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Automotive Header design?

  1. May 1, 2015 #1
    Hey guys I'm new here. I'm an ME student and I'm working on a project. I originally posted this on my350z forum and no one replied. Maybe it's too boring I don't know. If anyone has experience can you help me? Thanks.

    Hey guys, so I've been recently delving more and more into my z and wanting to design something myself for it. I decided that it would be kinda cool to design my own headers. I have the math down for it I just need help figuring out the length for the primary tubes.

    (Peak Torque RPM) = ((Primary Pipe Area)(88,200)/(1cyl displacement))

    I was looking for ~6k peak torque so I decided to plug in some different pipe areas to see which peak torque rpm I liked.

    1 3/4" pipe area = 2.41in^2
    1 cyl displacement = ((3500/6) * 0.0610237) = 35.597in^3/cylinder

    Plug into formula - Peak TQ RPM = ((2.41)(88200)/(35.597))
    Peak TQ RPM = 5,971 RPM

    This is the closest to what I was looking for.

    Now for a street car, a good multiplication factor for collector size is anywhere from 1.6 - 1.8. I want to be a little more aggressive so I'm going with 1.75.

    So to find it we take the OD of the primary pipe and multiply it times the 1.75 and we get 3.0625 so 3" pipe is almost perfect. I wouldn't want the collector size to be too long so I feel 5" should be good, depending on the fit with the cats. I just don't know how to calculate the tube length. I've searched far and wide but there's no definite answer. Anyone have any idea? Thanks for your time. -David

    EDIT: Ok, so I found a formula from A. Graham Bell's book that says header pipe length is ((850*ED)/RPM) - 3 with ED being 180 + degrees the exhaust valve opens before reaching BDC and RPM is the RPM you're tuning to.

    So I looked up stock cam specs and the EO is 52 degrees. So my length should be ((850*(180+52))/RPM) - 3 which would give me 32.86 inches. Can anyone run through these numbers with me? Any engineers confirm? Fluid dynamics gives me headaches.
     
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  3. May 2, 2015 #2

    jack action

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    There is no way to predict accurately exhaust geometry without a good computational engine simulation. Too much variation mainly due to the exhaust gas temperature varying from one end to the other. What you are trying to do is to size the area to achieve a certain flow velocity and a length to create a path for the pressure waves. These pressure waves travel at the speed of sound ... and speed of sound depends on the gas temperature ... which varies across the pipe length! So the more you simplify, the more assumptions have to be made.

    These rules of thumb are as good as you're gonna get. Le me explain to you where they come from, such that you understand them better.

    Basics

    When the valve opens, a huge pressure wave is sent down the exhaust pipe. Pressure waves are the way gas molecules «communicate» between each other. The message that is sent is «We have lot us coming thru, make some space». This message travels at the speed of sound in all directions. The higher the pressure inside the cylinder, the stronger the pressure wave. Once it reaches an area change, the pressure wave continues to go on, but at a weaker pressure. The larger the area change, the weaker the signal. But it also sends back another pressure wave, but this time a really weak one (think vacuum). This one goes back to the valve with the following message: «We found some free space, it's ok to come this way.» So when this «vacuum» arrives at the valve, it actually helps emptying the cylinder. If they arrive and the valve is close (whether weak or strong pressure waves), they just bounce back into the pipe with the same strength as they came, kind of saying: «There is no way thru here, don't come in here.» These pressure waves bounce back like that until the pressure equalizes across the pipe.

    Of course, because in an engine the exhaust valve keeps opening with fresh new pressurized gases, it never stabilizes. So the pressure waves are everywhere, crossing path with each other. Once in the collector, they go back into the other primary pipes (they look for an exit everywhere) and once they go back into the actual outlet pipe, once they reach another area change (cat, muffler, atmosphere. etc), they send back another weak pressure wave that will go back into every primary pipe (even weaker as it is divided by number of primary pipes). You can see the whole mathematical puzzle of keeping track of all of those pressure waves. And I haven't even got into the math, as when they cross, their pressure combines, which affects their local temperature, which affect their local speed of sound, which means that their actual speed slows down or speeds up while they cross. A big mess.

    Simplification

    primary pipe area

    What we know is that the smaller the pipe area, the stronger will be the pressure wave. But if the area is too small, the actual flow will choke. So it has been found that the port area that creates average port velocities of about 75 m/s performs best. So the volumetric flow rate [itex]\dot{V}[/itex] is equal to the port velocity [itex]v_p[/itex] times the port area [itex]A_p[/itex]:
    [tex]\dot{V} = v_p A_p[/tex]
    But the volumetric flow rate is also dependent on what the engine produces which is one cylinder volume [itex]V_{cyl}[/itex] per exhaust event duration [itex]\theta[/itex] at the given rpm [itex]\omega[/itex]:
    [tex]\dot{V} = \frac{V_{cyl} \omega}{\theta}[/tex]
    So:
    [tex]\omega = \frac{A_p \theta v_p}{V_{cyl}}[/tex]
    This is in SI units, so to convert angular velocity from rpm to rad/s, area from in² to , angle duration from deg to rad and volume from in³ to , we have to include the following conversion factor:
    [tex]\color{red}{\omega = 6.5617 \frac{A_p \theta v_p}{V_{cyl}}}[/tex]
    If we assume that the exhaust duration is one stroke, then [itex]\theta[/itex] = 180° and that [itex]v_p[/itex] = 75 m/s then:
    [tex]6.5617 (180) (75) = 88 583[/tex]
    or:
    [tex]\color{blue}{\omega = 88 583 \frac{A_p}{V_{cyl}}}[/tex]
    Which looks a lot like your equation.

    You referred to A. Graham Bell. His equation for the diameter of the exhaust pipe [itex]D_p[/itex], based on exhaust length [itex]L[/itex], is:
    [tex]D_p = 2.1\sqrt{\frac{V{cyl}}{25 (L+3)}}[/tex]
    Putting it for the area [itex]A_p[/itex] instead:
    [tex]A_p = \frac{\pi}{4}D_p^2 = \frac{\pi}{4}2.1^2\frac{V{cyl}}{25 (L+3)}[/tex]
    Replacing [itex]L[/itex] by his other equation you mentioned, we get:
    [tex]A_p = 3.46\frac{V{cyl}}{25 \left(\frac{850 \theta}{\omega}\right)} = \frac{V{cyl} \omega}{6141.6\theta}[/tex]
    But in his equation, [itex]V_{cyl}[/itex] is in cm³ so it has to be converted to in³ to be consistent with our other equation, so it becomes:
    [tex]A_p = \frac{V{cyl} \omega}{374.8\theta}[/tex]
    or:
    [tex]\color{green}{\omega = 374.8 \frac{A_p \theta}{V_{cyl}}}[/tex]
    Comparing to our theoretical equation (in red) [itex]374.8 = 6.5617 v_p[/itex] or [itex]v_p[/itex] = 57.11 m/s. This might seem different than the 75 m/s of the previous equation, but Bell doesn't assume the exhaust duration is 180°, he takes the 180° + the exhaust valve opening duration. If we assume an exhaust valve opening of 55°, then [itex]\theta[/itex] = 180 + 55:
    [tex]374.8 * (180 + 55) = 88 078[/tex]
    Which is close to the 88 200 that you find in the equation you gave in the OP.

    Conclusion:

    [itex]v_p[/itex] is found empirically. Bell took care of including a better approximation of exhaust duration in his equation to find it, so it might be a better assumption. But it is far from an exact science.

    Collector area

    You mentioned that the «multiplication factor for collector size is anywhere from 1.6 - 1.8». Where does it come from? I told you that it is the area change that counts when comes the time to reflect the strong pressure wave as a weak pressure wave. The largest area change is when the pipe exit directly to the atmosphere (straight pipe), in which case the area ratio is infinite. But it has been found that when the area ratio is larger than about 6, the effect increases very little, meaning that it is almost like being straight pipe. Here is how we calculate the collector's area ([itex]A_c[/itex]) vs the primary pipe's area([itex]A_p[/itex]). The primary pipe exits into the collector pipe, but also in the other primary pipes (say there are [itex]N[/itex] primary pipes in total), then if we want the area ratio to be greater than 6:
    [tex]6 < \frac{A_c + (N-1)A_p}{A_p}[/tex]
    Or:
    [tex]\color{red}{\frac{A_c}{A_p} > 7 - N}[/tex]
    So, if [itex]N[/itex] = 4, then [itex]\frac{A_c}{A_p} > 3[/itex] and since the area is proportional to the square of the diameter, [itex]\frac{D_c}{d_p} > \sqrt{3}[/itex] or 1.73. That is where 1.6-1.8 comes from. You have a V6 - so your collector have 3 primary pipes, not 4 - thus a diameter ratio of 2 would be best suited.

    Primary pipe length

    To define the length of the pipe, we know that a pressure wave has to travel the entire length of the primary pipe [itex]L[/itex], then comes back to the valve and all of that at the speed of sound [itex]v_s[/itex], so the time [itex]t[/itex] taken to do that is:
    [tex]t =\frac{2L}{v_s}[/tex]
    We usually want it to return when the valve closes to help empty the cylinder. The time between the opening of the exhaust valve (when the pressure begins its journey down the pipe) and the end of the exhaust stroke can be found with the crank angle duration [itex]\theta[/itex] and the rpm [itex]\omega[/itex]:
    [tex]t = \frac{\theta}{\omega}[/tex]
    Such that:
    [tex]\frac{2L}{v_s} = \frac{\theta}{\omega}[/tex]
    Or:
    [tex]L = \frac{v_s\theta}{2\omega}[/tex]
    Those are in SI units, so converting rad/s to rpm, rad to deg and m to in you have to include the following conversion factor:
    [tex]L = 3.28\frac{v_s\theta}{\omega}[/tex]
    you should already see a resemblance with Bell's equation:
    [tex]\color{red}{L = \frac{850\theta}{\omega} - 3}[/tex]
    Where [itex]\theta[/itex] is 180° + the exhaust valve opening duration.

    We already mentioned that the speed of sound varies a lot from one end of the pipe to the other, so we can't use the true speed of sound. But we can used some sort of «average» speed of sound for the pipe. This «average» speed is found empirically by testing different pipe length on the same engine. Bell's equation has [itex]850 = 3.28v_s[/itex] or [itex]v_s[/itex] = 259 m/s.

    Bell just subtracted 3 in from this calculated length to take into account the exhaust port length. Of course, you can adjust this number for your particular engine, but this equation is such a crude approximation that it probably wouldn't matter much anyway.

    Collector length

    For the collector length, if you have an area ratio greater than 6, it doesn't matter much: For the exhaust gases, it will feel like they reached the atmosphere. The greater the area ratio, the less significant is the length. Still, in one of my books, there is a «large ballpark» figure for a high-speed drag-racing engine which says that the collector's length should be half the primary pipe length. If the primary pipe length is a crude approximation, imagine how worst is this one.

    With a stock vehicle - with cat, muffler and exit pipe - these would also have an effect, but very little if you've managed to respect the area ratio greater than 6.
     
  4. May 2, 2015 #3
    I'm at a loss for words. That was possibly the best, most well-written, and intelligent reply to any question I've ever received. Thank you so much for clearing a lot up. Basically in the end, I would need to actually run tests and simulations to find the best length, and that's not even considering if I have enough room to fit it all. If only I had the money to do any of this. I figured they were approximations, just not that vague. I don't start any of my physics until this summer and engineering classes until this fall. I've been at a junior college getting a lot of core classes and all of my math out of the way. I'm in DE right now and it's running me over. I'm very excited to take these classes now after seeing just how much I can learn. I haven't been able to find any information like that anywhere and while it's hard to pick up from text, I am really grateful. Hopefully one day I'll be half as smart as you are haha. Thanks again! -David
     
  5. May 27, 2015 #4
    The reply by Jack is spot on. Well written and calls the right things. Supplementing that i would recommend for your project looming into the comparison of a true 4-1 header and a 4-2-1. Essentially the former will give you more Hp but it will be very high in the range of the engine. You cam beat that a little with longer tubes but that reduces high end. The latter (also known as a "tri-y" gives more gain in the low end then a flatter curve ending with slightly lower high end.

    The real fun of designing your own exhaust is in putting the gain where you will use it. Also Concider the rest of the path. A good header followed by a restrictive underbody shows very little real gain on the ground.
     
  6. Oct 19, 2016 #5
    Jack, your response is brilliantly clear. Thank you for it. How did you get this information? From reading A. Graham Bell he gives no clear description of the theoretical case.
    I would love to get hold of one of your books that you talk about this. Can you suggest one?
     
    Last edited: Oct 19, 2016
  7. Oct 19, 2016 #6

    Ranger Mike

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    As always..Jack is the best!!..
    ps
    i would go with tri-Y for the street!
    rm
     
  8. Oct 19, 2016 #7

    jack action

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    The books I used were:
    • DeskTop Dynos (ISBN 1-884089-23-2): Excellent to understand the basic principles, even with a high school degree. A slightly different version is found here.
    • Design and Simulation of Four-Stroke Engines (ISBN 0-7680-0440-3): Very math intensive, but this is basically a recipe to built your own simulation program. Prior knowledge of thermodynamics and gas dynamics is helpful before reading. The problem with this book is how the information is presented: All the equations are there, but trying to find how one links to the other is terribly difficult. I basically had to rewrite the book for myself to understand how to use these equations (thus why I could come up with post #2). But maybe it's just that my brain and the author's brain are connected very differently o0):woot:. There is a version for two-stroke engines also where a lot of the information overlaps.
     
  9. Oct 26, 2016 #8
    I must be wired the same way then as Ibeen reading Blair's book and really struggling. I could never have linked Bell's equations. Thank you once again!
     
  10. Mar 10, 2017 #9
    i don't understand this equation !
     
  11. Mar 10, 2017 #10

    jack action

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    By definition:
    [tex]\dot{V} = \frac{V_{cyl}}{t}[/tex]
    and:
    [tex]\omega = \frac{\theta}{t}[/tex]
    Where ##t## is the time for an exhaust event duration.
     
  12. Mar 11, 2017 #11
    Just adding more kudos for the excellent answer. Adding your thoughts on a writers ability to organize and present information was an added bonus to your answer. Technical publications are often difficult and it's due to no fault of the reader while unfortunately the reader is wondering why they aren't "getting it."
     
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