropriate precautionary measures.

What types of analysis can we obtained by solving such equations using Matlab?

1. Controlled System AnalysisIn our system if equation we have various control variables like Wcold, Cdand Watt.

We can solve all the system of variables by giving different values to these user

controlled variables and find out how can we obtaining a particular behavior of the

system according the ambient conditions.

2. Uncontrolled System AnalysisWe have solved the system of linear equations which in uncontrolled. Changes in all the

user controlled variables are assumed zero. We have analysed how the system behaves

when left open.


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3. Dependence of all the variables on each other.a. How will the enthalpy of the Superheater 1 change when its temperature is

changed?

Figure 5

Scale:

X-axis: 1 unit = 0.002 Degree Celsius

Y-axis: 1 unit = 0.01 Joules

b. How will the Water in drum level vary when the temperature and enthalpy ofthe Economiser is changed?

(Refer Figure 1)

c. When the temperature of the economiser is increased by say 100 C, after howmuch time the temperature in the evaporator will reach its maximum rated

value?


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4. Behaviour of a particular variable in the system:During perturbation in Case 4, changing the temperature from the steady state, the

temperature shoots up enormously and can be dangerous for the steam plant. This can

be anticipated using Matlab and precaution can be taken at the time of initiation.

(Refer Figure 4)

How can we conclude that our method is correct and all the results we have obtained are

accurate and match the original values?

Below is the graph of all the variables with respect to time when we have perturbed the

system by changing two values T1 and h1.

Each graph has 4 values for the same variables calculated by Runge-Kutta method by varying

the step size

1. h=0.01 variable a(10,501) with colour red2. h=0.05 variable e(10,101) with colour magenta3. h=0.1 variable b(10,51) with colour blue4. h=1 variable c(10,6) with colour green5. By using ode45 variable d with colour black

There are three graphs:

For the following graphs:

For a1, a2, a3, a4

Units: Degree Celsius

For a5, a6, a7, a8

Unit: Joules

For a9

Unit: Kg/m3

For a10

Unit: m3


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1. System left open for 5 seconds

FIGURE 6


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2. System left open for 15 seconds

FIGURE 7


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From the third graph we can conclude that for small interval after the system is left open say 40

seconds, Runge-Kutta method is accurate with the inbuilt function

After 40 seconds, our calculations are accurate as we can see the graphs using different sizes

coincide but it shows variations with the ode45 solutions.

BIBLIOGRAPHY:

PROJECT TAKEN FROM RESEARCH PAPER

BY: AMIT NATH PANDEY AND BHARAT PANDEY

MENTOR: PROF. MANMOHAN PANDEY
http://c/Users/yashshah/Downloads/paper171%20(2).pdfhttp://c/Users/yashshah/Downloads/paper171%20(2).pdfhttp://c/Users/yashshah/Downloads/paper171%20(2).pdf


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APPENDIX:

VARIABLES:

1. T1 =a12. T2 =a23. T3 =a34. T4 =a45. h5 =a56. h6 =a67. h7 =a78. h8 =a89. g =a910.V10out =a10

Matlab Program for the project (The following code gives the graph of FIGURE 6 ,

rest of the graphs are plotted accordingly by changing the initial values as per

mentioned below every graph. Also the function sole is made a separate

function [ da ] = sole( t,a )%UNTITLED Summary of this function goes here% Detailed explanation goes hereda(1)=-0.157* a(2) + 0.142*a(5);

da(5)=-0.00114-0.048*a(5)+0.051*a(1); da(2)=-0.090*a(2)+0.017*a(9);da(6)=-0.00312*a(6)+1.48*a(2)+0.00337*a(1)-0.279*a(9); da(10)=0.00411;da(9)=0.0038-0.395*a(9)+0.0374;da(3)=-0.037*a(3)+0.069*a(7)+0.00672*a(9); da(7)=-0.28*a(7)+0.024*a(6)-0.0134*a(5)+0.128*a(3)-0.01*a(9)+0.000642; da(4)=-0.037*a(4)+0.094*a(8)+0.0036*a(9); da(8)=0.032*a(7)-0.022*a(6)+0.012*a(5)-0.57*a(8)+0.212*a(4)-0.00638*a(9)+0.000382+0.000255;

da = da(:);

end


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p=zeros(10,501);q=zeros(10,501);r=zeros(10,501);s=zeros(10,501);a=zeros(10,501);

a(1,1)=10;a(2,1)=0;a(3,1)=0;a(4,1)=0;a(5,1)=10;a(6,1)=0;a(7,1)=0;a(8,1)=0;a(9,1)=0;a(10,1)=0;

for j = 1:500

s(1,j)=-0.157* a(2) + 0.142*a(5);r(1,j)=s(1,j);q(1,j)=r(1,j);p(1,j)=q(1,j);a(1,j+1)=a(1,j)+6*0.01/6*(p(1,j));

s(10,j)=0;r(10,j)=s(10,j);q(10,j)=r(10,j);p(10,j)=q(10,j);a(10,j+1)=a(10,j);

p(2,j)=-0.090*a(2)+0.017*a(9);q(2,j)=-0.090*(a(2)+p(2,j)*0.01/2)+0.017*a(9);

r(2,j)=-0.090*(a(2)+q(2,j)*0.01/2)+0.017*a(9); s(2,j)=-0.090*(a(2)+r(2,j)*0.01)+0.017*a(9); a(2,j+1)=a(2,j)+0.01/6*(p(2,j)+2*q(2,j)+2*r(2,j)+s(2,j));

p(3,j)=-0.037*a(3)+0.069*a(7)+0.00672*a(9); q(3,j)=-0.037*(a(3)+p(3,j)*0.01/2)+0.069*a(7)+0.00672*a(9); r(3,j)=-0.037*(a(3)+q(3,j)*0.01/2)+0.069*a(7)+0.00672*a(9); s(3,j)=-0.037*(a(3)+r(3,j)*0.01)+0.069*a(7)+0.00672*a(9); a(3,j+1)=a(3,j)+0.01/6*(p(3,j)+2*q(3,j)+2*r(3,j)+s(3,j));

p(4,j)=-0.037*a(4)+0.094*a(8)+0.0036*a(9);

q(4,j)=-0.037*(a(4)+p(4,j)*0.01/2)+0.094*a(8)+0.0036*a(9); r(4,j)=-0.037*(a(4)+q(4,j)*0.01/2)+0.094*a(8)+0.0036*a(9); s(4,j)=-0.037*(a(4)+r(4,j)*0.01)+0.094*a(8)+0.0036*a(9); a(4,j+1)=a(4,j)+0.01/6*(p(4,j)+2*q(4,j)+2*r(4,j)+s(4,j));

p(5,j)=-0.048*a(5)+0.051*a(1);q(5,j)=-0.048*(a(5)+p(5,j)*0.01/2)+0.051*a(1); r(5,j)=-0.048*(a(5)+q(5,j)*0.01/2)+0.051*a(1); s(5,j)=-0.048*(a(5)+r(5,j)*0.01)+0.051*a(1); a(5,j+1)=a(5,j) +0.01/6*(p(5,j)+2*q(5,j)+2*r(5,j)+s(5,j));

p(6,j)=-0.00312*a(6)+1.48*a(2)+0.00337*a(1)-0.279*a(9); q(6,j)=-0.00312*(a(6)+p(6,j)*0.01/2)+1.48*a(2)+0.00337*a(1)-

0.279*a(9);


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r(6,j)=-0.00312*(a(6)+q(6,j)*0.01/2)+1.48*a(2)+0.00337*a(1)-0.279*a(9);

s(6,j)=-0.00312*(a(6)+r(6,j)*0.01)+1.48*a(2)+0.00337*a(1)-0.279*a(9);

a(6,j+1)=a(6,j)+0.01/6*(p(6,j)+2*q(6,j)+2*r(6,j)+s(6,j));

p(7,j)=-0.28*a(7)+0.024*a(6)-0.0134*a(5)+0.128*a(3)-0.01*a(9); q(7,j)=-0.28*(a(7)+p(7,j)*0.01/2)+0.024*a(6)-0.0134*a(5)+0.128*a(3)-

0.01*a(9);r(7,j)=-0.28*(a(7)+q(7,j)*0.01/2)+0.024*a(6)-0.0134*a(5)+0.128*a(3)-

0.01*a(9);s(7,j)=-0.28*(a(7)+r(7,j)*0.01)+0.024*a(6)-0.0134*a(5)+0.128*a(3)-

0.01*a(9);a(7,j+1)=a(7,j)+0.01/6*(p(7,j)+2*q(7,j)+2*r(7,j)+s(7,j));

p(8,j)=0.032*a(7)-0.022*a(6)+0.012*a(5)-0.57*a(8)+0.212*a(4)-0.00638*a(9);

q(8,j)=0.032*a(7)-0.022*a(6)+0.012*a(5)-0.57*(a(8)+p(8,j)*0.01/2)+0.212*a(4)-0.00638*a(9);

r(8,j)=0.032*a(7)-0.022*a(6)+0.012*a(5)-0.57*(a(8)+q(8,j)*0.01/2)+0.212*a(4)-0.00638*a(9);

s(8,j)=0.032*a(7)-0.022*a(6)+0.012*a(5)-0.57*(a(8)+r(8,j)*0.01)+0.212*a(4)-0.00638*a(9);

a(8,j+1)=a(8,j)+0.01/6*(p(8,j)+2*q(8,j)+2*r(8,j)+s(8,j));

p(9,j)=-0.395*a(9);q(9,j)=-0.395*(a(9)+p(9,j)*0.01/2); r(9,j)=-0.395*(a(9)+q(9,j)*0.01/2); s(9,j)=-0.395*(a(9)+r(9,j)*0.01);a(9,j+1)=a(9,j)+ 0.01/6*(p(9,j)+2*q(9,j)+2*r(9,j)+s(9,j));

end

a

p=zeros(10,51);q=zeros(10,51);r=zeros(10,51);s=zeros(10,51);b=zeros(10,51);

b(1,1)=10;b(2,1)=0;b(3,1)=0;b(4,1)=0;b(5,1)=10;b(6,1)=0;b(7,1)=0;b(8,1)=0;b(9,1)=0;b(10,1)=0;


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for j = 1:50

s(1,j)=-0.157* b(2) + 0.142*b(5);r(1,j)=s(1,j);q(1,j)=r(1,j);p(1,j)=q(1,j);

b(1,j+1)=b(1,j)+6*0.1/6*(p(1,j));

s(10,j)=0;r(10,j)=s(10,j);q(10,j)=r(10,j);p(10,j)=q(10,j);b(10,j+1)=b(10,j);

p(2,j)=-0.090*b(2)+0.017*b(9);q(2,j)=-0.090*(b(2)+p(2,j)*0.1/2)+0.017*b(9); r(2,j)=-0.090*(b(2)+q(2,j)*0.1/2)+0.017*b(9); s(2,j)=-0.090*(b(2)+r(2,j)*0.1)+0.017*b(9); b(2,j+1)=b(2,j)+0.1/6*(p(2,j)+2*q(2,j)+2*r(2,j)+s(2,j));

p(3,j)=-0.037*b(3)+0.069*b(7)+0.00672*b(9); q(3,j)=-0.037*(b(3)+p(3,j)*0.1/2)+0.069*b(7)+0.00672*b(9); r(3,j)=-0.037*(b(3)+q(3,j)*0.1/2)+0.069*b(7)+0.00672*b(9); s(3,j)=-0.037*(b(3)+r(3,j)*0.1)+0.069*b(7)+0.00672*b(9); b(3,j+1)=b(3,j)+0.1/6*(p(3,j)+2*q(3,j)+2*r(3,j)+s(3,j));

p(4,j)=-0.037*b(4)+0.094*b(8)+0.0036*b(9); q(4,j)=-0.037*(b(4)+p(4,j)*0.1/2)+0.094*b(8)+0.0036*b(9); r(4,j)=-0.037*(b(4)+q(4,j)*0.1/2)+0.094*b(8)+0.0036*b(9); s(4,j)=-0.037*(b(4)+r(4,j)*0.1)+0.094*b(8)+0.0036*b(9); b(4,j+1)=b(4,j)+0.1/6*(p(4,j)+2*q(4,j)+2*r(4,j)+s(4,j));

p(5,j)=-0.048*b(5)+0.051*b(1);q(5,j)=-0.048*(b(5)+p(5,j)*0.1/2)+0.051*b(1); r(5,j)=-0.048*(b(5)+q(5,j)*0.1/2)+0.051*b(1); s(5,j)=-0.048*(b(5)+r(5,j)*0.1)+0.051*b(1); b(5,j+1)=b(5,j) +0.1/6*(p(5,j)+2*q(5,j)+2*r(5,j)+s(5,j));

p(6,j)=-0.00312*b(6)+1.48*b(2)+0.00337*b(1)-0.279*b(9); q(6,j)=-0.00312*(b(6)+p(6,j)*0.1/2)+1.48*b(2)+0.00337*b(1)-

0.279*b(9);r(6,j)=-0.00312*(b(6)+q(6,j)*0.1/2)+1.48*b(2)+0.00337*b(1)-

0.279*b(9);s(6,j)=-0.00312*(b(6)+r(6,j)*0.1)+1.48*b(2)+0.00337*b(1)-0.279*b(9);

b(6,j+1)=b(6,j)+0.1/6*(p(6,j)+2*q(6,j)+2*r(6,j)+s(6,j));

p(7,j)=-0.28*b(7)+0.024*b(6)-0.0134*b(5)+0.128*b(3)-0.01*b(9); q(7,j)=-0.28*(b(7)+p(7,j)*0.1/2)+0.024*b(6)-0.0134*b(5)+0.128*b(3)-

0.01*b(9);r(7,j)=-0.28*(b(7)+q(7,j)*0.1/2)+0.024*b(6)-0.0134*b(5)+0.128*b(3)-

0.01*b(9);s(7,j)=-0.28*(b(7)+r(7,j)*0.1)+0.024*b(6)-0.0134*b(5)+0.128*b(3)-

0.01*b(9);b(7,j+1)=b(7,j)+0.1/6*(p(7,j)+2*q(7,j)+2*r(7,j)+s(7,j));

p(8,j)=0.032*b(7)-0.022*b(6)+0.012*b(5)-0.57*b(8)+0.212*b(4)-0.00638*b(9);

q(8,j)=0.032*b(7)-0.022*b(6)+0.012*b(5)-0.57*(b(8)+p(8,j)*0.1/2)+0.212*b(4)-0.00638*b(9);


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r(8,j)=0.032*b(7)-0.022*b(6)+0.012*b(5)-0.57*(b(8)+q(8,j)*0.1/2)+0.212*b(4)-0.00638*b(9);

s(8,j)=0.032*b(7)-0.022*b(6)+0.012*b(5)-0.57*(b(8)+r(8,j)*0.1)+0.212*b(4)-0.00638*b(9);

b(8,j+1)=b(8,j)+0.1/6*(p(8,j)+2*q(8,j)+2*r(8,j)+s(8,j));

p(9,j)=-0.395*b(9);q(9,j)=-0.395*(b(9)+p(9,j)*0.1/2);r(9,j)=-0.395*(b(9)+q(9,j)*0.1/2);s(9,j)=-0.395*(b(9)+r(9,j)*0.1);b(9,j+1)=b(9,j)+ 0.1/6*(p(9,j)+2*q(9,j)+2*r(9,j)+s(9,j));

end

b

p=zeros(10,6);q=zeros(10,6);r=zeros(10,6);s=zeros(10,6);c=zeros(10,6);

c(1,1)=10;c(2,1)=0;c(3,1)=0;c(4,1)=0;c(5,1)=10;c(6,1)=0;c(7,1)=0;c(8,1)=0;c(9,1)=0;c(10,1)=0;

for j = 1:5

s(1,j)=-0.157* c(2) + 0.142*c(5);r(1,j)=s(1,j);q(1,j)=r(1,j);p(1,j)=q(1,j);c(1,j+1)=c(1,j)+4/6*(p(1,j));

s(10,j)=0;r(10,j)=s(10,j);q(10,j)=r(10,j);p(10,j)=q(10,j);

p(2,j)=-0.090*c(2)+0.017*c(9);

q(2,j)=-0.090*(c(2)+p(2,j)*1/2)+0.017*c(9); r(2,j)=-0.090*(c(2)+q(2,j)*1/2)+0.017*c(9); s(2,j)=-0.090*(c(2)+r(2,j)*1)+0.017*c(9);


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c(2,j+1)=c(2,j)+1/6*(p(2,j)+q(2,j)+r(2,j)+s(2,j));

p(3,j)=-0.037*c(3)+0.069*c(7)+0.00672*c(9); q(3,j)=-0.037*(c(3)+p(3,j)*1/2)+0.069*c(7)+0.00672*c(9); r(3,j)=-0.037*(c(3)+q(3,j)*1/2)+0.069*c(7)+0.00672*c(9); s(3,j)=-0.037*(c(3)+r(3,j)*1)+0.069*c(7)+0.00672*c(9);

c(3,j+1)=c(3,j)+1/6*(p(3,j)+q(3,j)+r(3,j)+s(3,j));

p(4,j)=-0.037*c(4)+0.094*c(8)+0.0036*c(9); q(4,j)=-0.037*(c(4)+p(4,j)*1/2)+0.094*c(8)+0.0036*c(9); r(4,j)=-0.037*(c(4)+q(4,j)*1/2)+0.094*c(8)+0.0036*c(9); s(4,j)=-0.037*(c(4)+r(4,j)*1)+0.094*c(8)+0.0036*c(9); c(4,j+1)=c(4,j)+1/6*(p(4,j)+q(4,j)+r(4,j)+s(4,j));

p(5,j)=-0.048*c(5)+0.051*c(1);q(5,j)=-0.048*(c(5)+p(5,j)*1/2)+0.051*c(1); r(5,j)=-0.048*(c(5)+q(5,j)*1/2)+0.051*c(1); s(5,j)=-0.048*(c(5)+r(5,j)*1)+0.051*c(1); c(5,j+1)=c(5,j) +1/6*(p(5,j)+q(5,j)+r(5,j)+s(5,j));

p(6,j)=-0.00312*c(6)+1.48*c(2)+0.00337*c(1)-0.279*c(9); q(6,j)=-0.00312*(c(6)+p(6,j)*1/2)+1.48*c(2)+0.00337*c(1)-0.279*c(9); r(6,j)=-0.00312*(c(6)+q(6,j)*1/2)+1.48*c(2)+0.00337*c(1)-0.279*c(9); s(6,j)=-0.00312*(c(6)+r(6,j)*1)+1.48*c(2)+0.00337*c(1)-0.279*c(9); c(6,j+1)=c(6,j)+1/6*(p(6,j)+q(6,j)+r(6,j)+s(6,j));

p(7,j)=-0.28*c(7)+0.024*c(6)-0.0134*c(5)+0.128*c(3)-0.01*c(9); q(7,j)=-0.28*(c(7)+p(7,j)*1/2)+0.024*c(6)-0.0134*c(5)+0.128*c(3)-

0.01*c(9);r(7,j)=-0.28*(c(7)+q(7,j)*1/2)+0.024*c(6)-0.0134*c(5)+0.128*c(3)-

0.01*c(9);s(7,j)=-0.28*(c(7)+r(7,j)*1)+0.024*c(6)-0.0134*c(5)+0.128*c(3)-

0.01*c(9);c(7,j+1)=c(7,j)+1/6*(p(7,j)+q(7,j)+r(7,j)+s(7,j));

p(8,j)=0.032*c(7)-0.022*c(6)+0.012*c(5)-0.57*c(8)+0.212*c(4)-0.00638*c(9);

q(8,j)=0.032*c(7)-0.022*c(6)+0.012*c(5)-0.57*(c(8)+p(8,j)*1/2)+0.212*c(4)-0.00638*c(9);

r(8,j)=0.032*c(7)-0.022*c(6)+0.012*c(5)-0.57*(c(8)+q(8,j)*1/2)+0.212*c(4)-0.00638*c(9);

s(8,j)=0.032*c(7)-0.022*c(6)+0.012*c(5)-0.57*(c(8)+r(8,j)*1)+0.212*c(4)-0.00638*c(9);

c(8,j+1)=c(8,j)+1/6*(p(8,j)+q(8,j)+r(8,j)+s(8,j));

p(9,j)=-0.395*c(9);q(9,j)=-0.395*(c(9)+p(9,j)*1/2);r(9,j)=-0.395*(c(9)+q(9,j)*1/2);s(9,j)=-0.395*(c(9)+r(9,j)*1);c(9,j+1)=c(9,j)+ 1/6*(p(9,j)+q(9,j)+r(9,j)+s(9,j));

end

t = 0:.01:5;t_1 = 0:.1:5;t_2 = 0:5;

c


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ic = [10 0 0 0 10 0 0 0 0 0];[T,d] = ode45(@sole,[0:1:5], ic)AF = d(6,:)d=d';

figure

for i= 1:10subplot(5,2,i)hold onplot(t,a(i,:),'r')plot(t_1,b(i,:),'b')plot(t_2,c(i,:),'g')plot(t_2,d(i,:),'k')xlabel('time(sec)')legend(strcat('a(',num2str(i),')'),strcat('b(',num2str(i),')'),strcat('c(',num2str(i),')'),strcat('d(',num2str(i),')'))hold off

end

numerical analysis term project_yash_ravi_final report

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