Honda EU10 generator - OK to use like this??

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ceejay22

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I have a Quick calorifier with a 1200W element, which I'd like to be able to run from my EU10 Honda generator. The generator has a 900VA continuous rating.

I'm considering inserting a series diode in the supply from the generator. This will cut the heating load to 600W by only using one half of the AC waveform. But it's an asymmetric load and I'm uncertain whether this would be a problem for the generator. I've asked our local Honda office but they were no help at all - I don't think the guy I spoke to even understood the question. Any ideas?
 
I really can't see that working!!!
The diode will cut the load to 600W but it will also mean that you can only use half the waveform output from the generator, ie 450VA. The other half of the generator output will get dissapated as heat by the diode - so you end up with half the output heating air and half heating the water!!!

Best solution is to source a lower wattage heating element, most marine calorifiers use standard domestic heating elements so a trip to the local plumbers merchants or an internet search should find a 600W element.
 
I would stay well away from the diode. The calorifier will still be wanting to draw 1200W but the generator will be limited to 450W. I would just connect it as normal. If the load is too great either the voltage will drop and you will only get 900W of heat or the overload protection will cut in and you will get 0.
 
Diodes in the reverse direction dissipate nothing. 0 amps x any voltage = 0 Watts

In the forward direction they dissipate about 0.6 volts x 4amps = 2.4 watts but only for 1/2 the time = 1.2 watts average

4 amps comes from 4amps x 240volts = 960 watts - near enough to 1000watts for this discussion.

For other EEs - I know that these are approximations, but plenty god enough for this discussion.

Since the heater element is a resistor it will not care about the waveform.
I think that these generators are a modern inverter type. The alternator or it's inverter may object to the half-wave load waveform but is unlikely to be damaged.
 
Three options:
1) Get a bigger generator
2) Change the element for a 500 or 600w one (http://www.heatrod.com/)
3) Suck it and see but don't come crying when you have wrecked your over priced Honda and voided the warranty.

The invertor will not be happy with you cancelling ½ the wave form nor will the DC side as it will be cycling to keep up with the imbalance loading. How do I know this? . . . . . . Guess :mad:
 
Dimmer switch?

I don’t think the diode will work as you will be at the full power of 1200 w during one half of the cycle.

What about a dimmer switch as used in lighting. The load from the heater is purely resistive and similar to incandescent lights.

JC
 
Jcorstorphine said:
I don’t think the diode will work as you will be at the full power of 1200 w during one half of the cycle.

What about a dimmer switch as used in lighting. The load from the heater is purely resistive and similar to incandescent lights.

JC
Click to expand...

Most are thyristors so in effect variable diode and in normal use reducing the power out, also most are only rated for 250 watts so may get warm at 900.

A lower output immersion is probably the cheapest option

I sized my suitcase generator to allow the immmersion and fridges and battery charger to run together
 
Jcorstorphine said:
1000 watt here

http://www.theinternetelectricalsto...dimmer-light-switch-matt-chrome-hs9-128-p.asp
Click to expand...

Found a better one.

http://www.ebay.co.uk/itm/POWER-CON...ial_Automation_Control_ET&hash=item2a2307dc84

The other option is to fit a high power resitor in series with the heater as follows

Volts Power Amps Resistance
240 1200 5 48
240 900 3.75 64

At the moment your heater has a resistance of 48 ohms so you need to increase the total resistance to 64 ohms by adding a 16 ohm (ish) resistor in series which would dissipate 225 watts so it needs to be a chunky unit.
 
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ytd said:
I would stay well away from the diode. The calorifier will still be wanting to draw 1200W but the generator will be limited to 450W. I would just connect it as normal. If the load is too great either the voltage will drop and you will only get 900W of heat or the overload protection will cut in and you will get 0.
Click to expand...

I tried that approach and it didnt work. The load on the genny slowed it down so that it wasnt outputting its rated output at all. Result it took ages to make the water at all warm.

This outfit do a 750 watt immersion heater element. Not that difficult to change since the boss is standard domestic.
 
Same generator on The Kipper, so changed the element to 750w which gives enough spare to run the battery charger. The water takes longer to heat up but no problem in a marina as it's always on, using the genny it's also OK as we need to charge the batteries.
 
Thanks for the replies.

Consensus seems to be the diode approach isn't worth the risk, given the cost of a new generator. I guess I just have to cough up for a new element. I've been quoted $140 for an 800W or 500W replacement. Can anyone can suggest an Australian supplier that will do it for less? All I've found so far from local domestic suppliers are 2-3kw units.
 
why is it that generators nowadays express their output as a va figure rather than the w/kw that we want to know about?
from my memory, rather than understanding, i was indoctrinated to believe that p=iv and v=ir, where i = current, yet it seems that va is greater than wattage.
does va not stand for volt x amp?
i was just wondering why the op has an inadequate generator and thought that the new output expression may have been a contributing factor.
this looks like a solution http://www.deluxeproducts.com.au/Enduro-VX7000i-Commercial-Power-4.0kva-Silent-Digital-AVR-Generator
 
I think the difference between VA and W relates to power factor, but others will be better able to explain than I.

My generator was originally bought for top-up battery charging in the fairly rare times where the solar panel can't cope. It works very well, hooked up to the shore supply which has a 50A charger. I don't need it for other AC purposes as I already have a 2kw inverter. I bought the lightest and most compact I could find as it spends most of its time at the bottom of a deep locker.

The water-heating role wasn't originally planned.
 
If an AC source (generator) is feeding a resistive load (immersion heater) then VA = Watts. In this case the current directly tracks the waveform of the voltage. The peak current happens at the same time as the peak voltage.

If the load is partly or completely inductive ( motor ) or capacitive (?) then the current waveform is either ahead of or behind the voltage waveform. Because the peak current and peak voltage no longer happen at the same time then the power (Watts) is lower. But VA stays the same. This is measured as power factor.

So for the immersion heater 250 volts x 4 Amps dissipates 1000 Watts

For a motor 250 volts x 4 Amps x power factor of 0.8 dissipates 800 Watts but has a VA of 1000.

The generator rating is for VA, you cannot cheat and get 1000W at PF=0.8 from a 1000VA generator.

PF in this text is purely for this description. In practice the PF of a motor will vary between start-up, light load and heavy load.
 
ceejay22 said:
Thanks for the replies.

Consensus seems to be the diode approach isn't worth the risk, given the cost of a new generator. I guess I just have to cough up for a new element. I've been quoted $140 for an 800W or 500W replacement. Can anyone can suggest an Australian supplier that will do it for less? All I've found so far from local domestic suppliers are 2-3kw units.
Click to expand...

Still think you should look at the Thyristor Controller on Ebay. The cost is £20 and they will ship to Oz for about £10. The unit can vary the power from 0 to 100% so you can tune the unit to suit the generator i.e. set it 0%, start the genny and slowly increase the % until the engins sound as if it is struggling, back it off (say 15%) Stop the genny and see if it will cope with a cold start.

Link to Thyristor supplier.

http://go.redirectingat.com/?id=635...://www.ybw.com/forums/showthread.php?t=331521
 
Jcorstorphine said:
Still think you should look at the Thyristor Controller on Ebay. The cost is £20 and they will ship to Oz for about £10. The unit can vary the power from 0 to 100% so you can tune the unit to suit the generator i.e. set it 0%, start the genny and slowly increase the % until the engins sound as if it is struggling, back it off (say 15%) Stop the genny and see if it will cope with a cold start.

Link to Thyristor supplier.

http://go.redirectingat.com/?id=635...://www.ybw.com/forums/showthread.php?t=331521
Click to expand...

He doesnt need to restrict the power of the generator so much as to restrict the demand of the calorifier element.
 
Chalker said:
If an AC source (generator) is feeding a resistive load (immersion heater) then VA = Watts. In this case the current directly tracks the waveform of the voltage. The peak current happens at the same time as the peak voltage.

If the load is partly or completely inductive ( motor ) or capacitive (?) then the current waveform is either ahead of or behind the voltage waveform. Because the peak current and peak voltage no longer happen at the same time then the power (Watts) is lower. But VA stays the same. This is measured as power factor.

So for the immersion heater 250 volts x 4 Amps dissipates 1000 Watts

For a motor 250 volts x 4 Amps x power factor of 0.8 dissipates 800 Watts but has a VA of 1000.

The generator rating is for VA, you cannot cheat and get 1000W at PF=0.8 from a 1000VA generator.

PF in this text is purely for this description. In practice the PF of a motor will vary between start-up, light load and heavy load.
Click to expand...

Thanks for that - I understand. But reversing your argument, doesnt that mean that with a VA of 1000 from the alternator in the Honda, he might well have wattage higher than 1000. In other words is the current and voltage in phase as they come out of the alternator? If not why not since an alternator and a motor are really the same thing.
 
When the voltage and current are in-phase watts and VA are the same.
When they are out of phase you have a PF of <1, Watts are less than VA
A PF>1 does not exist.

Watts = V x A x PF , PF can have a value between 0 and 1

The phase is normally measured at the generator output.
 
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