Good morning all....

In the graphs above, you will notice that between 22h00 and 00h00 last night we had load shedding. When the grid is on, we get, what looks like an average of 210/215 volts. When the grid goes off, and we are running on battery, I get a constant 230v from the inverter. (I see there is a setting for this on the inverter). As soon as load shedding starts, I can actually hear the room fans speed up. When the grid returns, I hear them slow down a bit. My question is: If I change my inverter setting to produce 215v instead of 230v, would my batteries last longer?

(Deye 8Kv inverter. 3 x Hubble AM2)

The difference will by minimal depending on the base load, but using a lower voltage means your Amps increase and your battery has X amount of Amp hours.
Lets assume your base load overnight is 500W.  If you are at 230V, then 500/230 = 2,17A.  If you are at 210V, then 500/210 = 2,38A
You can see with the lower voltage your Ah will increase not decrease, so it will have the opposite effect of what you wish to achieve.

Edited January 13, 20242 yr by zsde

Your supply Voltage is already on the lower of the scale end most of the time, and after loadshedding it drops out of range. You could rather consider setting the Voltage limits on the inverter to the standard 230V +/-10% to help protect your appliances so they don't draw excessive current.

 

18 minutes ago, GreenFields said:

Your supply Voltage is already on the lower of the scale end most of the time, and after loadshedding it drops out of range. You could rather consider setting the Voltage limits on the inverter to the standard 230V +/-10% to help protect your appliances so they don't draw excessive current.

 

The inverter is already set at 230V as can be seen when there is LS. Other times the grid voltage goes to loads as the inverter is on bypass. The inverter adds no contribution to steady voltage when the grid is on. 

My feeling is one can get about 3.5% longer run time when there is LS by reducing the voltage to 215V. Not worth it to reduce due to negatives already indicated. 

Having a too narrow grid healthy setting could mean more battery power will be used when the grid is on. 

Edited January 13, 20242 yr by Scorp007

4 hours ago, zsde said:

The difference will by minimal depending on the base load, but using a lower voltage means your Amps increase and your battery has X amount of Amp hours.
Lets assume your base load overnight is 500W.  If you are at 230V, then 500/230 = 2,17A.  If you are at 210V, then 500/210 = 2,38A
You can see with the lower voltage your Ah will increase not decrease, so it will have the opposite effect of what you wish to achieve.

I think we need to explore the half truth from social media/forums about the fact that motor loads always draw more current when you reduce voltage. 

I will try to explain this statement. 

If a 220W motor load is connected to 220V (for ease we ignore PF) we draw 1A.

Most household motor loads we will find only needs below 180W. In this case the current will not be higher than 1A.

If the above example needed the full 220W for the load(output power) then the current could go up to around 1.1A if the voltage drops to 200V.

Below is an example of the same load being a fridge with 110W motor but only needs around 70W for the output load. 

Reduced power used even when the voltage is only 2 volt lower and current remained the same. 

Motor loads vary and that I understand.
However when you have a device that draws a constant load, i.e. a globe and you reduce the voltage, then the Amps increase.
A good example are computer chips and specifically CPUs. Over the years the voltages for electronic components have been reduced where a typical CPU nowadays runs at 1,5V or less. The frightening part is when you see how the Amps have since shot up where a 100Amps has become quite normal just for the CPU or GPU for that matter.
So as the load increases so do the Amps.
If the formula of P= V X I is not valid anymore for constant currents then I may be living in the past and would like to be educated.
 

58 minutes ago, zsde said:

Motor loads vary and that I understand.
However when you have a device that draws a constant load, i.e. a globe and you reduce the voltage, then the Amps increase.
A good example are computer chips and specifically CPUs. Over the years the voltages for electronic components have been reduced where a typical CPU nowadays runs at 1,5V or less. The frightening part is when you see how the Amps have since shot up where a 100Amps has become quite normal just for the CPU or GPU for that matter.
So as the load increases so do the Amps.
If the formula of P= V X I is not valid anymore for constant currents then I may be living in the past and would like to be educated.
 

My post was about AC. Your examples are for DC circuits. 

I cannot educate but the impedance is the important factor for AC circuits. Z=root of Rsq+Induct sq. So the impedance Z remains the same in the circuit. Lower volts means lower current. All circuits are made up of inductance/capacitance as well as resistance. 

Thus if you lower the voltage there is already a reduced current due to the lower voltage with the same impedance that cannot be changed by the voltage you apply. 

I can't explain well and for that reason I supplied the actual measurements with the variable being the voltage. 

Power for a AC circuit was never P=VxI as the PF can be as low as 0.5 which will reduce the power to 50% at the same volts and current. 

It is for this reason that we find inverter specs at times showing VA as well as Watts and the values differ. 

We as small consumers are charged only for the active power Watts we use. Big consumers are charged for the VA used which is always higher and the VA dictates what amps will have to be provided or transmitted. The best example that I can think of is when you buy a transformer. It is never specced in Watts but VA. 

 

Edited January 13, 20242 yr by Scorp007

There is a category of appliances however that do reduce power when reducing voltage. That are the purely resistive ones, such as heaters, cooking top, oven, fryer, kettle, geyser, etc. But I think that are not the ones you want to run during load shedding. And take the kettle for example, it would take longer to boil the water, take less current, but at the end will have consumed the same amount of energy from the battery - that is the energy required to boil the given quantity of water.

1 minute ago, Beat said:

There is a category of appliances however that do reduce power when reducing voltage. That are the purely resistive ones, such as heaters, cooking top, oven, fryer, kettle, geyser, etc. But I think that are not the ones you want to run during load shedding. And take the kettle for example, it would take longer to boil the water, take less current, but at the end will have consumed the same amount of energy from the battery - that is the energy required to boil the given quantity of water.

On average I agree with you. But we just have to look around how many people install 2x8kW inverters and that indicates they want to run all their loads as normal during LS. Also look how many want to run their geysers from the inverter=always on. 

Over the last few days there was a question on how to run the geyser on essentials instead of non essentials. 

3 minutes ago, Scorp007 said:

On average I agree with you. But we just have to look around how many people install 2x8kW inverters and that indicates they want to run all their loads as normal during LS. Also look how many want to run their geysers from the inverter=always on. 

Agree - I'm also one of them with 2x5kW inverter. And what I wrote concerning the kettle also applies to the geyser. However I pay attention not to run the geyser during LS at night. At the end of the day it all boils down to how much energy you have stored in your battery. Therefor the more battery capacity the better.

I'll add that roughly matching your inverter's voltage with the typical utility voltage means that the changeover relays have an easier life and with load shedding, they will work hard.

I increased my output voltage from 230V to 240V for that reason. 

On 2024/01/14 at 6:37 AM, Coulomb said:

I'll add that roughly matching your inverter's voltage with the typical utility voltage means that the changeover relays have an easier life and with load shedding, they will work hard.

I increased my output voltage from 230V to 240V for that reason. 

Hi there @Coulomb - can you explain a little more around the "work hard"?

Our office site as an average grid voltage of 240v but the inverter is set to 230v.  Should we consider changing this to 240v?

4 hours ago, Douw G. Gerber said:

Our office site as an average grid voltage of 240v but the inverter is set to 230v.  Should we consider changing this to 240v?

I think you should consider it. 

The problem is when changing to and from bypass modes. If you have a model where this is infrequent, there is probably nothing to worry about.

When there are inductive loads, there will be arcing at the contacts, and I think that there may effectively be connection between AC-in and inverter-out when changing over. These are always in sync, which helps. But if the voltage differs, then there can be extra current in the arc, which would burn the contacts faster. This is just my impression of how the changeover takes place; I've not done any tests. Subjectively, the changeovers seem less noisy when the voltages are closer, but again I haven't done anything rigorous to check this.

Edited January 17, 20242 yr by Coulomb