Heads Up - Victron Isolation Transformer

[Dockhead]

Originally Posted by FlyingCloud1937

Are they really the same?

Next



Do you really believe that the blue terminals can stand up to the potential of 450v X 27a = 12,150 watt?

Lloyd

Lloyd, really.

I am the farthest thing in the world from an electrical engineer, but even I understand that a terminal rated like that is not intended to dissipate the full amount of power transmitted. It's job is to transmit the power, not dissipate it -- two entirely different things. It means the wire is big enough and the connectors have enough contact surface so that nothing has to be dissipated in the first place -- it means that this much power can go through it without any significant resistance, so it won't heat up in the first place. Come on now.

That's why they were laughing at you. I think they were unnecessarily rude, but in fairness you walked right into it.

[Dockhead]

Nick, I have a question for you, the answer to which may help other sufferers from this device.

If the breaker is disabled, don't we need to install an external one between the isolation transformer and the shore power cable?

The power in UK marinas is nominally 16 amps * 230 volts but the real voltage is 240 and the breakers on the pontoon power points don't seem to trip much before 20 or even 24 amps or more.

I can limit the power taken from the shore power connection using my Victron inverter/charger, but I'm afraid I will forget and fry the isolation transformer. What do you think? Or do you think that the isolation transformer should tolerate 20 -- 24 amps in case?

Naturally, all AC power runs through an RCD. But this is not intended to function as a circuit breaker, as far as I understand, and I think it's rated much higher, like 48 amps. It would need to be since my generator produces 6.5kW and so will put out more than 28 amps, which could be combined with inverter power for nearly double that, momentarily at least.

Advice?

I have a good ratchet crimper and good waterproof crimps for the jumpers; no soldering for me

[FlyingCloud1937]

Originally Posted by Dockhead

Lloyd, really.

I am the farthest thing in the world from an electrical engineer, but even I understand that a terminal rated like that is not intended to dissipate the full amount of power transmitted. It's job is to transmit the power, not dissipate it -- two entirely different things. It means the wire is big enough and the connectors have enough contact surface so that nothing has to be dissipated in the first place -- it means that this much power can go through it without any significant resistance, so it won't heat up in the first place. Come on now.

That's why they were laughing at you. I think they were unnecessarily rude, but in fairness you walked right into it.

DockHead,

Where did I say that the connectors had to dissipate the heat while conducting.

What Said is it has to stand up to the potential resistance of the circuit. In other words, it doesn't matter in where the circuit resistance is, the connector has to carry the heat.

Now the bigger worry is that the friction fit of the connector won't be able to carry the load, 1. because the thermal protection is gone by the bypass. 2. The friction fit is subject to heat from the resistance caused by the marine environment. 3. The unit is subject to vibration, 4. The friction fit is reduced...due to the annealing of the temper of the original design of the temper of the quick connect, that was in fact based on the temp of the connector.

So don't go down the same road as Jedi, in thinking that I said the connector has to dissipate the full resistance of the load. That is not what I said.

As an aside everyone can laugh at me...but that's not the point.

Have you ever seen a boat fire, a marina fire?

I have pulled a burning man from a building, So if by chance it can cause a fire, then yes I'm for taking precautions.

Lloyd

[mrm]

Originally Posted by FlyingCloud1937

What Said is it has to stand up to the potential resistance of the circuit. In other words, it doesn't matter in where the circuit resistance is, the connector has to carry the heat.

Would you care to elaborate on the 'anywhere resistance' part?

Another thing, let us assume a simple scenario: a socket, some wire, a spade connector, more wire, load (and same back to the socket). Let us asusme this setup creates a series of resistances: 0.03R(socket) + 0.1R(wire) + 0.01R(spade) + 0.1R(wire) + 26R(load) + 0.1R + 0.01R + 0.1R + 0.03R (BTW, this approximates a typical ~2kW heater). Let us further assume - for total simplicity - that our load is purely resistive and we apply 230V DC to the socket.

Can you, please, give me the numbers for the total power dissipated by each component in this setup?

1R == 1Ohm

Further, coming back to your 'anywhere resistance' statement. Please explain, how the heat (dissipated power) of the main load influences the spade connector located, say, a couple of meters distance from this load.

Marius

[FlyingCloud1937]

Originally Posted by mrm

Would you care to elaborate on the 'anywhere resistance' part?

Another thing, let us assume a simple scenario: a socket, some wire, a spade connector, more wire, load (and same back to the socket). Let us asusme this setup creates a series of resistances: 0.03R(socket) + 0.1R(wire) + 0.01R(spade) + 0.1R(wire) + 26R(load) + 0.1R + 0.01R + 0.1R + 0.03R (BTW, this approximates a typical ~2kW heater). Let us further assume - for total simplicity - that our load is purely resistive and we apply 230V DC to the socket.

Can you, please, give me the numbers for the total power dissipated by each component in this setup?

1R == 1Ohm

Further, coming back to your 'anywhere resistance' statement. Please explain, how the heat (dissipated power) of the main load influences the spade connector located, say, a couple of meters distance from this load.

Marius

ok fine you boxed me in

so what you are saying is I can now deduces from the equation of 250v X 16 a = 4000 watt / 12v =333a

that I can use a friction fit connector to carry the load?
bc the potential of the connector doesn't effect the transfer as long as the connector doesn't add any resistance.

and what connector would that be?

what happens when the connector develops resistance?

does the resistance begete more resistance?

my whole point is that,
the point where resistance develops is where fire begins

lloyd

or are you saying that it's all relative?

[goboatingnow]

Lloyd, the key to getting out of a hole is ...
To stop digging

" your not boxed in ", you quite frankly dont understand the theory. I squared R Lloyd look it up. ( joules law plus ohms law )

While inappropriate resistance in a high power circuit can create fires. Theres no statistical evidence to suggest spade connectors are a serious offender. The vast majority of problems are in the crimp. Friction fit connectors , off various types, are extremely common

Secondly such connectors or derivatives of them are used all over the truck, auto, and bus industry, where NVH issues are far far greater them boats ( which suffer comparatively little high frequency vibration. ).

Your understand of UL derating as applied to breakers is also flawed. A 18 amp breaker is not de -rated. What you can get is de rating by changing the breaker to comply with circuit de-rating ( its more complex then that )

The main thing Lloyd that you suffer from is pushing an argument in the face of poor understanding. My experience of installers in general ( and please this isn't about you) is that they often know what to do but don't understand really why they do it. You get quoted lots of code or practice but not the reason why. ( ie the theory) engineers on the other hand are supposed to be " creators " who " should" understand the theory.

The advantage of both is you can apply one set of understanding derived from one set of experiences to another circumstance

Dave

[mrm]

Originally Posted by FlyingCloud1937

ok fine you boxed me in

so what you are saying is I can now deduces from the equation of 250v X 16 a = 4000 watt / 12v =333a

that I can use a friction fit connector to carry the load?
bc the potential of the connector doesn't effect the transfer as long as the connector doesn't add any resistance.

and what connector would that be?

what happens when the connector develops resistance?

does the resistance begete more resistance?

my whole point is that,
the point where resistance develops is where fire begins

lloyd

or are you saying that it's all relative?

Lloyd, I am not trying to box you in. I am trying to understand your point and where you are coming from.

I notice, that you have an important quality of thought of a good field engineer, i.e. you worry about components being less than perfect and components deteriorating over time. And that you are aware of necessity for extra margin (safety factor) due to harsh environment. This is all good.

But... I guess what I am trying to say is.. everything is good in moderation.

I mean, yes to quality components, yes to safety factors etc., but there are limits.

For the spade connector in this example. It will add some resistance. In fact its inherent resistance is a very important factor for its rating specification... But, as long as this connector is used below its maximum continuous ratings (after including all appropriate deratings and factoring in the safety margin) it is all you - as an engineer - can do. The rest is care and maintenance. If an owner or a service tech is negligent.. well.. you can not protect stupid from there own..

Marius

[s/v Jedi]

Originally Posted by Dockhead

Nick, I have a question for you, the answer to which may help other sufferers from this device.

If the breaker is disabled, don't we need to install an external one between the isolation transformer and the shore power cable?

The power in UK marinas is nominally 16 amps * 230 volts but the real voltage is 240 and the breakers on the pontoon power points don't seem to trip much before 20 or even 24 amps or more.

I can limit the power taken from the shore power connection using my Victron inverter/charger, but I'm afraid I will forget and fry the isolation transformer. What do you think? Or do you think that the isolation transformer should tolerate 20 -- 24 amps in case?

Naturally, all AC power runs through an RCD. But this is not intended to function as a circuit breaker, as far as I understand, and I think it's rated much higher, like 48 amps. It would need to be since my generator produces 6.5kW and so will put out more than 28 amps, which could be combined with inverter power for nearly double that, momentarily at least.

Advice?

I have a good ratchet crimper and good waterproof crimps for the jumpers; no soldering for me

Hi Dockhead,

Normally, you should have a dual pole breaker between the inlet and the IT. It should be as close to the inlet as possible. The breaker ashore protects your shorepower cable: if your cable between inlet and IT is better or the same as the shorepower cable and if the shore breaker is in good condition, it is a redundant breaker... but that is two if's... in the US this breaker is mandatory IIRC.

The IT will take 20-24 amps for a short while. The torroidal transformer inside has an internal temperature sensor to protect the unit, which is separate from the two breakers concerned.

Keep in mind the original function of these two breakers: protect the primary transformer windings. If these are in parallel for 120V input, you have a 30A feed that splits over the two windings. For 240V service the windings (and breakers with them) are in series and the 16A feed first travels through winding 1, next through winding 2.

Now, if your wiring is for 30A as for lots of US boats, you will have a 30A breaker here. When you sail to the EU and find 230V, you re-jumper the IT for that and connect to 230V.... but you don't change this breaker to a 16A breaker.... which is okay, because the wiring can withstand that much current, the voltage is irrelevant regardless of what Loyd says. We must use wire and components rated for the voltage because of the insulation value only.

Tue other way around is trickier... The EU boat sails to the US and jumpers the IT for 120V. Now you get 30A instead of the 16A. Your breaker will flip quickly.... but before you replace it with a 30A version, you must carefully check if the wiring you have between dock-post and IT is rated for 30A !!

This sounds so obvious and even stupid to mention here..... but a walk along the dock here, I will see several use a 16A cord with blue EU plug into an adapter cable to 120V 30A outlet. So for some it is not obvious enough yet!

ciao!
Nick.

[Dockhead]

Originally Posted by s/v Jedi

Hi Dockhead,

Normally, you should have a dual pole breaker between the inlet and the IT. It should be as close to the inlet as possible. The breaker ashore protects your shorepower cable: if your cable between inlet and IT is better or the same as the shorepower cable and if the shore breaker is in good condition, it is a redundant breaker... but that is two if's... in the US this breaker is mandatory IIRC.

The IT will take 20-24 amps for a short while. The torroidal transformer inside has an internal temperature sensor to protect the unit, which is separate from the two breakers concerned.

Keep in mind the original function of these two breakers: protect the primary transformer windings. If these are in parallel for 120V input, you have a 30A feed that splits over the two windings. For 240V service the windings (and breakers with them) are in series and the 16A feed first travels through winding 1, next through winding 2.

. . .

ciao!
Nick.

Hmm, odd that my boat didn't have any breaker at the shore power inlet socket. The shore power inlet socket was connected to a breaker at the main electrical power, a kind of combination breaker/transfer switch. I guess UK rules are different.

So, to protect the isolation transformer windings I should install a breaker -- I guess 16 amps -- in a waterproof box between the isolation transformer (which is mounted in the lazarette very close to the shore power inlet socket) and the shore power inlet socket.

Well, I know what I'm going to be busy with next weekend.

[s/v Jedi]

Originally Posted by FlyingCloud1937

Now the bigger worry is that the friction fit of the connector won't be able to carry the load, 1. because the thermal protection is gone by the bypass. 2. The friction fit is subject to heat from the resistance caused by the marine environment. 3. The unit is subject to vibration, 4. The friction fit is reduced...due to the annealing of the temper of the original design of the temper of the quick connect, that was in fact based on the temp of the connector.

So don't go down the same road as Jedi, in thinking that I said the connector has to dissipate the full resistance of the load. That is not what I said.

1. The breakers that I jumpered and which Victron completely removed in current versions, do NOT provide thermal protection. Please read some source that you believe, that thermal breakers are called so because their internal functioning is based on warming up. They only protect against over-current. What differentiates them from magnetic breakers is that a thermal breaker can detect a slight but constant overload.

2. The friction fit might well develop a resistance due to corrosion. The same is true for every other type of connection with terminals.

3. You made that up. If not, post references.

4. You made that up too. Annealing of copper needs a red hot material. When this happens, it is because it failed already and the annealing is irrelevant.

What you said is that a faulty connection has he potential to have to dissipate the full load, 24 hours per day. Now, before that ever happens, you completely ignore the fact that the load of the IT is needed to complete a circuit to start with. Without a circuit, no current will flow. With that load, any heating by power transfer will split over the load carrying components in a ratio defined by their resisance because of I^2 x R. This means that in reality most of he power goes to the load, not the jumper. The full load to the jumper is only possible when all other components have zero resistance, so the load must be a short circuit... which it potentially can be, because one might also put a jumper in an outlet at the moment that the terminal becomes corroded.

Back to serious: yes, things can fail due to corrosion. If that happens, there are many many connections that will fail earlier than these disconnects. Like a 30A connector that sits in the boat inlet. This connection is the #1 point to go bad. The disconnect terminals will go bad before a soldered connection does, but just look at the pics I posted: they are everywhere, you can't eliminate them. And those really are the same Loyd, my pretty blue ones even cost extra.

ciao!
Nick.

[Dockhead]

Originally Posted by FlyingCloud1937

DockHead,

Where did I say that the connectors had to dissipate the heat while conducting.

Well, in Post #14 you wrote:

"Considering that at 230 volt, that connector at the rated voltage will need to support potentially 3680 watts heat dispation...ie 230 v X 16 amps? 24-7"

It's true you didn't say where the heat was supposed to be dissipated, or what you mean by "support". But the heat certainly wouldn't be dissipated at the connector -- no connector, or wire for that matter, is designed to dissipate all the energy which it might be capable of transmitting -- there are entirely different things.

You talked about watts of heat -- what is the relevance of that? At 1000 volts, a thin 16 gauge wire can carry enough current to produce 5 kilowatts of energy over 5 meters -- does the thin 16 gauge wire need to "support" potentially 5000 watts "heat dispation"? What in the world is this supposed to mean? In fact, for the circuit to be properly engineered, the wire (and all of the connectors) need to be rated to safely carry 5 amps of current. Watts, kilowatts, heat, and dissipation are all irrelevant. Heat appears only with resistance which only appears when something is improperly engineered, improperly constructed, or improperly maintained.

Originally Posted by FlyingCloud1937

What Said is it has to stand up to the potential resistance of the circuit. In other words, it doesn't matter in where the circuit resistance is, the connector has to carry the heat.

What?! What in the world do you mean "carry heat"? How does the connector carry the heat if the resistance is somewhere else in the circuit? It seems to me, in my humble non-engineer opinion, that this reflects a total misunderstanding of basic principles of electricity.

Let's say you have a circuit whose whole purpose is especially to produce heat through resistance. Your calorifier, let's say. Let's say it's 1000 watts. It is connected to 230v shore power through a 10 amp breaker and 12 gauge wire. The wire is crimp connected everywhere. How is the connector supposed to "carry the heat"? The wire is rated to carry 10 amps over up to 20 feet -- the real load will be about 4.5 amps. The connectors will be rated at 40 amps, probably. The breaker is set to trip at 10 amps. So 1000 watts of resistance is produced in the circuit, by design. But the whole system will not "carry heat" at all unless it is damaged or improperly crimped or installed. You cannot design a circuit, each part of which is capable of "carrying" the whole "heat" which the power carried by the circuit is capable of producing.

Originally Posted by FlyingCloud1937

Now the bigger worry is that the friction fit of the connector won't be able to carry the load, 1. because the thermal protection is gone by the bypass.

Um, what thermal protection? A thermal breaker doesn't provide thermal protection. It protects against over-current by using a bi-metallic strip which heats up and trips the breaker if the current limit is exceeded. "Thermal" is just the mechanism used.

Originally Posted by FlyingCloud1937

Have you ever seen a boat fire, a marina fire?

I have pulled a burning man from a building, So if by chance it can cause a fire, then yes I'm for taking precautions.

Lloyd

Everyone agrees that fires are horrible, and boat fire are doubly horrible. But it does not seem to me that we can reduce the number of boat fires by spreading basic misunderstandings about the design of circuits on board, and of the role of different protective mechanisms which are provided. On the contrary, very careful design of the circuits based on an exact understanding of the principles involved -- and preferably by professional electrical engineers, and not by amateurs like me or you -- is at the heart of safe electricity on board, in my opinion. Then quality installation, with quality maintenance of the systems.

[FlyingCloud1937]

What do we see here as the common denominator?

How many shore power cords are and receptacles are toasted each year, how many start boat fires?

In the 60's and 70's quick connects were common, then in the 80's they began to disappear.

Now I see them making a come back, especial in places they don't belong. On a boat is one of those places

They are used not because they are sufficient, but because it's a short cut in manufacturing both in costs and labor.

[colemj]

Originally Posted by FlyingCloud1937

What do we see here as the common denominator?

Difficult to say what the common denominator is because all we have is a picture with no information.

If these all came off the same boat at the same time, then I would say the common denominator was a serious problem on the boat unrelated to the connectors.

Mark

[FlyingCloud1937]

Originally Posted by colemj

Difficult to say what the common denominator is because all we have is a picture with no information.

If these all came off the same boat at the same time, then I would say the common denominator was a serious problem on the boat unrelated to the connectors.

Mark

Yes the came off the same boat.

There is a lot of information in the picture, just look at the labels. All of the loads that were of the restive type are burnt. Now look at the cable to the far right galley outlet.

Lloyd

[Dockhead]

Originally Posted by FlyingCloud1937

Yes the came off the same boat.

There is a lot of information in the picture, just look at the labels. All of the loads that were of the restive type are burnt. Now look at the cable to the far right galley outlet.

Lloyd

So, are you trying to say that because a circuit carries a resistance load somewhere, that the "heat" is magically transmitted to the connectors, and burn up the connectors? Sorry, an amp is an amp. If the connectors burned up, it means they were either not properly specified for the current, or not properly installed. I would suspect bad crimps. Dollars to doughnuts, there's nothing wrong with the spade connectors.