如何根据时间采样数据计算三相千瓦时 [英] How calculate three phase kilowatt hour from time sampled data

问题描述

我的问题是我想根据电流和电压的时间采样数据计算三相功率.

My problem is I want to calculate three phase power from time sampled data of current and voltages.

我的问题:

1. 如何根据时间采样数据计算能量(单位千瓦时)?是否有可用的方程式?

1. How can I calculate the energy (unit kilowatt hour) from time sampled data? Are any equations available?

是否需要考虑相移?(如何计算相移?如何将其与计算三相功率联系起来?)

Is it needed to take the phase shift in account? (How can I calculate the phase shift? How do I link this to calculating the three phase power?)

是否有更好的平台可以解决我的问题?

Is some better platform is available for solving my question?

我得到的是瞬时样本值(不是连续的).(我有一些传感器可以提供电流和电压 - 我将其转换为数字进行处理).每秒获取大约 50 个样本.(由于 120 的相移，当我们将三相的所有功率全部加满时，它是否为零?)如何从这些采样值计算三相总能量?我正在 Arduino 中处理我的数据.

I get the instantaneous sample value (not continuous). (I have some sensors that gives the current and voltage - I convert this to digital for processing). Around 50 samples are got per second. (Is it to be zero when we some up all the power of three phase - due to phase shift of 120?) How can I calculate total three phase energy from these sampled values? I am processing my data in Arduino.

(我不知道这是问我问题的地方(如果我能从其他地方获得更好的帮助，请建议我).)

(I don't know this is the place to ask my question (if I can get a better help from some where else please suggest me).)

推荐答案

数值演算来救援.

如果你有多个电压和电流样本，那么你也有很多瞬时功率样本:P(t) = U(t) * I(t).

If you have several samples of voltage and current, then you also have that many samples of momentary power: P(t) = U(t) * I(t).

现在你有权力也有时间，你可以整合 关于时间的力量.一个简单的数值方法是梯形规则.这个问题被标记为Arduino"，我对 C 相当了解，所以这里有一些说明该技术的伪 C:

Now you have power and you have time, you can integrate the power with respect to time. A simple numeric approach is the trapezoidal rule. This question is tagged "Arduino" and I know C reasonably well so here's some pseudo-C that illustrates the technique:

int n_samples = 1000; // or however many samples you have
double integral = 0.0;
for (int i = 0; i < n_samples - 1; i++) {
    integral += (samples[i] + samples[i + 1]) / 2;
}

integral *= (t_max - t_min) / n;

其中t_min和t_max分别是采样的开始和结束时间，n_samples是你得到的样本数，samples 是一个包含计算出的瞬时功率值的数组(大概是 double 左右).integral 将保存结果.

Where t_min and t_max are the beginning and ending time of the sampling, respectively, n_samples is the number of samples you got, samples is an array (presumably of double or so) that contains the calculated momentary power values. integral will hold the result.

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