Maximum Flow Through 3/4" Copper K?

[SimplyCivil]

Hi there, I'm a new EIT and starting to get more involved with small water systems. One question that keeps on cropping up is "What is the maximum flow possible through [X] diameter pipe?"

For a practical example, let's use a 6-inch watermain in the road (say @50-80psi Static) and a connection of 100' of 3/4" Copper K piping to serve a household. What is the maximum flow that can come out at the house?

Could someone break this problem down into the steps and calculations required (I'm not really looking for just "the answer", but more the steps/theory behind the calculation).

Thanks!!

 

[bimr]

This question has been answered on this site. For practical matters, copper is a soft material and the flow velocity must be limited to protect the copper.

"To avoid excessive system noise and the possibility of erosion-corrosion, the designer should not exceed flow velocities of 8 feet per second for cold water and 5 feet per second in hot water up to approximately 140°F."

http://www.copper.org/applications/plumbing/cth/design-installation/cth_3design_size.html

That will give you 10.9 gpm for the maximum velocity used for normal service. That velocity has a pressure loss of 0.126 psi per linear foot of tubing.

For fire flow, you can probably get 15 ft/sec for a short time. That will give you around 20 gpm but is only suitable for emergency conditions.

Pressure loss is on the attached chart. The chart is limited by the maximum velocity of 8 ft/sec.

http://www.copper.org/applications/plumbing/cth/technical-data/tables/cth_table6.html

 

[fel3]

There is no one answer to this question as posed, so we need to establish some constraints to put this question into a form (or forms) that can be analyzed. Also, you should convert pressures to heads for doing the calculations. That way, all parts of Bernoulli's Equation will have the same units (feet or meters).

Here are some things that must be determined before you can calculate. There may be a few others, but I'm running on inadequate sleep today.
(a) What is the elevation difference between the main in the road and point of discharge at the downstream end of the copper pipe? This elevation difference affects the maximum flow (see Bernoulli's Equation). It should be obvious that you will get less water out of the copper pipe if its point of discharge is higher than the main compared to if the point of discharge is lower than the main.
(b) What elements along the copper pipe exist that create minor head losses over and above the friction head loss due to the copper pipe itself? Every bend, tee, valve, etc. on the copper pipe adds minor head losses and thus reduces the maximum flow available through the copper pipe.
(c) Does the copper pipe branch and are those branches also drawing water? If so, then the flow into the copper pipe from the main will be more than the flow that comes out of the copper pipe at the point of discharge. This is a more complicated scenario, but it's not all that hard to solve. However, your question implies a homogeneous flow, so we will assume flow through only ONE pipe and no branches, or at least no active branches (i.e. faucets on the branches are closed).

Here are some reasonable assumptions we can make:
(a) The water main in the road is large enough compared to the copper pipe that we can treat it as a constant head reservoir. This means there is no need to investigate what happens upstream of the point of connection to the copper pipe. Assume a stating pressure in the water main (e.g. 50 psig; psig = psi gage), then convert it to a pressure head, then add the pressure head to the water main elevation to establish the hydraulic grade line (HGL).
(b) The point of discharge of the copper pipe is open to the atmosphere, which means that its pressure is about 14.7 psia (psia = psi absolute). This also means its pressure is ZERO psig and its HGL equals its elevation.
(c) Velocity heads will be small enough that we can safely ignore them. In any case, the velocity head at either end of a uniform pipe segment (same diameter and no branches) is identical and these will cancel out.

You need to pick an equation for that part of Bernoulli's Equation that deals with friction losses. For potable water, the two most common choices are Hazen-Williams and Darcy. If you use Hazen-Williams, you need to select a reasonable value for the pipe roughness coefficient, C. If you use Darcy, be prepared to iterate to find the friction factor, f. However, you can take a first cut at Darcy with an assumed f, just assume something that is reasonable for your particular situation. Here, the Moody Diagram is your friend.

Here are two scenarios to look at. The first is simple and idealized; the second is more realistic but more work to solve.
(1) The simplest scenario is a 3/4" copper pipe connected to the main at one end, open to the atmosphere at the other end (e.g. an open faucet or hose bibb), and NO minor losses. In this case, your headloss between the ends of the copper pipe is simply the HGL at the water main minus the HGL at the point of discharge. Insert either Hazen-Williams or Darcy into Bernoulli, rearrange and make substitution as needed (e.g. Q=A*V), then solve for Q.
(2) The more realistic scenario is like the first, but also includes all the minor losses. You will need to come up with minor loss coefficients for each offending bend and fitting (including inlet and outlet minor losses for the ends of the copper pipe), add them up, then fit that total into Bernoulli's Equation. This makes the most sense if you're using Darcy for piping head losses because the forms are similar. Alternatively, if you're using Hazen-Wiliams, you can find tables that provide "equivalent lengths" for bends and fittings (each bend or fitting will produce a minor loss that is equivalent to a length of pipe of the same size). This is a less accurate method, but it's easier to do. To use equivalent lengths, simply add them to the length of the copper pipe and do the calculation in (1) with a longer L.

I hope this helps without giving away the answer.

Fred

==========
"Is it the only lesson of history that mankind is unteachable?"
--Winston S. Churchill

 

[fel3]

I meant to add this, but bimr beat me to it. The procedure I outlined will calculate a theoretical maximum flow. There are some practical limitations, such as velocity, code requirements etc.

==========
"Is it the only lesson of history that mankind is unteachable?"
--Winston S. Churchill

 

[bimr]

http://www.eng-tips.com/viewthread.cfm?qid=276054

 

[SimplyCivil]

Thank you fel3 and bimr for your thoughtful input. I'm also glad to see others have had this question to what seems so "simple" a problem to the lay person. fel3 - I very much appreciate your detailed steps. I will "do my homework" and post back in a couple days or so. Hope you get some sleep

 

[rconner]

While it is a little more complicated than this, as an EIT exercise you may wish to go back and revisit the web or your hydraulic textbook(s), looking perhaps for discharge of water through smooth pipes, orifices, and fittings/valves etc. While there are many more considerations and limitations, the discharge of water at the same elevation is basically limited by entrance, loss of head due to friction, minor losses such as fittings and valves and loss of head due to velocity in the pipe. (When you do all the calculations, I guess I wouldn’t be surprised however if you came up with bimr's suggestion of about 20 gpm or so from your 50 psi main source, in WAO/wide open condition;>)