# Fluid on pipes systems

1. Dec 28, 2004

### lord_temple

Let's say that we have a 3/4" copper line to supply water. We have a volumetric flow of 20 gpm (0.0445 ft^3/s). With this, we have a velocity of 14.35 ft/s. Now, at the end of the copper line, I want to service two homes, so I split the service. Comming out of the 3/4" copper line, I have two 3/4" copper lines more.

How do I know how much pressure I'm going to loose for every branch that I add? Let say that later on I want to split the service into three or four 3/4" services, comming out of the same 3/4" "main line". I'm trying to apply Bernoulli's equation, but somehow I don't believe my number because, according to my interpretation of my results, the pressure on the branches is higher that the pressure on the "main" line, this is not possible, is it?

2. Dec 28, 2004

### sid_galt

Have you taken into account that the average density of water is decreasing in the branched pipes due to the amount of water being distributed in a larger amount of volume?

3. Dec 28, 2004

### Q_Goest

Bernoulli's is only part of the equation. You need to determine pressure drop due to friction. Darcy-Weisbach (sp?) is used for pressure drop due to friction for incompressible flow or for compressible flow with small pressure drops.

Crane paper #410 is the bible for calculating pressure drop for practicing engineers, though it does not cover two phase flow.

http://www.cranevalve.com/tech.htm

4. Dec 28, 2004

### lord_temple

Well... I'm taking water as an incompressible fluid, and density will be constant. I don't think density would affect a lot on the calculations, would it?

5. Dec 28, 2004

### Grogs

If you're looking at it from an ideal standpoint, couldn't you just start with the equation of continuity?

$$A_{1}v_{1}=A_{2}v_{2}$$

=> $$v_{2}=\frac{A_{1}v_{1}}{A_{2}}$$

You can then substitute into Bernoulli's equation to calculate the pressures.

BTW, is this just a thought problem, or are you actually looking at plumbing a house/houses?

6. Dec 28, 2004

### lord_temple

You're right.. but remember I have mass flow going in from one end, and mass flow going out from two ends, that will be:

A1v1=A2V2+A3V3 (all that goes in, goes out)

At the inlet, I have 20 gpm going in (0.0445 ft^3/s), and at a 3/4" that's a velocity of 14.35 ft/s.

Now, if I take into account that only half of the flow is going to one of the two branches (let say, 10 gpm of the 20 total are going through "A2"), and if I get the velocity from that:

Q=AV
V=Q/A
V=0.0222 ft^3/s (10 gpm)/0.0031 ft^2 (area of the tube)= 7.16 ft/s

I'm not sure if I can just plug in these velocities into Bernoulli's equation. If I plug in these info, I get a differential pressure (P1-P2)=-1.041 psi (density = 62.4 lb/ft^3)

That is:
P1-P2 = [(v1^2-v2^20)/2]*density

Am I doing something wrong? or what does the negative pressure means?

Oh, by the way, its just and "ideal scenario" I'm not lookint into doing something serious with it, since I'm doing quite a bit of assumptions. But I still want to have an idea of how much the pressure drops with each branch that I add.

Last edited: Dec 28, 2004
7. Dec 28, 2004

### Q_Goest

Bernoulli's equation is a conservation of energy equation. But for a fluid flowing through a pipe, energy is not conserved. That's not to say Bernoulli's is the wrong equation, it's just incomplete.

The energy which is not conserved can be seen as pressure drop due to frictional losses within the pipe. That pressure drop can be calculated using the Darcy Weisbach equation. Here's a calculator:
http://www.lmnoeng.com/darcy.htm

Bernoulli's is still used, for example if you have a drop in elevation, the pressure increases as the fluid goes to a lower point in the system, just as pressure increases as you go deeper in water. The reason is also the same.

Bernoulli's is also used to calculate what amounts to kinetic energy increases/decreases. This is what you've calculated so far, and is why the pressure mysteriously increases when you have numerous branches. The dynamic pressure of the water which has a high velocity is being converted to static pressure as the velocity drops. That's what you're calculating, and that's why you calculated a pressure rise. Note that this is a conservation of energy calculation, the kinetic energy of the moving water with high velocity is converted to static pressure as the water moves at a lower velocity. Note also this is mathematically identical to a diffuser where the flow area increases, velocity decreases, and static pressure increases.

Now if you want to model the system more accurately, you need to calculate the pressure drop due to friction using the Darcy Weisbach equation and subtract that from the pressure increase or decrease due to velocity changes and elevation changes. Since your system has a branch, you need to calculate the pressure drop for each section of pipe separately using the Darcy Weisbach equation.

The link I gave is a bit overly simplified though. It does not show you what to do about elbows, valves, and other restrictions to fluid flow. In short, what's done there is to come up with the "equivalent length" of each restriction and add it to the total actual length. For example, an elbow may be the equivalent of 12 times the diameter, so if you have 3 elbows in a 3/4" ID pipe, you would add 12*3*.75" = 27 inches to the overall length of the pipe.

Last edited: Dec 28, 2004
8. Dec 28, 2004

### lord_temple

Q_Goest,

Now I understand a lot more about Bernoulli's equations. I really appreciate it, and thanks for your time!

9. Dec 30, 2004

### FredGarvin

Lord Temple...If I'm not too late here's what's up with your calculation:

The negative dP you calculated is a result of using the most basic form of the Benoulli equation. Using this equation in it's basic state, you are making three assumptions about the flow:

1.) Incompressible (for water, a good assumption, especially at relatively low press.)
2.) Steady State (no transients...again a good assumption)
3.) Non-viscous flow (no frictional effects, i.e. no losses)

The way you are looking at this is as an energy balance, as stated in an earlier post, (pressure head + velocity head + gravitational head = constant). You say that the velocity reduces from point 1 to point 2. That means the velocity head reduces at point 2. For the energy balance to remain in check (assuming no elevation changes), the pressure term at point 2 must increase, making P2 larger than P1, hence the negative dP you calculated.

Q Goest had a great recommendation. Get Crane's tech. paper 410. It's worth it's weight in gold. At \$35 it's a steal when it comes to technical references.

10. Dec 30, 2004

### lord_temple

Yeah... I think my problem was too many assumptions, and trying to use Bernoulli's formula with very basic assumptions. I ran some numbers taking into account friction, length of pipe and some other stuff that I didn't take into account at the beginning, using the Darcy Weisbach equation. Now I'm getting some realistic numbers (to my point of view). I'm also cheking my calculations with WaterCad, which helps me a lot. I still have some minor doubts on details on my calculations, but I think I can figure those out. I'll get that tech paper as soon as I can.

It took me some time to undestand the meaning of the results, and I totally agree with your comments now that I can see it better... thanks FredGarvin.