50 Hz watt-hour meter monitoring 60 Hz

[dmsv10]

161216-2024 EST

No way for us to know.

But you can determine by experiment.

We have a local energy show every year sponsored by our mayor. This closes our main street for one whole evening. If you want to exhibit there are a number of conditions that you must agree to. One is that you will only use recycled paper, but a nonexistent condition is competence.

One year there was one exhibitor that was showing that a CFL was using about the same amount of power as was a standard incandescent with the same light output rating. Totally wrong, but his meter showed this and was marked in watts as a wattmeter. This exhibitor was a PE and he advertized that. But he was a civil engineer that did business with the city and clearly did not know much about electrical engineering.

Whatever this meter was it was probably just an ammeter calibrated in watts at 120 V with a resistive load.

A true wattage measurement will not care what the frequency is. However, there are restrictions to the statement.

Instantaneous power is simply instantaneous voltage times instantaneous current. Anything beyond this and you have to define average power and over what period the averaging is performed. Energy use is always instantaneous power integrated (summed) over some time period.

.

Do you remember what type of meter it was, handheld or otherwise i.e. was it built into a display panel?


Sent from my iPad using Tapatalk

 

[gar]

161221-0825 EST

dmsv10:

I don't know the brand of the meter that claimed to be a wattmeter. It plugged into a wall outlet just like the Kill-A-Watt does.

One can easily test an instrument to determine whether it is really a wattmeter. I will assume a 120 V circuit.

1. Connect the meter to your 120 V source. It should read zero power. If there are other measurements you can check these for likely correctness at zero load.

2 Connect a 100 W incandescent. Should read fairly close. PF close to 1.0 . At 124 V I read 107 W on a Kill-A-Watt.

3. In parallel with the bulb add a 30 ufd good quality capacitor (polypropylene). I read 106 W PF = 0.51 .

4. Turn off the bulb leaving only the 30 ufd. With one Kill-A-Watt I read 0.5 W, I = 1.4 A, and PF = 1.0 . Clearly some reading errors. On only one Kill-A-Watt does this test work well, can't find that instrument now.

5. Next I went to a smaller resistive load (higher resistance) so the phase shift is quite as close to 90 degrees. Used a Cree LED 9 W bulb. V = 124.4 V, I = 1.43 A, P = 10.4 W, VA = 178 VA, and PF = 0.05 .

A good quality capacitor can hold its charge for quite a while. A random turn off time can leave this voltage near 170 V from a 120 V source. Have a 100 W incandescent shunting the capacitor after power is removed. This will bleed the 30 ufd capacitor down rapidly. Time constant will range from about 30*100/10^6 = 0.03 seconds to 0.003 seconds near end of discharge. Bulb resistance changes with voltage.

.

 

[dmsv10]

I was a little confused or concerned, that the vendor was attempting to compare a 60 W incandescent light vs a 60 W CFL equivalent light in lumens, attempting to show someone that they use the same amount in power, or watts. ? I would be concerned as well.

Maybe the meter was connected to a Ct, or set of coiled up #14 gauge wire, to measure the total power the same way to measure the total power, including ballast and bulb, of electronic ballasts vs magnetic during lighting upgrades, since the total wattage was so small, we took a 10 turned coil used with an amp meter to measure, or in his case, the watt meters he was using could have been wired wrong or just not a real watt meter, last time recall that va in or va out or kva in = kva out, has never changed.


Sent from my iPad using Tapatalk

 

[gar]

1612224-0759 EST

dmsv10:

This incorrectly labeled meter was about the size of a Kill-A-Watt, and also used an integral plug. One really needs a short extension cord with these devices.

For a low cost device I would not expect a CT to be used and it doesn't matter anyway because of the setup of the demonstration. Note: the Kill-A-Watt uses an internal shunt resistor for current measurement. This is actually better than a CT because there is no phase shift. A Kill-A-Watt does a better job of measuring power than a TED System. TED uses CTs.

In the demonstration I previously described the same meter was used to measure the incandescent as was used with the CFL. So whether a CT or resistor was used to measure current would not make much difference.

.

 

[dmsv10]

Thanks and yes, the Kill-A-Watt I have used many over the years to measure smaller residential load and kVA totals for appliances and E.G. Power on boats, very consistent. The only issue I had with the 5 yr plus models was the ground plug on the device coming loose.


Sent from my iPad using Tapatalk

 

[gar]

170102-1059 EST

ike5547:

Have you yet run any experiments on your meter to determine how it performs on 60 Hz?

.

 

[NewtonLaw]

Simple Test

Simple Test

Customer provided a made-in-China small electronic watt-hour meter rated for 50 cycles only (though not expressed in those exact words -- just says 50 Hz and nothing else).

I'm concerned that we'll have misreadings when monitoring 60 Hz but am unable to explain how, why or in what way. I don't have the technical background.

This thing is pretty compact and pretty cheap so I don't think it can adapt to multiple input/output frequencies. Chinese-English documentation provides no info -- except 50 Hz.

Can you engineering guys enlighten me so I don't end up having to install (probably/possibly) the wrong equipment for this application?

You might just use a known load to test the meter. Say a hair dryer, 1200 watts, is plug into the unit. It should register about 1200 watts. If you own or have access to a DDM, measure the supply voltage to the hair dryer while it is running. Measure the current if possible. Multiply the voltage and current to get the running watts approximately. What does your watt meter read?

If you can not get a current reading but you are able to read the voltage while running the hair dryer, you could use the approximate hair dryer resistance, 1200 watts/120 volts = 10 ohms, then calculate the wattage of the hair dryer running by Watts = (Voltage while running)^2 / 10 ohms.

Hope this helps,

Newton Law