Comparative Safety: 12v v 24v v 48

[evm1024]

As both Nick and I noted the total current is limited by the total resistance. And thus the power drops.

In the 12 volt 1000 watt system we get 83.33 amps (1000/12). The current is limited to 83 amps by the loads resistance of 0.144 ohms (12/83.33).

Thus the total current and power through the circuit when we have a 0.144 ohm load and a 10 ohm series fault resistance is (12/10.144) 1.18 amps and (12*1.18) 14.16 watts.

Our 1000 watt load only gets 0.2 watts of that 14 watts with the rest dissipated in the 10 ohm fault.

But, as I noted the 10 ohm fault appears to be an unreal value. HRC (high resistance contact) values appear to be in the 0.25 ohm range. (citation needed)

Running that math for a more realistic faulty connection resistance of 0.25 ohms we get:

12 volt system - 231 watts in series fault, 133 watts to load for a total wattage of 365 watts in our nominal 1000 watt circuit.

48 volt system - 88 watts in series fault, 813 watts to load for a total wattage of 902 watts in our nominal 1000 watt circuit.

48 volts delivers more watts to the load with less watts to the fault in this fault condition.

Does that help?

[ramblinrod]

Originally Posted by CatNewBee

that is true, same power, same contact, higher system voltage leads to less voltage drop.

This is incorrect.

As can be seen in posts # 1 and 5.

One has to perform a complete circuit analysis of the load AND contact resistance.

Everything else equal, the voltage drop across the contact increases with resistance.

To determine whether the voltage drop across that contact will either increase or decrease with a change in system voltage, one has to perform a circuit analysis considering the load resistance AND contact resistance.

[Kelkara]

Originally Posted by Dockhead

Since physics are "the law", there must be a single right answer to these questions.

I'm sure you understand the "law" ... these graphs show how the power in the load and the resistive contact will vary with the resistance of the contact for 12V and 48V. As you can see the graphs are almost exactly the same, with the peak occurring where the resistance of the bad contact equals the resistance of the load ... The only difference is that the peak occurs at a 16x higher resistance ... but the shape of the graph is exactly the same, with the same power dissipated.

Since the discussion is about safety ... I'm going to ask the question from a different angle ... assuming that the equipment was properly installed and so working correctly originally (OK, I admit that this is a big ask) ... that means that our bad contact is deteriorating.

When are we going to notice that we have a problem? Either when we notice the contact getting hot, or when we notice the load is starved of power ... in both the 12V and 48V system it is the same ... the power dissipated by the bad contact will follow the same curve for the loss of power in the load. The only difference is that by this time the bad contact will have deteriorated 16x as much in the 48V system than the 12V system.

So which is safer? the 12V system where the problem becomes apparent earlier, or the 48V system where the early stages go unnoticed? Either way we have exactly the same problem to fix once we do notice it. Personally I think that the difference is in the original installation, which for such a high current load would be easier to get right with 48V.

[ramblinrod]

Originally Posted by evm1024

As both Nick and I noted the total current is limited by the total resistance. And thus the power drops.

Hmmm, whoda figured?

Glad to see ya comin' round EVM1024! ;-)

[transmitterdan]

Be careful when quoting Ohm’s law. It applies to a constant real resistance. Bad connections are not ever constant resistance. So we should not get too smug about how much we “know” about faults. Ohm’s law applies to a functioning system but is not really applicable when there are loose hot connections and arcs.

[ramblinrod]

Originally Posted by Dockhead

That seems very true!


But how then to explain this statement of yours:


[I]


Isn't that a contradiction?


Why would the power dissipated by the contact be more, if the power of the load is the same?

No contraction.

The power consumed by a load is based on its resistance, the corresponding voltage across it, and the current through it.

A 12 Vdc 1000 W heater, does not deliver 1000W of heat, if there is only 10 Vdc supplied.

If the voltage across the load is less, the current through it, and power consumed by it is less. Has to be. I = E/R and P = E*I.

As soon as we introduce contact resistance, we get less voltage across the load.

The higher the contact resistance, the lower the total circuit current, but the greater the voltage drop across the contact vs the load.

[CatNewBee]

Originally Posted by ramblinrod

No contraction.

The power consumed by a load is based on its resistance, the corresponding voltage across it, and the current through it.

A 12 Vdc 1000 W heater, does not deliver 1000W of heat, if there is only 10 Vdc supplied.

If the voltage across the load is less, the current through it, and power consumed by it is less. Has to be. I = E/R and P = E*I.

As soon as we introduce contact resistance, we get less voltage across the load.

The higher the contact resistance, the lower the total circuit current, but the greater the voltage drop across the contact vs the load.

Resistance does not change, nor does the system voltage, so for the same power same circuit, 4 times the voltage leads to 1/4 of the current and 1/4 of the voltage drop, causing 1/16 of the power loss.

[ramblinrod]

Originally Posted by transmitterdan

Be careful when quoting Ohm’s law. It applies to a constant real resistance. Bad connections are not ever constant resistance. So we should not get too smug about how much we “know” about faults. Ohm’s law applies to a functioning system but is not really applicable when there are loose hot connections and arcs.

When one applies Ohms law, it is by very nature for a fixed voltage, current, and resistance.

Ohm's law has to be re-applied and recalculated as any variable changes.

(Which is what is/was catching most people in this thread.)

If just an increase system voltage, the circuit current through the fixed load resistance would increase proportionally.

But for the example case, we are increasing system voltage while maintaining the same power dissipation at the load, so we have to increase it's resistance to reduce current.

Still all fine.

But as soon as we introduce an increasing contact resistance, now the load is receiving a reduced proportion of the supply voltage, the circuit current is decreasing, the power dissipated by the load is dropping, all the while voltage drop across the contact increases, to a point.

At this point, the voltage drop across the contact is max.

Any further increase in contact resistance decreases circuit current to the extent that the voltage drop across the contact falls.

This can all be seen in the table of post # 1 and 5.

[KTP]

He is right.

If you have a 1 ohm pure resistive load fed by 12V, you draw 12 amps from the battery. The load dissipates 12 watts. A good contact in this series circuit with 0.001 ohm resistance would dissipate a negligible 0.14 watts.

If the contact resistance in the series circuit to that 1 ohm load is also 1 ohm, you draw 6 amps from the battery. The load has 6V across it and the contact has 6 volts across it. Each dissipates 6 watts.

If the contact resistance in the series circuit was 100 ohms, then almost all the voltage would be across the contact. The load would draw 119mA. The contact would dissipate 1.4 watts.

[ramblinrod]

Originally Posted by CatNewBee

Resistance does not change, nor does the system voltage, so for the same power same circuit, 4 times the voltage leads to 1/4 of the current and 1/4 of the voltage drop, causing 1/16 of the power loss.

It may just be the way this is worded, and what resistance, voltage, current, voltage drop, and power loss you are referring to, but based on my interpretation, it doesn't make any sense.

Resistance does change.

If we change the system voltage, to have a load equal in power dissipation, we have to change the load resistance. If we didn't the load would consume more power proportional to the voltage increase.

If we have a contact resistance variable, that resistance does change, to what ever value we have and calculate for.

[ramblinrod]

Originally Posted by Dockhead

You lost me in the last paragraph.


So are you saying that in a lower voltage system, you get a greater voltage sag for a given amount of resistance? That seems logical, but what do you mean ". . . the greater the voltage drop across the contact vs the load"?

No.

This is covered in post # 1, but go to post # 5, its simpler and clearer.

Draw yourself a picture. On the left a battery, across this battery draw two resistors in series. Call the first resistor R contact, and the second resistor R load.

From post # 5....

For a 1000W load supplied by 12 Vdc, the power dissipated by the contact impedance, varies with the contact impedance, as follows:

For 0.001 ohms, 6.9W
For 0.01 ohms, 60.7 W
For 0.1 ohms, 242 W
For 1 ohm, 110 W
For 10 ohms, 14 W
For 100 ohms, 1.4W

Now if we make that a 1000 W load supplied by 48 Vdc, the following results occur...

For 0.001 ohms, 0.4 W
For 0.01 ohms, 4 W
For 0.1 ohms, 40 W
For 1 ohm, 211 W
For 10 ohms, 152 W
For 100 ohms, 22

<<<<<<<<< >>>>>>>>>>>

For the first example, R Load = 1/ P / E^2 = 1/ 1000W / (12 * 12)V = 1/1000W/144V = 0.14 ohms.

For each case of contact resistance, R total = R load + R contact.

For each case of contact resistance, I = E/R total.

For each case of contact resistance, P contact = I^2*Rcontact.

Or this could be calculated by, P contact = Econtact^2/R.

The same holds true for the 48 Vdc circuit, except we have modified the load to dissipate the same power as in the first circuit despite the higher supply voltage.

For the second example, R Load = 1 / P / E^2 = 1/ 1000 W / (48 * 48) V = 1 / 1000W / 2304 V = 2.304 ohms.

This now has to be used to calculate Rtotal, and all of the other calcs to derive P contact.

OK guys, that is all I can play today, I have to prepare a presentation I am delivering to a boating group tomorrow.

Fortunately it is not about 12 vs 24 vs 48 Vdc electrical systems....this time. ;-)

[Jammer]

Originally Posted by KTP

He is right.

If you have a 1 ohm pure resistive load fed by 12V, you draw 12 amps from the battery. The load dissipates 12 watts. A good contact in this series circuit with 0.001 ohm resistance would dissipate a negligible 0.14 watts.

The load dissipates 144 watts. 12v x 12a.

If the contact resistance in the series circuit to that 1 ohm load is also 1 ohm, you draw 6 amps from the battery. The load has 6V across it and the contact has 6 volts across it. Each dissipates 6 watts.

Each dissipates 36 watts, a quarter of the intended load. 6 volts x 6 amps. (This is the most that can be dissipated at the "contacts" since any other resistance chosen will dissipate less. You can try a bunch of values for yourself and see, or look at the math I linked upthread)


The 1/4 of the load is the maximum regardless of the voltage used.

If the contact resistance in the series circuit was 100 ohms, then almost all the voltage would be across the contact. The load would draw 119mA. The contact would dissipate 1.4 watts.

[Jammer]

Originally Posted by Dockhead

Since physics are "the law", there must be a single right answer to these questions.

From a standpoint of the physics of it, here's what happens if you choose to model the "flaky contact" as a fixed resistance, then see how much power the "flaky contact" dissipates in 12v vs. 48v, with the same power dissipated at the intended load :


You get a different answer depending on how your chosen fixed resistance for the "flaky contact" compares to the resistance of the intended load.



But it will never be more than 25% of the power the intended load is supposed to dissipate.

[evm1024]

I went back and re-read the original thread http://www.cruisersforum.com/forums/...ne-199962.html to refresh my memory as to why we are here.

Quoting Ocean Planet from that thread:

Originally Posted by OceanPlanet

MOST of the larger yacht systems we do are now 24V, not 12V Sometimes there is a small backup 12V battery to keep the 12V nav or lighting running in the case of a main bank LVC.

You will soon see more high-performance yachts going 48V (actual nominal for Li systems will be 51V) for the charging & batteries, as it is considerably more efficient and less weight. One example of the charging efficiency: A 28V x 185A American Power HPI alternator is cold-rated at 185A, and the same size unit in 56V is cold-rated at 150A. If hot outputs are 80% (usually are more, however for calculation purposes) that's 4.1kW for the 28V and 6.7kW for the 56V. Note that I'm using actual charging voltages vs. nominal.

Anyhow, that's at least 50% more output in kW, from the same size alternator, through cables that are 1/2 the weight. The inverter conversion efficiency to the AC loads is also more efficient at the higher DC voltage.

Note that residential energy storage is already going to much higher DC voltages (288V to 360V), for the same gains in efficiency. However that level would not be very safe on a boat...

We can see that the state of the art is advancing and that systems other than 12 volts are likely in our future with economy of scale bring the cost of "devices" down.

Right now on one hand we have those who say that 48 volts has enough advantages and few enough risks to be a viable option.

On the other hand we have those who say that 12 volts is much safer and much more common than 48 volts and thus there is no need to even consider the risks of 48 volts.

At the moment we are fixated on the heating and fire potential in a faulty connection that introduces a series resistance (HRC).

It is unclear how much resistance the typical faulty contact will have. Examples have been given with a resistance of around 10 ohms which show that the 48 volt systems dissipate hundreds of watts in the fault. And others where the fault resistance is 0.5 ohms where the 48 volt system dissipated less power than the 12 volt system.

It is noted that the resistance of a faulty connection is not a single value. The resistance will move around for a number of reasons both mechanical as well as material (e.g. material changes due to heating and arcing).

It appears that we are going in circles in a kind of he said she said battle of ohms law calculations.

One of the reasons that high resistance contact faults are a "hot topic' is that they are not able to trip circuit breakers commonly in use in our boats. One innovation that would be desirable as 48 volt systems become more common would be to include devices that do detect such restive faults and shut down the load in response.

Because the HRC resistance may be in the connection to the device we wish to protect it is likely that the HRC detection circuity be in the load its'self.

I actually have a few devices on my boat that do shutdown if there is a HRC fault.

[evm1024]

Originally Posted by Jammer

From a standpoint of the physics of it, here's what happens if you choose to model the "flaky contact" as a fixed resistance, then see how much power the "flaky contact" dissipates in 12v vs. 48v, with the same power dissipated at the intended load :


You get a different answer depending on how your chosen fixed resistance for the "flaky contact" compares to the resistance of the intended load.



But it will never be more than 25% of the power the intended load is supposed to dissipate.

I see the math for that. It works out.

So if we were to feed a 25 watt VHF with 48 volts we would not get more than 6 or so watts dissipated in the fault. Easy for most wiring to handle. (ignore the assumption of the actual power needed to transmit - it is not germane).

A 150 watt PEP SSB might peak out at 30 or 35 amps draw at 12 volts but because the duty cycle is less than 100% likely will draw less even when doing pactor transmissions. But still we end up with less than 90 watts in the fault.

Stick a 10 ohm fault in series with the SSB and we likely will get a low voltage shutdown. Sounds like it is protected to me.

So the assertion that 12 volts systems are safer than 48 volt systems might be losing steam when looked at with real world usage.