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Coffee Machine driven by Inverter

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Ben54b

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Big cables. I love them. I already have 0 guage to the rear wall of the ute. I could put the inverter there I suppose, but that would take up space recruited for amps.
 

vquigg

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I know this thread is dead, but I wanted to explain something for if anyone else comes along and reads it.

A sine wave inverter wattage is measured at the output. This means that a 1500w sinewave inverter, if running a 240V(in Australia) output would allow for 6.25A. Also, a pod coffee machine that runs at 1350-1450W pulls (1350/240=5.62) 5.62A-6.04A. You are all calculating wattage off the 12V DC voltage, which is extremely inaccurate.

If you don't believe me, think of a pod machine plugged into the wall at home. Nothing has changed with the pod machine, it still around off 240 and still pulls the same power. If it does that, at 160A or so, how can the wire from the wall to the machine be such a small cable?

Some pod machines, such as nespresso, have a microprocessor inside them that detects an unregulated power source and prevents too high of a current drawer to prevent cooking it by accident. But if you get a lower cost machine, such as an ALDI one, it should work fine. A deep cycle 100Ah battery will run one of them (if it's pulling water the whole time), in theory, or 20 hours before killing the battery.

Bare in mind, some inverters are different to others and you can't exactly calculate the power draw on the 12V side, but it wouldn't be 160A. If you want to know how much you are drawing, turn the machine on and feel how hot the cable gets, for every 20 degree increase in temp you are drawing double the amount of power. No hot = no low battery.
 

KevinE

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I know this thread is dead, but I wanted to explain something for if anyone else comes along and reads it.

A sine wave inverter wattage is measured at the output. This means that a 1500w sinewave inverter, if running a 240V(in Australia) output would allow for 6.25A. Also, a pod coffee machine that runs at 1350-1450W pulls (1350/240=5.62) 5.62A-6.04A. You are all calculating wattage off the 12V DC voltage, which is extremely inaccurate.

If you don't believe me, think of a pod machine plugged into the wall at home. Nothing has changed with the pod machine, it still around off 240 and still pulls the same power. If it does that, at 160A or so, how can the wire from the wall to the machine be such a small cable?

Some pod machines, such as nespresso, have a microprocessor inside them that detects an unregulated power source and prevents too high of a current drawer to prevent cooking it by accident. But if you get a lower cost machine, such as an ALDI one, it should work fine. A deep cycle 100Ah battery will run one of them (if it's pulling water the whole time), in theory, or 20 hours before killing the battery.

Bare in mind, some inverters are different to others and you can't exactly calculate the power draw on the 12V side, but it wouldn't be 160A. If you want to know how much you are drawing, turn the machine on and feel how hot the cable gets, for every 20 degree increase in temp you are drawing double the amount of power. No hot = no low battery.
Thank you for your very detailed post. It makes the subject much clearer! (y)

I bought a 1500W inverter about 5-6 years ago to run a microwave, kettle, or a toaster while on the road. It's very well travelled, but I've never used it other than to watch TV at home during the SA state wide blackout in 2016 (too scared of killing the battery!)

Thanks again! :)
 

Old.Tony

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You're correct that the current draw on the 12V side is not quite right, because little allowance for inverter loss has been made. However, using the 12V side (based on the wattage, we're not unfamiliar with how electrickery works) is vital to ensure that the 12V cabling is adequate. Let's get the details hashed out.

The better inverters lose around 10% (hardly ever see less than that, I've seen some that lose 20% (80% efficient). I am using a 1500W inverter in my tub with an 85% efficiency rating, but it only runs when the engine is going and I'm only powering 2 fridges (less than 100W total) so I really don't care about the loss.

So the math for the 12V side is a little off. Conservation of energy laws dictate that what comes out must be at least equal (and certainly no more than) what went in. So if an inverter is producing 1500W of output (watts is the measure of energy) and losing (say) 10%, the input is going to have to be (conservatively) 10% higher. Notice that I am only concerned with wattage here, not the amps.

So let's say the inverter is drawing 1600W at 12V to produce 1500W at 240V. 1600W at 12V is a tad over 133 amps (133A x 12V = 1596W). This satisfies the conservation of energy laws. The missing 100W is dissipated as heat (in the power transistors inside the inverter, which usually attach with some thermal paste to the sides of the inverter) and in operating the inverter's fans, lights and circuit.

If you're only going to use a small amount of power you can use lighter cables - like in my case, I'm drawing no more than 100W (it's actually closer to 84W peak, but on average it's less than 40W) so at most I'm not even reaching 10A - my cable is rated to allow 56A continuous (8Ga fig8 cable from Jaycar) - it's fit for purpose.

Calculating the current draw on the 12V side is vital to match cabling that won't start a fire.
 

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