What amp fuse for a fridge freezer

[JohnW2]

I don't see the problem. That's exactly what it means. Change it to rated power if it helps.

I'm not sure that "helps" is the word. As far as I am concerned, that change in wording would totally move the goalposts, and goes quite I way to explaining why I was getting confused by what you've been writing. "A given power output" is very largely dependent upon the load, and could be dramatically less than the "rated power" (which is the maximum output power it is designed to be able to provide.

However, I think I'm coming to understand what has caused all the confusion, even though it still leaves me with one big question.

As we know, the rise in motor speed when one increases voltage from, say,380V to 415V is trivial, typically 0.1% - 0.2% for all but the smallest motors - and if the load is constant and we assume that (actual) power output is roughly proportional to voltage squared, the increase in (actual) power output is also trivial. The situation is therefore that of an (actual) output power that remains essentially unchanged when voltage rises from 380V to 415V - and if we assume unchanged efficiency, that also means that the (actual) input power to the motor remains essentially unchanged when voltage increases from 380V to 415V.

If input power remains unchanged when voltage is increased, that obviously must mean that current (and/or PF) has reduced, and it is that reduction in current that you have been 'majoring' on.

As you've been saying, that reduction in current is 'anomalous' 'in terms of what one would expect (from Ohm's Law) with a resistive load, or even a 'simple inductive load'. What I am struggling with, in the case of a motor, is the mechanism of the reduction in current (and PF).

On the face of it, I would have thought that the only things which the motor 'knows about' are the load, its speed and the supplied voltage. If the first two remain essentially unchanged when voltage is increased, what is the mechanism whereby the current and PF reduce? I'm clearly 'missing something', but obviously don't know what!

There's no need to bring PF into it, but a reduction in PF causes the current to rise (for a given power).

PF has brought itself 'into it' to an appreciable extent. Just looking at one typical example from your catalogue (the "315 SMA" motor), when voltage increases from 380V to 415V (an increase of about 9.2%), motor speed increases from 1486 to 1488 rpm (and increase of about 0.13% -so' essentially unchanged'). With the increase in voltage, current decreases from 200A to 189A (a reduction of 5.5%), and PF reduces from 0.88 to 0.86 (a reduction of about 2.27%). It is the combination of that 5.5% reduction in current and ~2.3% reduction in PF that means that power remains essentially unchanged despite the (~9.2%) increase in voltage - with the figures given, the power input is 66,800 W with 380V and and 67,454 W with 415V.

However, as above, I still don't know enough to understand the mechanism of those decreases in current and PF, despite essentially constant load and speed.

Kind Regards, John

 

[fixitflav]

Increase the load and the current will increase, because the motor has to struggle more, to maintain sync.

Apart from all this being about async motors, I'm talking about load being constant, except for small changes due to change in motor speed. Of course, if the load varies otherwise, the current changes.

 

[JohnW2]

I've not been following this very closely, but you both seem to be saying the same thing - voltage increase, current reduction - assuming the both speed (synchronous) and load remain constant.

I would hope that no-one involved has ever doubted that.

The only reason why some people have perceived an apparent anomaly in terms of Ohm's Law is that we have been talking about a situation in which power is constrained to remain essentially constant (for a given load), even if supply voltage changes. If we had such a constraint in a purely resistive circuit (i.e. if the resistance automatically increased when voltage increased, such as to maintain the same power), then Ohm's Law would correctly indicate the reduction in current when voltage increased.

Interesting though this discussion has been (at least for me), we should perhaps remember because it all started with the ('silly' in my opinion) implication that one could get a reasonable estimate of the current drawn by a frisge/freezer by looming at the voltage and current on its reating plate

Kind Regards, John

 

[fixitflav]

I'm not sure that "helps" is the word. As far as I am concerned, that change in wording would totally move the goalposts, and goes quite I way to explaining why I was getting confused by what you've been writing. "A given power output" is very largely dependent upon the load, and could be dramatically less than the "rated power" (which is the maximum output power it is designed to be able to provide.

I've been talking about load being equal to motor rated power, but that's not essential. If the power was below rated, whatever the current was at one voltage, it would be lower at a higher voltage. Of course the whole basis is that the load is one machine with a fixed characteristic - pump, blower etc. If something different is driven, obviously things will change.

With the increase in voltage, current decreases from 200A to 189A (a reduction of 5.5%), and PF reduces from 0.88 to 0.86 (a reduction of about 2.27%).

You're overlooking the fact that lower PF means higher current. If the PF had not reduced the reduction in current would be greater.

It is the combination of that 5.5% reduction in current and ~2.3% reduction in PF that means that power remains essentially unchanged

There's no need for an explanation of why the power hasn't changed. You've agreed the increase in speed is negligible, so for a fixed load (eg pump) the change in power is also negligible.

However, as above, I still don't know enough to understand the mechanism of those decreases in current and PF, despite essentially constant load and speed.

I've tried, I don't know what more I can say. But it's been an interesting discussion

 

[fixitflav]

Sometime soon I will be visiting a friend with a workshop & 3ph generator (adjustable), I'll make an attempt to do some tests if I can, this could well be a month or so and very likely to forget by then.

In the meantime, will you explain how the speed control of a CH pump works, if it doesn't use electronics?

 

[JohnW2]

I've been talking about load being equal to motor rated power ...

So I now understand. However, as you say, that really does not affect our discussion, provided only that the output power (whatever it may be) remains essentially unchanged when voltage changes.

You're overlooking the fact that lower PF means higher current. If the PF had not reduced the reduction in current would be greater.

That's obviously true. As you say, if it weren't for the appreciable reduction in PF when voltage increased, current would have to fall even more to maintain a constant power.

I've tried, I don't know what more I can say. But it's been an interesting discussion

It's not your fault that I still don't understand 'the mechanism', but that remains the case!

Kind Regards, John

 

[SUNRAY]

I've not been following this very closely, but you both seem to be saying the same thing - voltage increase, current reduction - assuming the both speed (synchronous) and load remain constant.

Increase the load and the current will increase, because the motor has to struggle more, to maintain sync. Which is why at startup, such motors can draw quite large currents, until they gain sync..

We are both saying the same thing, except I'm saying it is not as predictable as Fixitflav seems to be saying, I have very different observations in real world experience. And I have most certainly known motors which work opposite to this.
Yes of course the current will rise when more mechanical load is applied, the additional energy requirement has to be found from somewhere.

Synchronous and asynchronous (induction) motors have significantly different characteristics, for a start a loaded synchronous motor will stall at the slightest opportunity, especially if running at the lowest 'functional' voltage whereas an asynchronous will lumber on, typically something with varying loads (such as a conveyor belt carrying varying loads) synchronous motors will not normally be used for such purposes. Asynchronous motors always run below any synchronous speed and are far more suited to VSDs, due to being able to follow the frequency changes without loss of control IIUIC.


As mentioned before most of my motor experience is in the control panel/wiring so do not claim to be an expert on them.

 

[SUNRAY]

In the meantime, will you explain how the speed control of a CH pump works, if it doesn't use electronics?

Different windings, although I haven't dismantled one for many years.
Some middle sized pumps have a reversible (2/3/4 position) plug mounted on them to alter the speed/power which doubles as a dedicated VSD connector.

 

[JohnW2]

Different windings, although I haven't dismantled one for many years. ... Some middle sized pumps have a reversible (2/3/4 position) plug mounted on them to alter the speed/power.

Last time I dismantled one (which, as with you, was a long time ago) it was as you describe - a 'switch' which selected different windings and/or a different 'arrangement' of the windings. Certainly no 'electronics'- although 'who knows?' in terms of more modern ones .

Kind Regards, John

 

[Harry Bloomfield]

In the meantime, will you explain how the speed control of a CH pump works, if it doesn't use electronics?

My own guess, plucked out of the air...

ac input, rectified then the dc is converted back to an ac waveform of a frequency to drive the pump at the desired speed.

 

[Harry Bloomfield]

Different windings, although I haven't dismantled one for many years.
Some middle sized pumps have a reversible (2/3/4 position) plug mounted on them to alter the speed/power.

Nope! My pump (around 10+ years old) has a continuously variable adjustment and I think includes an auto position. I assume the auto position in some way checks the pressure, to adjust itself to the demand.

 

[fixitflav]

Different windings, although I haven't dismantled one for many years.
Some middle sized pumps have a reversible (2/3/4 position) plug mounted on them to alter the speed/power which doubles as a dedicated VSD connector.

I don't know whether that's still the case with modern pumps, extract below from Grundfos UPS3 data sheet, but thanks anyway

Features
• Three constant curves/constant speed curves.
• Two proportional-pressure curves.
• Two constant-pressure curves.
• Speed control via a low-voltage PWM

 

[JohnW2]

I don't know whether that's still the case with modern pumps, extract below from Grundfos UPS3 data sheet, but thanks anyway

As I implied, it wouldn't surprise me if modern pumps do have much more sophisticated ('electronic' control - but, as I said, my dissection of CH pumps of old found the same (non-electronic) system as SUNRAY described.

In terms of the discussion we've been having, I imagine that 'all bets are off' as regards predicting how varying voltage will affect current when there is 'electronic control' as part of the equation?

Kind Regards, John

 

[fixitflav]

As I implied, it wouldn't surprise me if modern pumps do have much more sophisticated ('electronic' control - but, as I said, my dissection of CH pumps of old found the same (non-electronic) system as SUNRAY described.

In terms of the discussion we've been having, I imagine that 'all bets are off' as regards predicting how varying voltage will affect current when there is 'electronic control' as part of the equation?

Kind Regards, John

I suppose it depends what the electronic controls are doing. If it's just an inverter set to a fixed speed and voltage (other than 400V, 50Hz) I would guess the behaviour is similar.

 

[JohnW2]

I suppose it depends what the electronic controls are doing. If it's just an inverter set to a fixed speed and voltage (other than 400V, 50Hz) I would guess the behaviour is similar.

Maybe, but I suspect it could be appreciably more complicated than that. Don't forget that what you quoted included:

• Speed control via a low-voltage PWM

... in which case, as I said, I suspect the effect of varying voltage might be difficult to predict (without knowing a lot more). In fact, I wouldn't be surprised if (as with things supplied by SMPSUs - which also 'disobey' Ohm's Law in the manner you have been describing), for a given 'load' (flow rate and pressure) power would remain fairly constant over a wide range of supply voltages - so that, just as you've said of 'uncontrolled' motors, current would fall when voltage increased.

Kind Regards, John