vpasolve gives different answers for the same equation

Hi,

I want to use vpasolve to solve a nonlinear equation numerically. when I insert the coefficients manually, the result seems fine but when I use variables with the same value as I put manually the result changes. It doesn't seem logical to me at all. I would appreciate any help in this regard.

Here is the code (I have put both manual and variable case so you can compare the results) :

NA = 6.022*10^23; %1/mol

R = 8.3145*10^-6; %m^3*MPa/(K*mol)

M_ps = 329; %kg/mol

T_cent = 70; %C

P = 14.8; %MPa

T = T_cent + 273.15; %K

cT_ps = 739.9; %K

cP_ps = 387; %Mpa

crho_ps = 1108; %kg/m^3

r_ps = M_ps * cP_ps / (R * cT_ps * crho_ps)

tT_ps = T / cT_ps

tP_ps = P / cP_ps

syms x

vpasolve(x^2+ tT_ps* (log(1-x) + (1-1/r_ps)*x) + tP_ps,x)

vpasolve(x^2+ 0.4638* (log(1-x) + (1-1/18679)*x)+ 0.0382,x)

Answer 1

[Manan Mishra]

The difference in results is because of the values you have used in place of those variables.

The default output display format in MATLAB is "short" which gives 4 digits after the decimal point (the output gets rounded-off to 4 decimal places). To see a more precise value of those variables, you can change the output display format to "long" and then use those values in the "vpasolve" command.

>> format long

After executing this command, you can see the values with 15 digits after decimal places.

>> tT_ps

tT_ps =

     0.463778889039059

>> r_ps

r_ps =

       1.867918561070000e+04

>> tP_ps

tP_ps =

     0.038242894056848

When you give these values to "vpasolve" command, the results are same:

>> vpasolve(x^2+ tT_ps* (log(1-x) + (1-1/r_ps)*x) + tP_ps,x)

ans =

- 0.0047320390312815607624270500579465 - 0.22208533470256431164303895468295i

>> vpasolve(x^2+ 0.463778889039059* (log(1-x) + (1-1/18679.1856107)*x)+ 0.038242894056848,x)

ans =

- 0.0047320390312815607624270500579465 - 0.22208533470256431164303895468295i