Increase a variable by 0.1 each iteration
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I have created code to find Scrit and Dmax for a series of Dd values, but each iteration seems to increase by a random amount - I need it to increase by 0.1 each time? I believe I need to edit the line i=0.01:0.1:1?
I would also like to be able to save each iteration table in a new excel row, but at the moment the code just overwrites into the first row of my excel spreadsheet?
T=table(Dd, Scrit, Dmax)
filename='Dry diameter.xls';
writetable(T,filename)
- do I need to edit this to new column each run, if so how?
Here is the code:
% Using dry diameter in um
K=0.5;
sigma=0.072;
T=298.15;
Mw=18.01528;
Pw=0.997;
R=8.3143;
x=(4*sigma*Mw)/((R*T*Pw));
Dd=0.1
for i=0.01:0.1:1
Dd=Dd+i
D=linspace(Dd,2,10);
S=((D.^3-Dd^3)./(D.^3-Dd^3+Dd^3*K)).*exp(x./D)
Scrit=max(S)
Dmax=D(find(S==Scrit))
T=table(Dd, Scrit, Dmax)
filename='Dry diameter.xls';
writetable(T,filename)
end
采纳的回答
Star Strider
2019-3-4
You appear to be increasing ‘Dd’ in each iteration. It is increasing by adding whatever value of ‘i’ exists in each iteration to the previous value, so it only appears to be random. If you want ‘Dd’ to increase by 0.1 each time, try this:
Dd=Dd+0.1;
With respect to writing the ‘T’ table, a more efficient approach to your entire code would likely be something like this:
% Using dry diameter in um
K=0.5;
sigma=0.072;
T=298.15;
Mw=18.01528;
Pw=0.997;
R=8.3143;
x=(4*sigma*Mw)/((R*T*Pw));
Dd=0.1
i=0.01:0.1:1;
parmtrs = zeros(numel(i), 3); % Preallocate
for k1 = 1:numel(i)
Dd=Dd+0.1
D=linspace(Dd,2,10);
S=((D.^3-Dd^3)./(D.^3-Dd^3+Dd^3*K)).*exp(x./D)
Scrit=max(S)
Dmax=D(find(S==Scrit))
parmtrs(k1,:) = [Dd, Scrit, Dmax];
end
T=table(parmtrs(:,1), parmtrs(:,2),parmtrs(:,3), 'VariableNames',{'Dd', 'Scrit', 'Dmax'})
filename='Dry diameter.xls';
writetable(T,filename)
with ‘T’ now being:
T =
10×3 table
Dd Scrit Dmax
___ _______ ____
0.2 1.0005 2
0.3 0.99936 2
0.4 0.99703 2
0.5 0.99317 2
0.6 0.98735 2
0.7 0.97912 2
0.8 0.96796 2
0.9 0.95326 2
1 0.93431 2
1.1 0.91022 2
Experiment to get the result you want.
4 个评论
Alice Cozens
2019-3-4
Star Strider
2019-3-4
As always, my pleasure.
‘Could you just clarify what the purpose of the varible i is now?’
I kept ‘i’ because yoiu originally wanted to use it as a loop counter. It is now simply a vector, and it couold also be used as the independent variable if you wanted to plot ‘parmtrs’ as a function of it. It has no other purpose at present.
‘So if i wanted Dd to run from 0.01 to 5 in 0.01 increments?’i = 0.0
There are several options, the easiest being to specify ‘i’ as:
i = 0.01 : 0.01 : 5;
and changing ‘Dd’ to either:
Dd=Dd+0.01;
or:
Dd = i(k1);
‘Basically I don't understand what the three numbers for i are’
The definition for the ‘i’ vector are: start_value : step_value : finish_value. See the documentation on colon,: (link) for details.
Alice Cozens
2019-3-5
Star Strider
2019-3-5
As always, my pleasure!
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