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ShengWang-EE avatar ShengWang-EE commented on September 15, 2024

你好,感谢你对本论文的关注!

  1. 应该都是电力/电力的效率(即单位都是功率,例如MW)。baseMVA只是进行了折算(把单位从MW折算为100MW)(matpower利用这个变量进行标幺化加速计算),本质上不影响物理变量的含义。
  2. 我再推了一遍,的确好像是减号,感谢指正

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touchfishfieldofficer avatar touchfishfieldofficer commented on September 15, 2024

感谢您的解答,但是对于第一个问题我还是有一些小小的疑问,PTG的所消耗的电力是不是应该是乘以效率才等于他所产出的H2与CH4的总能量之和呀?以下是我的思考过程:
PTG将电能转化为H2,会存在PTG使用电能的效率,实际上只有所消耗的电能✖电力的效率的能量得到了利用,有效利用的电能=所有产出H2的能量;
H2转为CH4又存在一个甲烷化的效率,所以单位的CH4=(单位的CH4/甲烷化的效率)倍的H2;
综合以上信息,得出的结论是 电力✖电力的效率=单位时间H2的能量+单位时间CH4的能量/效率
而代码中是 电力/电力的效率=单位时间H2的能量+单位时间CH4的能量/效率

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ShengWang-EE avatar ShengWang-EE commented on September 15, 2024

你好,感谢你的问题!
代码中是
电力/电力的效率=单位时间H2的功率+单位时间CH4的功率/效率。
因为ptg制备的H2和CH4的流量的单位(包括全文中与气体流量相关的单位)都是Mm3/day。所以这是个流速的单位,因此该式子中是功率而不是能量。
祝好!

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touchfishfieldofficer avatar touchfishfieldofficer commented on September 15, 2024

好的,非常感谢您的解答

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touchfishfieldofficer avatar touchfishfieldofficer commented on September 15, 2024

可以再问一下您文章中caseA中的节点组成成分的相对误差和操作成本的相对误差是用什么衡量的吗?
QQ截图20231213152312

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ShengWang-EE avatar ShengWang-EE commented on September 15, 2024

你好,在这里例如methodA下的误差=2*abs(methodA的值-methodD的值)/(methodA的值+methodD的值)。

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touchfishfieldofficer avatar touchfishfieldofficer commented on September 15, 2024

谢谢,非常感谢您的解答。我想问一下气体成分的误差是指相对于上一次迭代过程中所产生的误差吗,操作成本的误差是指罚函数产生的误差吗?
另外,此处的公式是不是存在书写错误呀
image
微信图片_20231223192303

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