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Noob to solar having some issues


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I am very new to this solar stuff, hope somebody can advise on the issue I'm experiencing.

I have the following:

1200 W Pure sine wave inverter Make :Devel Model No : GD-6236
Solar controller 40A (Make : Easton)
100Ah (Allgrande Gel Battery VRLA Model : 6-CNFJ-100) + 120Ah ( G-energy Gel Battery Model : DG12-120)  wired in parallel (fully charged)
Heater with inductive load 650W
1 X Mono Solar panel 160W (Make : Oushang)


The 1200W inverter + Solar panel + batteries are all connected to the solar controller
The heater 650W is plugged into the inverter.
I turn on the heater and the inverter shuts down after a few seconds.

I assumed 1200 W pure sine inverter would be able to handle a 650 W heater. However it just shuts off after few seconds.

Any tips appreciated.

Thank you
 

 

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So this is a 12V system. 

According to this you have a 1000W inverter... but the image shows 1500W printed large on the inverter. And you say it is 1200W... These varying specifications indicate a cheap brand that will lie to get you to buy their product. But sometimes they can work ok. Lets give the inverter the benefit of the doubt.  Do you know what the low voltage cutoff of the inverter is?

When trying to draw 650W from the inverter, it will be drawing about 720W (including inefficiencies) from the battery - and at 12V, that is 60 Amp.

I hope you have very thick cables running from the batteries to the inverter.  These cables must also be as short as possible, since any voltage drop over the cables is a much more pronounced loss at 12V than a typical 48V system. You can measure the DC input of the inverter with a cheap digital multimeter while drawing a load to see how that voltage drops.

Also - Why do you say "inductive load"? Heaters are normally resistive loads.  Inductive loads are normally motors, compressors, airconditioners etc. A resistive heater should be an easier load to carry than an inductive load which could have a large startup "surge" current.  

Can you upload photos of your battery wiring, and heater load so we can get a better idea of your setup?

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11 hours ago, Ironman said:

So this is a 12V system. 

According to this you have a 1000W inverter... but the image shows 1500W printed large on the inverter. And you say it is 1200W... These varying specifications indicate a cheap brand that will lie to get you to buy their product. But sometimes they can work ok. Lets give the inverter the benefit of the doubt.  Do you know what the low voltage cutoff of the inverter is?

When trying to draw 650W from the inverter, it will be drawing about 720W (including inefficiencies) from the battery - and at 12V, that is 60 Amp.

I hope you have very thick cables running from the batteries to the inverter.  These cables must also be as short as possible, since any voltage drop over the cables is a much more pronounced loss at 12V than a typical 48V system. You can measure the DC input of the inverter with a cheap digital multimeter while drawing a load to see how that voltage drops.

Also - Why do you say "inductive load"? Heaters are normally resistive loads.  Inductive loads are normally motors, compressors, airconditioners etc. A resistive heater should be an easier load to carry than an inductive load which could have a large startup "surge" current.  

Can you upload photos of your battery wiring, and heater load so we can get a better idea of your setup?

Hi Ironman

Thank you for the detailed reply.

1) I don't know the voltage cutoff of the inverter, because the inverter shuts down as soon as I plug in the heater

2)  I have attached a picture of the battery with the cables

3) I made a mistake it should be resistive load, I was also testing an airconditioner (inductive) with the same result.

4) I have uploaded some pictures of my setup.

 

"When trying to draw 650W from the inverter, it will be drawing about 720W (including inefficiencies) from the battery - and at 12V, that is 60 Amp."

According to your point above I am guessing it is my solar controller 40A is too small to handle the 60Amp current and that is why it shuts down?

Does that sound right?

We bypassed the controller and the inverter does not shut down and powers the heater.

 

 

 

Regards

 

 

 

 

 

 

 

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25 minutes ago, Eastboy said:

We bypassed the controller and the inverter does not shut down and powers the heater.

There is your problem. The inverter needs to be connected to your battery directly. In the photo above I see it was connected to the Output/Load of the Charge Controller. Rule of thumb, if it is a 20Amp charge controller , the load connected to it should not exceed 20 amps. Same for a 30A and so on. 

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3 minutes ago, Jaco de Jongh said:

There is your problem. The inverter needs to be connected to your battery directly. In the photo above I see it was connected to the Output/Load of the Charge Controller. Rule of thumb, 

if it is a 20Amp charge controller , the load connected to it should not exceed 20 amps. Same for a 30A and so on. 

Thank you sir.

So should I not connect the inverter to the output/load of the charge controller? I assumed the controller offers more protection by monitoring the battery etc.

If I connect the solar panel and the battery to the solar controller, and then also connect the battery directly to the inverter, is this an acceptable setup?

Will the inverter switch off before the battery runs down and then the solar controller will recharge the batteries from the solar panels.

 

Hope I explained that clearly .

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25 minutes ago, Eastboy said:

I assumed the controller offers more protection by monitoring the battery etc.

It does, but will only work if your load is lower than the rating. 

26 minutes ago, Eastboy said:

and then also connect the battery directly to the inverter, is this an acceptable setup?

Only if there is a correctly rated fuse between them. In your case a 125 Amp should do. 

28 minutes ago, Eastboy said:

Will the inverter switch off before the battery runs down

Only if the inverter has its own low voltage cutoff protection. If you bypass the Charge controller, then all protection should come from the inverter itself. 

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9 minutes ago, Jaco de Jongh said:

It does, but will only work if your load is lower than the rating. 

Only if there is a correctly rated fuse between them. In your case a 125 Amp should do. 

Only if the inverter has its own low voltage cutoff protection. If you bypass the Charge controller, then all protection should come from the inverter itself. 

Understood. Thanks very much for your help

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On 2019/11/20 at 8:49 AM, Jaco de Jongh said:

Dex, a heater is resistive load, there is no inrush current. The "inrush" and "switch-off" current of a resistive load is equal to it's "steady-state" current value.

yeah he said inductive so i assumed it was not your regular heater type element but some funky space heater with a motor perhaps

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13 hours ago, Jaco de Jongh said:

Can you please help me understand the inrush current of a heater? 

The resistance of the element increases with temperature. It ends up glowing red hot, so the difference in temperature is extreme. Not quite as much as a light bulb, which go to yellow hot or beyond, but still the cold resistance is considerably less than when running. Hence, there is an inrush current of sorts, which might last for a minute or so. 

Edit: so it's nothing to do with inductance or capacitance.

Edited by Coulomb
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6 hours ago, Coulomb said:

there is an inrush current of sorts

I have tried on numerous occasions to measure this with normal testing equipment and could never see any significant inrush on a element, nothing compared to a motor starting for example. But o many people talk about inrush currents on resistive loads lately that I just had to ask.  

What I have learned in my studies years ago is that the "inrush" and "switch-off" current of a resistive load is equal to it's "steady-state" current value, but maybe that was wrong. 

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4 hours ago, Jaco de Jongh said:

I have tried on numerous occasions to measure this with normal testing equipment and could never see any significant inrush on a element, nothing compared to a motor starting for example. But o many people talk about inrush currents on resistive loads lately that I just had to ask.  

What I have learned in my studies years ago is that the "inrush" and "switch-off" current of a resistive load is equal to it's "steady-state" current value, but maybe that was wrong. 

There is a slight inrush current on most heating elements, because metals have a positive thermal coefficient and the resistance goes up with the temperature. This effect is even used in cheaper soldering irons: It is designed such that it is thermally balanced around 420°C without requiring any electronics to control the temperature.

I do not know how big this effect is. Like Jaco, my experience has always been pretty much that there is no inrush current to speak of. So I did what all armchair engineers do and looked it up on the interwebz. Wikipedia says this can be up to 14 times the normal current, but it lasts for milliseconds only. This very short period is likely why it's usually not even noticeable.

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12 hours ago, Coulomb said:

Not quite as much as a light bulb, which go to yellow hot or beyond, but still the cold resistance is considerably less than when running.

Previously I used incandescent lamps as a load to do load tests on switched mode power supplies & batteries etc. However their cold resistance can be such that the PSU can't get over the initial  current peak so not such a good idea after all!   

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1 hour ago, plonkster said:

Wikipedia says this can be up to 14 times the normal current, but it lasts for milliseconds only.

That's for lamps though, not heater elements. Try connecting two not quite identical 12 V headlights to a 24 V source and see what happens. Often, one will blow, as the one with the higher on-rush current over-voltages the other.

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