# Sizing of 1600' water pipe run

I'm looking at running water to a water trough that is 1600' away from my water source. I'll need about 5-10 gpm flow at the trough. My water source produces 40-60 psi from a bladder tank and has a 3/4" port. I am hoping to run a 3/4" PVC all the way but am concerned about flow loss from psi loss?

• Holy cow! How far up or down hill is this trough from the source? – ThreePhaseEel Jan 13 '17 at 4:52
• Hello, and welcome to Stack Exchange. Unfortunately, I'm voting to close this question as off-topic because it has nothing to do with Home Improvement. – Daniel Griscom Jan 13 '17 at 13:22
• @DanielGriscom I disagree. Just because you don't have water trough at your home, doesn't mean others don't. – Tester101 Jan 13 '17 at 14:13
• Looks like a DIY home project to me. What is the change in elevation is the trough higher, lower or at the same level. I voted to keep it open. – Ed Beal Jan 13 '17 at 14:13
• @DanielGriscom It's a plumbing question, which are on topic here. The OP is not asking about the trough, they're asking about getting water to it. Answers to this could be useful for sizing pipe of various lengths for various applications. If the OP was asking about sizing wire to a receptacle for a trough heater, would that be off topic? – Tester101 Jan 13 '17 at 15:05

Tl;dr

To get 5 gpm, you need a pressure head of about 85 ft. (i.e your inlet would have to be 85 ft above your outlet). Of course, this is what they make pumps for! To get 85 ft of pressure head, you need about a ~0.15-hp pump.

Detailed Answer Someone should definitely check my math on this :P

Assumptions and constraints:

1. No rise or drop in pipe (i.e. you're pumping across flat ground)
2. Minor losses (e.g. through unions, etc.) are negligible given the large length-to-diameter ratio of the tubing run.
3. ID of 3/4" PVC: 0.824 in or 68.7E-3 ft (based on OD nominal of 1.050 in and wall thickness nominal of 0.113 in)
4. Equivalent roughness (ε) of PVC pipe: 0.000005 ft (ref. Moody & Colebrook)
5. Density of water (ρ) at 60 °F: 1.94 slugs/ft3
6. Dynamic viscosity (μ) of water at 60 °F: 2.34E-5 lb-s/ft2

Solution:

Because the diameter is constant over the length of the pipe, inlet and outlet velocities are the same. We can assume pressures at the inlet and outlet are the same, too (large, open tanks). So if we modify the Bernoulli equation with p1=p2=V1=V2=0, we get:

hp = (f * l * V2) / (D * 2g )

where hp is the head pressure required to make the flow, f is the friction factor of the pipe, l is the length, V is the linear flow velocity, D is the diameter (ID), and g is acceleration due to gravity.

We find V by:

V = Q/A

where Q is the flow rate and A is the area of the pipe. Q is 5 gpm, or 11.14E-3 ft3/s (60 seconds in a minute, and 7.48 gallons per ft3). The area of a circle is π * (D/2)2, so our area is 3.71E-3 ft2. Thus, our velocity is 3.01 ft/s.

To find f, we need to know the ratio of equivalent roughness to diameter (ε/D) and the Reynolds number. ε/D is 72.8E-6. We find the Reynolds number, Re, using the equation

Re = (ρ * V * D) / μ

Thus, Re for our flow is 17.14E3. With Re and ε/D known, we determine f from the Moody Chart. f is approximately 0.0258.

Solving our original equation above for hp gives us hp = 84.5 ft

To find hp in terms of power, Php, we use the equation

Php = ρ * g * Q * hp

which yields Php of 58.8 ft-lbf/s. Converted to horsepower, this is 0.106 hp.