hingstonwm

iTrader: (2)

I performed a search but it didn't return and results, perhaps I didn't phrase my search correctly.

I have read several post where people have mentioned going to heavier flywheels to help with spool.?? I must admit that I am stumped by this statement. How does a heavy flywheel increase spool? Correct me if I am wrong since I am a noob in FI theory, spool is a function of exhaust gases spinning the turbine to a point sufficient enough to raise intake tract pressure, thus allowing more fuel to be introduced to the combustion chamber, and as a result more power?

It seems to me anything that slows the engines acceleration would reduce spool. Slower, or less RPM,h means less exhaust gas exiting the cylinder and less exhaust gas to spin the turbine and create boost. Am I missing some load consideration?

Braineack

iTrader: (62)

A heavier flywheel does not cause a turbocharger to reach it's boost threshold any earlier than a light flywheel does.

thagr81 us

iTrader: (5)

Me thinks someone was ******* with you...

18psi

iTrader: (76)

I "THINK" the theory they're applying to that is that the heavier flywheel creates more unsprung weight causing a higher load on the engine causing more exhaust gasses and therefore decreasing spool time.

A couple bags of sand or a passenger would probably create a "better" spool increase though LOL

Braineack

iTrader: (62)

loading is the same, flywheel is there to maintain inertia, so it doesn't slow down. If it can rev quicker, it can load quicker... something like that load is from EGTs and crap.

18psi

iTrader: (76)

I say just run 2 flywheels. Full boost by 200rpm

Braineack

iTrader: (62)

except it will take longer to reach 200rpm. however you will be able to run a saw blade a lot easier.


its just like when you use 2- stage rev limiter. hold rpms at 2k and reduce the timing significantly and the turbo will spool like crazy...

shifty35

Flywheels suck.

Joe Perez

iTrader: (8)

Depleted uranium flywheels, anyone?

E-NA6CE

iTrader: (2)

A lighter flywheel only shortens your turbo's threshold in a zero-load circumstance. While in gear and under load threshold remains the same. The only way it makes a difference is in TIME. Sure, you can spool faster, but you can rev faster. You will hit boost and reach peak boost sooner (in a time sense) but they will still happen at the same RPM, obviously dependent on load.

The only argument to the fact is that a heavier flywheel does increase load, which increases EGT's, which increases the fluid's volume, which increases exhaust pressure, which theoretically can increase spool.

fooger03

iTrader: (2)

• Heavier flywheel will show boost coming on at an insignificantly lower RPM (likely only noticable if you're datalogging)
• Lighter flywheel will show boost at a higher RPM
• Lighter flywheel will reach the higher RPM faster than the heavier flywheel will reach the lower RPM

So:
Heavier flywheel may cross boost threshold at a lower RPM, but the otherwise identical car with the lighter flywheel is faster 100% of the time.

Alternatively:
If you're running a time-sweep dyno as opposed to a fixed load dyno, the ONLY difference you will notice is that the lighter flywheel will show a higher TQ number.

Braineack

iTrader: (62)

I've have a stock, 8lb and not 1.8L flywheel on my turbo setups. could not notice a difference in either... only the lightened makes it easy to blip throttle and stall in traffic.

i noticed no spool chane with 1.8L flywheel. I datalog almost every time id rive my car.

E-NA6CE

iTrader: (2)

Word. Zero change in spool under load, super easy to blip... and stalling. Bah ha ha ha ha. You don't even know you stall sometimes because you don't get the slight shake the OEM **** gives you.

miatauser884

iTrader: (11)

I'm jealous of your new-found ability to datalog via sim card.

I agree with the spool faster as a function of time not as function of rpm

hingstonwm

iTrader: (2)

Thanks for the responses! You guys confirmed what I was thinking.

Fireindc

iTrader: (13)

Would sure make clutch jobs more hazardous.

E-NA6CE

iTrader: (2)

Wouldn't the heat caused by friction also cause some concern? Then again, I'm no terrorist so I have no idea...

thagr81 us

iTrader: (5)

Shouldn't be a problem as heat doesn't do anything to uranium. However, achieving critical mass is a WHOLE different story....

lordrigamus

iTrader: (14)

Only if it was this kind of friction:

E-NA6CE

iTrader: (2)

Here's my Out Of The *** post on Critical Mass.

Uranium-233 is a fissile artificial isotope of uranium, part of the thorium fuel cycle which has been used in a few nuclear reactors and has been proposed for much wider use as a nuclear fuel. It has a half-life of 160,000 years.

Uranium-233 is produced by the neutron irradiation of thorium-232. When thorium-232 absorbs a neutron, it becomes thorium-233, which has a half-life of only 22 minutes. Thorium-233 decays into protactinium-233 through beta decay. Protactinium-233 has a half-life of 27 days and beta decays into uranium-233; some proposed molten salt reactor designs attempt to physically isolate the protactinium from further neutron capture before beta decay can occur.

233U usually fissions on neutron absorption but sometimes retains the neutron, becoming uranium-234, although the proportion of nonfissions is smaller than for the other common fission fuels, uranium-235, plutonium-239, and plutonium-241. It is slightly smaller at all neutron energies.

The fission of one atom of U-233 generates 197.9 MeV = 3.171 × 10−11 J, i.e. 19.09 TJ/mol = 81.95 TJ/kg.

Critical Mass is roughly 15kg.

The critical mass for lower-grade uranium depends strongly on the grade: with 20% U-235 it is over 400 kg; with 15% U-235, it is well over 600 kg.

The critical mass is inversely proportional to the square of the density. If the density is 1% more and the mass 2% less, then the volume is 3% less and the diameter 1% less. The probability for a neutron per cm travelled to hit a nucleus is proportional to the density. It follows that 1% greater density means that that the distance travelled before leaving the system is 1% less. This is something that must be taken into consideration when attempting more precise estimates of critical masses of plutonium isotopes than the approximate values given above, because plutonium metal has a large number of different crystal phases which can have widely varying densities.

Note that not all neutrons contribute to the chain reaction. Some escape and others undergo radiative capture.

Let q denote the probability that a given neutron induces fission in a nucleus. Let us consider only prompt neutrons, and let ν denote the number of prompt neutrons generated in a nuclear fission. For example, ν ≈ 2.5 for uranium-235. Then, criticality occurs when ν·q = 1. The dependence of this upon geometry, mass, and density appears through the factor q.

Given a total interaction cross section σ (typically measured in barns), the mean free path of a prompt neutron is \ell^{-1} = n \sigma where n is the nuclear number density. Most interactions are scattering events, so that a given neutron obeys a random walk until it either escapes from the medium or causes a fission reaction. So long as other loss mechanisms are not significant, then, the radius of a spherical critical mass is rather roughly given by the product of the mean free path \ell and the square root of one plus the number of scattering events per fission event (call this s), since the net distance travelled in a random walk is proportional to the square root of the number of steps:

R_c \simeq \ell \sqrt{s} \simeq \frac{\sqrt{s}}{n \sigma}

Note again, however, that this is only a rough estimate.

In terms of the total mass M, the nuclear mass m, the density ρ, and a fudge factor f which takes into account geometrical and other effects, criticality corresponds to

1 = \frac{f \sigma}{m \sqrt{s}} \rho^{2/3} M^{1/3}

which clearly recovers the aforementioned result that critical mass depends inversely on the square of the density.

Alternatively, one may restate this more succinctly in terms of the areal density of mass, Σ:

1 = \frac{f' \sigma}{m \sqrt{s}} \Sigma

where the factor f has been rewritten as f' to account for the fact that the two values may differ depending upon geometrical effects and how one defines Σ. For example, for a bare solid sphere of Pu-239 criticality is at 320 kg/m2, regardless of density, and for U-235 at 550 kg/m2. In any case, criticality then depends upon a typical neutron "seeing" an amount of nuclei around it such that the areal density of nuclei exceeds a certain threshold.

This is applied in implosion-type nuclear weapons where a spherical mass of fissile material that is substantially less than a critical mass is made supercritical by very rapidly increasing ρ (and thus Σ as well) (see below). Indeed, sophisticated nuclear weapons programs can make a functional device from less material than more primitive weapons programs require.

Aside from the math, there is a simple physical analog that helps explain this result. Consider diesel fumes belched from an exhaust pipe. Initially the fumes appear black, then gradually you are able to see through them without any trouble. This is not because the total scattering cross section of all the soot particles has changed, but because the soot has dispersed. If we consider a transparent cube of length L on a side, filled with soot, then the optical depth of this medium is inversely proportional to the square of L, and therefore proportional to the areal density of soot particles: we can make it easier to see through the imaginary cube just by making the cube larger.

Several uncertainties contribute to the determination of a precise value for critical masses, including (1) detailed knowledge of cross sections, (2) calculation of geometric effects. This latter problem provided significant motivation for the development of the Monte Carlo method in computational physics by Nicholas Metropolis and Stanislaw Ulam. In fact, even for a homogeneous solid sphere, the exact calculation is by no means trivial. Finally note that the calculation can also be performed by assuming a continuum approximation for the neutron transport. This reduces it to a diffusion problem. However, as the typical linear dimensions are not significantly larger than the mean free path, such an approximation is only marginally applicable.

Finally, note that for some idealized geometries, the critical mass might formally be infinite, and other parameters are used to describe criticality. For example, consider an infinite sheet of fissionable material. For any finite thickness, this corresponds to an infinite mass. However, criticality is only achieved once the thickness of this slab exceeds a critical value.