Sesshoumaru

iTrader: (1)

velocity is distance/time

i pulled this from a website



once calculated the velocity you can use

Q = AV

Where:

Q = Quantity of flow in cubic feet per minute.
A = Cross sectional area of duct in square feet.
V = Average velocity in feet per minute.

the formula they use for density does not depend on the ideal gas law.

Static Pressure is usually stated either in inches of water (H2O) or in millimeters of water (mmH2O). It is essentially a measure of the differential air pressure between the air pressures inside an application vs ambient air pressure outside of an application, which for airflow calculation purposes is usually 0 (zero).

In practical situations, the velocity of the air stream is not uniform across the cross section of a duct. Friction slows the air moving close to the walls, so the velocity is greater in the center of the duct

getting some of those readings would be hard but i think the main idea here is how u see they are related.

is this closer?

Detbyron

cool. Now we're getting closer. Where did you pull this from?

I am curious about hv (velocity pressure). What is the definition of it and can it be related to pressure.

paw140

The ideal gas law still applies, even though there is flow. Regardless of the fact that we don't have a closed system (which the ideal gas law does not require, by the way), the relationship is still the same, pv=nrt. No matter how you look at it, in the system we are talking about (constant V and P), increasing T is going to decrease n. Thus a higher temp will result in a lower mass of air.

The ideal gas law applies perfectly in this situation.

DaedelGT

What the hell major were/are you?

Mass can't change ummm... ever?

The ideal gas laws ONLY apply at high temperatures and pressures anyway. And no, the temperatures and pressures we're talking about don't count (excpept maybe at combustion).

Time for me to go read a bit in this thread to get you guys back on track.

DaedelGT

Ideal gas laws apply to flow systems as well. If you guys want proof, I can take pics of my thermodynamics book.

The fact remains that those laws don't work for turbochargers.

The reason higher temperature gasses do not create as much energy, is because the enthalpy of said gasses is higher entering the combustion chamber, leaving less energy to create the mechanical motion. That works on the difference in energys.

paw140

You're way way off. First, volumetric flow = cross sectional area * velocity. Density doesn't matter. If you are talking about mass flow rate, then you are correct. Now, I don't know where you got number of moles (n) = density?? That's quite a leap, especially considering moles is a unitless number, and density is mass per volume.

paw140

Are you talking to me, or the other guys? The ideal gas law applies to gasses with low densities (i.e., high temperature and low pressure). I just pulled that from my Thermodynamics book. I don't know if it applies exactly to the system we are talking about, but regardless, the PVT relationship is still the same.

paw140

I have a BS in Plastics Engineering (just like ME, only for plastics) and am getting my PhD in Polymer Science.

Regarding the mass changing, I'm not sure if you are referring to my comment, but yeah, if P and V are constant, and T is increased, n is going to decrease.

DaedelGT

Where is the mass going if n is decreasing?

paw140

Well, it's not just disappearing

Think back to our example, the FD's induction system. Volume cannot change (unless the inake hoses stretch, but disregard that). We increase the temperature of the intake charge (simulating the difference between a single and twins). What is going to happen to the gas inside the intake system? If you increase the temperature, the gas wants to expand... it can't because the volume doesn't change. Thus it wants exert a higher pressure... but we already stated that we are also keeping pressure constant. The only other variable to change is the mass of the gas, represented by n, the number of moles. Therefore, the intake system can hold less mass of air when it is hot than when it is cold.

Think about it another way. Take a soda bottle at room temperature, open to the atmosphere, and then seal it. Heat the bottle up, and then open up the cap. Air will escape, right? Seal it back up. There is less air mass in the hot bottle than the room temperature bottle, while both have the same volume and pressure.

I apologize for probably not explaining it too well, but there it is.

DaedelGT

Ooooooh!!!

I kept telling my room mate I'm going to sound like an ******* when I finally find out that you guys are talking about something other than what I'm thinking about... I understand exactly what you're saying. I was too lazy to go back and read enough to matter.


*twiddling thumbs*

Detbyron

Like you said, if we keep the pressure and volume the same and temperature goes up, that means that density decreases.
Or to make it time dependant:
keep pressure and cross-sectional area of the TB the same... increase temperature so the density decreases, and thus mass flow rate (not volumetric flow rate like I stated before in error)... assuming that the average velocity remains constant. But I don't know if it would go up or down.

By the way, I majored in ME, which is an embaressment to the school.

Sesshoumaru

iTrader: (1)

well i'm a electrical engineer.

and i only learned the basics.

You ME's figure it out and just tell me. This is getting way out of league.

can't remeber where i pulled it from.

It was on an air conidtioning site or someting (calculating duct width or something)

ReDLiNe@t9

so i guess its worth it to have one

Spool Up

You guys keep wondering where that mass goes....it's cause you can't use the ideal gas law in the sense that you're trying to with that equation. It wasn't designed for solving steady state steady flow (SSSF) problems that this is more atuned to. This is something more along the First Law of Thermodynamics which states that
[Qin + Win + massin(h + PE + KE)] = [Qout + Wout + massout(h + PE + KE)].
Q = enery transfer (mainly in the form of heat here)
W = work done by/on the system
h = enthalpy
PE = potential energy
KE = kinetic energy
I know that it's not, but if we assume that the tubing and manifolds are adiabatic, we will loose the Q terms. Assuming there's no flex in the lines like somebody mentioned earlier, we lose our work terms as well. We can also ditch the PE and KE terms in the equation which leaves us left with it just being a matter of enthalpy and mass. Assuming it's SSSF like we said, then mass remains constant and it's just a deal of enthalpy. This is why DaedelGT said that it's the enthalpy that causes a larger turbo to produce more power. Enthalpy is just the internal energy of the air (u) plus the work done by the air (Pdv) which we said was zero. Since the internal energy of hotter air (stock twins) is much higher than cooler air (single), it's just like DaedelGT said.......i think this might explain why a little better though.

You can probably use the ideal gas law to find certain parts of the equation (mainly the ethalpy at a given temp and pressure), but you can't really solve anything of this sort with it. This situations involves mass flow rates, so when you try to do your ideal gas law, you're not necessarily even comparing the same air on both sides of the equation.

paw140

If you understood my posts, you would see that I was using the ideal gas relationship as a qualitative means of comparing the amount of air in a hot versus a cold system. A hot system will have less mass of air in it than a cold system. Less mass of air means less power. It's as simple as that.

The mass *does not* remain constant. You are making this way more complicated than it is. The equation your threw out doesn't even apply because were are looking at two separate systems, not a steady state flow problem.

Oh, and why are you bringing this thread back from the dead?

wiblergt

Think of it this way...

small turbos = small diameter pipe (straw)
large turbos = larger diameter pipe (storm drain)

10psi of water flowing through a straw gives a small amount of water.

10psi of water flowing through a storm drain yealds A FLOOD!

I don't care if the water/air is hot or cold. That is a separate equation/problem that can be "fixed" with a larger IC, aquamist, or a liquid nitrogen setup. You are going to get more FLOW with a larger TURBO.

the "MODS" we do to the forced air system (hard pipes, cooling procedures, polishing, y-pipes) gets rid of, or decreases, all the other variables in other equations.

Try not to make every situation ONE EQUATION... break it down! Even though your Professor may tell you different, or want you to do the work different, it will help with ALL your Physics, Thermo, or whatever classes!!! TRUST ME... that's how i passed!!


Big Will

Spool Up

Sorry for ressurecting an old post, I was searching for something and saw this so thought i'd leave a comment. when i said that the mass doesn't change, it was in referral to the SSSF at any given instant in time (not over time). I know I might have complicated things, but in that sense so does using any other equation stated before if all that had to be said was cold air's denser that hot.

wilbergt....but as soon as get to the manifolds, your straws become the same size. I'm not trying to solve anything here, I just saw an interesting debate over some thermo-type problems. It's not often you see something like this on the boards, just thought it was fun. Haha...i definately agree w/ what you said about the one equation stuff though, especiall when they plug everything in all at once.

wiblergt

Spool Up... EXACTLY!!! The Thermo debate is very impressive!

when the air gets to the manifold, nothing else matters except HOW HARD IT IS BEING FORCED THROUGH. This is where a BIG TURBO will make a difference...

AND... back to the original topic...

this is when you need an upgraded BOV.

THANK YOU... I'LL BE HERE ALL WEEK!!

BIG WILL

paw140

We've been through this before, dammit What you said is totally irrelevant to this discussion.

This is getting really old, actually. Did you even read the thread?

Sesshoumaru

iTrader: (1)

yeah i was surprised to see this one up from the grave.

It's true though, not many indepth discussions.

wiblergt

YES... PROBABLY 3 TIMES.

the thread keeps evolving into slightly different discussions.

You seem to know your stuff... you are correct.

the reason people keep "bringing this thread back from the dead" is because it is INTERESTING. most people don't discuss, or CAN'T discuss this stuff.

it's a discussion FORUM... any 2 cents should be appreciated!

REX-722JC

As for the octane question, theres 93 octane in the east coast. I'm originally from Jersey and moved to Cali cuz of the military. All I've seen is 91 rarelly I see 92. Cali sucks for our rides cuz smog,cheap gas quality, and very expensive gas almost 2.50 for 91. As for the bov, this is to reduce compressor surge and recirculate the air in to the intake. The stock ecu was meant to read it that way. It gets fooled by thinking its getting more air than it should. When u decide to boost higher i would be good to change your bov,more boost-more air-bigger surge, u r just going to shorten the life of your turbo. Its also a good idea to get rid of your stock ecu if u going to do anything that changes air flow on your engine. Almost any change you do to the car is like a slap in the face to your stock ecu.....

Fred Sickert

So if you have a really big turbo, would it run at a lower psi at the outlet than at the throttle body ?

RX7UP

Uh oh!....... Here we go again......