# Equivalent Volumetric Flow (with NPS & friction)

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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.  Likewise, it is sometimes important to know how many smaller pipes can be fed from a larger pipe so that the pressure in the smaller pipes is maintained.

# 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)

Additional fluid mechanics analysis provides a better estimate of the volumetric flow equivalent that includes frictional loss at the walls using the formula:

[11] N = sqrt("diameter"_"(larger)"^5/ "diameter"_"(smaller)"^5)

And so finally, instead of substitute [1] & [2] into [10], we substitute [1] & [2] into [11] to give us a better estimate of the Equivalent Volumetric Flow.

[11] N = sqrt("ID"_"(lg)"^5 / ("ID"_"(sm)"^5) ) = sqrt(("OD"_"(lg)" - 2 * "WT"_"(lg)")^5 / ("OD"_"(sm)" - 2 * "WT"_"(sm)")^5)

As you can see the flow of some number N of small tubes is increased in equation [11], since frictional force decreases the flow more in the smaller tubes.

# Usage

To estimate the number of smaller pipe that would provide a volume flow equivalent to a larger pipe, we enter the NPS and schedules for both pipe sizes into this vCalc equation, which gives us information to compute the inner diameter of the two pipes.  This calculation is a rough estimate because the outer diameter and wall thickness of NPS scheduled pipes have a manufacturing tolerance that could cause the inner diameter to vary slightly.

This version of the calculation should do a better job of estimating the frictional loss of liquid flow due to liquid contact with the walls of the pipes than the simpler Equivalent Volumetric Flow (with NPS Choices) equation. 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 14.54  10.66 of the smaller pipes will have approximately the same volumetric flow as the larger pipe.

Compare this with the same calculation performed with the Equivalent Volumetric Flow (with NPS Choices) equation which shows the volumetric flow equivalent of a 4 inch pipe is approximately 10.66 1-1/4 inch pipes. The Equivalent Volumetric Flow (with NPS Choices) equation is derived from a purely geometric model of the volume flow and ignores wall friction.

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 14  1-14 inch pipes to any 4 inch pipe.

Example 2   If you're 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 5.12 of the smaller pipes has equivalent volumetric flow to the larger pipe, which means at most five of the smaller pipes can feed the NPS = 6 inch pipe without having the flow restricted at the 6 inch pipe.