0
\$\begingroup\$

I am trying to parse the ECO dataset, which includes 1-second electric data collected from residential houses. Link to dataset: http://vs.inf.ethz.ch/res/show.html?what=eco-data

But the relationship between the power angle and the real and reactive power do not always make sense to me. Here is an example of 3 rows, each representing a second:

                               57168     57169     57170
--------------------------  --------  --------  --------
powerallphases               19.506    19.724    42.936
powerl1                      19.506    19.724    42.936
powerl2                       0         0         0
powerl3                       0         0         0
currentneutral                0.1127    0.1125    0.3519
currentl1                     0.2039    0.2046    0.4279
currentl2                     0.1061    0.1072    0.1084
currentl3                     0.0645    0.0645    0.0645
voltagel1                   239.6     239.48    239.48
voltagel2                   239       239       239
voltagel3                   240       240       240
phaseanglevoltagel2l1       240       240       240
phaseanglevoltagel3l1       120       120       120
phaseanglecurrentvoltagel1  300       300       326
phaseanglecurrentvoltagel2  270       270       270
phaseanglecurrentvoltagel3  270       270       270

I am interested only in phase 1, so let's look only at rows that are related to it:

                               57168     57169     57170
--------------------------  --------  --------  --------
powerl1                      19.506    19.724    42.936
currentl1                     0.2039    0.2046    0.4279
voltagel1                   239.6     239.48    239.48
phaseanglecurrentvoltagel1  300       300       326

Here is the meaning of the fields:

* powerl1: Real power phase 1
* currentl1: Current phase 1
* voltagel1: Voltage phase 1
* phaseanglecurrentvoltagel1: Phase shift between current/voltage on phase 1

Apparent power is current * voltage.
However, I would expect that apparent power * cos(power angle) = real power But if I add it to the table I get:

                               57168     57169     57170
--------------------------  --------  --------  --------
powerl1                      19.506    19.724    42.936
currentl1                     0.2039    0.2046    0.4279
voltagel1                   239.6     239.48    239.48
phaseanglecurrentvoltagel1  300       300       326

apparentPower               48.8544   48.9976  102.473
cos(θ) * apparentPower      24.42     24.4988  84.953     

In the first two rows, apparent power * cos(power angle) is close enough to the real power to be explained by a measurement error. But in the last row, it is completely out of scale. And it's not a single measurement, all the following rows keep being in error:

                               57170     57171     57172     57173
--------------------------  --------  --------  --------  --------
powerl1                      42.936    30.469    36.452    42.268
currentl1                     0.4279    0.2309    0.2607    0.2856
voltagel1                   239.48    239.48    239.48    239.48
phaseanglecurrentvoltagel1  326       318       323       328
apparentPower               102.473    55.2959   62.4324   68.3955
cos(θ) * apparentPower      84.953     41.092   49.860     58.002

Am I missing anything in my calculation? Thanks in advance!

\$\endgroup\$
2
  • \$\begingroup\$ Questions on the use of electronic devices are off-topic as this site is intended specifically for questions on electronics design. This is a site rule but your device is also not supported by any data sheet so how can anyone trust it. \$\endgroup\$
    – Andy aka
    Mar 30, 2020 at 7:36
  • \$\begingroup\$ Hi, my question is mostly about the mathematical relation between power angle, apparent power and real power, I believe the kind of device is irrelevant \$\endgroup\$ Mar 30, 2020 at 8:14

2 Answers 2

2
\$\begingroup\$

In the case with distorted currents and voltages: $$V_{rms}^2I_{rms}^2 = S^2 = D^2 +\sum_h P_h^2+\sum_h Q_h^2$$.

Most likely the phase angle reported between \$V_h\$ and \$I_h\$ is that of the fundamental (\$h=1\$) component. There is also the active- and reactive- power for each harmonic \$h\$ for all \$h>1\$, as well as a pure distortion component \$D\$ that you have to cater for with distorted waves.

In this application it will be best to approximate and use:

$$V_{rms}^2I_{rms}^2 = S^2 = D_{thd}^2 +P_1^2+Q_1^2$$ where only the fundamental powers are used and all the harmonic and other distortion are lumped into \$D_{thd}\$.

For a simpler view of the analysis have a look at this slideshow explaining AC power and energy from first principles.
(Use Acrobat instead of Google viewer to display the animations.)

In concrete terms reactive power does not exist. Only active- and apparent- powers exist. Maybe the best way to explain reactive power, \$Q\$ is:

$$Q= \sqrt{V_{rms}^2I_{rms}^2-P^2}$$

This means that \$Q\$ is a imaginary value to express the difference between the apparent power and the active power in an alternating current circuit if the values are not distorted.

If the currents or voltages are distorted you will have an active- and reactive- power for each harmonic component. Further to that, there will also be distortion that only show up in the current or voltage RMS as described above.

\$\endgroup\$
1
\$\begingroup\$

Hi, my question is mostly about the mathematical relation between power angle, apparent power and real power, I believe the kind of device is irrelevant

Real devices and loads can take highly non-sinusoidal currents and hence phase angle measurements are irrelevant. The power triangle of apparent, real and reactive powers as shown below is irrelevant when the load is non-linear.

enter image description here

The power triangle above assumes that loads are linear. A typical appliance current might look like this: -

enter image description here

How would you measure or determine a meaningful phase angle from the current waveform above?

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.