Port Diameter

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sreten

R.I.P.
Joined 2003
Hi,

the minimum port diameters are determined by the
SPL capability and the maximum air velocity allowed.

Edit :
rereading you post seems this is not what you are asking.
Possibly you are not considering a ports length is not its effective length.
The effective length is somewhat longer depending on dimensions.

:)/sreten.
 
Ron E

Ron E

Member
Joined 2002
jwatts said:
Does any body know how the minimum port diameter equations are derived. I have derived a formula based on the pressure exerted on the port and it gives good numbers but they are slightly smaller than what is typically used by the norm. Could some tell how the norm is usually derived.
Click to expand...

They are usually derived by calculating the air velocity through the port. It depends on which minimum port diameter equations you mean. Post a couple.

While not impossible, I doubt you have come up with a correct formula without knowing how the derivation is performed - but who knows? While the pressure in a box created by a driver does have something to do with the flow through the port and the sound output of the system, there are other things to be considered.

Post your formula or derivation and people can pick it apart.

I think I have posted my formula here, which gives air velocity at Fb given SPL and tuning frequency and port diameter. Do a search and use it to check your results. My derivation merely assumes that all sound at Fb comes from the port, and treats the port as a rigid piston.
 
Notax

Notax

Member
Joined 2004
Xc - calculated excursion

k1=1/(Qts+0.68)*0.68+0.28

Xc=k1*srq(Re)*Sd/BL

Fb=h*Fs
h=0.382/Qts+0.14*Qts

Rt=Dt/2=0.501*sqr(Fb*Xc*Sd)

Its minimum Dt(Rt) for not to come to situation that air velocity in port is greater than 1/10 of C (speed of sound), or 34 m/s.

This is some of my formulas (corections of T/S) for calculations of bass-reflex enclosures.
 
Ron E

Ron E

Member
Joined 2002
I'm not sure that's te hway I would caclulate excursion, but apart from your Xc and h business, the formula:

Dmin=0.5*sqrt(Xmax*Sd*Fb)

is well known. I played around with it some time ago and (from memory) it doesn't seem to correlate all that well with port air velocity. Perhaps I mixed up the units??

That is why I came up with my own formula, which matches my actual simulations of port velocity. I like being able to specify SPL and port velocity and have it spit out a diameter.
 
Notax

Notax

Member
Joined 2004
Ron E said:
I'm not sure that's te hway I would caclulate excursion, but apart from your Xc and h business, the formula:

Dmin=0.5*sqrt(Xmax*Sd*Fb)

is well known. I played around with it some time ago and (from memory) it doesn't seem to correlate all that well with port air velocity. Perhaps I mixed up the units??

That is why I came up with my own formula, which matches my actual simulations of port velocity. I like being able to specify SPL and port velocity and have it spit out a diameter.
Click to expand...

I wrote "corections of T/S". I use some of known formulas, some that I corected, and some of mine formulas or correction coeficients.

Xc is NOT Xmax. Formula for Xc looks simle, but I got it from more complicated calculations with SPL, Sd and other T/S parameters.
Xc is calculated real excursion around Fb, and because of that I got near exact result for Rt(Dt) as in simulation program for port and air velocity.

Give me few examples (with all T/S parameters).
 
Ron E

Ron E

Member
Joined 2002
AndrewT said:
none of you have considered pressure.

But, for a vented cabinet, calculating the pressure might be more than a little difficult.

Has he been particularly clever in deriving an alternative method of analysis?
Click to expand...

I considered pressure in my response.

Actually the sound power output from either a vented or sealed box is equal to the power required to compress the air in the box, but in a vented box it is more than just diaphragm displacement, there is a resonance going on. If you ignore diaphragm output and assume it all comes from the port - things simplify and you get my relations.

Now for Notax.
Consider:
Fs 22 Hz
Qts 0.38
Qes 0.4
Vas 144 Liters
Re 2.7 Ohms
Dia 24.5 cm
Xmax 15 mm
Pe_max 300 Watts
Le 0 Henrys
P_input 90 Watts@8ohms
Box parameters
Vb 100 Liters
Fb 22 Hz

You don't define units, so I assume SI.
From your relation I seem to get 42mm diameter
From the 0.5*sqrt(Vd*Fb) I get 62mm

My equation says, for 107dB and 17m/s at Fb you would need a 135mm diameter vent. For 34m/s you would need 0.7 times that.

From my box simulator calculations I get 80m/s at 90Watts input with a 62mm diameter port and 170m/s at 90 watts input with a 42mm diameter port. So in my opinion both of these calculations give ports that are too small.

Obviously those are linear predictions and port compression would reduce actual velocity.

BTW, when I say 90 watts I actually mean the voltage that would cause 90 watts to be dissipated in an 8 ohm resistor.
V=sqrt(90*8)
 
Ron E

Ron E

Member
Joined 2002
pinkmouse said:
There's some good stuff in this white paper from the guys behind Soundeasy that makes interesting reading.
Click to expand...

Yup. Bohdan's a smart guy. So what? Are you suggesting we all go out and buy soundeasy becasue we cannot possibly cope without it?

My formula is not a nonlinear port velocity simulation, it is a linear simulation based on T/S acoustic theory - but it is still closer than the other two formulas we have seen here.

I modeled with my formula using the woofer and box specs in the article you cite. Mine suggests ~106mm by inputting 104dB and 17 m/s.

Both Notax's and the "conventional formula" give ~56mm diameter

This particular alignment is one where velocity below Fb is much more than at Fb - If that is taken into account, my full sim software recommends a diameter of 120mm. This port would need to be more than 36cm (14") long.

In practice, port design is more an exercise in what will fit, rather than what is technically best.
 
pinkmouse

pinkmouse

Ex-Moderator
Joined 2002
Ron E said:
Yup. Bohdan's a smart guy. So what? Are you suggesting we all go out and buy soundeasy becasue we cannot possibly cope without it?
Click to expand...

No, of course not, I just pointed it out as interesting reading. I also have some links to white papers by the guys behind the Audio Precision testing kit. I can't afford that, but it doesn't mean to say I can't learn things from them.
 
Notax

Notax

Member
Joined 2004
Ron E said:



Now for Notax.
Consider:
Fs 22 Hz
Qts 0.38
Qes 0.4
Vas 144 Liters
Re 2.7 Ohms
Dia 24.5 cm
Xmax 15 mm
Pe_max 300 Watts
Le 0 Henrys
P_input 90 Watts@8ohms
Box parameters
Vb 100 Liters
Fb 22 Hz

You don't define units, so I assume SI.
From your relation I seem to get 42mm diameter
From the 0.5*sqrt(Vd*Fb) I get 62mm

My equation says, for 107dB and 17m/s at Fb you would need a 135mm diameter vent. For 34m/s you would need 0.7 times that.

From my box simulator calculations I get 80m/s at 90Watts input with a 62mm diameter port and 170m/s at 90 watts input with a 42mm diameter port. So in my opinion both of these calculations give ports that are too small.

Obviously those are linear predictions and port compression would reduce actual velocity.

BTW, when I say 90 watts I actually mean the voltage that would cause 90 watts to be dissipated in an 8 ohm resistor.
V=sqrt(90*8)
Click to expand...



1. You didnt give all data.
2. Several times I wrote its HALF-DIAMETAR - Rt in my formula.

From data you gave I reconstruct Sd and BL that miss.

0,062=0,5*sqr(Vd*Fb) its Vd=698,9 cm3 and Sd=466 cm2 and Cms=0,473644 mm/N and Mms=110,5 gr and BL=10,154 N/A

Sd=466 cm2, BL=10,154 N/A

k=0,9215
Xc=k*Sd*sqr(Re)/BL=0,9215*0,0466*1,643/10,154=0,00695 m (6,95 mm)
h=0,382/Qts+0,14*Qts=0,382/0,38+0,14*0,38=1,011
Fb=h*Fs=22,242 Hz
Rt=0,501*sqr(Sd*Fb*Xc)=0,501*sqr(0,0466*22,242*0,00695)=0,0425 m

HALF-DIAMETAR is 42,5 mm
DIAMETAR is 85 mm

And it works for every common driver unit (as minimal for velocity of air in port not become greater then 34 m/s).

In BassBox Pro I got 34,5 m/s with Dt=85 mm.
 
J

jwatts

Member
Joined 2004
quadcenterprise.com
Cool

Sorry for such a late return. I am entirely to busy. It looks like there are several different views on how to do it. I have changed my mind again on what to do. Here is the basic idea. Assume the force exerted on the cross sectional area of the port has to be equal to or greater than that produced by the driver. I am assuming no nonlinearity and the Reynolds number is less than
2000(Laminar Flow).
F=k*x

F(port)=k(port)*x(vent length)
F(driver)=k(driver)*x(Xmax)

k(port)=(1/Cmb)
k(driver)=(1/Cms)

x(vent length)=Lv

(1/Cmb)*Lv=(1/Cms)*Xmax

Lv=((1/Cms)*Xmax)/(1/Cmb)

alpha=Cms/Cmb=Vas/Vb

Xmax/alpha=Lv

Vent length equation below (no flare correction)

Lv=(c^2*D^2)/(4*pi*Fb^2*Vb)

Substitute and solve for D

D=sqrt((Xmax*4*pi*Fb^2*Vb)/(alpha*c^2))

An example result would produce 2.03 inches when other popular equation typically produce 3.7 to 6.7 inches.
 
A

AndrewT

R.I.P.
Joined 2004
Hi,
Assume the force exerted on the cross sectional area of the port has to be equal to or greater than that produced by the driver
Click to expand...
why should we or you assume this?
Where is the correlation between force on the driver (from the voicecoil) and force (pressure) on the column of air in the port?
Laminar Flow). F=k*x
Click to expand...
where does this come from?

Why should { force = length of air column times constant k }
 
Notax

Notax

Member
Joined 2004
Re: Cool

jwatts said:
Sorry for such a late return. I am entirely to busy. It looks like there are several different views on how to do it. I have changed my mind again on what to do. Here is the basic idea. Assume the force exerted on the cross sectional area of the port has to be equal to or greater than that produced by the driver. I am assuming no nonlinearity and the Reynolds number is less than
2000(Laminar Flow).
F=k*x

F(port)=k(port)*x(vent length)
F(driver)=k(driver)*x(Xmax)

k(port)=(1/Cmb)
k(driver)=(1/Cms)

x(vent length)=Lv

(1/Cmb)*Lv=(1/Cms)*Xmax

Lv=((1/Cms)*Xmax)/(1/Cmb)

alpha=Cms/Cmb=Vas/Vb

Xmax/alpha=Lv

Vent length equation below (no flare correction)

Lv=(c^2*D^2)/(4*pi*Fb^2*Vb)

Substitute and solve for D

D=sqrt((Xmax*4*pi*Fb^2*Vb)/(alpha*c^2))

An example result would produce 2.03 inches when other popular equation typically produce 3.7 to 6.7 inches.
Click to expand...


You assume that F of driver and port are same. I am not sure it is like that.
Second, how you get Xmax, or what value you use for Xmax to calculate?
 
A

AndrewT

R.I.P.
Joined 2004
Hi,
well, I have to admit to looking up Hooke's Law.

What has
Laminar Flow). F=k*x
Click to expand...
got to due with stress and strain and their proportionality?

Since we are into compressible fluids in this example you might have done better trying to convince us that Boyle's law is appropriate, or adiabatic expansion, or some other fluid mechanics jargon.

I just cannot see what "F", the force coming from the voice coil of the driver, has to do with "x", the length of the column of air in the port.
 
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