Question What is an electrical bus exactly?

[C3PO]

Is there (I assume there is) a way to compute the potential magnetic force of such an event based on the anticipated peak flow (which of course would be much higher than normal)?

Of course. Every electrician here has done that during training. In reality square copper bars aren't used "below" distribution switch boards. Most switch boards use copper profiles specifically to save on material.

 

[Proximus]

On the machines we design at work we usually use a terminal system.
DC Power supply (or mains AC) to a set of jumper-ed terminals to allows for easy expand ability or just adding additional hardware.

As stated previously the supply must be short protected as well as each device individually.

A typical system would have a mains bus, and one or two DC buses.
ABB/entrelec have an excelent selection of jumper terminals and sensor terminals.

 

[jedidia]

you can 'tie' two (or more) compatible buses such that the voltage equalizes

Could you explain this point a bit better? I'm really a bit of a noob when it comes to electricity. It is my understanding that I'd need a transformer to hook two buses of different voltages together? What exactly do you mean by "the voltage equalises"?

 

[Urwumpe]

Could you explain this point a bit better? I'm really a bit of a noob when it comes to electricity. It is my understanding that I'd need a transformer to hook two buses of different voltages together? What exactly do you mean by "the voltage equalises"?

When you tie two busses, the voltage flows from the bus with the higher voltage to the bus with the lower voltage, until the sum of power generation and voltage added and subtracted is equal on both busses. For example, when you have batteries powering the busses, the battery with the higher voltage will get drained faster than the battery with the lower voltage (because they are switched parallel).

 

[Linguofreak]

Could you explain this point a bit better? I'm really a bit of a noob when it comes to electricity. It is my understanding that I'd need a transformer to hook two buses of different voltages together? What exactly do you mean by "the voltage equalises"?

You use a transformer when your power supply and load need different voltages. You use a bus tie when the power supply on one bus isn't keeping up with the load and the voltage drops as a result, and the other bus has enough reserve power to bring the voltage into the correct range (and the two buses use more or less the same design voltage).

 

[Zatnikitelman]

Actually it can, but only for a very short time. Circuit breakers have to be quick to protect the system. BTW a million amps is a million amps no matter what the voltage.
Fun fact: One of the reason for the bulky copper bars is to withstand the magnetic forces during a short circuit.

What I meant by that is high voltage + high current potential = scary. I'll happily walk up and touch a bus bar with 1 volt potential otherwise carrying a million amps. That 1 volts isn't enough to drive enough current through me to kill me. On the flip side, I have no problem touch a million volts as long as the current is limited (think electric gag-buzzers). But my point is I doubt those copper bars pictured only sit at 1 volt, or only carry < 1mA, therefore, they're a little scary.

 

[jedidia]

Welp, that was confusing. Still is somewhat.
Here's the two things I thought I knew:

1. The voltage in a parallel curcuit is the same at any point.
2. A bus is a parallel curcuit.

So I started looking up what would happen if you place two power supplies with different voltages together, since that seems essentially the same problem. From what I've learned, it's very highly not recommended, but generally works with similar enough voltages and a whole lot of backfeed protection, but it seems to me that the arrangement of plugging a bus into another bus (which would at that point essentially become another power supply) would bring the voltage down on the bus supplying the initial power so that the voltage in the overall system is again the same at every point.

I guess that's what thorsten meant by equalising, but would that not essentially lead to undervoltage on everything that is connected to the supplying bus and relies on its voltage?

In any case, I'm beginning to form a picture here of how things are supposed to work, so thanks everyone for their explanations. You've been a great help!

Now, let me see if that picture of mine is correct by a more practical example. Back to the spaceship!

Let's say I have an electric engine, so I have a quite a strong generator. On the one side, that thing hooks probably pretty much directly into the engine, since it was probably designed for it.
But when the engine's not running, I'd also like to power the rest of the ship with it, of course. So I put in a transformer and feed an ISS style 160 volts main bus, because high voltage is preferable for getting electricity from point a to point b.

Now I'm hooking cables into that bus, probably one to each module. In those modules I'm most probably going to need something like a 26 volt bus, as that seems a very common thing in avionics or spaceflight. Maybe I'll put in a 9-volt bus in my cabin to hook up all the guitar effects without needing a dedicated power supply for every single one :shifty:

In any case, that would essentially mean hooking that cable up to a transformer that brings the voltage down to 26 volts to feed my 26-volt internal module bus. In my cabin I'd hook up another transformer to that 26V bus to power my private 9V bus. And similar to that in every module.

Does that seem a plausible way of doing things? I know I'm ignoring AC/DC here (always liked Maiden better anyways), and messy stuff like phases, but I really don't want to go this complex. All I want is a reasonably involved power distribution system that is still abstracted enough to be used by most people willing to read two pages of manual.

That 1 volts isn't enough to drive enough current through me to kill me.

No, but it might burn your fingers of... :shifty:

 

[InjectorPlate]

A bus in electrical system terms is can be a data bus, a power distribution bus (be it A/C D/C), an audio bus/mixer, or what have you. How you actually connect to the bus is not an issue. In engineering we used to call it "your either on the bus or off the bus".

A typical defined bus has a standard in engineering as in specifications for a power distribution bus such as a defined voltage/amperage rating, conductor size for the former, physical requirements (bus bar, transfer conductors/cable), etc.

It's simple really if you look at it. Do not open a capsule, aircraft and wonder what all the wiring does, unless your like me when I was young! Main cables from your car battery to your fuse panel, and of course to ground, could be call the BATT A bus. All a bus does is take a source and "offers up what it carries to those that are invited".

---

 

[paddy]

In short a power system can be designed that can take in 415v 3 phase, 240vac, 220vac, 110vac and 12v DC. It could give out 240ac ( for my kettle) 110ac for my US made PC and 12DCv (for my Car radio) and 5vDC for all my USB goodies.

If you ask for it .... I build it... If you say its available... Its there!
If one input fails i.e. no solar power, the other work!
If the PC blows up I can still make tea.

It is all connected to, comes from, goes via, a BUS

Draw a Square Box, call it a power bus , connect EVERYthing to it, drink tea, job done.

 

[Linguofreak]

I guess that's what thorsten meant by equalising, but would that not essentially lead to undervoltage on everything that is connected to the supplying bus and relies on its voltage?

Yes, the voltage will sag. Whether that continues, and whether it can be considered an under-voltage depends on whether the power supply for the bus has any reserve power left (if, say, the power supply is a diesel generator, it might be running at half throttle, and then start throttling itself up when the bus tie switch is thrown and the voltage starts to sag) and what the tolerances are for the equipment on the bus. Can it take an error of 1 percent? 10 percent? 30 percent?

 

[JonnyBGoode]

 

[paddy]

ok you win

 

[Thorsten]

Yes, the voltage will sag. Whether that continues, and whether it can be considered an under-voltage depends on whether the power supply for the bus has any reserve power left

If you want to model this, one way to do it is to characterize your power source by a table of voltage the source can keep up in the bus vs. power demand.

What then happens is that if you demand too much power, the operating voltage drops to a level that equipment fails.

Below is such a table for a Shuttle fuel cell I've been using (which inputs the relative load, i.e. power demand scaled by the max. power the cell can supply and outputs fuel cell voltage - bus voltage is marginally lower).

		<fcs_function name="systems/electrical/fc/voltage">
		<function>
		<product>
		<table>
        		<independentVar lookup="row">systems/electrical/fc/rel-load</independentVar>
        		<tableData>
          	 		0.00	32.5
				1.00	27.5
				1.33	24.5
				1.41	23.5
				10.0	1.0
        		</tableData>
      		</table>
		<property>systems/electrical/fc/fc-ready-for-load</property>
		<property>systems/failures/fc1-condition</property>
		</product>
		</function>
		</fcs_function>

So if you ask for nothing, you'll measure 32.5 volts if the cell is sound, if you ask for full load (I think 12 kW) the voltage drops to 27.5, if you ask a third more than that, voltage is down to 24.5 which is below the 26 V operating voltage of most equipment, so you'll get a bunch of failures.

 

[jedidia]

What then happens is that if you demand too much power, the operating voltage drops to a level that equipment fails.

Interesting... I always thought drawing too much power burns up the power supply/conductors because you're shoving more amps through them than they are designed for. I wasn't aware that the voltage drops, but then again it makes perfect sense since P = IV, so something's got to give when the supply just doesn't have enough P.

I take it that the additional amps will still heat up and eventually fry the supply though, right?

 

[Thorsten]

I take it that the additional amps will still heat up and eventually fry the supply though, right?

I believe the drop in operating voltage before you reach maximal power output is characteristic of an electrochemical source like a battery of fuel cell (in the event, I have the numbers for the fuel cell from the Shuttle Manual) - a Diesel generator might well be able to keep voltage constant till it reaches maximal throttle.

From the point of maximal power, you can use P = UI with the operating voltage to get the current through the circuitry and compare with the maximal current in whatever the choke-point is - and whenever that exceeds the limit, it will fry.

Hopefully it's a circuit-breaker.

Also note that currents won't grow beyond bound, because the power source also has something called 'inner resistance' (I think that's the correct English term, it's been a while...) which limits the current even if there's zero resistance in the circuitry otherwise.

 

[Notebook]

it gets worse...

Internal resistance

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

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

Theory[edit]
Impedance is the opposition by a system to the flow of energy from a source. For constant signals, this impedance can also be constant. For varying signals, it usually changes with frequency. The energy involved can be electrical, mechanical, acoustic, magnetic, or thermal. The concept of electrical impedance is perhaps the most commonly known. Electrical impedance, like electrical resistance, is measured in ohms. In general, impedance has a complex value; this means that loads generally have a resistance component (symbol: R) which forms the real part of Z and a reactance component (symbol: X) which forms the imaginary part of Z.
In simple cases (such as low-frequency or direct-current power transmission) the reactance may be negligible or zero; the impedance can be considered a pure resistance, expressed as a real number. In the following summary we will consider the general case when resistance and reactance are both significant, and the special case in which the reactance is negligible.

I know of this slightly as I used to work with various bits of equipment...

Nothing has changed, Electrical Engineering is still as it was when Mr Edison did things.

N.

 

[Linguofreak]

Interesting... I always thought drawing too much power burns up the power supply/conductors because you're shoving more amps through them than they are designed for. I wasn't aware that the voltage drops, but then again it makes perfect sense since P = IV, so something's got to give when the supply just doesn't have enough P.

I take it that the additional amps will still heat up and eventually fry the supply though, right?

Well, if you've got two buses that are designed to be tied together, presumably the power supplies and conductors will be designed to handle the current draw of all of the equipment on both buses (or, at least, all of the equipment on each bus that is not redundant on the other bus).

Whether the power supply itself burns in conditions where the connected equipment will draw large amounts of current will depend somewhat on the power supply. The energy that's dissipated by whatever burns out in the process if burning out has to come from somewhere. For a generator driven by an internal combustion engine, assuming all energy supplied to the circuit is used immediately and none is stored by capacitors or whatever, the maximum power that any part of the circuit can dissipate as heat is the full-throttle output of the engine driving the generator. If the generator can dissipate that much power as heat, assuming that it is the only power source on the circuit, I don't think it will generally be possible for the generator to blow (the engine might blow if it is run too long at full throttle, but that's a mechanical problem, not electrical).

For a battery, the voltage is determined by battery chemistry and number of cells, and all the energy that will ever be available to the circuit (assuming no recharge) is already stored chemically in the battery. Unlimited current draw from attached equipment will cause current to increase until either internal resistance stops further current growth, or until the power dissipated by internal resistance heats the battery enough that it catches fire.

 

[jedidia]

The energy that's dissipated by whatever burns out in the process if burning out has to come from somewhere.

I'm under the impression that in a conductor, it's the amps generating the heat, not the volts. So if too many amps are drawn, the voltage drops to accommodate more amps, hence you get more heat at the same power.
I might have a severe misconception here, but I thought this was the reason why long-range power transfers are conducted at as high a voltage as possible.

 

[Xyon]

I thought this was the reason why long-range power transfers are conducted at as high a voltage as possible.

I might chime in with my own misconception here, but I thought this was more to do with minimising losses in transport.

 

[tl8]

I might chime in with my own misconception here, but I thought this was more to do with minimising losses in transport.

High voltage is used to minimise losses caused by high current flow (I2R Losses, IE cable resistance)). By going to a high voltage, but keep the power the same, you reduce the amount of current and therefore losses due to cable resistance.

---------- Post added at 11:55 AM ---------- Previous post was at 11:50 AM ----------

I'm under the impression that in a conductor, it's the amps generating the heat, not the volts. So if too many amps are drawn, the voltage drops to accommodate more amps, hence you get more heat at the same power.
I might have a severe misconception here, but I thought this was the reason why long-range power transfers are conducted at as high a voltage as possible.

Amps and the resistance of the cable generate the heat. Of course resistance is affected by heat and there is a nice feedback loop. Generally we specific parts so we can ignore this.

When the voltage drops, there is too much load (resistance) on the system. By dropping the voltage, the current also drops which reduces the load.