Why does this for loop fail?

6 Ansichten (letzte 30 Tage)
cmcelm
cmcelm am 10 Nov. 2019
Kommentiert: cmcelm am 10 Nov. 2019
I'm trying to create a for loop that will assign a solved element of a function to a blank vector for every value of a given vector. However, when I try to run this code it gives "Error: Assignment has more non-singleton rhs dimensions than non-singleton subscripts" and stops on about the 44th value or so. I made the two vectors the same dimensions, so why am I getting dimensional errors?
syms T V
Ttp = 89.4
% function from earlier
PRfxn = matlabFunction(PREos)
% will use later
dPdV = diff(PREos, V)
dPdT = diff(PREos, T)
CpMinusCv = (-T*(dPdT)^2)/dPdV
CpCvfxn = matlabFunction(CpMinusCv)
% trying to get my T and V for a given P == Pc (variable defined earlier) variable Tc is also defined earlier
Trange = [Ttp:10:2.5*Tc]
Vrange = zeros(1, length(Trange))
i = 1;
for T = Trange
Vrange(1, i) = vpasolve(PRfxn(Trange(1, i), V) == Pc, V, [0 inf])
i = i+1;
end

Akzeptierte Antwort

Stephan
Stephan am 10 Nov. 2019
Bearbeitet: Stephan am 10 Nov. 2019
vpasolve may find more than one solution. Try to save your results in a cell array:
...
Vrange = cell(1, length(Trange))
...
Vrange{i} = vpasolve(PRfxn(Trange(1, i), V) == Pc, V, [0 inf])
  1 Kommentar
cmcelm
cmcelm am 10 Nov. 2019
Thanks! I tested vpasolve and found that it was returning two roots. I'll fix the dimensions on my vpasolve because I only need one root.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Mehr zu Symbolic Math Toolbox finden Sie in Help Center und File Exchange

Produkte


Version

R2017b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by