关于FLUX计算的电磁转矩

giuseppe

很疑惑FLUX计算的电磁转矩是否包含损耗, 大家来讨论一下

理论计算的电磁转矩, 一般可由这个公式得到, Te=Pe/wm=3*E*I*cos(gama)/wm
这样得到的转矩其实是包含了铁损和永磁损的(SPM电机), 但不包含铜损, 因为它假设所有的电磁功率(总输入功率-铜损)都转化为转矩了, 实际上轴上的转矩要比这个值小

FLUX的计算方法为: Te=dWm/dtheta; dWm=integrate(H)dB, 这样计算得到的电磁转矩是不是就是不包含任何损耗的纯的电磁转矩了?

这涉及到计算电机效率时, 第一种方法为 (Te*wm-铁损-永磁损)/(Te*wm+铜损)
而第二种方法为 Te*wm/(Te*wm+铜损+铁损+永磁损)
查看FLUX11手册, SPM教程, 可以看到它是按第二种方法来计算效率的, 说明FLUX计算的电磁转矩是不包含任何损耗的

cym

請問版主是否可加我的MSN, 有些問題想特定請教
cym1013@msn.com

lijo

Your reasoning is correct. The iron losses can be included in a resistor in parallel with the FEM coils also. In my opinion the Flux examples are not rigorous, they are examples for beginners, with approximations, not valid in all cases.

giuseppe

回复 2# cym

msn我很少登陆, 方便的话请加我的qq或者gmail讨论

giuseppe

回复 3# lijo


    I was wondering in which way the FLUX could calculate the pure eletromagnetic torque. I ever knew there are several FE analytic method to calculate electromagnetic torque like maxwell stress or something alike. Do you have any hints from the FLUX manual?

Anyway, in another hand, it doesn't deteriorate efficiency calculation much even if the torque calculated by FLUX is not taken as a pure eletromagnetic torque. But I'm just curious.

lijo

Cem = K *( B_stator  ∧ B_rotor)

from the F2D76_VOLUME3_EN.pdf user guide:
-------------------------------------------------------------------------------------------------------------------
'Cm is the electromagnetic torque obtained by the magnetic calculation (virtual works
method)'

Fm the calculated electromagnetic force:
- on the whole of the moving part if the region selected in PROPHY is magnetic or comprises
a magnet (virtual works method),
- on the region selected in PROPHY if it is not magnetic (Laplace's method J ∧ B )

------------------------------------------------------------------------------------------------------------------- Note:
The electromagnetic force (or the torque) is calculated when the modeled part of the device
is solved. This quantity is stored at each time step in the *.RES_n results file.

-------------------------------------------------------------------------------------------------------------------


In the new Flux:




AT THE MOMEMT NOT AT ALL CLEAR "...the computation of the torque using the power sums is more precise."

woodylacurs

我觉得新版本既然强调了是电磁转矩而不是旧版本中后处理用的轴上转矩，就是想凸出概念。而根据电机学中的定义电磁转矩本身就是不包含铜损，但包含铁损和机械损耗。而在计算的时候是否赋予电导率和损耗去参数并不影响计算结果（ansoft貌似有个可以考虑损耗影响的，但是flux似乎没有），所以在不考虑机械损耗的时候电磁功率是要包含输出功率（轴功率）和铁损以及机械损耗的。那效率的计算公示就是（Tem*w-pfe)/(Tem*w+pcu).
不知是否有误，请您赐教。

giuseppe

回复 7# woodylacurs


   Te*w包含铁损, 那包不包括永磁损耗那?

woodylacurs

回复 8# giuseppe

我认为是要包含的，并且永磁损耗本来就是广义的铁损的一部分。在单独计算有限元结果的时候，是按照虚功原理计算的转矩，在一次侧功率没有增加的同时，他包含了所有需要扣除的损耗（定子铜耗除外）；在考虑损耗之后，这部分功率是要扣除的。当然，如果有软件能够考各种虑损耗和转矩之间的耦合关系，我认为那计算出来的就是实际损耗，但是flux似乎没有这个功能。

但是异步电机的电磁关系就不同了，很多书上都是将异步电机铁耗归算到一次侧的，这个是根据等效电路图来的，我没有设计过异步电机，不知道根据实测结果是否会包含在有限元计算的电磁功率中。还望各位指教。

海的背影

应该是包括铁耗的吧，至少flux10.3电磁转矩是考虑了涡流损耗的（软磁材料的铁耗）。铜耗就不确定了，不过应该没有，因为如果建立的是nonmesh的线圈，电磁转矩是不计算在内的。呵呵

e=mc^2

从能量传递的角度来说（电动机）（不计杂散损耗_稳态场）

总的有功：定子绕组吸收有功功率P1

定子上消耗的功率：铜耗Pcu+铁耗Pfe     （由电磁能量转换产生的功率：Pe=P1-Pcu）

由气隙传递到转子侧的有功功率P1-Pcu-Pfe

在转子上消耗的有功：Pal（P永磁）

最终转化为机械能的功率：P2+P机械（风磨等）=P1-Pcu-Pfe-Pal(P永磁)


在所有有限元软件中，由于磁滞而产生的铁耗（一般都不计算硅钢片上的涡流）是不在有限元处理过程中的，此时，电磁功率Pe仍然是P1-Pcu，而由气隙传递到转子端的功率则由变成了(P1-Pcu)即电磁功率，此时电磁转矩Te是没有考虑到铁耗的影响的（若考虑铁耗，则Te值会变小），故而为得到实际的电磁转矩，我们需要再减去Pfe的影响：Te‘=Te-Pfe/omgs。

但是对于永磁电机来说，转子上的永磁损耗是不对电磁转矩有影响的，它影响了输出转矩T2的值 T2=（P1-PCU-Pfe-P永磁-P机械）/omgs

lijo

for induction motors


ELECTROMAGNETIC_TORQUE = SYNCHRONOUS_POWER / SYNCHRONOUS_SPEED

SYNCHRONOUS_SPEED = ( 1 - S ) / MOTOR_SPEED

e=mc^2

lijo 发表于 2012-11-22 23:57

                               
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for induction motors

1、In Steady state AC Magnetic, the stray load losses is not included in computational process

2、stray load losses occured both in stator and rotor

3、I want to know the definition of synchronous power，and the difference with electromagnetic power


thanks

lijo

We are modelling.
Stray losses how you say can not be modeled in stedy state AC magnetic but I can include it in the equivalent circuit:



the same for Pfe+Pfw

If you into the above FEM model put Radd=1e-16 Rfe,fw=1e6 you get the same Electromagnetic Torque but different Pabs and Iabs

e=mc^2

lijo 发表于 2012-11-24 03:45

                               
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We are modelling.
Stray losses how you say can not be modeled in stedy state AC magnetic but I can  ...

you are right

By Circuit coupling,you can take every kind of loss into consideration.

But what I mean, the equivalent resistance of Pstray and Pfe, is  calculated by experimental method or other equivalent computation method(In FEA,the computation of iron loss  is not precise enough,and is calculated in postprocessor).But in pre-processor of stedy state AC Magnetic,Pstray is not modeled in maxwell's equations of each finite element.


If I make any mistakes, please point them out.  

by the way,how is your FLUX3D? can you help me with " h t t p://bbs.simol.cn/thread-117525-1-1.html "


thanks

lijo

OK. As you well say, Pstray or supplementary are given by experimental tests and Pfe and Pfrictional&windage by Non-Load Test FEM simulation at AC SS with slip=0.00001 for example, then Pstray acting as a certain kind of balancing term.
The motor manufacturers do not know how to calculate or mesure the stray losses.


Sorry, I have only experience with FluxSkewed, 2D1/2 Not 3D for the time being, 3D is very hard time computation. Today my main job is in optimization with Flux2D and GOT-It, very hard time computation too.

Anyway I will examine your 3D project and try to help you.