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发表于 2019-6-19 13:48
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来自: 中国湖北武汉
关于Maxwell中计算视在电感和增量电感的计算原理,跟您讨论一下:
首先是视在电感,您提到的是两步法:1.冻结磁导率,2.单相加1A电流,磁链除电流。
存疑处:思路肯定是对的,但帮助文档中,静态场时貌似是能量除电流平方,瞬态场中却又是磁链除电流,不知为何?
其次是增量电感,也是两步法:1.冻结磁导率,2.XXX.。
存疑处:既然磁导率已被冻结,那无论第2步怎么处理,得到的都不是增量电感吧?
所以我认为您想表达的意思是:1.加三相电流激励,算负载场,保存磁链数据,不冻结磁导率;2.仍加三相电流激励,但其中一相电流增加△I,算负载场,得到磁链数据,由磁链差/电流差得到增量电感。不知我的理解是否正确?
另外帮助文档中讲瞬态电感计算时提到:After the FEA normal solution is completed at each time step, the permeability associated with apparent inductance, or the differential permeability associated with incremental inductance, of each element, is frozen.
也就是说,不管求什么电感,都是分两步:1.冻结磁导率(求视在电感就冻结绝对磁导率,求增量电感就冻结切线磁导率);2.单相加1A电流(求增量电感也是加1A,非△I),磁链除电流。您提到的求增量电感的方法是否就是这种冻结切线磁导率的方法?
附帮助文档摘抄:
Because both 2D and 3D transient solutions are nonlinear, for inductance computation, Maxwell
provides two options: apparent inductance and incremental inductance. After the FEA normal
solution is completed at each time step, the permeability associated with apparent inductance, or the
differential permeability associated with incremental inductance, of each element, is frozen – which
is equivalent to freezing the coefficient matrix (the left side of the equation to be solved). In order to
obtain winding self-inductance and mutual inductance, the solver sets an excitation current of 1A in
the first winding, while setting all other winding currents to zero (permanent magnet effects are
excluded). This excitation assignment corresponds to one source vector on the right-hand side of the
equation to be solved. As a result, the calculated flux linkage provides the self-inductance for the first
winding with 1A excitation current, and the calculated flux linkages represent mutual inductance for
all other windings with zero current. Next, the solver excites the second winding with a current of 1A,
while setting all other winding currents to zero (permanent magnet effects are excluded). This
excitation assignment corresponds to another source vector on the right-hand side of the equation to
be solved. The solver continues this process until all windings have been assigned 1A current in turn. |
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