如何计算三相千瓦时从时间采样数据 [英] How calculate three phase kilowatt hour from time sampled data

问题描述

我的问题是我想从电压的当前时间采样数据并计算三相电源。

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

我的问题：

1. 我如何从时间采样数据计算出的能量（单位千瓦时）？是否有任何可用的公式？

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

是不是需要采取帐户相移？ （如何计算的相移？如何链接到这个计算三相电源？）

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?)

是一些更好的平台可为解决我的问题？

Is some better platform is available for solving my question?

我得到的样本瞬时值（不连续）。 （我有一些传感器，使该电流和电压 - 我将其转换为数字进行处理）。大约50样本每秒得到。 （难道是零，当我们有些三个阶段的一切权力 - 由于120相移）我怎样才能从这些采样值计算总的三相能量？我在Arduino的处理我的数据。

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.

（我不知道这是要问我的问题（如果我能得到一些别的地方，请建议我）一个更好的帮助的地方。）

(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

数值演算救援。

如果您有电压和电流的几个样品，那么你也有一个瞬时电源的许多样品： 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，你有时间，可以 整合 相对于时间的功率。一个简单的数值方法是<一个href=\"http://en.wikipedia.org/wiki/Numerical_integration#Quadrature_rules_based_on_interpolating_functions\"相对=nofollow>梯形规则。这个问题被标记的Arduino我知道的C相当好所以这里的一些伪-C是说明了该技术：

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 的开始和结束采样时间，分别为 N_SAMPLES次是你得到的样本数，样品是一个数组（$ p $ 双击左右），包含计算的瞬时功率值。 积分将持有的结果。

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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