8
\$\begingroup\$

The dielectric constant of air is approximately 3 kV/mm. The typical space between car spark plugs is between 0.6 and 1.8 mm. By multiplying those numbers in mm by the dielectric constant of air, we figure that a spark plug would work for between 1800 V and 5400 V.

Why is it that cars require between 12000 and 25000 V to fire properly? I feel like I am missing something important with these equations. Why am I so far off with these calculations? As I said, 1800-5400 V would work.

\$\endgroup\$
8
  • 3
    \$\begingroup\$ Are you taking into consideration that the spark needs to ignite the air/fuel while on a compression stroke, so there is a high pressure in the cylinder? \$\endgroup\$
    – HandyHowie
    Commented Dec 15, 2022 at 12:45
  • 1
    \$\begingroup\$ 3 kV at atmospheric pressure. With say 11:1 compression and an atmosphere of fuel vapors, you need higher voltage. \$\endgroup\$
    – winny
    Commented Dec 15, 2022 at 12:46
  • \$\begingroup\$ @winny That would make more sense. I didn't think of that. Could you answer this question 11:1 compression and atmosphere of fuel vapors? I'm just curious how this would work mathematically speaking. \$\endgroup\$ Commented Dec 15, 2022 at 12:46
  • 1
    \$\begingroup\$ I'll try to dig up more background information than a one sentence answer. \$\endgroup\$
    – winny
    Commented Dec 15, 2022 at 12:49
  • 1
    \$\begingroup\$ The relationship is not linear. If it was, it would be impossible to have voltage in vacuum. I'll try to dig up a curve for you. \$\endgroup\$
    – winny
    Commented Dec 15, 2022 at 13:10

3 Answers 3

10
\$\begingroup\$

Compression is key here as Paschen's law is in effect.

https://en.wikipedia.org/wiki/Paschen%27s_law

Except for close to vacuum, it's a near linear relationship between pressure and dielectric breakdown for most gases. You can substitute air for pure nitrogen for the sake of argument here.

enter image description here

On top of that, if an auto manufacturer finds that raising the voltage even higher gives 0.1 % fuel saving due to some property of the combustion, they will.

\$\endgroup\$
5
  • \$\begingroup\$ So the curve will be slightly different as it is an atomised fuel / air mix and it is not homogenous... \$\endgroup\$
    – Solar Mike
    Commented Dec 15, 2022 at 14:14
  • \$\begingroup\$ @SolarMike Probably. The auto industry will probably compensate by going even higher in voltage to overcome that too, along with swirl in the intake and piston dish. \$\endgroup\$
    – winny
    Commented Dec 15, 2022 at 14:29
  • \$\begingroup\$ yes, they have done all of those - some interesting shaped piston crowns... \$\endgroup\$
    – Solar Mike
    Commented Dec 15, 2022 at 14:45
  • \$\begingroup\$ Do you know what the units are on the x-axis? I'm just curious now. The way I am interpreting it is that at 760 Torr (atmospheric pressure) along the x-axis, it would be a little more than 20,000 volts for nitrogen which doesn't make sense to me as air is almost all nitrogen. I would have expected it to be much lower. \$\endgroup\$ Commented Dec 15, 2022 at 14:51
  • 4
    \$\begingroup\$ @WilliamGarske it's the product of pressure and gap distance, that's why the unit is "Torr cm". So 760 on the X-axis corresponds to 760 Torr at 1cm, or 7600 Torr at 1mm, etc. \$\endgroup\$
    – hobbs
    Commented Dec 15, 2022 at 22:35
9
\$\begingroup\$

So, you are missing several factors:

  1. compression ratio means that the pressure is above atmospheric - probably what you assumed with your calculation. The pressure AND temperature is also increased - something used in diesel engines so they don't need sparks...

  2. the spark needs to be "big & fat" ie just breaking down to have a feeble spark is not good enough,

  3. the mechanics of the combustion process also come into play here, as the initial combustion from the spark has a delay time then the flame front will progress through the fuel charge, that is why there is spark advance so that the fuel charge has sufficient time to burn during the power stroke.

Champion used to make sparkplug testers, which had a chamber that the sparkplug to be tested was screwed into. There was a viewing port and a spark generator. The test involved watching how the spark behaved as one increased the air pressure (workshops have compressed air). As the pressure increased then if the insulation - the cone - was cracked or contaminated then the spark would deviate from just between the terminals. Good kit and solved many misfire issues...

Denso has some good information (should do they make sparkplugs...):

enter link description here

enter image description here

\$\endgroup\$
2
  • 1
    \$\begingroup\$ This was very helpful too. I very much appreciate it \$\endgroup\$ Commented Dec 15, 2022 at 14:29
  • \$\begingroup\$ The big and fat spark is rather important for the proper burn timing. Weak spark makes the forming of the flame front and thus the delay somewhat unpredictable. \$\endgroup\$
    – fraxinus
    Commented Dec 16, 2022 at 8:26
1
\$\begingroup\$

Short answer: the dielectric strength of the dense air-fuel mixture encountered at the top of the compression stroke is much higher compared to free air (it will have a higher 'strike voltage'.) The higher spark voltage overcomes that.

The strike voltage will vary based on the engine's operating state. At high load (high BMEP) that voltage increases considerably: an engine that misfires will usually do so under wide-open throttle if the spark is too weak.

Knowing this, tuners building high-performance engines will thus use even higher spark voltages - 50kV or more - to avoid misfires. Aftermarket ignition suppliers like Mallory, MSD, Accel and others offer whole systems based on this need.

On the other hand, is there a downside to having less (just-adequate) spark voltage? I once worked on a system for V-twins that varied the spark energy based on engine load. The idea was to use a lighter spark at light load and thus use less energy. In retrospect it didn't make much sense since having a hot spark at light load does no harm (this observation, and other reasons led me to not waste a lot of time on it and instead walk away from the project.)

Nonetheless you can buy aftermarket systems that do this from an outfit called Jacobs Electronics, who make lots of claims about increased performance / mileage / time between tune-ups, etc. Personally I think it's a lot of baloney.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.