基于Simulink的发电厂接入大电网的仿真模型，求助各位帮忙

shanby

ShortCurrentFor300MW.rar (9.3 KB, 下载次数: 5)

2014-12-21 21:17 上传

点击文件名下载附件

, V' l3 z6 ~+ O接线情况：模拟2台150MW，35kV的发电机分别接入2台分裂升压变压器，再由220kV接入大电网。在35kV母线接入3相短路模块，求解短路电流。模型见附件。参数设置可能不合理。仿真时Simulink提示错误信息为：
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The following two blocks cannot be connected in series:

: a% v7 ?* i/ F: Y, ^Block 1: 'A: Simplified Synchronous Machine SI Units1'
" T: m, X( x/ p0 q$ {Block 2: 'transfo_1_winding_2: Split winding type transformer1'
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( Q: R4 o4 K5 Y+ o8 `The first block, modeled as a current source, cannot be connected in series with the inductive element of the second block.
# [! \5 l0 }7 O1 l+ C0 p% d, lAdd a high-value resistance in parallel with one of the two block.
You can also specify high-value resistive snubbers if the blocks have a snubber device.

5 l7 |5 B: n+ n$ J* u$ V意思是名字叫Simplified Synchronous Machine SI Units1的发电机是一个电流源，不能与感性元件（变压器，名字为transfo_1_winding_2: Split winding type transformer1）连接。需要与发电机并联一个高值电阻。请问大家我错在哪里？参数改如何设置。有好心人麻烦就自己理解修改这个模型的参数上传，万分感谢。
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300MW short current diagram

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tommy_jiang

说得已经很清楚了，让你在发电机机端并联一个电阻。

wangwei9911

我并了一个大电阻，没有数据啊。你们呢？

cgtaoy1234

并联一个三相并联RLC负荷应该就可以了

shanby

接线情况：模拟2台150MW，35kV的发电机分别接入2台分裂升压变压器，再由220kV接入大电网。在35kV母线接 ...
shanby 发表于 2014-12-21 21:06

                               
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   查看如下帮助文档：Assumptions
( `1 e! y9 J7 L, tThe electrical system of the Simplified Synchronous Machineblock consists solely of a voltage source behind a synchronous reactanceand resistance. All the other self- and magnetizing inductances ofthe armature, field, and damping windings are neglected. The effectof damper windings is approximated by the damping factor Kd. The threevoltage sources and RL impedance branches are Y-connected (three wiresor four wires). The load might or might not be balanced.Limitations
3 g' D9 Y  f: T/ i. @4 XWhen you use Simplified Synchronous Machine blocks in discretesystems, you might have to use a small parasitic resistive load, connectedat the machine terminals, in order to avoid numerical oscillations.Large sample times require larger loads. The minimum resistive loadis proportional to the sample time. As a rule of thumb, remember thatwith a 25 μs time step on a 60 Hz system, the minimum load isapproximately 2.5% of the machine nominal power. For example, a 200MVA simplified synchronous machine in a power system discretized witha 50 μs sample time requires approximately 5% of resistive loador 10 MW. If the sample time is reduced to 20 μs, a resistive loadof 4 MW should be sufficient.
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得出：
1.简单同步发电机模型采用带电阻和电抗、独立电压源，三相Y连接的电气模型；
2.使用边界限制：如果是采用离散系统仿真，则需要在发电机端子连接寄生小电阻，其阻值或功率与离散系统的采样时间成比例（举例有计算实例）

试错结论：
2 b' D( x  ^. e3 ]) |1.错误提示说简单同步发电机是电流源，需要并联高值电阻。这与帮助文档电压源的说法不一致（尽管电压源和电流源和也互相变换）
8 O" m( y3 L6 W& J0 O) [! h2.将powergui模块参数设置为continous，仿真报错与设置为discrete的报错信息一样。
3.将powergui模块参数设置为discrete， 按错误提示边栏电阻，则仿真不报错（虽然结果不是预期）。
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1 \9 o; z0 R1 G1 Q0 d$ y ShortCurrentFor300MW2.rar (9.71 KB, 下载次数: 1)

2014-12-31 11:44 上传

点击文件名下载附件
修改后

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电压波形

: h4 t; `/ A/ c! w疑问：
) J, F) J- w3 x7 n4 \1.如何解释模型搭建、报错和试错结论？
3 K8 T7 E: d$ N2.Simulink仿真的逻辑？
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redplum

为什么接两个transformer呢

redplum

而且还要从一个驱动器出来吗

dc慕晗

把变压器的电感设置为零，就可以解决了