1FASTLS

Can the solenoids get to hot and not work, they are on a braket mounted to intake bolt which transfering heat, seams to be working intermittantly.

Matt@HSW

iTrader: (14)

Yes, they could. It could also be an electrical issue (bad ground or connection) or if it's just the nitrous solenoid high bottle pressure will keep them from opening.

Matt

terry s

iTrader: (5)

I am concerned about how hot they will get when mounted to the engine. I am almost finished making a mounting bracket that mounts to the front of the head. I am using silicone insulators which will significantly reduce the temp. of the solenoids.

cantdrv65

iTrader: (3)

They will probably be ok as long as you keep them away from the headers.

Robert56

iTrader: (2)

Yes, heat equals resistance, and once you excede your noids amp rating failure is next. To make it more clear lets say you have a noid that pulls 7amps at 13 volts, now heat that bad boy up big time and resistance increases and you will now be pulling in excess of the 7 amps, This is not good as windings and external wiring are set-up for a 7 amp load, and failure is just around the corner.
With this said I have had mine mounted on 1' spacers off the head for a while and recently did like Terry (his idea)and put the silcone spacers in for peace od mind.
Robert

JoeyGC5

iTrader: (24)

Robert, I've never had to correct you till now Voltage = current times resistance. If voltage is constant and resistance increases, then current decreases. However when a solenoid heats up, it creates more interanl "short" circuits, so to speak. The hysterisis losses (heated internal wiring of the solenoid), then add to the increased current consumption. Causing solenoid failure and erratic operation.

Robert56

iTrader: (2)

Is that not what I said? What about Ohms Law takes 1-volt to push one amp through 1-ohm of resistance. You say current decreases, yes to a point but amp load is still trying to be met, through increased voltage demand? and then over whelming circuit?
Example: 12volts / 7amps = 1.71ohms, or E / I = R, so with my example we have approx 1.71ohms of resistance.
Then we get some heat and resistance goes up to 2.5ohms,
I x R = E, or 7amps x 2.5ohms = 17.5volts
Now you've over loaded the circuit and it still trys to perform and then the internal short you suggest and failure?

Now I am not an Electrical engineer, so I can take my lumps if I am off base.
Robert

JoeyGC5

iTrader: (24)

I was just busting your chops on the wording .

Insanekamel

Please rewrite all of that in ENGLISH.. j/k nice to see some of us know what they are talking about