# Calculating features of a 3 phase power timeseries

I have data from a number of high frequency data capture devices connected to generators on an electricity grid. These meters collect data in ~1 second "bursts" at ~1.25ms frequency, ie. fast enough to actually see the waveform.

The meters are collecting voltage and current from each of the 3 phases. An example of the data (plot and tabular) is shown below, with one phase shown in each colour.

I want to roll this waveform data up to some summary statistics at a lower frequency (20ms). Specifically, I am looking to calculate:

• Active power, reactive power and power factor
• The grid frequency as it changes over time

Apologies but I'm a mechanical engineer and this is not my strong suit! All of the references I can find refer to idealised situations, where the phase angles etc are pre defined. I could fit idealised sin curves to each of the timeseries, but I feel there is a better solution. Are there any simple techniques to calculate the above directly from the timeseries?

Here's a "toy" data set of the first few waves of one voltage phase as a pandas Series for those who are interested:

import pandas as pd, datetime as dt
import pandas as pd, datetime as dt
ds_waveform = pd.Series(
index = pd.date_range('2020-08-23 12:35:37.017625', '2020-08-23 12:35:37.142212890', periods=100),
data = [  -9982., -110097., -113600.,  -91812.,  -48691.,  -17532.,
24452.,   75533.,  103644.,  110967.,  114652.,   92864.,
49697.,   18402.,  -23309.,  -74481., -103047., -110461.,
-113964.,  -92130.,  -49373.,  -18351.,   24042.,   75033.,
103644.,  111286.,  115061.,   81628.,   61614.,   19039.,
-34408.,  -62428., -103002., -110734., -114237.,  -92858.,
-49919.,  -19124.,   23542.,   74987.,  103644.,  111877.,
115379.,   82720.,   62251.,   19949.,  -33953.,  -62382.,
-102820., -111053., -114555.,  -81941.,  -62564.,  -19579.,
34459.,   62706.,  103325.,  111877.,  115698.,   83084.,
62888.,   20949.,  -33362.,  -61791., -102547., -111053.,
-114919.,  -82805.,  -62882.,  -20261.,   33777.,   62479.,
103189.,  112195.,  116380.,   83630.,   63843.,   21586.,
-32543.,  -61427., -102410., -111553., -115374.,  -83442.,
-63565.,  -21217.,   33276.,   62024.,  103007.,  112468.,
116471.,   84631.,   64707.,   22405.,  -31952.,  -61108.,
-101955., -111780., -115647.,  -84261.])

• if you add up the 3 phases you get a single sine wave. – Tiger Guy Sep 1 '20 at 19:51
• no, try it. We derived 3-phase power in the Navy like this. The peak is less than the opposing 2 phases that are opposite it. – Tiger Guy Sep 1 '20 at 20:00
• $\vec{V_A} + \vec{V_B} + \vec{V_C} = 0$ That's what defines 3-phase @TigerGuy – StainlessSteelRat Sep 1 '20 at 20:05
• @StainlessSteelRat well my memory is proved shot. Maybe equals zero is how we generated the waves? But clearly it is zero which is why it works. – Tiger Guy Sep 2 '20 at 18:41

## regarding Total, active, reactive Power and $$\cos \phi$$

1. Estimate the energy in a timeperiod T for every timestep ($$t \in [t_0,t_0+T]$$) by using the formula

$$P = \frac{1}{T}\sum_{t=i} I_{A,i} V_{A,i}dt + \frac{1}{T}\sum_{t=i} I_{B,i} V_{B,i}dt + \frac{1}{T}\sum_{t=i} I_{C,i} V_{C,i}dt$$

where:

• $$dt$$: is the timestep (e.g. 1.25ms)
• $$T$$ is the integration period
• $$\frac{1}{T}\sum_{t=i} I_{k,i} V_{k,i}dt$$: the active power of each line k $$\in \{A,B,C\}$$
1. Find the RMS values for Voltage and Current ($$V_{k, RMS}, I_{k, RMS}$$) for each line (A,B,C) in the time period.

The total power for each line is $$P_{k,T} = V_{RMS} * I_{RMS}$$

make sure you don't confuse line with phase (see star and delta configurations)

1. The reactive power will be the difference between the two (Total-Active).

2. if you want the $$cos \phi$$ for each line then just do

$$\phi_k = \arccos \left(\frac{P_{k,Active}}{P_{k,Total}}\right)$$

Caveat: This is a numerical estimation. Depending on the duration of integration, you might get some strange results (e.g. Total Power less than Active). That is why you should prefer complete periods if possible only one at a time (longer time periods will tend to average too much the data).

## Regarding the grid frequency:

what you should do is perform an fft and find the peak frequencies. For a signal like that, you would need to add a window (usually Hanning, or Hamming) and also get a longer time period (e.g. 10 or more standard grid frequency periods).

## Python Code

I am also adding some python code, just for verification of the above:

#%%
import numpy as np
import matplotlib.pyplot as plt
# %%

Tmax = 1/50 #integratin period
f_g = 50  # grid frequency
dt = 0.00125

ts = np.arange(0, Tmax, step=dt)

# %%  Plot 3 line voltages
Vmax = 10
V_As = Vmax*np.cos(2*np.pi*f_g*ts)
V_Bs = Vmax*np.cos(2*np.pi*(f_g*ts + 1/3))
V_Cs = Vmax*np.cos(2*np.pi*(f_g*ts + 2/3))

plt.plot(ts, V_As, label='A')
plt.plot(ts, V_Bs, label='B')
plt.plot(ts, V_Cs, label='C')
plt.legend()

# %%  Estimation of Power in line A, for a given impedance R_A (complex)
R_A = 2+1/2*np.pi*1j
angle = np.angle(R_A)  # actual cos phi
print(angle)
I_As  = np.real(Vmax/np.abs(R_A) * np.exp(1j* (2*np.pi*f_g*ts + np.angle(R_A))))

# plots V and I for verification
plt.plot(ts, V_As, label='V_A')
plt.plot(ts, I_As, label='I_A')
plt.legend()
plt.grid()
# %%
def calc(Vs,Is):
P_activ = np.sum(Vs*Is)*dt/Tmax
Vrms = np.std(Vs)
Irms = np.std(Is)
P_total = Irms*Vrms
phi = np.arccos(P_activ/P_total)
print ('Active :{:.3f}'.format(P_activ))
print ('Vrms  :{:.3f}'.format(Vrms))
print ('Irms  :{:.3f}'.format(Irms))
print ('P_total  :{:.3f}'.format(P_total))
print ('phi  :{:.3f}[rad] = {:.3f}'.format(phi, phi*180/np.pi))

calc(V_As, I_As)


• Thank you NMech for the high quality and comprehensive response :) – DaveB Sep 1 '20 at 21:23
• I made a small correction to the formula to account for dt, and T. Also I added some python code, which you seem comfortable with. – NMech Sep 1 '20 at 22:09
• Hi NMech - thank you very much for the Python code, that is very useful. I have implemented my own version of the frequency calculation just using zero-passes which seemed to work quite well. Working through your power calculations now; I really appreciate your help with this. – DaveB Sep 3 '20 at 6:53
• I'm glad my wasted youth helped someone :-). I'm also a mechanical engineer and I had to do measurements on a small wind turbine project. The amount of headache that I had to go through at least is helping someone else. – NMech Sep 3 '20 at 7:08