如何在 Modelica 中描述 dy/dx 的导数?

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问题描述

我正在研究一个包含 dy/dx 导数的简单模型，但在 Modelica 中，我不能直接写出这个方程，我可以使用 x=time 和 der(y)，但我认为这是一种妥协，因为 Modelica 语言的限制.

I am working on a simple model which includes a derivative of dy/dx, but in Modelica, I can't write this equation directly, I could use the combination of x=timeand der(y), but I think this is a compromise because of limitation of Modelica language.

我的问题是:有没有其他更好的方法来描述 Modelica 中的导数?

My question is:Is there another better method to describe derivative in Modelica?

代码如下:

model HowToExpressDerivative "dy/dx=5, how to describe this equation in Modelica?"
  Real x,y;
equation 
  x = time;
  der(y) = 5;
end HowToExpressDerivative;

我也试过用der(y)/der(x)来表达dy/dx，但是x等于time时出现错误^2.

I also tried to use der(y)/der(x) to express dy/dx, but there is an error when x equals time^2.

model HowToExpressDerivative "dy/dx=5, how to describe this equation in Modelica?"
  Real x,y;
equation 
  x=time^2;
  der(y)/der(x)=5;
end HowToExpressDerivative;

Error: The following error was detected at time: 0

Model error - division by zero: (1.0) / (der(x)) = (1) / (0)

Error: Integrator failed to start model.
... "HowToExpressDerivative.mat" creating (simulation result file)

ERROR: The simulation of HowToExpressDerivative FAILED

推荐答案

给定焓 h 和曲柄角 phi 你可以替换 dh/dphi=... 作者:

Given enthalpy h and crank-angle phi you could replace dh/dphi=... by:

  der(h)/der(phi)=...

但是，即使正确该公式也会在引擎静止时失效(der(phi)=0)，因此并不理想.

However, even if correct that formula will break down when the engine is standing still (der(phi)=0), so it is not ideal.

另一种方法是重写公式.仔细观察公式似乎是:

An alternative would be to rewrite the formulas. Looking more closely the formula seems to be:

  dh/dphi=(\partial a/\partial T)*dT/dphi+...

这表明它们可以重写为:

which suggests that they could be rewritten as:

  der(h)=(\partial a/\partial T)*der(T)+...

这篇关于如何在 Modelica 中描述 dy/dx 的导数?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！

DER