Equivalent Volumetric Flow (with NPS Choices)

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Equation / Last modified by mike on 2015/07/29 07:14
`"Equivalent Number of Pipes" = `
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MichaelBartmess.Equivalent Volumetric Flow (with NPS Choices)
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This equation computes the number of smaller pipes that can be supported by an equivalent volumetric flow to a larger pipe. It is often important in plumbing to understand the combined volumetric flow equivalent of some number of smaller pipes flowing into a larger pipe to ensure the flow into the large pipe is unrestricted flow.

Inputs

The pipe's dimensions are specified by selecting the following inputs for a chosen smaller and larger pipe:

  • Smaller Pipe's NPS - the Nominal Pipe Size (NPS) designator for the pipe (this is not the same as the outer diameter)
  • Smaller Pipe's Schedule - this selects a schedule defining the standard wall thickness
  • Larger Pipe's NPS - the Nominal Pipe Size (NPS) designator for the pipe (this is not the same as the outer diameter)
  • Larger Pipe's Schedule - this selects a schedule defining the standard wall thickness

Derivation

Inner Diameters

The outer diameter (OD) of the pipe is determined from a look-up function based on the pipe's Nominal Pipe Size (NPS).

The wall thickness (WT) of the pipe is determined from a look-up function based on the pipe's Nominal Pipe Size (NPS) and the pipe's Schedule.

Once the OD and WT are determined for the two pipe sizes, the inner diameter (ID) is computed for each pipe as follows:

     [1] `"ID"_"(sm)" = "OD"_"(sm)" - 2 * "WT"_"(sm)"`

     [2] `"ID"_"(lg)" = "OD"_"(lg)" - 2 * "WT"_"(lg)"`

Next, we find a unit volume in the pipe:

     [3] `"Volume" = "area"_"(cross sectional)" * "length"`,

then volumetric flow is just the unit volume that flows through the pipe per unit time.

     [4] `"Volumetric Flow" = "Volume"/"Time"  = ("area"_"(cross sectional)" * "length")/"Time"`

Since area is: `"area"_"(cross sectional)" = pi*"radius"^2 = pi* (1/2 * "diameter")^2`, we can re-write the Volumetric Flow as

      [5] `"Volumetric Flow" = (pi * (1/2 *"diameter")^2 * "length")/"Time"`

And simplifying [5]:

     [6] `"Volumetric Flow" = (pi/(4 * "Time")) * "diameter"^2`

To find out when the volumetric flow through the large pipe equals the volumetric flow through some number of smaller pipes, we write the equation:

     [7] `N * "Volumetric Flow"_"(smaller)" = "Volumetric Flow"_"(larger)"`, letting N be the number of smaller pipes providing equivalent volumetric flow to the one larger pipe.

Solving for N, we get:

     [8] `N = "Volumetric Flow"_"(larger)" / "Volumetric Flow"_"(smaller)"`

And substituting in equation [6] we get:

     [9] `N = (pi/(4 * "Time") * "diameter"_"(larger)"^2) / (pi/(4 * "Time") * "diameter"_"(smaller)"^2)`

The constant terms `pi/(4*Time)` are equivalent in the numerator and the denominator, so they cancel out giving us the equation for N in terms of the two diameters of the two sizes of pipes:

     [10] `N = "diameter"_"(larger)"^2 / ("diameter"_"(smaller)"^2)`

And finally, the diameter through which the fluid flows is the inner diameter of each pipe, so we substitute [1] & [2] into [10].

     [11] `N = "ID"_"(lg)"^2 / ("ID"_"(sm)"^2) = ("OD"_"(lg)" - 2 * "WT"_"(lg)")^2 / ("OD"_"(sm)" - 2 * "WT"_"(sm)")^2`

UsagePlumbing - equivalent volume flow.png

To estimate the number of smaller pipe that would provide a volume flow equivalent to a larger pipe, we simply take the ratio of the square of the larger pipe's diameter to the square of the smaller pipe's diameter.  This calculation is a rough estimate because the outer diameter and wall thickness of NPS scheduled pipes have a tolerance that could cause the inner diameter to vary slightly. This calculation also neglects the frictional loss of volumetric flow due to liquid contact with the walls of the pipes. Some number of smaller pipes will have a larger wall surface total than the larger pipe.

If the volumetric flow is equivalent, the pressure in the larger pipe should not exceed the pressure of any of the smaller pipes.

Example 1    As an example, if you have want to know how many NPS 1-1/4 inch pipes from the Schedule 40s & Std will have an equivalent volumetric flow to an NPS 4 inch pipe from the Schedule 40s & Std, this equation computes that  10.66 of the smaller pipes will have approximately the same volumetric flow as the larger pipe -- again, ignoring wall friction and pipe manufacturing tolerances.  So, if you were creating a sprinkler system where a 4 inch pipe was going to feed a number of 1-1/4 inch pipe, you could keep maximum pressure in the smaller pipes while optimizing the number of smaller pipes that can be fed by one larger pipe by connecting a maximum of 10 1-14 inch pipes to any 4 inch pipe.

Example 2   If your plumbing a closed water system of NPS = 3 inch pipes feeding a drain pipe that is NPS = 6 inches, and both are from Schedule 5, then this equation tells you that approximately 3.69 of the smaller pipes has equivalent volumetric flow to the larger pipe, which means at most three of the smaller pipes can feed the NPS = 6 inch pipe without having the flow restricted at the 6 inch pipe. 

Note that given that there will be slightly less flow in four 3 inch pipes due to larger surface area friction in the smaller pipes, drainage of the four NPS = 3 inch pipes might be accommodated by the 6 inch pipe but we cannot verify this without a better model of the fluid flow in the pipes to at least consider friction with the pipe walls.