Info
이 질문은 마감되었습니다. 편집하거나 답변을 올리려면 질문을 다시 여십시오.
variable solve for T help
조회 수: 1 (최근 30일)
이전 댓글 표시
molar_composition=[30.4/100,43.1/100,21.5/100,3.1/100,1.9/100]
Feed_stream=100
A=[6.8760,6.8968,6.9187,7.1161,6.9509]
B=[1171.1700,1264.9000,1351.9900,1444.5900,1342.3100]
C=[224.4080,216.5440,209.1550,240.1840,219.1870]
P=5625.42
syms T
5625.42==(molar_composition(1))*10.^(A(1)-(B(1)/T+C(1)))
%+(molar_composition(2)/100)*10.^(A(2)-(B(2)/T+C(2)))+(molar_composition(3)/100)*10.^(A(3)-(B(3)/T+C(3)))+(molar_composition(4)/100)*10.^(A(4)-(B(4)/T+C(4)))+(molar_composition(5)/100)*10.^(A(5)-(B(5)/T+C(5)))
solve(T)
댓글 수: 0
답변 (1개)
KALYAN ACHARJYA
2019년 10월 22일
편집: KALYAN ACHARJYA
2019년 10월 22일
molar_composition=[30.4/100,43.1/100,21.5/100,3.1/100,1.9/100];
Feed_stream=100;
A=[6.8760,6.8968,6.9187,7.1161,6.9509];
B=[1171.1700,1264.9000,1351.9900,1444.5900,1342.3100];
C=[224.4080,216.5440,209.1550,240.1840,219.1870];
P=5625.42;
syms T
fun=(molar_composition(1))*10.^(A(1)-(B(1)/T+C(1)))==5625.42;
%+(molar_composition(2)/100)*10.^(A(2)-(B(2)/T+C(2)))+(molar_composition(3)/100)*10.^(A(3)-(B(3)/T+C(3)))+(molar_composition(4)/100)*10.^(A(4)-(B(4)/T+C(4)))+(molar_composition(5)/100)*10.^(A(5)-(B(5)/T+C(5)))
solve(fun,T)
Result:
ans =
-1/((549755813888*log(20346100990538375/1099511627776))/(643857516551209*log(10)) + 14948685213335552/80482189568901125)
댓글 수: 5
KALYAN ACHARJYA
2019년 10월 22일
편집: KALYAN ACHARJYA
2019년 10월 22일
clc;
clear;
syms x t
molar_composition=[30.4/100,43.1/100,21.5/100,3.1/100,1.9/100];
Feed_stream=100;
A=[6.8760,6.8968,6.9187,7.1161,6.9509];
B=[1171.1700,1264.9000,1351.9900,1444.5900,1342.3100];
C=[224.4080,216.5440,209.1550,240.1840,219.1870];
P=5625.42;
syms T
fun=(molar_composition(1))*10.^(A(1)-(B(1)/T+C(1)))==5625.42;
%+(molar_composition(2)/100)*10.^(A(2)-(B(2)/T+C(2)))+(molar_composition(3)/100)*10.^(A(3)-(B(3)/T+C(3)))+(molar_composition(4)/100)*10.^(A(4)-(B(4)/T+C(4)))+(molar_composition(5)/100)*10.^(A(5)-(B(5)/T+C(5)))
result=vpa(solve(fun,T));
fprintf('The solution is %.2f\n',result);
참고 항목
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!