Gymnasieopgave: Varmeledning
> | restart;interface(warnlevel=0):with(plots): |
1. Differentialligningen
> | lign:=C*diff(T(t),t)=P-U*A*(T(t)-T0); |
Ligningen er en 1. ordens lineær differentialligning. Vi bruger MAPLE
> | los:=dsolve({lign,T(0)=T0},T(t)); |
Heraf ses at for
gælder
> | Tu:=P/U/A+T0; |
Måles Tu - T0 = P/UA kan U bestemmes.
Talværdier:
> | V:=0.5^3; |
> | A:=6*0.5^2; |
> | lambda:=4.1*10^(-2); |
> | L:=2.0*10^(-2); |
> | U:=lambda/L; |
> | rho:=1.20; |
> | cp:=1.01*10^3; |
> | m:=rho*V; |
> | C:=cp*m; |
> | T0:=20.0; |
> | P:=60.0; |
> | tau:=C/U/A; |
> | cp := 1010.00; |
> | los:=dsolve({lign,T(0)=T0},T(t)); |
> | los1:=subs(t=t*60,rhs(los));evalf(los1); |
> | s0:=textplot([[0.75,41.5,"t"],[2.9,24," t = 0.82"]],font=[SYMBOL]): |
> | s:=textplot([[1.5,32,"Temperaturkurve for flamingokasse"],[1.5,24,"Tidskonstant"],[3.4,3,"[min] , tid "],[0.1,45.0,"[C] , temperatur"],[3.4,24,"min"]],align={RIGHT}): |
> | p2:=plot(20.0+19.512*60*t/tau,t=0..0.85,0..45,linestyle=2): |
> | p0:=plot(40.0,t=0..4,0..40,linestyle=4): |
> | p1:=plot(los1,t=0..4.5,0..40): |
> | display(s0,s,p1,p2,p0); |
> |