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Wikipedia's Components of Jet Engines has this to say on the igniters:

A high voltage spark is used to ignite the gases. The voltage is stored up from a low voltage (usually 28 V DC) supply provided by the aircraft batteries. It builds up to the right value in the ignition exciters (similar to automotive ignition coils) and is then released as a high energy spark. ...

Most modern ignition systems provide enough energy (20–40 kV) to be a lethal hazard should a person be in contact with the electrical lead when the system is activated, so team communication is vital when working on these systems.

I bolded that part I'm interested in. 20 to 40 kV? That is a huge, huge amount of voltage. Why on Earth would a jet engine need that much voltage to produce a spark? I thought kerosene was relatively easy to ignite.

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    $\begingroup$ Whoever wrote "energy (20-40 kV)" does not know what energy is. $\endgroup$ – Sanchises Jan 19 at 14:09
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In short, 40 kV isn't that much voltage for applications that are intentionally creating an electrical arc. Car spark plugs also use voltages in the tens of thousands of volts for the same reason, for example.

As for why that is:

In general, air acts as an electrical insulator. That is, electricity won't pass through air at normal voltages. Which is good because otherwise you'd have a constant arc to ground through the air from any exposed hot conductor and that would cause lots of problems.

As with any insulator, air has a dielectric strength, measured in volts per unit distance. For a given length of air (such as the spark gap in a spark plug,) there exists a breakdown voltage at which point the voltage differential per unit distance exceeds the dielectric strength. When a voltage equal to or greater than the breakdown voltage is applied to conductors on either side of said length of air, the air will suddenly ionize, creating a channel of plasma through the air. The plasma channel now has a much, much, much lower resistance than air normally has, allowing current to flow across the gap and quickly equalizing the potential (i.e. voltage) on each side of the gap.

As you're likely aware, plasma tends to be quite hot. The plasma channel created by the above process is the visible arc that you'll see between the conductors in a spark plug.

For air at normal pressures, the dielectric strength is about 3 MV/m. That is, if you want to create a 1-meter arc, you'd need 3 megavolts of potential across it. Or, in units more suited to creating ignition sparks (which are hopefully not a meter in length,) you need around 3 kV per millimeter. So, for example, in the winter when you take off a fleece coat and then experience a shock the next time you get close to a grounded conductor, the static potential that has built up on you is about 3 kV per millimeter (i.e. 30 kV/cm or around 75 kV/inch) of the maximum length of the arc you can create to such a conductor.

So, in summary, 40 kV really isn't all that much when you're trying to create arcs through air, especially not highly pressurized air, such as in a piston or jet engine. Even relatively small distances require quite high voltages in order to arc. Which is a good thing so that exposed conductors don't just constantly create arcs to ground. Or to grounded people, electronics, etc.

By the way, this ability of an electric field to ionize air is why switches for very high voltage transmission lines have to have much, much longer poles than what you'll find on normal switches. On a 138 kV line, for example, the electricity would simply arc across a switch that created a gap of less than couple of inches or so when open. Perhaps an even larger gap when humidity is high. And these switches do indeed arc while opening and closing. They can also create quite long arcs while opening, since the plasma channel is already in place. Here's an example of what that can look like at 138 kV:

Of course, in this case, you have a more-or-less continuous voltage source, not just a stored-up charge that is suddenly released, so the arc won't just be a brief flash, but will continue until the distance between the conductors becomes too high to maintain the arc.

And, of course, when the potential between the base of a cloud and the ground gets too high, you get lightning. And, yes, that's a very, very high voltage to arc across a kilometer or more. According to the U.S. National Weather Service, a typical lightning bolt involves potentials of around 300 MV. While that actually seems a bit low to arc across the typical distance of a lightning bolt, lightning normally occurs in thunderstorms where the atmosphere is already very humid and, thus, more conductive than normal.


As far as the ease of igniting jet fuel, while that's mostly unrelated to the voltage required to create a spark across a given distance of air, it's actually not that easy to ignite. At standard air pressure, you can drop a lit match in a bucket of jet fuel and it will just extinguish the match. You can also create a raging fire in the fuselage of a Boeing 777 and the fuel remaining in the tanks of said 777 won't burn. This is by design and it's one of the great things about jet fuel for aviation. While the energy density and specific energy of petroleum fuels is very high (by chemical reaction standards,) it actually requires quite high temperatures and/or pressures to ignite jet fuel, making it an ideal fuel for providing lots of energy per unit weight while not tending to randomly burn or explode at inconvenient times.

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    $\begingroup$ This is the best answer here, even briefly addressing the relative difficulty of igniting jet fuel. Thank you. $\endgroup$ – DrZ214 Jan 19 at 20:57
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    $\begingroup$ @Fattie Hmm... the answer states, "So, in summary, 40 kV really isn't all that much when you're trying to create arcs through air, especially not highly pressurized air, such as in a piston or jet engine." Do you think it would be less confusing without the transmission line and lightning comparisons? I just added those because I thought they'd be useful as a point of comparison for voltages required to arc across air gaps. $\endgroup$ – reirab Jan 19 at 22:22
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    $\begingroup$ @Fattie It's true that the voltage needed to arc across any air gap is a bit extreme compared to voltages normally used on wires (because we normally don't want conductors to be able to arc across air gaps,) but 40 kV certainly isn't unusual for applications where you actually do want an arc. $\endgroup$ – reirab Jan 19 at 22:25
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    $\begingroup$ @reirab - I'm all for incredibly simple answers. I would have said up front "30k is the norm for all sparks, such as in a car." That's the whole answer to this question per se. Then - by all means - if I was able to write an answer this good I would then say "Here's some more info on sparks." And then had the answer as seen. $\endgroup$ – Fattie Jan 19 at 23:32
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    $\begingroup$ @reirab - don't get me wrong. Every writer should follow their own lights. But then, every a-hole can criticize from a sofa in comments :) :) $\endgroup$ – Fattie Jan 19 at 23:47
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That's nothing special. Car ignition coils also put out 40kV (try grabbing a spark plug wire on your car, that is leaking spark due to insulation breakdown, with your hand while it's running; FUN FUN FUN!). You need that voltage to jump an air gap reliably in a piston, or jet engine's, combustion chamber. That high a voltage allows fairly large air gaps to be used, making the spark large and hot, all good things to have when you are waiting for the engine to relight after flaming out due to water ingestion while flying in heavy rain.

The 40kV coils in cars are the reason auto plugs operate with very large gaps, like .040", and will run fine until the electrodes are down to little nubs.

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    $\begingroup$ @DrZ214 From this answer on Physics.SE: you need at least 10kV to create a 1cm spark in air. If you want the spark to be produced reliably, more voltage is better. $\endgroup$ – Bianfable Jan 19 at 8:17
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    $\begingroup$ @DrZ214 Its the difference between potentially and reliably. An inerting system is about reducing the minimal chance of arc, that would still happen at least a couple of times in a million hours of flights. And a couple of times is too much for the risk of fire. Where as in an ignitor, having an ignitor that only worked 2 times in a million would be unacceptable, so you need a voltage with an extremely high likelihood of arcing. $\endgroup$ – user1937198 Jan 19 at 14:16
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    $\begingroup$ @DrZ214 one thing to keep in mind is pressure. The insulating quality of the air increases with pressure, so in a piston engine's combustion chamber or a jet engine's burner can, the high pressures require a higher voltage for a given gap. The opposite case is a problem for high voltage components in ambient air when you go to altitutude. Magnetos at high altitude have a problem of high voltage current internal leakage and have to either have large internal clearances to make large air gaps between current carrying parts, or if it's still a problem, the mags are actually pressurized. $\endgroup$ – John K Jan 19 at 15:49
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    $\begingroup$ @DrZ214 The reason for inerting the fuel tanks is if the insulation on the wires comes off and the conductors actually touch. That will produce quite a nice spark. It won't be able to arc between the conductors at any meaningfully-non-zero distance at those voltages, but that doesn't mean it won't produce quite the spark if they touch. I'm not sure if you've ever seen what happens when a 120 or 240 VAC hot wire touches a ground/neutral wire, but it produces a quite large and extremely hot spark for the hopefully very short duration until the circuit breaker trips. $\endgroup$ – reirab Jan 19 at 17:27
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    $\begingroup$ I appreciate the levity, but obviously readers should NOT try grabbing live ignition parts while the car is running. :) Also, it's better not to let the spark plug electrodes wear down to little nubs, as this will increase the required spark voltage and stress your (far more expensive) ignition coil(s), possibly to the point of misfires, rogue arcing and failure. $\endgroup$ – screwtop Jan 20 at 0:37
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40kV is typical for a spark igniter. Air, especially dry air, is a pretty good insulator, and it takes that level of voltage to cause a breakdown. An ignition spark is effectively a miniature lightning strike and propagates through air in much the same way; consequently it needs much the same electrical field strength (in volts per metre) to get started.

But it is wrong to talk of the kV level as "energy". It is more akin to pressure. Static electricity can build up to these voltage levels, but if you walk across a plastic floor in plastic shoes you will not die when you grasp a metal door handle and zap yourself. The energy of a discharge manifests as the current times the time it lasts, and the current from such a simple static discharge is very short-lived.

An ignition system suitable for starting a big engine will be designed to reliably produce a nice fat spark - that is to say, a higher current for longer - and is therefore that much more dangerous.

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  • $\begingroup$ Large trucks with compression-ignition engines will usually use a glow plug to start, not a spark plug. $\endgroup$ – forest Jan 21 at 2:43
  • $\begingroup$ @forest True. Large petrol (gasoline) engines do exist but are not common. I have rephrased my answer accordingly. $\endgroup$ – Guy Inchbald Jan 21 at 10:29
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If one passes current through a coil at low voltage and disconnects it suddenly, it will produce a voltage which is orders of magnitude higher unless current starts flowing again (the only reason voltage is limited at all is that if it gets high enough, current will start flowing again, somewhere). If one uses two inter-wound coils with different numbers of turns and interrupts a current in the coil with fewer turns, the coil with the larger number of turns will reach a higher voltage yet unless or until current starts flowing through one of the coils. It is thus simple to have a small assembly turn a low voltage into spurts of high voltage very near the location where the high voltage is needed. Further, it's possible to use a moving magnet in the vicinity of the coil to induce the initial current in the coil without having to deliver any sort of voltage to it. This is the principle upon which magnetos work.

Note that such devices do not continuously produce high voltage, but instead store up some energy when current is driven at a low voltage (or using a moving magnet), and then release it all quickly at high voltage. Devices may pose danger if the amount of energy that they store at once is sufficiently large, if they cycle at a fast enough rate to deliver a substantial average current, or if they cycle it at a rate which can interfere with nerve impulses in the body.

I don't think the text is intended to suggest that the ignition systems could be used as a source of "reliably lethal" voltage, but rather to warn that even inadvertent incidental contact could, under just the "right" circumstances, be dangerous. Even if the risk of death from an accidental shock would only be 1 in 10,000, a prudent person should not take such risks needlessly.

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    $\begingroup$ While this is useful information, it doesn't answer the question at all. $\endgroup$ – Phil Jan 21 at 1:55
  • $\begingroup$ You could add to this by prepending to this answer that the voltage is what it is because it couldn't be more, or less. The voltage will keep building in the ignition coil until it's able to bridge the gap between the spark plug, after which it discharges immediately. $\endgroup$ – forest Jan 21 at 2:45
  • $\begingroup$ @forest: My point was that while 30,000VDC may sound like a lot, or would pose an ignition danger, or it would be hard to produce, it's fairly easy to produce, and the dangers can be minimized by generating such voltages near the point of use. $\endgroup$ – supercat Jan 21 at 6:34

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