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Voltage, current and wattage


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On 2022/04/20 at 10:05 AM, Leshen said:

Because the higher the voltage, the lower the current for the same wattage which makes motors run cooler. 

Motors don't have switch mode supplies in them - they will 100% run hotter with higher voltages. (Unless you are talking about BLDC motors - like inverter aircons; they will not be affected the driver will be).

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31 minutes ago, P1000 said:

Motors don't have switch mode supplies in them - they will 100% run hotter with higher voltages. (Unless you are talking about BLDC motors - like inverter aircons; they will not be affected the driver will be).

I’m not referring to running it at a voltage beyond its design limit. Read the context also. It was a change from 220V to 240V. 

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18 minutes ago, Leshen said:

I’m not referring to running it at a voltage beyond its design limit. Read the context also. It was a change from 220V to 240V. 

I know. Upping the voltage to a motor will also up current (and watts).

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Just now, Leshen said:

And what does lowering voltage do?

It lowers the current. Think of an ohmic load like a resistor - more voltage = more current = more power. Now an induction motor is not exactly an ohmic load, but some big parts of it is. Lowering the voltage will decrease the field strength until it is to weak to hold the rotor in sync (will depend on the load). A universal motor will spin faster with more voltage (while using more power).

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3 minutes ago, P1000 said:

It lowers the current. Think of an ohmic load like a resistor - more voltage = more current = more power. Now an induction motor is not exactly an ohmic load, but some big parts of it is. Lowering the voltage will decrease the field strength until it is to weak to hold the rotor in sync (will depend on the load). A universal motor will spin faster with more voltage (while using more power).

You are going to have many electricians disagree with you. Have a read 

 

Effects of low voltage

When you subject a motor to voltages below the nameplate rating, some of the motor's characteristics will change slightly and others will change dramatically. To drive a fixed mechanical load connected to the shaft, a motor must draw a fixed amount of power from the line. The amount of power the motor draws has a rough correlation to the voltage x current (amps). Thus, when the voltage gets low, the current must increase to provide the same amount of power. An increase in current is a danger to the motor only if that current exceeds the motor's nameplate current rating. When amps go above the nameplate rating, heat begins to build up in the motor. Without a timely correction, this heat will damage the motor. The more heat and the longer the exposure to it, the more damage to the motor.

The existing load is a major factor in determining how much of a decrease in supply voltage a motor can handle. For example, let's look at a motor that carries a light load. If the voltage decreases, the current will increase in roughly the same proportion that the voltage decreases. In other words, a 10% voltage decrease would cause a 10% amperage increase. This would not damage the motor if the current stays below the nameplate value.

Now, what if that motor has a heavy load? In this case, you already have a high current draw, so the voltage is already lower than it would be without the load. You may even be close to the nameplate's lower limit for voltage. When you have a voltage reduction, the current would rise to a new value, which may exceed the full-load rated amps.

Low voltage can lead to overheating, shortened life, reduced starting ability, and reduced pull-up and pullout torque. The starting torque, pull-up torque, and pullout torque of induction motors all change, based on the applied voltage squared. Thus, a 10% reduction from nameplate voltage (100% to 90%, 230V to 207V) would reduce the starting torque, pull-up torque, and pullout torque by a factor of 0.9 x 0.9. The resulting values would be 81% of the full voltage values. At 80% voltage, the result would be 0.8 x 0.8, or a value of 64% of the full voltage value. What does this translate to in real life? Well, you can now see why it's difficult to start "hard-to-start" loads if the voltage happens to be low. Similarly, the motor's pullout torque would be much lower than it would be under normal voltage conditions.

On lightly loaded motors with easy-to-start loads, reducing the voltage will not have any appreciable effect, except that it might help reduce the light load losses and improve the efficiency under this condition. This is the principle behind some add-on equipment whose purpose is to improve efficiency.

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2 minutes ago, Leshen said:

You are going to have many electricians disagree with you. Have a read 

 

Ok, I think both my explanation and the quoted bits are oversimplified too much for the way the conversation is headed. In the ranges we are talking about; for the motors we are talking about (fridges/pool pumps); it will not really matter - efficiency is quite low in these motors. Furthermore measured current will likely not change much at all, but power factor might. Anyway, I think we have derailed this thread way too much now.

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26 minutes ago, P1000 said:

Ok, I think both my explanation and the quoted bits are oversimplified too much for the way the conversation is headed. In the ranges we are talking about; for the motors we are talking about (fridges/pool pumps); it will not really matter - efficiency is quite low in these motors. Furthermore measured current will likely not change much at all, but power factor might. Anyway, I think we have derailed this thread way too much now.

Maybe you can explain on a different thread why example an electric lawnmower requires a certain length of lead, at a certain rating to avoid a volt drop and the reasoning behind it. I’m sure that would give perspective. 
 

Regards

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7 hours ago, Leshen said:

Maybe you can explain on a different thread why example an electric lawnmower requires a certain length of lead, at a certain rating to avoid a volt drop and the reasoning behind it. I’m sure that would give perspective. 
 

Regards

Ok, so a lawn mower is a bit special in the sense that it starts a lot - and it takes a lot of current to start. Mine takes in excess of 30A rms for more than 1s on startup. If your lead resistance is too high and the voltage drop too high it might never get up to speed and draw >30A continuously while trying to get up to speed. Obviously, the rest of the system is not designed for this and hopefully the breaker will trip before things start to melt.

So after reading your quote above I ran a couple of simulations for a single phase squirrel cage motor in PSIM and the current is lower for lower voltages, but it does take longer to start up. I am by no means an expert in power electronics, and took motor parameters from values I could find online, so my simulation is definitely not perfect. That said, I ran the simulation for a couple of different "motors" and it seems to always come to the same conclusion.

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Let me jump in here with some first hand experience of fridge / freezer,  pool pump motors, etc.

In the KZN midlands the power grid is atrocious, besides multiple power failures a week, (Sometimes in one day) we also get the dreaded low voltage. When someone notices a voltage of say 128V they plaster this on the local group very quickly, as many people in our area have lost appliances due to this.

I currently have a LV / HV protection because of this very problem.

https://banggood.onelink.me/zMT7/3c4cr445

Not very expensive, but has saved my equipment  many times, before the inverter was installed.

Now it just cuts power to the inverter, and we run on solar or batteries.:)👍

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