KETA01 Tenta 2012-03-08 Carmen Arévalo
Contents
Uppfift 1
Ainv=[0.074505 0 0 0 0.081619 0.081619 0 0 0.080795 0.080795 0.080795 0 0 0 0 0.084767]; A=inv(Ainv)
A =
13.422 0 0 0
-13.422 12.252 0 0
0 -12.252 12.377 0
0 0 0 11.797
The inverse cannot correspond to the figure because: According to this inverse, reservoir 4 is only affected by inputs to reservoir 4, but we see in the figure that it is affected by inputs in all other reservoirs
b = [750; 150; 50; 30]; c=Ainv*b
c =
55.879
73.457
76.755
2.543
Uppgift 2
t=[0 1 4 6 8 12 16 20]; c=[12 22 32 45 58 75 70 48]; Q=18;
The coefficients in order of decreasing degree are:
p=polyfit(t,c,3) tplot=linspace(0,20); cplot=polyval(p,tplot); plot(t,c,'o',tplot,cplot) legend('data','cubic polynomial','location','best') xlabel('time') ylabel('concentration') conc=@(t)p(1)*t.^3+p(2)*t.^2+p(3)*t+p(4); format shortg M=Q*[quad(conc,0,5);quad(conc,0,10);quad(conc,0,15);quad(conc,0,20)]
p =
-0.023301 0.34267 4.1382 13.888
M =
2372.5
7231.7
13760
19569
Uppgift 3
V=500; Cin=0.5; F=40; k=5e-3;K=0.1; fun=@(C)V/F+(K/k)*log(C/Cin)-(C-Cin)/k; fun10=@(C)V/F+(K/k)*log10(C/Cin)-(C-Cin)/k; Cplot=linspace(0.0001,6); funplot=fun(Cplot); plot(Cplot,funplot) xlabel('Concentration') ylabel('function') grid Cout=fzero(fun,0.5)
Cout =
0.57679
Other alternatives
Cout10=fzero(fun10,0.5) C=fsolve(fun,0.001) fu=@(C)(-K/k*log(C/Cin)+1/k*(C-Cin))*F/V-1; Cu=fzero(fu,0.5)
Cout10 =
0.56804
Equation solved.
fsolve completed because the vector of function values is near zero
as measured by the default value of the function tolerance, and
the problem appears regular as measured by the gradient.
C =
0.0018367
Cu =
0.57679