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Perforated Pipe Distributor Sizing Calculations




Perforated Pipe Distributor Sizing Calculations Perforated Pipe Distributors have been discussed many times on “Cheresources”. How to design a perforated pipe distributor also known as a sparger has been a frequent question on the forum. Some general replies without providing in-depth methodology of sizing a distributor can be seen in the queries raised.

Perry’s Handbook (8th Edition) has a brief sub-section on perforated pipe distributors in Section 6, Fluid & Particle Dynamics which I had gone through critically but which was not to my satisfaction.
 
Since I do have a penchant of finding and developing engineering calculations, I was continuously on the look-out for something related to perforated pipe distributors. Last week I struck gold, when I found a company design manual providing detailed calculations for perforated pipe distributors. I had a critical look at it and found it to be better than anything I had found and read earlier related to perforated pipe distributors. I could even develop an excel workbook for perforated pipe distributor from this very informative design procedure.
 
Today’s blog entry is meant to provide the stepwise calculations for a perforated pipe distributor including the design equations. Both liquid and gas pipe distributors are covered. Gas pipe distributors follow the same design steps as that for liquid pipe distributors except for some minor variations in the design procedure. Both Metric units and USC units have been provided in the methodology.
 
Let us straight away get to the steps for sizing a Perforated Pipe Liquid distributor:
 
Step 1:
Initially set the pipe size of the pipe distributor, same as the pipe size feeding the pipe distributor
 
Step 2:
Calculate the Reynolds number, Rei, of the inlet stream to the pipe distributor using the following equation:
 
Metric Units:
Rei = 1.27*Q*ρ / (d*μ)
 
USC Units:
Rei = 50.6*Q*ρ / (d*μ)
 
where:
Q = Volumetric flow Rate, liters/s (gpm)
ρ = Liquid Density, kg/m3 (lb/ft3)
d = inside diameter of pipe, mm (inch)
μ = Liquid Viscosity, Pa.s (cP)
 
Step 3:
Find the fanning friction factor f (dimensionless), for the given flow regime. For turbulent flow some values for fanning friction factor based on pipe size are provided in the attached table.
Attached Image
 
Step 4:
Calculate the Kinetic Energy per unit volume of the inlet stream, Ek, kPa (psi) from the following equations:
 
Metric Units:
Ek = 810*α*ρ*Q2 / (d4)
 
USC Units:
Ek = 1.8*10-5*α*ρ*Q2 / (d4)
 
where:
Ek = Kinetic energy per unit volume of the inlet stream, kpa (psi)
α = velocity correction factor (use α = 1.1 for turbulent flow and 2.0 for laminar flow)
 
Step 5:
Calculate the pressure change ΔPp along the distributor pipe due to friction and momentum recovery from the following equations:
 
Metric Units:
ΔPp = ((4000*f*L*J/α*d) – 1)*Ek
 
USC Units:
ΔPp = ((48.0*f*L*J/α*d) – 1)*Ek
 
where:
f = fanning friction factor, dimensionless
L = Length of perforated distributor pipe, m (ft)
J = dimensionless factor (Use J = 0.35 as an initial value)
 
Step 6:
Find the required pressure drop, ΔPo across the distributor holes by multiplying the greater of Ek or ΔPp by 10. If the calculated value of ΔPo is less than 1.75 kPa (0.25 psi) make it equal to at least 1.75 kPa (0.25 psi).   
 
Step 7:
Calculate the required total area of the pipe distributor holes, Ao, using the following equations:
 
Metric Units:
Ao = 22.3*(Q / C)*sqroot (ρ/ΔPo)
 
USC Units:
Ao = 3.32*10-3*(Q / C)*sqroot (ρ/ΔPo)
 
where:
Ao = Total required hole area, mm2 (in2)
C = flow coefficient, dimensionless (as a first value consider C = 0.60)
 
Step 8 (Guidelines for choosing hole diameter and hole-to-hole linear distance):
a. Minimum hole diameter ≈ 1/2-in. (13 mm) to avoid plugging and to limit the number of holes to a reasonable value. In very clean service, smaller holes may be considered, but in severely fouling service, 1/2-in. (13 mm) holes may be too small.
b. Maximum hole diameter = 0.2 times inside diameter of distributor.
c. The ratio of hole diameter, do, to inside pipe diameter, di, should be between 0.15 and 0.20 when the criterion ΔPo = 10 Ek is used. If it is necessary to use do /di < 0.10, then make ΔPo = 100 Ek.
d. To provide sufficient pipe strength, the minimum distance (edge-to-edge) between adjacent holes should approximately equal the hole diameter.
e. Within the limitations imposed by the above requirements, a larger number of small holes are preferred over a smaller number of large holes.
f. If slots are used instead of holes, the slot width should be at least 1/2-in. (13 mm).
 
Step 9:
The value of Rei / n should be greater than 4000. If it is not, then a new flow coefficient value "C" should be calculated from the attached chart. Instead of Re shown on the x-axis of the chart use Rei / n for reading the flow coefficient of the chart.
Attached Image
 
Step 10:
Using the calculated number of holes, find the factor "J" from attached chart and compare this with the assumed value of 0.35. If this revised value of J affects the value of ΔPo by more than 10%, substitute the revised value of J in equation given in Step 5 and repeat the steps starting from calculation of ΔPp.
Attached Image
 
Perforated Pipe Gas Distributor:
 
Step 1:
Initially set the pipe size of the pipe distributor, same as the pipe size feeding the pipe distributor
 
Step 2:
Calculate the Reynolds number, Rei, of the inlet gas stream to the pipe distributor using the following equation:
 
Metric Units:
Re = 1270*W*ρ / (d*μ)
 
USC Units:
Re = 6310*W*ρ / (d*μ)
 
where:
W = gas flow Rate, kg/s (lb/h)
ρ = Liquid Density, kg/m3 (lb/ft3)
d = inside diameter of pipe, mm (inch)
μ = Liquid Viscosity, Pa.s (cP)
 
Step 3:
Find the fanning friction factor f (dimensionless), for the given flow regime. For turbulent flow some values for fanning friction factor based on pipe size are provided in the attached table.
Attached Image
 
Step 4:
Calculate the Kinetic Energy per unit volume of the inlet stream, Ek, kPa (psi) from the following equations:
 
Metric Units:
Ek = 8.10*108*α*W2 / (ρ*d4)
 
USC Units:
Ek = 0.28*α*W2 / (ρ*d4)
 
where:
Ek = Kinetic energy per unit volume of the inlet stream, kpa (psi)
α = velocity correction factor (use α = 1.1 for turbulent flow and 2.0 for laminar flow)
 
Step 5:
Calculate the pressure change ΔPp along the distributor pipe due to friction and momentum recovery from the following equations:
 
Metric Units:
ΔPp = ((4000*f*L*J/α*d) – 1)*Ek
 
USC Units:
ΔPp = ((48.0*f*L*J/α*d) – 1)*Ek
 
where:
f = fanning friction factor, dimensionless
L = Length of perforated distributor pipe, m (ft)
J = dimensionless factor (Use J = 0.35 as an initial value)
 
Step 6:
Find the required pressure drop, ΔPo across the distributor holes by multiplying the greater of Ek or ΔPp by 10. If the calculated value of ΔPo is less than 1.75 kPa (0.25 psi) make it equal to at least 1.75 kPa (0.25 psi).
 
Step 7:
Calculate the required total area of the pipe distributor holes, Ao, using the following equations:
 
Metric Units:
Ao = 2.24*104*(W/C*Y)*(1 / sqroot (ρ*ΔPo))
 
USC Units:
Ao = 2.24*104*(W/C*Y)*(1 / sqroot (ρ*ΔPo))
 
where:
Ao = Total required hole area, mm2 (in2)
C = flow coefficient, dimensionless (as a first value consider C = 0.60)
Y = Gas expansion factor, dimensionless
 
Y is calculated as follows:
 
Y = 1 - (0.41 + 0.35*β4)*ΔPo/(k*P) for ΔPo/P <0.37 -------(i)
 
Y = Y0.37 - 0.37*( (ΔPo/P) - 0.37) for ΔPo/P >0.37 --------(ii)
 
where:
P = Pressure at the inlet of the gas distributor, kPa (abs) (psia)
k = ratio of specific heats, Cp / Cv
β = ratio of hole to the perforated pipe inside diameter (specified earlier as a value between 0.15 to 0.20)
Y0.37 = value of Y calculated using eqn (i) with ΔPo / P equal to 0.37
 
Steps 8,9 & 10: These remain the same as for Perforated Pipe Liquid Distributors.
 
This has been a rather detailed description on how to size a perforated pipe distributor and I am hoping that readers and members of "Cheresources" find it useful and can build a excel workbook using the equations and method provided. Please do refer the attachments while reading the text of this blog entry.
 
Looking forward to comments from the knowledgeable forum members.
 
Regards,
Ankur.




 

 

 

Dear Ankur,

Thank you very much for your valuable design calculation  method. Actually I am designing a perforated gas distributor for a vaccum vessel in which the outlet perforation area is exposed to a vaccum of 1 Pascal.In the above explained calculations we are getting a perforation area which is lesser than the inlet area of the tube.That doesn't satisfy the conductance principle for a vaccum vessel in which the perforation area of the tube should be atleast equal to the inlet area.So does this calculation is valid for this type of design....??

Please consider minimum distributor hole size as 13 mm (1/2"). Based  on the length of the distributor considering limitations of arranging the distributor pipe in the vessel, provide equidistant holes of 13 mm along the length of the distributor pipe. Since this is a vacuum service the chances of flow maldistribution are negligible.

 

Regards,

Ankur.

 

Dear Sir,

Since the flow "mal distribution" is negligible.So it is not necessary to maintain Rei/n of 4000 as per your specification mentioned above right?? and also whether our outlet perforation area can be higher than the inlet tube area for the vaccum system or it can be less also..Please suggest..Thanks in advance..

 

For gas distributors mal-distribution is negligible. It is more applicable to liquid distributors. I don't see any specific concern for the perforated area being more than the total area of the distributor pipe. Please provide some back-up where it mentions that the perforation area should be less than the total area of the inlet pipe.

 

Regards,

Ankur.

 

Dear Sir,

As per your calculations for our pipe.I am getting 

perforated area=0.2 times of total inlet area

But As per the need for the vaccum system in our design

We need the perforated area=1.25 times the total inlet area so that there won't                                                    be any back pressure at the outlet.

In this aspect what can be done to achieve this area for the perforations.

 

Regards,

 

Sharjun

Pressure drops or backpressure in gas distribution systems are negligible. In your case, you are sparging  into a vacuum vessel. With the kind of pressure ratio (Phole / Pvessel), it is very likely that you may have very high velocities (check if it is sonic velocity) at the vena contracta leading to very high turbulence and excellent mixing and distribution of the sparged gas in the vessel.

 

I would suggest you post your query in the discussion forums for others to review and provide their opinion and insight.

 

Regards,

Ankur.

Dear Ankur,

 

I would like to construct a gasfeed with square sectional area. I need 2mm holes (outlets) because smaller than that it can be covered and larger than that can localize plasma inside. 

I would like to achieve equal feed distribution through the outlets (2mm holes).

 

I´m working with low pressure (1Pa) and N2. 

 

Do you have any hints about how can I calculate this? I would like to associate the gas distribution with the amount of holes, the distance between the holes and the volume flow rate in the inlet.

 

Thanks

 

Regards

 

Gustavo

Gustavo,

 

The calculation procedure for perforated pipe distributor is quite explicit. However. I have never come across a pipe distributor with "1 Pa" gas pressure.

 

Your question is too general. If you have any specific doubts on the procedure and steps given in the article I can try providing an explanation.

 

Regards,

Ankur.

Hi Ankur,

 

thank you for your answer. The doubt I have is because I cannot go forward from the step 6 because I have very low pressure drop (3,72e-5 Pa) and your explanation works if ΔPo is at least 1.75kPa. Do you have any suggestion for my case? 

 

Just to make a little bit more clear:

1-I have set the hole diameter as 2mm 

2- Now I´m looking forward to find the ideal amount of holesthe ideal space between them and the ideal pipe diameter in order to have equal gas feed through the holes.

 

any suggestion would be very appreciated

 

Regards

 

Gustavo

Gustavo,

 

Keep the pipe diameter as:

 

do / di = 0.15 to 0.20 where:

 

do = hole diameter and

di = pipe inside diameter

 

Providing more number of holes will give better gas distribution. You can try an "edge-to-edge" distance of 10mm between any two holes considering mechanical strength of low diameter pipe and ease of punching holes in the pipe.

 

Hope this helps.

 

Regards

Ankur.

Dear Ankur,

 

for sure it will help. But what modifications should I do in the steps you showed in your article, specially from Step5.  I´m afraid I change something and because of that maybe the equation should be adpated, otherwise the results will be wrong.

 

thanks and Best Regards

Gustavo,

 

Not every design has to be as per equations. Some practical experience and common sense can be used to develop a workable design. Maybe the first time it may not be a perfect design but you can always do modifications to improve a design.

 

Why don't you try out the suggestions I have given? At least you will have something to start with and later you can do some modifications for improvement.

 

I do not have theoretical equations to offer for your case, so you will have to perform some trial and error design based on what I have suggested.

 

Regards,

Ankur.

Dear Ankur,

 

can the fanning friction factor be much greater than 1? I have laminar flow, with very low Reynolds(3,8e-04). So calculating f=16/RE I will find a very high f ~ 4,21e04. 

 

Regards

 

Gustavo

Dear Sir,

  Thanks a lot for this detailed design. It helped me a lot to design a sparger.

Regards

Amit

Hello Ankur

Thank you  for your guidance.

I need to design a perforated pipe for washing e Vessel content. Its configuration is a ring with piping like a plus shape in it.

What is the difference between this and a straight pipe design?

And one more question.  The minimum of Re/n ??

 

Regards

mrym

Hello Ankur

Thank you  for your guidance.

I need to design a perforated pipe for washing e Vessel content. Its configuration is a ring with piping like a plus shape in it.

What is the difference between this and a straight pipe design?

And one more question.  The minimum of Re/n ??

 

Regards

mrym

mrym,

 

Yes, you can use it for your plus shaped  pipe distributor.

 

Rei / n is Reynolds number divided by the number of holes and it should be greater than 4000.

 

Regards,

Ankur.

Hi Ankur,

 

I just want to check my understanding of your calculation.

 

For Liquid Distributor, I interpet determining ΔPo as follows:

 

Step 6   ΔPo = MAX(10*Ek, 10*ΔPp, 1.75)           for 0.15< do / di < 0.20

Step 7c ΔPo = 100*Ek                                         for di/do < 0.10

 

Is ΔPo simply a function of only Ek for do / di < 0.10??

 

For Gas Distributor, I interpet determining ΔPo as follows:

 

Step 6   ΔPo = MAX(10*Ek, 10*ΔPp, 1.75)           for all do / di

 

Or does the same criteria apply for 0.15< do / di < 0.20

and for di/do < 0.10 as in the Liquid Distributor case??

 

Thank you Brian

Hi Ankur,
 
I just want to check my understanding of your calculation.
 
For Liquid Distributor, I interpet determining ΔPo as follows:
 
Step 6   ΔPo = MAX(10*Ek, 10*ΔPp, 1.75)           for 0.15< do / di < 0.20
Step 7c ΔPo = 100*Ek                                         for di/do < 0.10
 
Is ΔPo simply a function of only Ek for do / di < 0.10??
 
For Gas Distributor, I interpet determining ΔPo as follows:
 
Step 6   ΔPo = MAX(10*Ek, 10*ΔPp, 1.75)           for all do / di
 
Or does the same criteria apply for 0.15< do / di < 0.20
and for di/do < 0.10 as in the Liquid Distributor case??
 
Thank you Brian

Brian,

Liquid Distiributor:
That is correct. For the case do / di <0.10 ΔPo is only a function of Ek. But make sure that when using in formula the absolute value of ΔPp is used because many times the calculated ΔPp value is negative.

 

Gas Distributor: Same criteria applies for gas distributor also as for liquid distributor

 

However, it is important to note that gas pressure drops are negligible and thus allow the use of more number of holes with smaller diameter unless the gas is dirty with entrained solids or has entrained liquid. Entrained solids can plug up small holes and also cause erosion while entrained liquid droplets are more likely to cause erosion of the hole. Having more holes of smaller diameter keeping in mind the constraints mentioned above naturally gives better distribution of the gas in applications where intimate mixing of the sparged gas is desirable.

 

Regards,

Ankur.

Hi Ankur,

 

Thank for the procedure of designing perforated pipes. I have applied it several times during different projects.

However, in step 5 to calculate the pressure drop it seems there is a mistake in the formula or units:

Is it  dp=((4000*f*L*J/α*d) – 1)*Ek or ((4000*f*L*J/(α*d)) – 1)*Ek

Most of the time it is negative value. what is the unit for d? is it mm or meter?

 

Regards

 

Mahdi

 

 
 

Hi Mahdi,

 

It is not J/α but ((α*d) -1) as you have rightly mentioned. Yes, ΔPp is calculated with a minus sign but you need to consider an absolute value. The higher of Ek or ΔPp is to be considered and multiplied by 10 to arrive at ΔPo.

Also check if calculated pressure drop is greater than 1.75 kPa. If not, a minimum value of 1.75 kPa needs to be considered.

 

Also note that d is in mm in Metric Units and inch in USC units.

 

Regards,

Ankur

Hi,

 

How do you determine the number of holes needed after you have calculated the total required hole area?

 

Total hole area is that for just one hole or all holes?

 

Here I have converted L to inches since it was given in feet.  dp=((4000*f*L*J/(α*d) – 1)*E

 

Any help on this is much appreciated.

How would this effect the calculations if I wanted x amount of holes every 6 inches along a 10ft pipe. Would this calculation be the right one to use for this case in finding pressure loss along the pipe?

Hi,

 

How do you determine the number of holes needed after you have calculated the total required hole area?

Step 8 of the procedure is self-explanatory

Total hole area is that for just one hole or all holes?

Step 7 clearly mentions that it is total area of all holes

Here I have converted L to inches since it was given in feet.  dp=((4000*f*L*J/(α*d) – 1)*E

 

Any help on this is much appreciated.

How would this effect the calculations if I wanted x amount of holes every 6 inches along a 10ft pipe. Would this calculation be the right one to use for this case in finding pressure loss along the pipe?

Hole-to-hole distance is supposed to be equi-distant for all holes and not random to prevent flow maldistribution.See 8d for minimum (edge-to-edge) distance between adjacent holes.

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