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Help calculating combined size of holes in a pipe

bigone5500

Apr 9, 2014
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I have a pipe which has 20 1/8" holes drilled into the side. I want to figure the single hole equivalent of all these holes added together. What I am coming up with seems wrong. Does anyone know what the correct formula is?
 

KrisBlueNZ

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What do you mean "the single hole equivalent"? Do you mean a hole with the same area? In that case you need to calculate the area of each hole, multiply by 20, and convert that area back into a diameter. Area of a circle is pi r2 IIRC!

BTW I've fixed your thread subject. "Heeeeeelp!" isn't very informative and just wastes users' time.
 

hevans1944

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Jun 21, 2012
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Following @KrisBlueNZ suggestion, I get an equivalent diameter of about 0.559 inches to equal the area of twenty 1/8" holes. What did you get?
 

bigone5500

Apr 9, 2014
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I was getting about the equivalent to a quarter inch hole. I'm usually pretty good figuring this stuff out. A co-worker helped out and got about 2 1/2" equivalent. I think what I was doing wrong was not converting back to diameter.

Sorry for the stupid title. That is just how I was feeling at that time.

Thanks guys.
 

Gryd3

Jun 25, 2014
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I was getting about the equivalent to a quarter inch hole. I'm usually pretty good figuring this stuff out. A co-worker helped out and got about 2 1/2" equivalent. I think what I was doing wrong was not converting back to diameter.

Sorry for the stupid title. That is just how I was feeling at that time.

Thanks guys.
I got a little over a quarter inch (0.2795") as well. Assuming the 1/8" holes are diameter and not radius.
(Note that this value is a Radius Though! So you would be using a half inch drill for this. 0.559")

@hevans1944 I think you did PI x 1/8"^2 ?
The formula calls for radius, so I used PI x 1/16"^2

*Facepalm... 0.559 and 0.279 are both correct... that's what happens when you ignore units!
 
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hevans1944

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Okay, let’s start with diameters, stick with diameters, and finish with diameters.

The area of a circle is A = (π) (D²) / 4. Where π = 3.14159 (approximately)
You could use A = (π) r² but since D = 2r then r = D/2 and r² = D²/4 so we can work with diameter instead of radius. Most drills I have used are specified by diameter, not radius.

We start with a 1/8-inch hole diameter, or D = 0.125 inches.

The area of that one hole is A = (π) (D²) / 4 = 0.01227 square inches (approximately). Then twenty of these holes would occupy an area of (20) (0.01227) = 0.2454 square inches (approximately). So, the area of a round hole with that area would have a squared diameter D² = (4) (0.2454) / (π) = 0.3125. Taking the square root of this, D = 0.5590.


If we work the problem backwards, from an assumed solution of 2 ½ inches diameter, then the area of that hole would be (π) (2.5)² / 4 = 4.9087 square inches. Divide this by 20 to obtain the area equivalent of twenty holes: 0.2454 square inches per hole. Solve for the squared diameter of each hole, D² = (4) (0.2454) / (π) = 0.3125. Taking the square root, obtain 0.5590 inches diameter.


Clearly, this does not equal 0.125 inches diameter as originally specified, so the assumed solution of 2 ½ inches diameter is wrong.
 
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KrisBlueNZ

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I went via millimetres, and got the same result: 0.559 inches diameter.
 

duke37

Jan 9, 2011
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The calculation equates the area of the holes. It does not equate the effect of the holes. A big hole will provide much less friction to a fluid than the same area of small holes. It is not clear what bigone5500 wants.
 

shrtrnd

Jan 15, 2010
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Yeah, I know from my evaporative cooler, that the placement of the holes along the pipe will affect he flow of the liquid from the drilled holes.
The pressure/flow out the holes closest to the in-flow of liquid will be greater than the flow at the 20th hole at the end of the pipe.
If you need a consistent flow from the pipe over the length of the 20 holes, you'll need smaller holes at the input to the pipe, and larger holes
headed toward the end of the pipe. Just something to consider with whatever your project is.
 
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