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Which is the correct approach to determine the total power loss in a wire with a given harmonic spectrum:

Approach A:

$$ P = \sum_{i=0}^n (i_{n, RMS}^2 \times R_{n})$$

Approach B:

$$ P = i_{total, RMS}^{2} \times R_{weighted} = \frac 1 {2\pi} \int_{0}^{{2\pi}} (i(t)_{1} + i(t)_{2} + ... + i(t)_{n})^{2}dt \times R_{weighted} $$

where

$$ R_{n} \equiv resistance \space of \space wire \space for \space harmonics \space order \space n $$ $$ R_{weighted} \equiv weighted \space wire \space resistance \space for \space all \space harmonic \space orders $$ $$ i_{n, RMS} \equiv root \space mean \space square \space for \space harmonics \space order \space n $$ $$ i(t)_{n} \equiv sinusoidal \space function \space for \space current \space for \space harmonics \space order \space n $$

I've noticed that approach A is the accepted way to calculate the total power loss from harmonics, however, it seems very wrong. What it's doing is finding the power loss for each frequency individually and then adding them. The problem I see with this is that the total RMS value of current is not the sum of individual RMS current values, but rather the RMS of the sum of individual currents. The RMS of a sinusoidal function is:

$$ i_{RMS} = \sqrt { \frac 1 {2\pi} \int_{0}^{{2\pi}} i(t)^{2}dt } $$

And therefore the RMS of a distorted current would be:

$$ i_{RMS} = \sqrt { \frac 1 {2\pi} \int_{0}^{{2\pi}} (i(t)_{1} + i(t)_{2} + ... + i(t)_{n})^{2}dt } $$

Is approach A used because it's simpler and conservative? It will always return a larger RMS current than approach B. am I missing something here? Every resource, including research papers, use approach A. I'm baffled by this...

Edit

I built a python script of a simple harmonic spectrum. The fundamental frequency is 1 Hz @ 1 A magnitude (blue) and the 2nd order harmonic is 2 Hz @ 0.5 A magnitude (orange). The summation of these two frequencies is the green wave. The script plots the waves, then does a discretized integration to get the RMS value. See the script and results below.

Simple 1st and 2nd Order Harmonic & Summation


import numpy
import matplotlib.pyplot as plt


def func(t, w, a):
    return a*numpy.sin(w*t)


t1 = 0
t2 = 2*numpy.pi
f = 1/(t2 - t1)

# array of time values (from 0 to 2*pi)
t = numpy.linspace(t1, t2, 10000)

# current values # 1 (amplitude = 1, angular freq = 1)
w = 2*numpy.pi*f
a = 1.0
i1 = func(t, w, a)
plt.plot(t, i1)

# current values # 2 (amplitude = 0.5, angular freq = 2) [2nd order harmonic @ 50% fundamental]
w *= 2
a /= 2
i2 = func(t, w, a)
plt.plot(t, i2)

plt.plot(t, i1 + i2)

# dt (time step)
step = (t2 - t1)/10000

# integral of # 1 current squared
int1 = numpy.sum(numpy.power(i1, 2))*step
# RMS of # 1 current 
rms1 = numpy.sqrt(int1/(t2 - t1))
print(f'Calc. # 1 current RMS: {rms1:.4f} ... Analytical # 1 current RMS: {1/numpy.sqrt(2):.4f}')


# integral of # 2 current squared
int2 = numpy.sum(numpy.power(i2, 2))*step
# RMS of # 2 current 
rms2 = numpy.sqrt(int2/(t2 - t1))
print(f'Calc. # 2 current RMS: {rms2:.4f} ... Analytical # 2 current RMS: {0.5/numpy.sqrt(2):.4f}')


int3 = numpy.sum(numpy.power(i1 + i2, 2))*step
rms3 = numpy.sqrt(int3/(t2 - t1))

rms = numpy.sqrt(numpy.power(rms1, 2) + numpy.power(rms2, 2))
print(f'Calc. total current RMS: {rms3:.4f} ... Analytical total current RMS: {rms:.4f}')

plt.show()

The output:

Calc. # 1 current RMS: 0.7071 ... Analytical # 1 current RMS: 0.7071
Calc. # 2 current RMS: 0.3535 ... Analytical # 2 current RMS: 0.3536
Calc. total current RMS: 0.7905 ... Analytical total current RMS: 0.7905

The calculated RMS currents are equal to the analytical solutions. So it turns out they are both correct. The math works out such that

$$ \sum_{i=0}^{n} (i_{n, RMS}^2) = \frac 1 {T} \int_{0}^{{T}} (i(t)_{1} + i(t)_{2} + ... + i(t)_{n})^{2}dt $$

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2 Answers 2

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Approach A is the same as B, only A is to the point. It only makes sense if you consider that A uses magnitudes (e.g. \$I_k\$ in \$I_k\sin(\omega_k t)\$) and -- since those magnitudes act as samples -- the discrete integration is a sum. So B is just a flamboyant way of computing the instantaneous values, which are pretty meaningless, since you are interested in a final number, not a continuously variable number. Plus, at the end of the integration cycle, you'll get the same values as the A method. But, if you really want to get busy, there's no one stopping you.

Just to be clear: the RMS waveform is calculated with the help of the integral (or a sum, for a sampled waveform). But, the distortion, (power factor) is a number pertaining to the whole waveform, it's not continuous, since it's evaluated for a whole interval (not necessarily a whole period).

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I've noticed that approach A is the accepted way to calculate the total power loss from harmonics, however, it seems very wrong. What it's doing is finding the power loss for each frequency individually and then adding them.

That happens to be the correct way.

The problem I see with this is that the total RMS value of current is not the sum of individual RMS current values

That's only a problem if you incorrectly believe that to be true. It isn't; the total RMS value of a current (or a voltage) is the square root of all the individual harmonic RMS values squared i.e. this: -

$$I_{RMS-TOTAL} = \sqrt{A^2 + B^2 + C^2 + \text{etc}}$$

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  • \$\begingroup\$ I added more info to back up my point. I still don't see how the "summation of RMS currents" is the same as the "RMS of summation of currents". \$\endgroup\$
    – alvrm
    Commented Nov 12, 2022 at 20:10
  • \$\begingroup\$ Neither are right or relevant and your addition is just perpetuating your problem. The total RMS current is the square root of the sum of the squares of each individual harmonic. \$\endgroup\$
    – Andy aka
    Commented Nov 12, 2022 at 20:48
  • \$\begingroup\$ They are equal, I had a problem in my script where I forgot to take the sqrt of the squared RMS currents. Both approaches work. \$\endgroup\$
    – alvrm
    Commented Nov 12, 2022 at 21:16
  • \$\begingroup\$ They may be equal but have no relevance. \$\endgroup\$
    – Andy aka
    Commented Nov 12, 2022 at 22:05
  • \$\begingroup\$ They are equal because one is derived from the other. Approach A is the mathematically simplified version of approach B. \$\endgroup\$
    – alvrm
    Commented Nov 13, 2022 at 14:27

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