Comparative Safety: 12v v 24v v 48

[Jammer]

Originally Posted by evm1024

Back to our 12 volt and 48 volt examples.

In the case of the 48 volt 10 ohm HRC fault the power dissipation is 152 watts which creates a very high temperature in a very short time. This high temp is more likely to arc, spark and burn itself into an open circuit is a very short time.

In the case of the 12 volt 10 ohm HRC fault the power dissipation is 14 watts which creates a high temperature but because of the much lower energy does not "blow" itself to bits in short order and thus has time, in some cases a lot of time to catch your boat on fire.

All this really begs the question - is a faulty 12 volt system safer than a faulty 48 volt system? I answer hardly.

I don't agree with this analysis on several points. In practice, the risks are about the same. Here's why.



In a DC circuit with a power source (battery) and a resistive load, a series fault can, at most, dissipate 1/4 of the power the load draws. There's something called the Maximum Power Transfer Theorem that covers this, and it is explained here better than I can do it. (The explanation covers AC systems but also holds true at DC)


The important point is that the series (HRC) fault can deliver exactly the same amount of power regardless of the supply voltage, given a load of equal wattage.


I've dealt with series (HRC) faults, in practice, and they start to become a frightening thing when the load is in the several hundred watts range. Below that they are usually more of a nuisance because the heat gets conducted away fast enough that nothing lights up.


So again it is the large loads you have to worry about and the advantage of the higher voltage is that you can keep the wire sizes manageable and your likelihood is better of having a high-quality, low-resistance connection -- one that stays that way given the realities of the environment aboard.

[ramblinrod]

[QUOTE=evm1024;2827845]Well, at least we are making progress. Rod had come to understand that a 24 volt system may not be less safe.

Error 1.

I made no such statement.

The statement I clearly made was...

"The 24 Vdc system is not "safer" than the 12 Vdc system. In fact, it may be "less safe", for the other reasons previously stated."

Error 2.

I have not just "come to understand" this; I have known it for about 45 years.

Please try to do a much better job presenting the positions of others.

IMHO, to not take proper care in this regard is highly disrespectful.

[evm1024]

[QUOTE=ramblinrod;2828884]

Originally Posted by evm1024

Well, at least we are making progress. Rod had come to understand that a 24 volt system may not be less safe.

Error 1.

I made no such statement.

The statement I clearly made was...

"The 24 Vdc system is not "safer" than the 12 Vdc system. In fact, it may be "less safe", for the other reasons previously stated."

Error 2.

I have not just "come to understand" this; I have known it for about 45 years.

Please try to do a much better job presenting the positions of others.

IMHO, to not take proper care in this regard is highly disrespectful.

Ad hominem attacks are not really necessary. Let's confine ourselves to the ideas and leave the ego and bluster out of the conversation. I pledge to do so. And you?

You have clearly stated that higher voltage systems are less safe than 12 volt systems. Two of the higher voltage systems discussed on this thread are 24 volts and 48 volts.

You have stated that they are not safe so much that at least one person stated that you expressed your views 20 times.

you stated and have quoted yourself above:

Originally Posted by RR

"The 24 Vdc system is not "safer" than the 12 Vdc system. In fact, it may be "less safe", for the other reasons previously stated."

Your first sentence states that 24 volts systems are not safer than 12 volt systems. And in plain English this means that 24 volts systems may be as safe as 12 volt systems.

Further in your second quoted sentence you state that 24 volts systems MAY be less safe than 12 volt systems. And in plain English this means that 24 volt systems could be just as safe as 12 volt systems.

Further you state:

Originally Posted by RR

I have not just "come to understand" this; I have known it for about 45 years.

This seams rather at odds with your prior statement. First you say that I mis-stated your position (saying that 24 volt systems may be as safe as 12 volt systems) then you say that you have known this for 45 years.

Lastly, my statement which you disagree with is:

Originally Posted by evm1024

Well, at least we are making progress. Rod had come to understand that a 24 volt system may not be less safe.

And I do think it is progress that you are now willing to concede that 24 volt systems MAY be as safe as 12 volt systems. MAY is a far cry from NOT.

[evm1024]

I note that the examples given to show the effects of various circuit faults in 12, 24 and 48 volt systems used a number of different loads along with a few in series fault resistances.

In my case I used a 1000 Watt windless. In another case a 1000 watt heater was used.

I know of many boats that have windless in the 1000 watt range. But, I cannot for the life of me think of any boat that has a 1000 watt heater.

A quick google search produced a number of 12 volt heaters in the 600 watt range. These heaters are in the 10,000 BTU range and as you might guess are power hogs. Running one off batteries runs them flat in just a few hours. In addition it does not make sense to run them off of an engine or generator based power source.

In the case of an engine based power source (alternator) you would be loading up the alternator for very little heat (10,000 btu). Also, of course with the engine running you would normally use the coolant loop for heat.

In the case of a generator I think we all know that they are in almost every case designed to produce AC in the voltage of our mains. Thus we would be using 120 volt or 240 volt heater if we were going to use a heater at all.

Why you might ask am I going on about this? The answer lies in time.

In the case of a 1000 watt windless should you have a high resistance connection somewhere in the circuit the windless will not run or would run very poorly. Because a windless is under direct manual control the operator would note the failure right away and say "Oh Crap" I have a fault. Of course they would push the button a few more times but still they would know that there is a problem and go looking for the cause of the fault. And thus there is a fairly small chance that the faulty connection would burn up and start a fire.

In the case of a 1000 watt heater one would be likely to turn it on and just forget about it for some time. Most would not even notice that is was putting out no or much lower heat than normal. The faulty connection would go undetected for some time and has a much greater chance of starting a fire.

But as noted, there really are not any 1000 watt 12 volt heaters that I could source. And given the limited 12v (24, 48) power on a boat even if one were able to source such a heater there are many many more options that provide more heat in a much more realistic way.

Also I might add that due to its continuous operating nature a 1000 watt 12 volt heater is much more like a propulsion motor that a windless.

And I do not think that anyone is willing to argue that a 12 volt sailboat propulsion motor is safer than a 48 volt propulsion motor. (4.8 kW at 36 volt in Sailing Uma's case)

Clearly, the case of any continuous high wattage device such as a heater or propulsion motor is safer and best served with higher voltages that 12 volts.

[CatNewBee]

12V systems are preferred because of the available batteries, originaly 6V out of 3 cells were used in cars start batteries, but with larger engines 12V was needed to keep the current under control.

Large series of cells have their specific problems with balancing and overcharging, in trucks 24V are common because of the higher current demand.

In e-mobility much higher voltages are common.

in the bigger rest of the world 240V/380V are common installations in households, even boatsnhave 110V/240V installations.

Nothing wrong with 48V, beside there is very little gear available for this voltages. My solar panels yeld 52V Vmp / 60V Voc, great for MPPT controller.
You can touch the wire and you'll feel the current, but the chock is not serious. 240V feels different.

Another remark regarding DC current and AC inverter. RMS is not higher onbthe DC side, it is lower or equal to the VA on the inverter and higher then the Wattage due to conversion losses. Inverter use high frequency switching ro transform the voltage and then create the sine wave. On the input side they use capacitors, that compensate the peaks. you can subtract 20% of the AC VA rating to calculate effective wattage, 5000VA inverter yelds 4000W DC equivqlent power and would draw 350A DC (4000/12)/0.95 because of 95% efficiency of the inverter.

[Dockhead]

Originally Posted by evm1024

. . . Clearly, the case of any continuous high wattage device such as a heater or propulsion motor is safer and best served with higher voltages that 12 volts.

Why "clearly"?


As a layman, I can't get any clear picture of the objective reality of the question, from what I've read in this and the other thread.


Someone said that the energy released in a high resistance connection is a function of the square of the voltage, so higher voltage means much greater risk in case of a high resistance connection.


Someone else said it's actually the square of the current, so on the contrary, the higher voltage system will be safer in case of a high resistance connection fault.


Then Jammer appeared to tell us that because of the Maximum Power Transfer Theorem, the energy released in a bad connection is always a function of the POWER of the load, so the voltage of the system doesn't make any difference.




These theses can't all be correct! So what is it? Enquiring minds would like to know.



If Jammer is right (and he sounds pretty convincing to me, but what do I know), then having a lower system voltage would not seem to have any advantage at all, other than perhaps less shock risk, and on the contrary, a higher system voltage should mean that it is easier to produce robust and reliable connections, because the wires are smaller and easier to secure, and currents handled are less.

[CatNewBee]

Yes, there is a lot of fake news inside this thread.

Here the facts:

Voltage alone is irrelevant for the resistance, it is always the Voltage / Current ratio. R = U / I and R = constant on a given connection of wires and linear elements.

Now to the other side P = U * I, with P = power, U = Voltage and I = Current, what means for a constant power output, if you double the voltage, the current is half.

Now to the voltage drop on a connection. Assuming resistance of 10 mOhm results in 1V/100A or 1V voltage drop at 100A current. (R=U/I)

Assuming we have a Load of 1000W or 1kW we want to operate. (P=U*I)

We start at 10V, I= P/U = 100A for the load, that also floats through our line resistance and causes a Voltage drop of 1 V at 100A, right?, so the load receives 9V at 100A = 900W and the line burns 1V*100A in heat = 100W

Now we use 100V, same setup. I= P/U = 10A, according to the resistance U = R*I = 0.01 Ohm* 10A = 0.1V voltage drop.
The load receives 99.9V at 10 A and yields 999W, the connection burns 0.1V * 10A = 1W

Lets do it for 1000V. I=P/U = 1A, Voltage drop R*I = 0.01Ohm * 1A = 0.01V, Power at the unit = 999.99V * 1A = 999.99W, power loss = 0.01V * 1A = 0.01W

You got the point? It is always the current that makes trouble along the line.

U = R*I is the voltage drop, Power loss = U * I (voltage drop * current), by replacing the voltage drop with R*I, the power loss

P = (R*I) * I = R * I^2

So the power loss is a function of the current I^2, because R is constant.


I hope, this explains it mathematically and sorts out errors.
You want the currents as low as feasible without risking a shock, so the maximum considered safe limit is 50V DC, above that you need better isolation and touch protection / residual current circuit breakers (RCCB) and the like.

We can also use the two formulae to substitute the current and show the relation of power loss to the voltage drop:

P = U*I and R = U/I or I = U/R
P = U * (U/R) = U^2 / R

So yes, there is also a U^2 relation, but this is not the system voltage, it is the voltage drop over the line and it relates only to the current and is decreasing in SQR to the higher system voltage, because of proportional decreasing of the system current.

[Dockhead]

Originally Posted by CatNewBee

Yes, there is a lot of fake news inside this thread.

Here the facts:

Voltage alone is irrelevant for the resistance, it is always the Voltage / Current ratio. R = U / I and R = constant on a given connection of wires and linear elements.

Now to the other side P = U * I, with P = power, U = Voltage and I = Current, what means for a constant power output, if you double the voltage, the current is half.

Now to the voltage drop on a connection. Assuming resistance of 10 mOhm results in 1V/100A or 1V voltage drop at 100A current. (R=U/I)

Assuming we have a Load of 1000W or 1kW we want to operate. (P=U*I)

We start at 10V, I= P/U = 100A for the load, that also floats through our line resistance and causes a Voltage drop of 1 V at 100A, right?, so the load receives 9V at 100A = 900W and the line burns 1V*100A in heat = 100W

Now we use 100V, same setup. I= P/U = 10A, according to the resistance U = R*I = 0.01 Ohm* 10A = 0.1V voltage drop.
The load receives 99.9V at 10 A and yields 999W, the connection burns 0.1V * 10A = 1W

Lets do it for 1000V. I=P/U = 1A, Voltage drop R*I = 0.01Ohm * 1A = 0.01V, Power at the unit = 999.99V * 1A = 999.99W, power loss = 0.01V * 1A = 0.01W

You got the point? It is always the current that makes trouble along the line.

U = R*I is the voltage drop, Power loss = U * I (voltage drop * current), by replacing the voltage drop with R*I, the power loss

P = (R*I) * I = R * I^2

So the power loss is a function of the current I^2, because R is constant.


I hope, this explains it mathematically and sorts out errors.
You want the currents as low as feasible without risking a shock, so the maximum considered safe limit is 50V DC, above that you need better isolation and touch protection / residual current circuit breakers (RCCB) and the like.

OK, so you're in the "the danger is a square of the current" camp, so the higher the voltage, the safer, but for the risk of shock.


So why is Jammer not right?


Or maybe you're both right? He's describing what happens at the LIMIT, and you're describing what happens before you get there. He says the power is limited to 1/4 of the power of the load, and all of your examples are less than that.

[CatNewBee]

Define: DANGER.

1. DANGER of Power loss: YES, higher voltage leads to lower current and to better, loss-less transfer to the loads, so better efficiency.

2. DANGER Risk of fire: Higher voltages lead to lower power loss and to less heat at connectors, wires etc., so safer installations / or cheaper components usable.

3. DANGER Risk of shock: Higher voltages cause higher currents, it is the current that kills you in the end, so yes, risk for lethal touching life-wires rises, 50V cannot drive lethal currents in a healthy body, but higher voltages can, so there is a limit by our biology (our body conductivity).

[Dockhead]

Originally Posted by CatNewBee

. . .

2. DANGER Risk of fire: Higher voltages lead to lower power loss and to less heat at connectors, wires etc., so safer installations / or cheaper components usable.


. . . .

This is the one which is controversial.


But you are assuming that the resistance in the connector is the same in all cases. But in reality, the higher voltage system will have smaller connectors, won't it? Because it CAN?


I guess (assuming you are right about the basic principles, and other information posted on here is wrong) that higher system voltage will allow you to more easily oversize and overengineer the connectors, so that they are safer. But comparing standard systems with standard solutions, wouldn't there be less of an advantage than you are stating, for higher voltage systems?

[transmitterdan]

The maximum power transfer theorem is correct. It states that if you have a voltage source with a fixed constant internal resistance and you wish to obtain the maximum power from said source then the optimum load resistance is equal to the aforementioned internal resistance.

It means that as series resistance increases due to a wiring fault the power wasted in the fault will increase until it is burning 1/4 the power of the original load and the original load will also be at 1/4 power. After that any further increase in series fault resistance will decrease the power in the fault resistance.

But this theorem does not work in reverse. I often trip up new engineers seeking employment by asking the question in reverse. If we have a given load resistance what should we choose for the internal resistance of the voltage source to achieve maximum power transfer? About 50% of respondents say make the internal resistance of the voltage source equal to the load. But that is a terribly wrong answer.

Consider that the maximum power transfer theorem means that at maximum power, equal internal and external resistances demand that the system efficiency is 50%. So if a series high resistance should become equal to the load resistance the power to the load will drop to 1/4 the normal power. Most high current loads will automatically shut off long before the power drops that much. Inverters and windlass motors will cut off long before that. So all of this to say the MPT theorem is of limited applicability in determining what voltage is "safest".

[Dockhead]

Originally Posted by transmitterdan

. . . Consider that the maximum power transfer theorem means that at maximum power, equal internal and external resistances demand that the system efficiency is 50%. So if a series high resistance should become equal to the load resistance the power to the load will drop to 1/4 the normal power. Most high current loads will automatically shut off long before the power drops that much. Inverters and windlass motors will cut off long before that. So all of this to say the MPT theorem is of limited applicability in determining what voltage is "safest".

OK, but by the time that happens as the result of a connection fault, the boat will already be in flames, no? I wouldn't think it would take many watts of power to combust a connector weighing a couple of grams -- no?


All this is making me think that it really makes sense to use large and hefty connectors, with large contact surface and significant mass. And higher system voltage? Do you have a view on this, the main topic of this thread?




Bit of thread drift, but all this also makes me think that the UK domestic power plugs may be a really good thing -- they are massive, and with a really large contact area compared to the Continental round pin Shuko plugs or the ultra cheap, ultra flimsy American ones. I used to think the UK ones were quite clunky . . .

[CatNewBee]

Originally Posted by Dockhead

This is the one which is controversial.


But you are assuming that the resistance in the connector is the same in all cases. But in reality, the higher voltage system will have smaller connectors, won't it? Because it CAN?


I guess (assuming you are right about the basic principles, and other information posted on here is wrong) that higher system voltage will allow you to more easily oversize and overengineer the connectors, so that they are safer. But comparing standard systems with standard solutions, wouldn't there be less of an advantage than you are stating, for higher voltage systems?

Not necessarily.

You can make every system safe, regardless of the voltage, but with higher currents you need more tough gear. Because the exponential (sqr) effect of the current, it means the lower the voltage, the bigger the cross-section for anything.

On the other hand, from 1000V on there is a risk of sparks across wires and with rising voltages, the distance and isolation must be better to prevent unwanted currents across the air or through the isolation. So this defines the upper limit.

And there is the human body with its conductivity depending on moisture and other factors. You do not want a current through your heart of over 0.1A. So assuming you are average and have a resistance of 2kOhm,
U = R*I leads to a Voltage of 2000Ohm * 0.1A of 200V, if your skin is wet and your body a good conductor, you may have 1kOhm resistance, and then 1000Ohm * 0.1A = 100V may kill you. It always depends where the current goes through, what the resistance is and if the heart is in the pass.

50 V would induce in the last case 0.05A, in the first case 0.025A, so more or less a painful chock, but nothing serious, except you are soaking wet, and grab a wire in each hand, so you have the perfect pass through, or you connect the wires on your ears.




https://en.wikipedia.org/wiki/Electrical_injury

[CatNewBee]

Originally Posted by Dockhead

OK, but by the time that happens as the result of a connection fault, the boat will already be in flames, no? I wouldn't think it would take many watts of power to combust a connector weighing a couple of grams -- no?


All this is making me think that it really makes sense to use large and hefty connectors, with large contact surface and significant mass. And higher system voltage? Do you have a view on this, the main topic of this thread?




Bit of thread drift, but all this also makes me think that the UK domestic power plugs may be a really good thing -- they are massive, and with a really large contact area compared to the Continental round pin Shuko plugs or the ultra cheap, ultra flimsy American ones. I used to think the UK ones were quite clunky . . .

You can start a fire by applying pressure on a crystal, see electric lighters. No batteries whatsoever. You can ignite things with a 1.5V battery. Even wth a flint-stone you can start a fire, or chemically by using substances that cause exothermic reactions.

Life is dangerous. For any power range and load there are metrics to run the gear safely. As long as you stick on the safe side when dimension you circuits, all will be OK, same for Isolation.

Regarding the plugs. I think UK and US plugs are less safe because of the exposed posts when sticking into the outlet, the (current) EU plugs prevent exposure of metal when the plug hits the contact.

Regarding the loads / currents, all this connectors are dimensioned to manage a specific current, it is not only the size of the posts, but also the contact surface and the cleaning effect, that scrubs the contacts clean and free of oxidation when plugging in, also the used contact materials and the contact to the wires on the back. There are good ones and poor ones in all systems.

BTW, I assume you even have a safe installation on several thousand volts on board, in your outboard engine between the coil and the spark plug, same in your car if you have a gas engine. If you have xenon headlights, there are also high voltage wiring's. You can handle any voltage when you do it right.

[GordMay]

Some of the 42V Engineering Challenges faced by Automakers in the 1990s:
- Some fuse panel and harness makers have found that common 14V mini- and maxi-fuses do not behave properly at 42V. They can fail to interrupt excessive currents properly, causing serious overload conditions.
- At 42V and higher power levels, many components, such as wires and relays, experience electrical stress that is three times higher than before. With higher stress, components tend to break down more often.
- Recent research shows that 42V arc energy is 50 to 100 times higher than in a 14V system. Such arcing can generate temperatures up to 1800F, ignite fuel vapors, start a fire in plastic insulation, and even melt metal. Simply redesigning relays, switches, and fuses for higher voltage and using flame-retardant materials is not a total solution; these component designs should suppress arcs.
- Electromechanical components such as alternators, motors, and starters may require more time on field coil winding machines to get the same number of ampere-turns (given that the current and wire gauge will be one-third of what it was for 14V devices).
- Additional production testing may be required to verify arc suppression and EMI/EMC compliance.