Richard Houlgate Posted December 13, 2023 Author Report Share Posted December 13, 2023 Power Factor - ouch! This seems an easy explanation http://www.smartgauge.co.uk/watts_4.html So if the PF is 0.9 and the cleaner is rated at 1500va it would actually take 1667 watts of power. Or did I misunderstand. (Thinks: A level Physics was 55 years ago!) Link to comment Share on other sites More sharing options...
Jen-in-Wellies Posted December 13, 2023 Report Share Posted December 13, 2023 3 minutes ago, Richard Houlgate said: Power Factor - ouch! This seems an easy explanation http://www.smartgauge.co.uk/watts_4.html So if the PF is 0.9 and the cleaner is rated at 1500va it would actually take 1667 watts of power. Or did I misunderstand. (Thinks: A level Physics was 55 years ago!) That's a good explanation. It makes my brain hurt too, even though my A level Physics was over a decade closer to the present day. Link to comment Share on other sites More sharing options...
nicknorman Posted December 13, 2023 Report Share Posted December 13, 2023 (edited) 39 minutes ago, Richard Houlgate said: One potential installer is keen to use the Victron Easy Plus Compact 12/1600/70, I suspect because he understands it as he has one on his boat. The Technical data gives So I suppose that when on shore power it would transmit 3000w. When on battery it would give a continuous 1600w, but peak would go up to 3000w (depending on the batteries). Or am I wrong? No the 3000w is the short time inverter overload power (a second or two, probably). And 1600va continuous. Not watts! The continuous power is 1300w (or 1200w if it’s hot). When on shore power the transfer limit is specified in amps, not power. And is normally more than 16A 11 minutes ago, Richard Houlgate said: Power Factor - ouch! This seems an easy explanation http://www.smartgauge.co.uk/watts_4.html So if the PF is 0.9 and the cleaner is rated at 1500va it would actually take 1667 watts of power. Or did I misunderstand. (Thinks: A level Physics was 55 years ago!) No the power is the power. 1300w continuous. If the PF is 0.9 then the current is more than might be expected for that power. As the power factor gets worse the output current increases without any increase in output power. So when you multiply the current by the voltage you might get up to 1500. 1500va, not 1500w! If the PF is even worse then the 1500va becomes the limiting factor and max continuous output power will be less than 1300w. Edited December 13, 2023 by nicknorman Link to comment Share on other sites More sharing options...
Ronaldo47 Posted December 13, 2023 Report Share Posted December 13, 2023 (edited) It's the other way round. VA can only equal W for unity power factor (purely resistive loads) . For power factors less than than unity, VA will always be greater than W. Domestic customers normally only pay for actual power, so PF doesn't concern them. Industrial customers usually pay for VA, so it is in their financial interest to get their PF as close to unity as possible. Edited December 13, 2023 by Ronaldo47 typos Link to comment Share on other sites More sharing options...
nicknorman Posted December 13, 2023 Report Share Posted December 13, 2023 31 minutes ago, Ronaldo47 said: It's the other way round. VA can only equal W for unity power factor (purely resistive loads) . For power factors less than than unity, VA will always be greater than W. Domestic customers normally only pay for actual power, so PF doesn't concern them. Industrial customers usually pay for VA, so it is in their financial interest to get their PF as close to unity as possible. Not too sure who you are disgreeing with! But anyway whilst what you say in second para is true for on shore systems, for boats it is more about the maximum current that the inverter can chuck out (internal resistance of the wiring, semiconductors etc) that can be limiting with a poor power factor. I think these days low PF is only likely to occur transiently, eg a vacuum cleaner starting up. Link to comment Share on other sites More sharing options...
Ronaldo47 Posted December 13, 2023 Report Share Posted December 13, 2023 1500W would be 1667 VA at 0.9 PF. Calculating PF is more complex these days with modern switch-mode power supplies that do not draw sinusoidal current. Link to comment Share on other sites More sharing options...
Richard Houlgate Posted December 13, 2023 Author Report Share Posted December 13, 2023 So if I understand Wikipedia correctly a Switch Mode Power supply is one that alters the voltage by switching on and off very rapidly, so it doesn't necessarily take the whole sinusoidal wave form making the idea of power factor irrelevant. Or am I wrong? Not that it matters though, but it's interesting to learn new things. https://en.wikipedia.org/wiki/Switched-mode_power_supply Link to comment Share on other sites More sharing options...
Richard Houlgate Posted January 13 Author Report Share Posted January 13 So to hopefully end the saga, the Heart stopped working and has been replaced by a Sterling Pro Combi https://sterling-power.com/products/pro-combi-s-pure-sine-wave-inverter. Why Sterling when everyone champions Victron? Because the local canal boat electrician prefers Sterling and was dragging her feet over the Victron and I wanted to get the job done. Joy of joys, the thing works! We now have 240v at the plugs again. What to do with the old Heart Freedom Combi 20? If someone wants it for spares and can pick it up from Shrewsbury, or the marina near Nantwich, they are welcome. Otherwise it's going down the tip. 1 Link to comment Share on other sites More sharing options...
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