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Posted: Mon Nov 20, 2006 9:24 am
by ChemCool
Correct.
Posted: Mon Nov 20, 2006 9:27 am
by Draco
So can you work it out and let me know what the best would be Gary..please
Posted: Mon Nov 20, 2006 9:48 am
by Gary57
What boost you running at?? What rev does your turbo peak at??
Posted: Mon Nov 20, 2006 9:49 am
by ChemCool
conservation of mass law

Bernoulli's equation.

Posted: Mon Nov 20, 2006 10:06 am
by Draco
I'm running 1bar, at 5500 revs.
Posted: Mon Nov 20, 2006 10:07 am
by Gary57
That is correct.
Posted: Mon Nov 20, 2006 10:11 am
by ChemCool
And then I hae noticed the following two equations:The first equation describes the overall heat transfer that occurs.
Q = U x A x DTlm and also:We also have an equation for checking the amount of heat lost or gained by the fluid on one side of the heat exchanger (ie, just the turbo air or just the outside air):
Q = m x Cp x DT
Too much for me thank you. Easy out for me , I have bypassed the intercooler.
Posted: Mon Nov 20, 2006 10:14 am
by ChemCool
Posted: Mon Nov 20, 2006 10:18 am
by Gary57
63mm in and 76mm out would be good. Inlet speed would be around 200km/h and exit around 135km/h which is pretty good.
Posted: Mon Nov 20, 2006 10:19 am
by Gary57
Why did you bypass the intercooler??
Posted: Mon Nov 20, 2006 10:21 am
by ChemCool
Gary, give us that formula. Would like to see it. Seriously very interesting and hope to learn from that.
Posted: Mon Nov 20, 2006 10:22 am
by Draco
So back to my intercooler thing Gary, have you worked it out? Freddys Water meth works well got alot of power out of it. looks cool as well.
Posted: Mon Nov 20, 2006 10:26 am
by ChemCool
Gary I bypassed the intercooler because I inject water/meths. I dont exceed 1.3 Bar. Up to that level, more than enough cooling. If you go dual nozzels, then perhaps more boost. I only use one nozzle 175ml/min before TB. Your next question going to be that the intercooler will assist in the water/meth effort. Yes, with higher boost perhaps. I dont need that.
No lag for me in 2nd , 3rd and so forth.
Posted: Mon Nov 20, 2006 10:42 am
by Gary57
1. find the amount of air the motor is consuming at peak rev or boost.
1800cc = 0.0018m3
5500rpm = 91.667rps
1Bar boost = 2 x atmospheric intake
SO... (0.0018/2) x 91.667 x 2 = 0.165m3/s
2. now work out the speed in the pipes.
m/s = m3/s / area of pipes
m/s = 0.165/???
3. now convert from m/s to km/h
km/h = 3.6 x m/s
Guys this is the very basic of equation, disregarding friction. If you want to work it out correctly you need to know the radius of the bend, pipe material, etc and that takes alot of time and serious thinking.
Posted: Fri Nov 24, 2006 1:13 pm
by Draco
Guys the cooler is in, my boost is at about 0.5 bar in 2nd at 3500 revs and then I am not pushing it so it is about the same as it was normal, if I put foot it will defenately boost full at this revs because previosly it did the same, will let you know what the dyno improovement is as soon as I have it so then we can see if the cooler really works and how ifficient it is.