Archived Forum

Which country are you in, and what is the mains supply voltage? Do you have an earthed centre-tapped supply (i.e as in the USA, 120 - 0 - 120 V)?

I think you are not measuring the correct current and/or voltage. Or it is possible that you have a serious fault in your house wiring, and part of the neutral current is being diverted past the CT.

I'm in germany, so we got 230V... We got the electricity meter in the basement and the fusebox in the kitchen. It's a bit messy inside, but I added the CT to the biggest blue cable I could find. I'll add a Photo later.

Could a possible confusion with the direction of the CT and/or AC-AC be the cause of the problem?

"Could a possible confusion with the direction of the CT and/or AC-AC be the cause of the problem?"

No, the two must be relatively correct (both one way round or both the other way round) so that Power is the correct sign, as you found. (Negative power means voltage is positive while current is negative, then a half-cycle later they are reversed). That does not explain why you only read half the power.

How many wires lead from the meter in to the fusebox? If you have a 3-phase system, it is possible that the Schlafzimmer is fed by a different phase to the kitchen. If that is the case, you need 3 CTs, one on each phase (and not one on the neutral, because the neutral current is ideally zero!).

This closeup is the main-fuse. Does this mean I got a 3-phase system?

Do I really need 3 CTs then? I have no Idea on how to fit them into the box... 

Yes,  you have a three-phase supply. The main circuit breaker has 3 linked poles and the cables into it should be brown, black & grey.  If you want to measure the total power coming in to your house, you need 3 CTs. The CTs will go on the 3 cables into terminals 2, 4 & 6 on the Siemens 3-pole MCB. These cables go across the bottom of the box - will the CTs fit there? Or can you fit 2 CTs close to terminals 2 & 4 and the third at the bottom?

When you fit the CTs, make sure they all point in the same direction.

Bad news then ^^

At the moment I only got 2 CTs. The Problem with the cables is, they are so thick and stiff, I can barely move them. I wonder how they were installed there in the first place...

But I will figure something out, thank you so far!

I'm an Electrical Engineer. We train gorillas as electricians to install cables like that.

But they are only 16 / 25 mm² ?  The biggest cable I have ever used was 630 mm².

(Take care when working in there. All those exposed live terminals would not be acceptable in the UK).

 Today I got my third CT and a gorilla to bend the cables for me. Now I'm looking forward to get some proper readings.

Is this 

openenergymonitor.blogspot.de/2012/02/if-you-are-active-member-of.html

the most recent info about the 3 phase system?

I would read that post with much caution. He states that the three phase voltages are always equal and always 120° displaced. This is true in theory and when you have a perfectly balanced system, but in practice not. All the conductors have an impedance - principally resistive - and the loads on the phases can vary greatly, thus the neutral will have a voltage with respect to earth, and the phase voltages will be unequal, and displaced if there are reactive components to the loads. Therefore you need to remember that measuring one voltage and assuming the other two will be the same will be wrong, but it is probably the best easy solution.

The standard emonTx voltage and power sketch measures about 100 samples per cycle at 50 Hz. If you read voltage and 3 currents in sequence, you will get 25 readings per cycle. That is probably enough to give reasonable accuracy. What I would do is save the voltage reading into an array (of not more than 25 elements - 16 or 17 is probably enough - you need readings for 240°), read the first current and calculate the instantaneous power and accumulate the answer (in order to calculate the average later), then read the second current and multiply by the voltage 8 or 9 samples earlier (or you may need to interpolate between 2 samples), calculate power the same way and then repeat this for the third current with the 16 or 17-samples delayed voltage. Do this for say 10 cycles of mains as the example sketch, then calculate the real and apparent powers, currents and rms voltage. You should be able to do all this in the emonTx.

If you need greater accuracy, the emonTx has 2 spare analogue inputs. It would be possible to add (on a separate circuit board) the additional components for 2 voltage inputs and have 3 transformers truly measuring the 3 phase voltages. I do not know whether anyone has done this.

Now I tested current measurement again with the three CTs. Still something strange here...

I connected the water heater in the kitchen and in my room and this is what I got:

Irms1 Irms2 Irms3 Vrms

my room:
2.42 0.82 0.02 235.28
2.41 0.77 0.03 235.44
2.42 0.79 0.04 235.22

kitchen:
0.77 0.33 7.60 233.34
0.77 0.32 7.55 233.61
0.76 0.33 7.54 233.74

When connecting the heater in the kitchen, the current is still higher than when connected in my room.

I also switched on the oven and it got me this:

0.27 5.82 0.57 235.67
0.27 5.77 0.59 235.46
0.27 5.75 0.59 235.61

Am I still missing something? I put the CTs to the three wires to the Siemens 3-pole MCB, as you said. And the measurements in the kitchen make sense, but in my room?

 Did you upload the new sketch to emonTx to support 3 phases?

 Which sketch are you refering to? I'm using this at the moment:

github.com/openenergymonitor/emonTxFirmware/tree/master/emonTx_CT123_Voltage

I only adjusted the calibration and use this:

Serial.print(" "); Serial.print(ct3.Irms);

to get the current on each CT

You have set the correct calibration constants for the 3 c.t's?

And the 3 wires with the c.t's are the ones coming from the meter in the basement?

The 7.6 A reading in the kitchen on L3 is correct for a 1800 W load. What power is the oven?

Can you use a multimeter and measure some voltages (TAKE CARE):

L1 (Breaker Terminal 2)  - N

L2 (Breaker Terminal 4)  - N

L3 (Breaker Terminal 6)  - N

L1 (Breaker Terminal 2)  - L2 (Breaker Terminal 4)

L2 (Breaker Terminal 4)  - L3 (Breaker Terminal 6)

L3 (Breaker Terminal 6)  - L1 (Breaker Terminal 2)

N - E

 Yeah! I think, I got the problem! The voltages are all as they should be:

L1 - N = ~233
L2 - N = ~233
L3 - N = ~233

L1 - L2 = ~410
L2 - L3 = ~410
L3 - L1 = ~410

N - E = 0

BUT: One of the CTs opened a little, as soon as I closed the fusebox. I did not notice at first, but with the box open, all readings are okai, and as soon as I closed it, one current dropped. So now I rearranged a little, and i seems to be working!

Now I'm have to concentrate on the Code. I'm not sure, if got this right: I put the voltages into an array(buffer) and then I multiply the first current with the most recent voltage. Then I multiply the second current with the voltage 7 samples earlier, and so the third. Then I got the power for each 'channel' and add them up? What do I have to average here? How do I know which one is the _first_ current? By measuring and testing only?

I could possibly answer these questions myself, but I haven't found good information about the 3phase system until now. So maybe you have a link or book or something? I really appreciate your help, thank you! :-)

Sebastian

Yes, a partly open c.t. will give very wrong readings. (I thought maybe the system was not what I thought it was - that was the reason for asking you to measure the voltages. I am happpy now).

I think you are heading along the correct path. You need to use the same general methods that are in the example sketch https://github.com/openenergymonitor/emonTxFirmware/tree/master/emonTx_C... and in EmonLib (https://github.com/openenergymonitor/EmonLib).

You read voltage, current L1, current L2, current L3. First you filter the readings. Then you calculate the instantanous power for each set of readings using the 3 currents and the voltage 3 times (now, delayed once and delayed twice), you add the powers separately and count the readings, then divide by the count of readings to get the average power per phase over 10 (? - your choice) cycles. Then you can add the powers to get the whole house consumption, or you can see the power on each phase separately - your choice again.

(But note - you need to delay adding up the powers until after the first cycle because you will not have a full set of voltage readings until 2/3 cycle has passed).

If you associate the wrong c.t. with the voltage, the power will be clearly incorrect - half the actual value and negative - because of course the phase angle will be 120°.

Here's a quick guide to the 3-phase system. [WARNING: Beware the Wikipedia article "Three-phase electric power" - large parts describe American practice which is totally inapplicable in the EU].

Take a bicycle wheel and put 3 labels on the tyre 120° apart: 'L1', 'L2', 'L3' (or 'Brown', 'Black', 'Grey', or 'Red', 'Yellow', 'Blue' according to how you label your 3 phases). Spin the wheel slowly and look at it from in front - the three labels appear to be going up and down. The up and down movement represents the three voltages, the axle is the neutral voltage (0).

For most well-behaved loads, the flags could equally represent current - near the centre equals small current, near the rim represents a large current. Now here's the clever bit: if the three phase currents are balanced, at any instant they add up to zero. That means no neutral current. In reality, that is rarely the case.

What you will do in your sketch is read one voltage, then wait until the next flag comes round (which represents the next voltage) before you use your reading of voltage in the calculation with the next current.

I don't have access to a 3 phase supply, so I cannot run any tests. I shall write/modify a sketch for you over the next few days. (The only way I will know it is working is if it gives the correct wrong results    ).

 Thanks to Robert Wall my system is now running! :)

Just one more thing: Whenever the emontx does restart (battery low...etc) the first ~10 Readings are way too high! I think it's because of the first reading being to high due to some weird power on strange thing... and then the filter kicks in. Is this known? And is there a better way other than to throw away (eg. not transmit) the first few readings?

That problem is due to the settling time of the high pass filter in the software. It sees a step of d.c. input at power-up and takes time to remove it. There ought to be a way to pre-load the filter, but I think it might imply waiting to start the readings at a special point in the cycle. That would be worth looking in to.

As you say, the easy answer is to throw away the first few readings.