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고체 세제의 녹는 속도와 최적의 크기 [옥타브 코드]

작성자
cfdkim
작성일
2021-10-06 13:12
조회
472
clear;clc;clf;

V = 100; c = 4*pi/3;

for n=1:50

V1 = V/n;

r0(n) = (V1/c)^(1/3); time(n) = r0(n)*n;

end

[~,h1,h2]=plotyy(1:n,time,1:n,r0);axis tight

set(h1, "color", "b");set(h2, "color", "r");







clear;clc;clf;

V = 100;

c = 4*pi/3;

n=1;V1 = V/n; r0(n) = (V1/c)^(1/3);

while r0(n)>1

time(n) = (r0(n)-1)*n + 1 ;

n=n+1; V1 = V/n; r0(n) = (V1/c)^(1/3);

end

[~,h1,h2]=plotyy(1:n-1,time,1:n-1,r0(1:n-1));


axis tight;set(h1,"color","b");set(h2,"color","r");





clear;clc

V = 100;

for n=1:6

switch n

case 1

c1 = sqrt(2)/12; c2 = sqrt(3);

case 2

c1 = 1; c2 = 6;

case 3

c1 = sqrt(2)/3; c2 = 2*sqrt(3);

case 4

c1 = (15+7*sqrt(5))/4; c2 = 15*tan(3*pi/10);

case 5

c1 = 5*(3+sqrt(5))/12; c2 = 5*sqrt(3);

case 6

c1 = 4*pi/3; c2 = 4*pi;

end

a = (V/c1)^(1/3); S = c2*a^2;

ss = sprintf('a = %.4f\nS = %.4f\n',a,S);

fprintf(ss)


end





clear;clc;clf;

V = 100;

c = 4*pi/3;

n=1;V1 = V/n; r0(n) = (V1/c)^(1/3);

while r0(n)>1

time(n) = (r0(n)-1)*n + 1 ;

n=n+1; V1 = V/n; r0(n) = (V1/c)^(1/3);

end

[~,h1,h2]=plotyy(1:n-1,time,1:n-1,r0(1:n-1));

axis tight;


set(h1,"color","b"); set(h2,"color","r");
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