How do fix this code to give me the error?

2022 10 17
2 个回答
更新时间：2022 10 17
2 次查看（30 天）
Ran in:

0 个投票

Ran in:
%% dt=10
h=10;
t=[100:h:1500];
Cp=(.00000000747989.*t.^3)-(.000028903.*t.^2)+.0423484.*t+36.1064;
dH=trapz(t,Cp)
dH = 7.4939e+04
for i=1:5
h=h*10;
tt=100:h:1500;
CcPp=(.00000000747989.*tt.^3)-(.000028903.*tt.^2)+.0423484.*tt+36.1064;
aa=trapz(tt,CcPp);
dHH(i)=aa;
error(i)=abs((dH-dHH(i))/dH);
hh(i)=h;
end
Error using matlab.internal.math.getdimarg
Dimension argument must be a positive integer scalar within indexing range.

Error in trapz>getDimArg (line 90)
dim = matlab.internal.math.getdimarg(dim);

Error in trapz (line 44)
dim = min(ndims(y)+1, getDimArg(dim));
[10 dH 0;hh' dHH' error']
plot(log10(hh),log10(error),'o')
xlabel('Log_{10} (\Delta h)')
ylabel('Log_{10} \epsilon')
a=polyfit(log10(hh),log10(error),1);
f=@(x) a(1)*x+a(2);
hold on
fplot(f,[-5 -1])
hold off
a
%% dt=20
t=[100:20:1500];
Cp=(.00000000747989.*t.^3)-(.000028903.*t.^2)+.0423484.*t+36.1064;
enthalpy_change=trapz(t,Cp)
%% dt=50
t=[100:50:1500];
Cp=(.00000000747989.*t.^3)-(.000028903.*t.^2)+.0423484.*t+36.1064;
enthalpy_change=trapz(t,Cp)
%% dt=100
t=[100:100:1500];
Cp=(.00000000747989.*t.^3)-(.000028903.*t.^2)+.0423484.*t+36.1064;
enthalpy_change=trapz(t,Cp)
%% dt=200
t=[100:200:1500];
Cp=(.00000000747989.*t.^3)-(.000028903.*t.^2)+.0423484.*t+36.1064;
enthalpy_change=trapz(t,Cp)

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回答（2 个）

Torsten
Torsten 2022-10-17
编辑：Torsten 2022-10-17
Ran in:

0 个投票

Ran in:
Cp = @(T) .00000000747989.*T.^3-.000028903.*T.^2+.0423484.*T+36.1064;
H = @(T) 1/4*.00000000747989.*T.^4-1/3*.000028903.*T.^3+1/2*.0423484.*T.^2+36.1064*T;
deltaT = [10 20 50 100 200];
Tstart = 100;
Tend = 1500;
deltaH_true = H(Tend)-H(Tstart);
for i = 1:numel(deltaT)
dT = deltaT(i);
T = Tstart:dT:Tend;
deltaH(i) = trapz(T,Cp(T));
error(i) = abs((deltaH(i)-deltaH_true)/deltaH_true);
end
format long
error
error = 1×5
0.000003409811578 0.000013639246313 0.000085245289456 0.000340981157826 0.001363924631303
p = polyfit(log10(deltaT),log10(error),1)
p = 1×2
1.999999999992101 -7.467269618875773
Walter Roberson
Walter Roberson 2022-10-17

0 个投票

h=h*10;
tt=100:h:1500;
CcPp=(.00000000747989.*tt.^3)-(.000028903.*tt.^2)+.0423484.*tt+36.1064;
aa=trapz(tt,CcPp);
What is happening is that your increment gets large enough that eventually tt becomes a scalar. But when you pass a scalar to the second parameter of trapz then it is interpreted as a dimension number rather than as a coordinate. (Note that trapz() of a scalar is 0)

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