![Proper Orthogonal Decomposition Framework for the Explicit ... › ~anil-lab › others › lectures › CMCE › 49-POD-Ceccato.pdf(DMD) [37,38]. Let us consider the equations of](https://static.fdocuments.in/doc/165x107/60d2fb8d998bab547864a5f1/proper-orthogonal-decomposition-framework-for-the-explicit-a-anil-lab-a.jpg)
cmce
8
Avmunj Avneesh123 ASSIGNMENT-2 Program-1 #include<iostream.h> #include<conio.h> #include<stdio.h> #include<math.h> void main() { clrscr(); float ta=30.0, tb=100.0, ts=150.0; float a=0.036, rm=0.0125, tt, lhs; float f[10], d[10], n[10]; n[0]=0; lhs=(ts-tb)/(ts-ta); for(int i=0;i<10;i++) { f[i]=0.692*exp(-5.78*n[i])+0.131*exp(-30.5*n[i])+0.0534*exp(-74.9*n[i])-lhs; d[i]=-3.9976*exp(-5.78*n[i])-3.9955*exp(-30.5*n[i])-3.99966*exp(-74.9*n[i]); n[i+1]=n[i]-f[i]/d[i]; } i=0; do { if(n[i+1]-n[i]<0.001) { cout<<"The time is :"<<n[i+1]*rm*rm*3600/a<<"seconds"<<endl; i=-1; } else i++; }while(i>=0); getch(); }
-
Upload
tanavi-rana -
Category
Documents
-
view
217 -
download
0
Transcript of cmce
8/3/2019 cmce
http://slidepdf.com/reader/full/cmce 1/8
Avmunj
Avneesh123
ASSIGNMENT-2
Program-1
#include<iostream.h>#include<conio.h>
#include<stdio.h>
#include<math.h>
void main()
{
clrscr();
float ta=30.0, tb=100.0, ts=150.0;
float a=0.036, rm=0.0125, tt, lhs;
float f[10], d[10], n[10];
n[0]=0;lhs=(ts-tb)/(ts-ta);
for(int i=0;i<10;i++)
{
f[i]=0.692*exp(-5.78*n[i])+0.131*exp(-30.5*n[i])+0.0534*exp(-74.9*n[i])-lhs;
d[i]=-3.9976*exp(-5.78*n[i])-3.9955*exp(-30.5*n[i])-3.99966*exp(-74.9*n[i]);
n[i+1]=n[i]-f[i]/d[i];
}
i=0;
do
{if(n[i+1]-n[i]<0.001)
{
cout<<"The time is :"<<n[i+1]*rm*rm*3600/a<<"seconds"<<endl;
i=-1;
}
else
i++;
}while(i>=0);
getch();
}
8/3/2019 cmce
http://slidepdf.com/reader/full/cmce 2/8
Output:
The time is :1.42489seconds
Program-2
#include<iostream.h>
#include<conio.h>
#include<stdio.h.>
#include<math.h>
void main(){
float A=2.6606, B=7.1819, b=0.125, tc=369.9,e;
int p;
float t=50.0, to;
for(p=750;p<=10000;p+=250)
{
do
{
to=t;
e=B-exp(-20*((to/tc)-b));t=(A*tc)/(e-log(p)/2.303);
}while(to!=t);
cout<<"The Boiling point at pressure "<<p<<" = "<<to<<endl;
}
getch();
}
Output:
The Boiling point at pressure 750 = 228.485
The Boiling point at pressure 1000 = 235.309
The Boiling point at pressure 1250 = 240.889
The Boiling point at pressure 1500 = 245.648
The Boiling point at pressure 1750 = 249.822
The Boiling point at pressure 2000 = 253.554
The Boiling point at pressure 2250 = 256.939
The Boiling point at pressure 2500 = 260.045
The Boiling point at pressure 2750 = 262.92
The Boiling point at pressure 3000 = 265.601The Boiling point at pressure 3250 = 268.115
8/3/2019 cmce
http://slidepdf.com/reader/full/cmce 3/8
The Boiling point at pressure 3500 = 270.487
The Boiling point at pressure 3750 = 272.732
The Boiling point at pressure 4000 = 274.867
The Boiling point at pressure 4250 = 276.902
The Boiling point at pressure 4500 = 278.85
The Boiling point at pressure 4750 = 280.717
The Boiling point at pressure 5000 = 282.512
The Boiling point at pressure 5250 = 284.24
The Boiling point at pressure 5500 = 285.908
The Boiling point at pressure 5750 = 287.52
The Boiling point at pressure 6000 = 289.081
The Boiling point at pressure 6250 = 290.594
The Boiling point at pressure 6500 = 292.063
The Boiling point at pressure 6750 = 293.49
The Boiling point at pressure 7000 = 294.879
The Boiling point at pressure 7250 = 296.231The Boiling point at pressure 7500 = 297.55
The Boiling point at pressure 7750 = 298.836
The Boiling point at pressure 8000 = 300.092
The Boiling point at pressure 8250 = 301.32
The Boiling point at pressure 8500 = 302.52
The Boiling point at pressure 8750 = 303.695
The Boiling point at pressure 9000 = 304.846
The Boiling point at pressure 9250 = 305.974
The Boiling point at pressure 9500 = 307.079
The Boiling point at pressure 9750 = 308.164The Boiling point at pressure 10000 = 309.228
Program-3
#include<iostream.h>
#include<stdio.h>
#include<math.h>#include<conio.h>
void main()
{
float ts=120,tb=100,ta=20,alpha=0.036,s=0.0125;
float e,a,p,t=0.001,to,eo;
int i=3;
a=pow(3.142,2)/4;
p=alpha/(s*s);
e=(ts-tb)/(ts-ta);
do{
8/3/2019 cmce
http://slidepdf.com/reader/full/cmce 4/8
to=t;
while(i<20)
{
eo=((1/(i*i))*(exp(-(i*i)*a*p*to)));
e=e-eo;
i=i+2;
}
t=(log(e))/(-p*alpha);
}while(to<t);
cout<<"The time of heating ="<<(to*3600)<<" seconds"<<endl;
getch();
}
Output :
The time of heating =698.541 seconds
Program-4
#include<iostream.h>
#include<conio.h>
#include<math.h>
int main()
{
float
Ao=6.87225,Bo=0.097313,Co=508256.0,a=0.9477,b=0.225,c=129000.0,alp=.00607175,gam=
0.022,r=0.08205,p,t,v,vo,e1,e2,e3,e4,e5,e6,e;
cout<<"ENTER PRESSURE(atm):";
cin>>p;
cout<<"\nENTER TEMPERATURE(K):";
cin>>t;
v=r*t/p;
do{
vo=v;
e1=(((Bo*r*t)-Ao-Co)/(t*t))/(vo*vo);
e2=(b*r*t-a)/(vo*vo*vo);
e3=(a*alp)/pow(vo,6);
e4=c/((pow(vo,4))*(pow(t,2)));
e5=1+(gam/(vo*vo));
e6=exp(-gam/vo);
e=e4/(e5*e6);
v=r*t/(p-e-e1-e2-e3);}
8/3/2019 cmce
http://slidepdf.com/reader/full/cmce 5/8
while(vo!=v);
cout<<"\nMOLAR VOLUME(Lit):"<<v;
getch();
return 0;
}
Output:
ENTER PRESSURE(atm):5
ENTER TEMPERATURE(K):300
MOLAR VOLUME(Lit):4.7279
Program-5
#include<iostream.h>
#include<conio.h>
#include<stdio.h>
#include<math.h>
void main()
{
clrscr();
float fo,nre,f=0.01,e;
for( nre=5000;nre<=100000;nre+=5000)
{
do
{
fo=f;
e=4.06*log(nre/sqrt(f))-0.6;
f=1/(e*e);
}while(fo!=f);
cout<<"Nre="<<nre<<" Friction factor="<<f<<endl;
}
getch();
}
Output :
Nre=5000 Friction factor=0.000402403Nre=10000 Friction factor=0.000357262
8/3/2019 cmce
http://slidepdf.com/reader/full/cmce 6/8
Nre=15000 Friction factor=0.000334376
Nre=20000 Friction factor=0.000319479
Nre=25000 Friction factor=0.000308615
Nre=30000 Friction factor=0.000300155
Nre=35000 Friction factor=0.000293277
Nre=40000 Friction factor=0.000287513
Nre=45000 Friction factor=0.000282572
Nre=50000 Friction factor=0.000278262
Nre=55000 Friction factor=0.000274449
Nre=60000 Friction factor=0.000271038
Nre=65000 Friction factor=0.000267957
Nre=70000 Friction factor=0.000265153
Nre=75000 Friction factor=0.000262582
Nre=80000 Friction factor=0.000260211
Nre=85000 Friction factor=0.000258015
Nre=90000 Friction factor=0.000255969Nre=95000 Friction factor=0.000254058
Nre=100000 Friction factor=0.000252264
Program-6
#include<iostream.h>
#include<conio.h>#include<stdio.h>
#include<math.h>
void main()
{
clrscr();
float Ao=1.3012, Bo=0.4611, r=0.08205, a=0.01931, b=-0.011101, c=43400.0;
float vo,e1,e2,e3,e,v,p,t;
cout<<"P T Molal Volume"<<endl;
for(p=10;p<=50;p+=5)
{for(t=173;t<=373;t+=50)
{
v=r*t/p;
do
{
vo=v;
e1=r*t*(1-c/(vo*t*t*t));
e2=vo+(Bo*(1-(b/vo)));
e3=Ao*(1-a/vo);
e=((e1*e2)-e3)/p;v=sqrt(e);
8/3/2019 cmce
http://slidepdf.com/reader/full/cmce 7/8
}while(vo!=v);
cout<<p<<" "<<t<<" "<<v<<endl;
}
}
getch();
}
Output :
10 173 1.71904
10 223 2.15877
10 273 2.58861
10 323 3.01307
10 373 3.43419
15 173 1.22671
15 223 1.53004
15 273 1.82389
15 323 2.11248
15 373 2.39775
20 173 0.974824
20 223 1.20974
20 273 1.43558
20 323 1.65633
20 373 1.8738
25 173 0.820224
25 223 1.0139
25 273 1.19886
25 323 1.37889
25 373 1.55572
30 173 0.714869
30 223 0.88088
30 273 1.03851
30 323 1.19136
30 373 1.34108
35 173 0.63801335 223 0.784131
35 273 0.922153
35 323 1.05555
35 373 1.1859
40 173 0.579197
40 223 0.710286
40 273 0.833537
40 323 0.952298
40 373 1.06809
45 173 0.53256145 223 0.651871
8/3/2019 cmce
http://slidepdf.com/reader/full/cmce 8/8
45 273 0.763575
45 323 0.870916
45 373 0.97536
50 173 0.494557
50 223 0.60437
50 273 0.706787
50 323 0.804958
50 373 0.9003
![FEM analysis of plates and shellsjpamin/dyd/CMCE/CMCE_lecture8_JP.pdf · FEM approximation [3] Element tangent stiffness k e T = k 0 + k e u + ke σ ke 0 = Z Z Ae B nTD B 0 0 BmTDmBm](https://static.fdocuments.in/doc/165x107/60b599060959c219dc2536fc/fem-analysis-of-plates-and-shells-jpamindydcmcecmcelecture8jppdf-fem-approximation.jpg)
