Oval pipe flow as good as round?
Oval pipe flow as good as round? For example 4inch being 2 3/4 tall and 4 3/8 wide or 3.5inch being 2 1/2 tall and 4 wide? Would it flow the same as a true round pipe with the same overall diameter?
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Joined: Apr 2006
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From: Naperville, IL
A perfect circle will outflow a non-perfect circle with the same circumference if that's what you're refering to, due to possessing more area. Get the exact sizes of the pipes you're comparing, and from there it's just a little calculus to determine the area inside each pipe. The bigger number will flow more.
Well in a nutshell the flat pipe should flow faster at a higher velocity. I have a set of Micron Serpent headers on my bike and they have a flattened section in them that prevents reversion by keeping the velocity up.
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Joined: Apr 2006
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From: Naperville, IL
Years ago, one of the car mags tested this on a flow bench. IIRC, they found that given the same cross sectional area, the oval pipe flowed SLIGHTLY better than the round pipe.
That's been at least 15 years ago... Damn I'm getting old.
That's been at least 15 years ago... Damn I'm getting old.
It's been a while since I've done any fluid dynamics, but here's a brief rundown for you:
Flow is based on a number of factors, but the primary one dealing with shape is Hydraulic Radius, which is calculated by Rh = dA/dP.
Given the Area of a Circle = pi*r^2, Perimeter = 2*pi*r
And the Area of an Ellipse = pi*a*b, Perimeter = pi(a+b)K, where a and b are the long and short radii, and K = (1 + 1/2m^2+1/64m^4...), where m=(a-b)/(a+b)
Set the perimeters equal (same size initial pipe to be used for circular and elliptical cross sections), so 2*pi*r = pi(a+b)K, which reduces to d = (a+b)K.
Therefore Rhc (circle) = (pi*r^2)/(2*pi*r) = r/2 or d/4
Rhe (ellipse) = (pi*a*b)/(pi(a+b)K) = a*b/d (remember our nice substitution earlier)
If we look at the numbers proposed:
4" circular - Rhc = 4/4 = 1
2 3/4 tall and 4 3/8 wide - Rhe = (1 3/8 * 2 3/16)/(2d) = .7519
3.5" circular - Rhc = .875
2 1/2 tall and 4 wide - Rhe = .625
In both of these cases, the hydraulic radii are decreased by going to an elliptical shape, which would, all else being equal, mean the elliptical pipes flow less than their circular counterpart. This, however, is best determined experimentally, as a number of other factors, such as temperature and pressure differentials, turbulent or laminar flow, friction of the pipe, fittings, etc. all greatly impact the real world dynamics of fluids in a pipe. In other words, depending on other factors, the advantage may go to the elliptical pipe.
Ultimately - in the situation you're looking at, either pipe would provide ample area for the exhaust of an LS1, so I wouldn't worry much about it. Just use what you've got that works.
Flow is based on a number of factors, but the primary one dealing with shape is Hydraulic Radius, which is calculated by Rh = dA/dP.
Given the Area of a Circle = pi*r^2, Perimeter = 2*pi*r
And the Area of an Ellipse = pi*a*b, Perimeter = pi(a+b)K, where a and b are the long and short radii, and K = (1 + 1/2m^2+1/64m^4...), where m=(a-b)/(a+b)
Set the perimeters equal (same size initial pipe to be used for circular and elliptical cross sections), so 2*pi*r = pi(a+b)K, which reduces to d = (a+b)K.
Therefore Rhc (circle) = (pi*r^2)/(2*pi*r) = r/2 or d/4
Rhe (ellipse) = (pi*a*b)/(pi(a+b)K) = a*b/d (remember our nice substitution earlier)
If we look at the numbers proposed:
4" circular - Rhc = 4/4 = 1
2 3/4 tall and 4 3/8 wide - Rhe = (1 3/8 * 2 3/16)/(2d) = .7519
3.5" circular - Rhc = .875
2 1/2 tall and 4 wide - Rhe = .625
In both of these cases, the hydraulic radii are decreased by going to an elliptical shape, which would, all else being equal, mean the elliptical pipes flow less than their circular counterpart. This, however, is best determined experimentally, as a number of other factors, such as temperature and pressure differentials, turbulent or laminar flow, friction of the pipe, fittings, etc. all greatly impact the real world dynamics of fluids in a pipe. In other words, depending on other factors, the advantage may go to the elliptical pipe.
Ultimately - in the situation you're looking at, either pipe would provide ample area for the exhaust of an LS1, so I wouldn't worry much about it. Just use what you've got that works.
Last edited by 2002BlackSS; Feb 29, 2008 at 06:51 PM.
Staging Lane
Joined: Mar 2015
Posts: 87
Likes: 0
From: Texas
It's been a while since I've done any fluid dynamics, but here's a brief rundown for you:
Flow is based on a number of factors, but the primary one dealing with shape is Hydraulic Radius, which is calculated by Rh = dA/dP.
Given the Area of a Circle = pi*r^2, Perimeter = 2*pi*r
And the Area of an Ellipse = pi*a*b, Perimeter = pi(a+b)K, where a and b are the long and short radii, and K = (1 + 1/2m^2+1/64m^4...), where m=(a-b)/(a+b)
Set the perimeters equal (same size initial pipe to be used for circular and elliptical cross sections), so 2*pi*r = pi(a+b)K, which reduces to d = (a+b)K.
Therefore Rhc (circle) = (pi*r^2)/(2*pi*r) = r/2 or d/4
Rhe (ellipse) = (pi*a*b)/(pi(a+b)K) = a*b/d (remember our nice substitution earlier)
If we look at the numbers proposed:
4" circular - Rhc = 4/4 = 1
2 3/4 tall and 4 3/8 wide - Rhe = (1 3/8 * 2 3/16)/(2d) = .7519
3.5" circular - Rhc = .875
2 1/2 tall and 4 wide - Rhe = .625
In both of these cases, the hydraulic radii are decreased by going to an elliptical shape, which would, all else being equal, mean the elliptical pipes flow less than their circular counterpart. This, however, is best determined experimentally, as a number of other factors, such as temperature and pressure differentials, turbulent or laminar flow, friction of the pipe, fittings, etc. all greatly impact the real world dynamics of fluids in a pipe. In other words, depending on other factors, the advantage may go to the elliptical pipe.
Ultimately - in the situation you're looking at, either pipe would provide ample area for the exhaust of an LS1, so I wouldn't worry much about it. Just use what you've got that works.
Flow is based on a number of factors, but the primary one dealing with shape is Hydraulic Radius, which is calculated by Rh = dA/dP.
Given the Area of a Circle = pi*r^2, Perimeter = 2*pi*r
And the Area of an Ellipse = pi*a*b, Perimeter = pi(a+b)K, where a and b are the long and short radii, and K = (1 + 1/2m^2+1/64m^4...), where m=(a-b)/(a+b)
Set the perimeters equal (same size initial pipe to be used for circular and elliptical cross sections), so 2*pi*r = pi(a+b)K, which reduces to d = (a+b)K.
Therefore Rhc (circle) = (pi*r^2)/(2*pi*r) = r/2 or d/4
Rhe (ellipse) = (pi*a*b)/(pi(a+b)K) = a*b/d (remember our nice substitution earlier)
If we look at the numbers proposed:
4" circular - Rhc = 4/4 = 1
2 3/4 tall and 4 3/8 wide - Rhe = (1 3/8 * 2 3/16)/(2d) = .7519
3.5" circular - Rhc = .875
2 1/2 tall and 4 wide - Rhe = .625
In both of these cases, the hydraulic radii are decreased by going to an elliptical shape, which would, all else being equal, mean the elliptical pipes flow less than their circular counterpart. This, however, is best determined experimentally, as a number of other factors, such as temperature and pressure differentials, turbulent or laminar flow, friction of the pipe, fittings, etc. all greatly impact the real world dynamics of fluids in a pipe. In other words, depending on other factors, the advantage may go to the elliptical pipe.
Ultimately - in the situation you're looking at, either pipe would provide ample area for the exhaust of an LS1, so I wouldn't worry much about it. Just use what you've got that works.
