Stanley

That's exactly what I'm saying should happen since it only measures current, but @bobmabenahas the opposite. His CoCT bill is much higher than what the Efergy reads.

I agree, but not taking power factor into account should mean that the Efergy meter reads more than your actual consumption, not less. Which makes the discrepancy between the CoCT reading and the Efergy meter even worse.

It will work, but car batteries are designed slightly differently to deep cycle batteries. They are not meant to be discharged very deeply or very often, so if you do cycle a car battery it will probably not last very long.

That depends if the more expensive hair dryer uses a diode for the low power setting or not. My wife has a "Dyson Supersonic Hair Dryer" which is quite expensive and our inverter has no problem with it on any setting. It does have more than two temperature and speed settings which is also a giveaway that it is not just using a diode. Edit: You could score some serious points with the wife by buying her one (J ust don't tell her it's to sort out the inverter issue XD )

I think you need to remove the k from all the kW in your post. Otherwise they are 1000 times too cheap. i.e. It should be R4.73/W

LEM makes a very large range of hall-effect sensors with different measuring ranges and supply voltages etc. https://www.lem.com/en R.S Components does stock quite a few of their products (for some reason they put them in the same category as the current transformers). Here is a link to one of them as an example: https://za.rs-online.com/web/p/current-transformers/0497182/

I don't think you can clear that error. Blue Nova use a little PCB on each cell with a microcontroller that measures the cell voltage and temperature etc. That error means that one or more of the little PCBs is not communicating with the BMS. I think you need to contact Blue Nova for support.

If they are using CTs, then if you know the CT ratio you can easily calculate the current in the secondary for a particular primary current.

CTs and Hall-effect sensors are 2 completely different things, although both are used for measuring current. CTs are passive devices (they really are just transformers) where the current in the secondary (the only winding on the CT) is proportional to the current in the primary (the conductor through the middle). This makes the CT a current source. In order to measure the current, the measuring device will usually have a burden resistor at the measuring end so that the voltage across the burden resistor is proportional to the current and will not be affected in any way by cable length etc. as any other voltage drops or resistance in the circuit will have no effect on the measurement. The burden resistor is usually quite a low resistance (often a fraction of an ohm) meaning it is very difficult for noise to cause a measurable voltage across the burden resistor. Of course the design of the measuring circuit is also important to make sure it get's rid of / can handle common mode noise. Putting the burden resistor at the CT instead of in the inverter / measuring device would of course negate the benefit of the CT being a current source. Hall-effect sensors are active devices (they have electronics in them that need to be powered) that measure current using the hall-effect, which means that they can measure both AC and DC current (unlike CTs which can only measure AC). Since they are active devices, their output depends on the circuit design. You can get these with a whole range of output options including current outputs (although this is most likely done using an opamp in a current source configuration). So I guess it depends on what the Sunsynk uses to measure the current. Is it a CT or a hall-effect sensor?

CTs are current sources and therefore the signal is very immune to noise. You shouldn't see any interference if you run the cables in the same conduit as the AC.

The switch with feedback to the inverter is the internal relay or contactor that the inverter uses to connect the load to the grid when in standby mode. (This is only applicable for Passive standby UPS used as off-grid SSEG) The inverter must know the state of this switch (not just based on what it wants the switch to do, because switches can become stuck or welded closed) so that it can never try to power the load from the inverter while the load is powered by the grid. The Hager changeover switch is just a manual external changeover switch (It is good to have this in your drawing when submitting your application as the people approving them don't know the difference, they just look for a changeover so make sure to label it as such in the drawing)

Thank you, that is the most important bit of information you have provided so far. I might expect that when trying to run a 1200W vacuum cleaner if it is a cheap inverter (square wave or quasi-sine output), but I wouldn't expect that from a resistive load like a kettle. Do you have a link to the product from loot so we can check the specs?

So, when you tested it by trying to boil a 2kW kettle was it connected to the 1500Ah battery bank or a different bank? What batteries are you using? To get 1500Ah at 12V from 6 x 6V batteries, each battery would have to be 500Ah What size and length cables did you use between the battery bank and the inverter? What failed when you tried to boil the kettle? Did the inverter shut down on low battery or overload?

You posted this exact question already. Now you have made another thread without answering any of the questions posed to you in the previous thread.

What exactly happened when you tried to boil the kettle? Did the Inverter shut down on battery low or was it an overload?

I'm assuming it's a single-phase installation. This works well and complies with the CoCT regulations: http://www.em.co.za/SF263.html

Generally you will have at least 2 Neutral points (more if you have more than one earth leakage protection device). One for the neutral before the earth leakage protection and one for after. If you mix them up you will have your earth leakage breaker constantly tripping.

Yes, if it isn't popped out then the circuit breaker has probably not tripped. This means that your problem must be elsewhere. Unfortunately I can't really help since I don't know much about that particular inverter, but I'm sure there are others on the forum who can give you some advice.

I don't know that particular inverter but all the circuit breakers I know of that look like that pop out when they trip and you push them back in to reset them.

I have been using this type of pre-charge for years. It is cheap and works really well. The nice part is that you only need to use it very rarely. After the initial startup when you install the inverter, the only time you would need to pre-charge it again is if you had to completely shut the system down for some reason and disconnect the battery, so it isn't worthwhile trying to find a fancy solution. (A fancy solution would be something like a contactor with the pre-charge resistor wired across it, and a circuit to close the contactor when the voltage on the inverter side of the contactor is high enough, but this could be very expensive since the contactor would need to be able to carry the full DC current, which could be hundreds of amps depending on the inverter size.)

Are you sure the surge current that trips the BMS isn't just when you connect the inverter to the battery for the first time? The inverter will have a fairly large DC bus capacitance which will cause a big current spike when you connect it to the battery. Turning pylontech batteries on with the Inverter connected will also cause a trip no-matter how many batteries you have because the Master battery turns on first and will trip before the others have started. The solution to this problem is to add a pre-charge circuit. The simplest way is to use a removable fuse between the battery and the Inverter (you must have a fuse anyway, so why not use a cartridge fuse or similar fuse that is easy to remove and insert) and then wire an normally-open momentary contact push button in series with a resistor (I would say 47 Ohms, 20W or so should be ok) across the fuse holder. Then to start the inverter, begin with the fuse open, turn the battery on, then press and hold the push button for 10 seconds or so before closing the fuse. This will eliminate most of the current surge and prevent tripping the BMS.

Hi Henk, I don't know which cells exactly (I only ever looked at the BN52V-77-4k and a big custom built battery that was a high voltage and a few hundred kWh) but BlueNova use Winston cells (you can check the link I posted above to the 60Ah cells). In the large custom bank, they used bigger Winston cells (I don't remember which ones exactly) and still paralleled 2 cells (Then they had lots of series connected pairs of cells). The last I knew they were still using the 123Smart BMS, but that was maybe 2 years ago now. They did say at the time that they had developed their own bespoke BMS, but when asked if they had a battery with the new BMS they said they didn't have one available yet. I have never seen a BlueNova battery with anything other than the 123Smart BMS (but then I don't really see a lot of their batteries anymore since I prefer other brands), so I don't know if they did end up using their own BMS or not.

Yes, 4kWh (or 3.2kWh in reality) is way to small for your inverter size. I know you may actually be able to get quite a bit more than the 'rated' energy from a new cell, but I still have to wonder about the fact that the manufacturer rates these as 60Ah cells. Surely they would rate them higher if they could. Here is a picture of a cell from a BlueNova 77Ah / 4kWh battery There are 16 of these cells in the battery. Each cell is clearly printed as 60Ah / 198Wh. 198Wh x 16 = 3.168kWh http://en.winston-battery.com/index.php/products/power-battery/item/wb-lyp60aha?category_id=176

Since that is supposedly a 4kWh battery, then to use 25% (down to 75% SOC) is only 1kWh. So if you use between 500W and 1kW then it will take 1 - 2Hours to go down to 75% SOC. Although in reality the "77Ah / 4kWh" BlueNova battery is actually a 60Ah 3.12kWh battery (You can open it up and check for yourself. The capacity of the cells is printed on them by the manufacturer). So 25% of the real capacity is actually only 0.78kWh

Actually, you could do it with just 1 diode to the battery. The Geyser element would just need to get it's power from the PV side of the diode. If you give some specs of the PV array I can probably find you a suitable diode at RS components. Also, how are you charging the battery? I assume the PV is not connected directly to the battery. You should have some sort of charge controller.