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

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/ ^. @2 _9 S+ z8 g3 d  x3 T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：

. N* {, A6 D: I4 b: PThe following two blocks cannot be connected in series:
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Block 1: 'A: Simplified Synchronous Machine SI Units1'
; S4 J0 j' V) l. a; x0 ]Block 2: 'transfo_1_winding_2: Split winding type transformer1'

) @0 U, ~, s) O0 h8 b6 q$ jThe first block, modeled as a current source, cannot be connected in series with the inductive element of the second block.
/ \5 R1 T" C) m4 Q1 y( KAdd a high-value resistance in parallel with one of the two block.
. m9 b8 }; a: j0 T" ZYou can also specify high-value resistive snubbers if the blocks have a snubber device.
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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

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

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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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9 v$ n+ j9 {) N; |   查看如下帮助文档：Assumptions
The 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
. e5 @" a% X& FWhen 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.

" w1 }- ^- W  g2 s, a$ g: j8 R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；
8 i$ q. j# ?: m5 k: g* _; l8 B2.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）
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试错结论：
0 \4 Q) T2 \2 ^: ~4 }% x8 z1.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）
2.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。
( V" |' c0 t% r: i: ?( l, A3 g3.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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2 ^8 i, l0 j9 g1 W3 V ShortCurrentFor300MW2.rar (9.71 KB, 下载次数: 1)

2014-12-31 11:44 上传

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修改后

电压波形

疑问：
1.如何解释模型搭建、报错和试错结论？
9 ~7 P+ Y2 k. A* @$ g/ b2.Simulink仿真的逻辑？

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