# Hydrodynamic Design, Part 3: Flows Through Connectors

**By C.F. ‘Chubb’ Michaud**

In Part 3 of this series, we look at how pressure drop relates to the number and types of fittings and how we measure that drop. Pressure drop due to length and rate of travel through pipes, through fittings and with elevation changes, can all be equated to a common unit of measure. This will help the reader understand the total energy requirement for moving water from point A to point B.

*Pressure drop in connectors*

The convention of relating the pressure drop through pipe as ‘feet of head per 100 feet of pipe’ is one of convenience. In calculating the total available pressure at a point of use, the engineer simply looks at the size of the pipe, the length of the run and change in elevation and adds it all up. In addition to the length, he must also figure in the changes in direction via elbows and tees and going through valves. It takes energy to push water through a pipe. It takes even more energy to make it change direction. Any time water changes its equilibrium, it uses energy. If you run a small pipe into a larger one, you will use more energy. If you go from big to small: more energy. For convenience, all fittings are equated to energy loss in terms of an *equivalent length of pipe*. If you count the elbows and tees and multiply by the equivalent length of pipe, you can calculate the total pressure loss in a system in feet of pipe and pull the ∆P right off Figure 2, Part 2. Table 3 gives the ∆P in fittings in terms of length of pipe. The smaller pipe sizes are not included. Note that larger pipe has larger losses due to the total mass flow being larger. It takes a lot of energy to turn corners.

*Calculating pressure demand*

Pipe length is measured in feet. The pressure losses through fittings is measured in pipe length equivalents and pressure losses due to elevation are measured in feet. We can add up the total resistance in equivalent feet of pipe to determine both pipe size and pressure requirements. As an example, let’s use a new vacation home, supplied by a well and situated at the top of a 30-foot hill. It is a two-story home, so the shower is situated approximately 15 feet above the base of the house and is about 40 linear feet from the point of entry. From the well house, a copper supply pipe is buried below grade with a diagonal run of 125 feet. The five-foot riser pipe from the supply line is a three-eighths-inch line. In laying out the plumbing design, we used three tees and 11 elbows, all square (to save space). How much pressure loss are we to expect by the time water reaches the dual shower head with a flow demand of five gpm with a 15 psi pressure drop through the head?

Length of pipe = 125 + 40 + 5 = 170 feet

Fittings: using 3/4-inch copper, 3 tees x 5 ft equiv = 15 feet

11 ells x 4 ft equiv = 44 feet

Not counting any filtration equipment, we have an equivalent total of 229 feet of pipe. At five gpm through three-quarter-inch pipe, we determine our linear flow at 3.65 ft/sec and the corresponding pressure drop (see Figure 2, Part 2) of 7.5 psi/100 ft. Since we have 229 feet of pipe, we multiply by 2.29 and find a 17.2 psi drop in available pressure at our shower. Right? Not quite. We also have the 30-foot hill and the 15-foot-climb to the shower. That’s another 45 feet of head loss or 19.5 psi. Our net dynamic pressure loss is 17.2 + 19.5 = 36.7 psi. If our well tank is set to kick the well pump on at 40 psi, you won’t be using your shower much at that point. If you max out at 60 psi, you can start your shower with a net pressure of 23.3 psi. You will lose another 1 psi coming up through the three eighths-inch riser so let’s say your net is 22 psi less the 15 psi to flow through the head. Will the operating pressure of only (23.3 – 15 = 8.3 psi) deliver five gpm to your new shower heads?

Using the pressure/flow Equation 1 from Part 2, we can calculate that the flow (q) will be the product of 20 x (3/8)^{2} x 8.31/2 (the square root) or 20 x 0.14 x 2.9 = 8.1 gpm. Enjoy your shower. But make it a quick one because by the time your well tank is down to 41 psi, your available pressure will not operate your shower head with more than a trickle.

**Equation 1 (Part 2).**

q = 20d2 p^{1/2}

Where:

q = rate of flow at the outlet (in gpm)

d = actual inside diameter (ID) of outlet (in inches)

p = flow pressure (in psi)

**Example 1.**

Assume a hose bib with a three-quarter-inch supply line and the flow pressure is 9 psi. Calculate the expected flowrate.

q = 20 x (3/4)2 x (9)½

q = 20 x 9/16 x 3 q = 33.8 gpm

**Example 2.**

Assume a faucet connected to a three-eighths-inch line and a flow pressure of 16 psi

q = 20 x (3/8)2 x (16)½

q = 20 x 9/64 x 4

q = 11.25 gpm

*Converting linear flow to volumetric*

Linear flowrates (as shown in Figure 2, Part 2) can be converted to volumetric flowrate as follows. First, you calculate the cross sectional area of your pipe:

A = πr^{2}

Where A = area in in^{2}

π = 3.14 r = 1/2 D (inside diameter of pipe)

Table 2, Part 2 shows the area of common sizes of pipe but this is with a given ID and not a measured ID. While one-inch copper Type L runs 1.025 in, the lighter-weight Type M is 1.055 in, one-inch PVC in Sch 80 is only 0.957 in ID and Sch 40 is 1.05-inch ID. For these examples, I am using a true one-inch ID.

With the cross-sectional area of the pipe, you can go two ways. You can take a given volumetric flow in gpm (1 US gallon = 231 in3) and determine how many linear feet of pipe is represented by that volume.

Example: Determine the linear flow of 15 gpm through oneinch ID pipe. Fifteen gallons = (15 x 231) 3,465 in^{3} per minute. A one-inch pipe has an area of 0.78 in2. So the total linear pipe equivalent is (3,465/0.78 =) 4,442 inches in 60 seconds or 6.17 ft/ sec. From Part 2, Figure 2 we can estimate the ∆P at about 8 psi.

OR, you can determine the maximum volumetric flow tolerated to a given psi ∆P.

Example: Determine the maximum volume of water that can be pushed through a one-inch pipe to a 20 psi ∆P. From Figure 2, Part 2 this is determined to be 11.6 ft/sec, which is the equivalent of 11.6 x 12 = 139.2 in/sec = 8,352 in/min. This is 8,352 in x 0.78 in^{2} (cross-sectional area of a one-inch pipe) = 6,515 in^{3}/min which (/231 in^{3}/US gal) = 28.2 US gpm.

**Summary**

In the above example, had we used sweeping ells instead of short, square ells, we might have saved the equivalent of 30 feet of pipe or about 2.25 psi. Had we plumbed in one-inch copper instead of three-quarter-inch, however, we could have saved an additional 5.75 psi in pressure losses from the well head to the upstairs bath. This may have made the difference between a satisfied homeowner and a problematic one.

*References*

- Michaud, C. F. ‘Chubb’. “Hydrodynamic Design, Part I: Pressure— Defined,”
*WC&P*, January 2013 - Michaud, C. F. ‘Chubb’. “Hydrodynamic Design, Part II: Flows Through a Pipe,”
*WC&P*, February 2013 - Cameron Hydraulic Data, Ingersoll-Rand Co, New York, NY, 14th Edition, 1965

**About the author**

*C.F. ’Chubb’ Michaud is the Technical Director and CEO of Systematix Company of Buena Park, CA, which he founded in 1982. He has served as chair of several sections, committees and task forces with WQA, is a Past Director and Governor of WQA and currently serves on the PWQA Board, chairing the Technical and Education Committees. Michaud is a past recipient of the WQA Award of Merit, PWQA Robert Gans Award and a member of the PWQA Hall of Fame. He can n be reached at (714) 522-5453 or via email at AskChubb@aol.com*