Bernoulli's principle
Thread Starter
|
Tuned
Joined: Feb 2007
Posts: 434
Likes: 0
Bernoulli's principle
If you aren't familiar with the principle it states (in my own words): when a fluid/gas is flowing through a pipe when the diameter of the pipe decreases the velocity increases and pressure decreases. and when a fluid/gas is flowing through a pipe and the diameter increases, velocity decreases and pressure increases.
now putting this theory into exhaust, wouold it be correct to say that if you started out with a 3" pipe, went a couple feet and then stepped it down to 2 3/4" then went a couple feet and stepped it down again to 2 1/2" that you would have better exhaust flow and ultimately increase power?
And going from 3" pipe and stepping up would decrease flow and increase preassure untimately decreasing power?
if this is a repeat subject just let me know, but what do you guys think? i know we have some intelligent people in here.
now putting this theory into exhaust, wouold it be correct to say that if you started out with a 3" pipe, went a couple feet and then stepped it down to 2 3/4" then went a couple feet and stepped it down again to 2 1/2" that you would have better exhaust flow and ultimately increase power?
And going from 3" pipe and stepping up would decrease flow and increase preassure untimately decreasing power?
if this is a repeat subject just let me know, but what do you guys think? i know we have some intelligent people in here.
New User
Joined: Apr 2006
Posts: 12
Likes: 0
Don't forget that in that principle, mass flow rate is constant.
The gasses would be moving faster, but you won't be making any more power.
Every see any exhaust that does this? Maybe someone tried it already.....
Greg
The gasses would be moving faster, but you won't be making any more power.
Every see any exhaust that does this? Maybe someone tried it already.....
Greg
Post Fiend
Joined: May 2006
Posts: 10,007
Likes: 4
From: Central Florida
1/2qv(squared)+p+qU= constant
q is density, v is velocity, p is pressure, U is potential energy
Is this the one you mean?
I don't know the specs, but p should be relative and not absolute. q I just don't know about, because we are not dealing with an uncompress-able liquid, density of air will vary with pressure.
However, when you change the pressure or density, you will affect velocity.
Any how you also have some other issues related directly to pipes and flow. The closer you measure velocity to the side wall, the closer velocity will read to 0. Presure will also increase the longer the pipe is. And you don't want to just deal with steady flow, the gasses in the exhaust are actually pulsing for some lenght. The Idea I use is you don't want the pulses to slow before exiting the pipe, but you don't want them to colide either.
q is density, v is velocity, p is pressure, U is potential energy
Is this the one you mean?
I don't know the specs, but p should be relative and not absolute. q I just don't know about, because we are not dealing with an uncompress-able liquid, density of air will vary with pressure.
However, when you change the pressure or density, you will affect velocity.
Any how you also have some other issues related directly to pipes and flow. The closer you measure velocity to the side wall, the closer velocity will read to 0. Presure will also increase the longer the pipe is. And you don't want to just deal with steady flow, the gasses in the exhaust are actually pulsing for some lenght. The Idea I use is you don't want the pulses to slow before exiting the pipe, but you don't want them to colide either.
Last edited by ReAX; Aug 10, 2007 at 07:37 PM.
Thread Starter
|
Tuned
Joined: Feb 2007
Posts: 434
Likes: 0
Don't forget that in that principle, mass flow rate is constant.
The gasses would be moving faster, but you won't be making any more power.
[/QUOTE]
Well jet engines use this theory to make more power, and be more efficient in all the sections of the engines, so im thinkin you would definitely be making more power by incorporating this into a cars exhaust system.
The gasses would be moving faster, but you won't be making any more power.
[/QUOTE]
Well jet engines use this theory to make more power, and be more efficient in all the sections of the engines, so im thinkin you would definitely be making more power by incorporating this into a cars exhaust system.
Post Fiend
Joined: May 2006
Posts: 10,007
Likes: 4
From: Central Florida
As I understand it though, a jet engine has no cylinder or seal combustion chamber. Just a venturi to accelerate the air so when the fuel is ignited, the expanding heated gases only travel in one direction.
To use your theory, you will want to create a mild vacuum behind the tip. Hopefully that will increase the headers/manifolds scavaging action and suck all of the gases out of the cylinders. Yes, if you can completely evacuate the cylinders, you could replace the burnt gas with fresh a/f.
To use your theory, you will want to create a mild vacuum behind the tip. Hopefully that will increase the headers/manifolds scavaging action and suck all of the gases out of the cylinders. Yes, if you can completely evacuate the cylinders, you could replace the burnt gas with fresh a/f.
Trending Topics
Elder User
Joined: Apr 2006
Posts: 804
Likes: 0
From: Edmonton, AB
You are not considereing that pressure in front of the step change would be higher because the smaller pipe downstream is more restrictive.
Two throw some food for thought back to you:
Why not just run the same smaller pipe all the way through?
Where is the smallest diameter in the system right now??
Just to parallel to real world why your idea does not work, Nascar v8s start small and take 2 steps larger before the collector.
Two throw some food for thought back to you:
Why not just run the same smaller pipe all the way through?
Where is the smallest diameter in the system right now??
Just to parallel to real world why your idea does not work, Nascar v8s start small and take 2 steps larger before the collector.
FTE Stories
Ford Trucks for Ford Truck Enthusiasts
AEV FXL Super Duty - the Super Duty Raptor Ford Doesn't Make
Brett Foote
Lobo Vs Lobo: Proof the F-150 Lobo Should Be Even Lower!
Michael S. Palmer
Ford's 2001 Explorer Sportsman Concept Looks For a New Home
Verdad Gallardo
10 Best Ford Truck Engines We Miss the Most!
Joe Kucinski
2026 Shelby F-150 Off-Road: Better Than a Raptor R?
Brett Foote
2027 Super Duty Carhartt Package First Look: 12 Things You NEED to Know!
Michael S. Palmer
10 Most Surprising 2026 Ford Truck Features!
Joe Kucinski
Top 10 Ford Trucks Coming to Mecum Indy 2026
Brett Foote
5 Best / 5 Worst Ford Truck Wheels of All Time
Joe Kucinski
Cross-Country
Joined: Apr 2004
Posts: 71
Likes: 0
From: Leesburg, VA
Originally Posted by duffman77
Just to parallel to real world why your idea does not work, Nascar v8s start small and take 2 steps larger before the collector.
Post Fiend
Joined: May 2006
Posts: 10,007
Likes: 4
From: Central Florida
I think this is the info you are looking for.
2" pipe head loss, 1' is 15.0gpm and 10' is 52gpm
2.25 1' is 21.0gpm and 72.8gpm at 10'
2.5 1' is 28.3 gpm and 98.1 at 10'
3 47.1 at 1' and 163 at 10'
references on 0.083 pipe and an non-compressable liquid.
I have never tried to use this data, so I am not 100% on how to apply it. However, you have a loss due to pipe size, it appears that the larger the pipe, the more loss. however, these are probably at constant pressures so they can be applied to the rest of the equation.
2" pipe head loss, 1' is 15.0gpm and 10' is 52gpm
2.25 1' is 21.0gpm and 72.8gpm at 10'
2.5 1' is 28.3 gpm and 98.1 at 10'
3 47.1 at 1' and 163 at 10'
references on 0.083 pipe and an non-compressable liquid.
I have never tried to use this data, so I am not 100% on how to apply it. However, you have a loss due to pipe size, it appears that the larger the pipe, the more loss. however, these are probably at constant pressures so they can be applied to the rest of the equation.
Elder User
Joined: Apr 2006
Posts: 804
Likes: 0
From: Edmonton, AB
Originally Posted by bad4dr
NASCAR engines are designed to run at 7,000+ RPM for hundreds of miles. They need peak HP gains at the very top of the powerband. No real world correlation whatsoever. Street engines spend most of their lives between idle and 5,500 RPM, especially truck engines. Their requirements are very different.
Originally Posted by bad4dr
A good, reliable street exhaust system will have 2 1/2" tubes into the muffler, and 2 1/4" tubes from the muffler back. The added backpressure of the smaller diameter pipe will create a torque advantage over larger tubes. CAR CRAFT magazine did a comparison of a full 3" exhaust vs. a 2 1/2" exhaust. The 3" system made more power upstairs (and made more noise), but it lost torque points to the smaller tubes. Fight the urge to run sewer pipe exhaust tubing on a truck unless it's a mudbogger or drag truck. You'll spend more money and have nothing to show for it but ringing ears.
Elder User
Joined: Apr 2006
Posts: 804
Likes: 0
From: Edmonton, AB
Sorry just rereading this post, my bad, cant read, full exhaust vs full exhaust. I dont change my opinion on moving to a smaller pipe downstream though and what I posted is still not wrong, they just moved past optimum as I stated.
Elder User
Joined: Jul 2004
Posts: 811
Likes: 0
995.4SD you have it a bit backwards. Think of it like this: no pipe=no backpressure, more pipe or smaller pipe = more backpressure. The is an optimum backpressure for each engine, that is the headers job. Beyond the collector the less you have, the better.
Elder User
Joined: Apr 2004
Posts: 797
Likes: 0
From: Wickenburg, AZ
Wow! All the old engineering stuff. I think we wrote it as delta P to signify that it was a change in pressure across the direction of flow. The pressure at the exhaust tip is going to be atmospheric. As you shrink the diameter of the pipe, you're going to need an ever greater amount of pressure at the exhaust manifold to force the exhaust out.
In optimizing flow, you're seeking to drive the required pressure at the manifold down, so you don't have to have the engine work so hard pushing out exhaust ("pump loss"). The reason why you have these "optimums" is that you start encountering turbulent flow in the pipes. When you hit turbulent flow, it's going to require more energy to move the exhaust through, translating back as more backpressure. Turbulent flow starts when the Reynolds Number hits 2100 or so. Reynolds# = Density*Velocity*Diameter of pipe/(fluid viscosity). That's the way I learned it, but judging from the posts here, we've definitely got some Engineers here that will take it well beyond what I can give.
In optimizing flow, you're seeking to drive the required pressure at the manifold down, so you don't have to have the engine work so hard pushing out exhaust ("pump loss"). The reason why you have these "optimums" is that you start encountering turbulent flow in the pipes. When you hit turbulent flow, it's going to require more energy to move the exhaust through, translating back as more backpressure. Turbulent flow starts when the Reynolds Number hits 2100 or so. Reynolds# = Density*Velocity*Diameter of pipe/(fluid viscosity). That's the way I learned it, but judging from the posts here, we've definitely got some Engineers here that will take it well beyond what I can give.