Help measuring power consumption on a fridge with pulsed DC current input

[Jman99]

I have a small Engel compressor fridge and are using a basic plug n play inline DC wattmeter to get an idea of the power consumption. The problem is it reads 2.8amps/37watts draw average ( which I verified with a clamp meter and handheld dmm) and the compressor is on continuously due to the summer heat, so in each hour I recorded the amp hours and watt hours, yet instead of getting 2.8ah/37wh in an given hour, the meter always reads about 1.7ah/22wh.

These fridges from what I have heard convert the DC to ac 20v square wave to run the compressor. So I then ran the fridge on 240vac with a mains inline wattmeter and got 44wh in any hour.

So very confused, which power consumption is accurate? The DC or ac meter?
Just trying to know what my fridge draws and what's happening here.

Thanks

 

[KlausST]

Hi,

basically both could be right.
* The AC measurement for AC input
* and the DC measurement for DC input.
Both don´t necessarily need to give the same results. It´s quite expectable that you have different power at differnt operation modes.

Measurement on DC sources are more simple and thus more axact.
You need a high quality AC power meter to get accurate results.

I´m not surprised by 2.8Ah/37Wh versus 1.7Ah/22Wh. It´s quite expectable that the power consumptionn is not constant.

Klaus

 

[Jman99]

Hi,

basically both could be right.
* The AC measurement for AC input
* and the DC measurement for DC input.
Both don´t necessarily need to give the same results. It´s quite expectable that you have different power at differnt operation modes.

Measurement on DC sources are more simple and thus more axact.
You need a high quality AC power meter to get accurate results.

I´m not surprised by 2.8Ah/37Wh versus 1.7Ah/22Wh. It´s quite expectable that the power consumptionn is not constant.

Klaus

Thanks Klaus.
Also my concern was the pulsing is happening too fast for the DC wattmeter so it's amphour reading would be quite poor accuracy.

 

[KlausST]

Hi,

pulsing with DC ... is not a big issue. Just use an appropriate RC filter and it should work.
Even a cheap DC meter should work. (for sure there ar bad ones)

Klaus

 

[albbg]

Sorry, I didn't understand exactly how did you measured the consumption under the various configuration. I mean

1. Istantaneous DC measurement. You measured 2.8A/37W (so you supplied it with about 13V) by means of both an inline wattmeter and clamp+dmm. Could you please let's know the model ?

2. DC measurement over 1h. You measured 1.7A/22W by means of what instrument ? Could you please let's know the model ?

3. AC measurement over 1h. You measured 44W by means of a mains inline wattmeter. Is it a professional instrument ? Could you please let's know the model ?

If measurements 2. and 3. are correct, than there is a loss of efficiency of about 50%. Too much for an AC/DC converter.
If the correct measurements are 1. and 3., instead, the efficiency is about 85% that could be acceptable.

Last question: the power consumption of the fridge is not declared by the manufacturer ?

 

[danadakk]

Your AC meter is true rms capable ? That way non resistive loads, like compressors,
read real power.

What is true-RMS?

The need for true-RMS meters has grown as the possibility of non-sinusoidal waves in circuits has greatly increased in recent years. Some examples include variable speed motor drives, electronic ballasts, computers, HVAC, and solid-state environments.

www.fluke.com

AN012 - Peak, RMS and averaging circuits

ESP Application Notes - AN012 - Peak, RMS and averaging circuits

sound-au.com

https://www.analog.com/media/en/technical-documentation/application-notes/an106f.pdf

Regards, Dana.

 

[KlausST]

Hi,

my pedantic comment. ;-)

For true power measurement one does not need true RMS.

True_power is the integral over time of (V x I) / t. It needs to measure (and multiply) the instantaneous values of V and I.

No "square" and no "root" is invloved so far.

****
On the other hand often good "true RMS meters" also provide accurate true power results (or vice versa).
From V_RMS and I_RMS they calculate apparent_power... and use it to calculate the power_factor:
power_factor = true_power / apparent_power.

Klaus

 

[danadakk]

Good DSO scopes or power analyzers or DMMs or VNAs handle this quite easily now..

Or the ancient heating approach + a mercury thermometer + a known mass (specific heat
known), get rid of all this electronic crap.

Back to reality.

assets.testequity.com/te1/Documents/pdf/power-measurements_AC-DC-an.pdf

Fundamentals of AC Power Measurements

Power analysis involves some measurements, terms and calculations that may be new and possibly confusing to engineers and technicians who are new to this discipline. And today's power

www.tek.com

Make Better AC RMS Measurements with Your Digital Multimeter

Learn about the compexity of ac rms measurements and how to deal with them.

www.keysight.com

Regards, Dana.

 

[dick_freebird]

A cheapo non-true-RMS meter might give a bogus reading on
a square wave power source if it just "scales for sine equiv"
rather than doing the homework.

The DC draw is all the power there is, and easy to measure.
Of course you do not get to know what losses are in the
chopper and what in the compressor, without digging deeper.
But given that the compressor and chopper are a "package
deal" maybe you do not care.

Nameplate values are a "not to exceed" for purposes of sizing
wire, socket and circuit breaker (and absolving mfr for responsibility,
re same). The real draw should be well less, as observed.

 

[KlausST]

Hi,

O.K. an example.

You have an expensive, high quality true RMS meter.
It shows perfect 230V AC RMS and perfect 2.0A AC RMS.
What is the true power? and what is the energy after 2 hours?

Answer: the true power is in the range of 0W and 460W depending on power factor. It could be 0.5W, 4W, 72W, 432W. Nobody can tell, even on perfect RMS values. No matter whether you have perfect clean sine or distorted sine.

...and the energy is somewhere in the range of 0 to 0.92Wh per 2 hours depending on power factor.

--> even perfect RMS measurements can not give any information about true power or energy.
Multiply true RMS voltage by true RMS current to get apparent power in VA or kVA. Still no information about the true power nor how high the bill will be you have to pay. (energy)

There is "true RMS voltage measurement", there is "true RMS current measurement",
but there is nothing like "true RMS power measurement", because RMS means Root Means Square.
For a power measurement, one does not need Root, nor Square.

*******

With this I just want to say that "true RMS" is no guarantee for good true power measurements.
It just may be a "hint" for a high qualitiy meter.

Klaus