Hello!
I need some help plotting the following:
kjb.PNG
This is what I have:
syms x
kw = 91371;
CaO= 1.81*10^(-4);
FaO= 0.1233;
K = 1.29;
fun =FaO*((1+K*CaO*(1-x))/(kw*CaO*(1-x)))
W=int(fun);
fplot(W,[0 1
But nothing comes out on the graph. I want to plot W on the Y-axis and x on the X-axis, between 0 and 1 (because it's a conversion in chemistry).
Does anyone know what I'm doing wrong?

5 件のコメント

KALYAN ACHARJYA
KALYAN ACHARJYA 2019 年 6 月 15 日
編集済み: KALYAN ACHARJYA 2019 年 6 月 15 日
Marina Gonzalez
Marina Gonzalez 2019 年 6 月 15 日
Same question because it doesn't work.
I obviously forgot the final parenthesis when I copied it.
Why doesn't it show on the graph? This is what I get when I plot it:
5646541.PNG
KALYAN ACHARJYA
KALYAN ACHARJYA 2019 年 6 月 15 日
Actually, at present I dont have Matlab,therfore unable test it pls show int(fun) expression?
Marina Gonzalez
Marina Gonzalez 2019 年 6 月 15 日
Sure!
W =
(10621383255117913671*x)/6101500778987273584640000 - (2711395674095616*log(x - 1))/363677786528305625
Walter Roberson
Walter Roberson 2019 年 6 月 15 日
In this calculation W is imaginary below 2.0007245246191665436401515314702

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 採用された回答

Star Strider
Star Strider 2019 年 6 月 15 日

2 投票

But nothing comes out on the graph.
The result of the int function is a constant. At best, you will get a straight line.
I am not certain what you want. You can get it to plot as a function of the upper limit of integration by creating a function from it:
syms x a
kw = 91371;
CaO= 1.81*10^(-4);
FaO= 0.1233;
K = 1.29;
fun = FaO*((1+K*CaO*(1-x))/(kw*CaO*(1-x)));
W(a) = int(fun, x, 0, a);
fplot(W,[0 1])

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