Maple Solutions to the Chemical Engineering Problem Set

ToKelvin:=x->x+273.15;
T(K) := map(ToKelvin,T(C));
ToKelvin := x → x + 273.15
TK ( ) := [ 236.45, 253.55, 261.65, 270.55, 280.75, 288.55, 299.25, 315.35, 333.75, 353.25 ]
We must take the reciprocal of these values
> r := x-> 1/x;
Tover := map(r,T(K));
1
r := x →
x
Tover := [ .004229223937 , .003943995267 , .003821899484 , .003696174459 , .003561887801 , .003465603881
,
.003341687552 , .003171079753 , .002996254682 , .002830856334
]
We are now ready to fit the transformed data
> CCfit:=fit[leastsquare[[Tau,rho], CC2, {A,B}]]([Tover,lnpdata]): CCfit;
ρ = 8.752009662 − 2035.330939 Τ
This result can be expressed in terms of our orginal variables
> CC3:=subs(v1,v2,CCfit): CC3;
2035.330939
log10( P )
= 8.752009662 −
T + 273.15
We make a function of the right hand side for plotting purposes
> CCF:=unapply(10^rhs(CC3),T);
⎛
2035.330939 ⎞
⎜8.752009662
−
⎟
⎝
T + 273.15 ⎠
CCF := T → 10
and display this new result along with the polynomial fits and the data.
> n:='n':
p1:=plots[logplot]([seq(rhs(eqn[n]),n=2..5),CCF(T)],T=-40..100,color=[red,yellow,green,black,blue,purple]):
> plots[display]({p1,lplot2,lplot1});
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