Determining Pressure in Pipe System

[Steel Worker]

There are (5) 3/8" diameter pipes supplying water at 3500 psi and 4.5 GPM each from 5 separate pumps. These (5) lines connect to a main 1.5" diameter pipe. What is the water pressure in the main pipe?

 

[tiny-tim]

Welcome to PF!

HI Steel Worker! Welcome to PF!

Show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[Steel Worker]

Neglecting friction and head losses, I elected to use Bernoulli's equation:

P1 + p(V1^2)/2 = P2 + p(V2^2)

Since I know Q1 = 5(4.5 GPM) = Q2 = 22.5 GPM or 86.6 in^3/sec
ID of 3/8" Sch. 80 pipe = 0.423 inch
A1 = 0.141 in^2
A2 = 1.77 in^2 (ID = 1.5 inch)

Q=AV, V=Q/A, V1 = 123.4 in/sec, V2 = 49.03 in/sec

Plugging into Bernoulli's equation, I get P2 = 88902 PSI
This seems a very high pressure. I actually put this into an excel speadsheet and when i made the 1.5" pipe smaller the pressure decreased also. The formula was right, but what am I doing wrong here? I appreciate any help.

 

[gneill]

Your value for V1 looks suspect; I see something somewhat higher. And even with your value for V1, the pressure shouldn't come out that high. Perhaps there's a unit conversion issue biting you somewhere?

If you solve Bernoulli for P2, you should be able to arrive at a form like:

P2 = P1 + \frac{1}{2} \rho \left(v1^2 - v2^2\right)

The second term shouldn't be very large (in psi) compared to P1.

 

[Steel Worker]

Q1 = 4.5 GPM, There are 5 inlet pipes of this same flow rate. Converting Q1 to cu.in/sec =
4.5 Gallon/min x (1min/60sec) x (0.1337 cu.ft/1 gallon) x (1728 cu.in/1 cu.ft) = 17.32 cu.in/sec.
To get the Velocity, V=Q/A, which is 17.32 cu.in/sec/0.141 sq.in = 123.3 in/sec.

I found the velocity for the outlet the same way, where V2 = 49 in/sec.

5(P1 + (pV1^2)/2) = P2 + (pV2^2)/2
P2 = 5P1 + 5(pV1^2/2) - (pV2^2)/2
P2 = 5(3500psi) + 5(1.94*123.3^2/2) - 1.94*49^2/2
P2 = 88905 psi

 

[SteamKing]

I see a couple of problems in your calculations.

1. You have 5 separate lines coming into one large line. You can write separate Bernoulli equations for each of the smaller lines. I don't think you can write one Bernoulli equation and multiply by 5 though. From a hydrostatic standpoint, if I had 5 lines pressurized at 3500 psi each, the pressure in the larger line would be just 3500 psi, not 5 x 3500 psi.

2. In calculating rho for water, the value 1.94 has units of slugs/cu.ft. Since your flow velocities are calculated in in/s, then some additional unit conversions are in order. Remember, g = 386 in/s^2.

 

[gneill]

Check your value for A1.

Why are you multiplying P1 by five? These pressures are essentially in parallel. All five feeder pipes have the same pressure and velocity. What adds is the total volume supplied.

You could replace the five feeders with a single pipe of 5x the area, the same pressure, and 5X the net flow.

 

[Steel Worker]

Why wouldn't I multiply each of the supply pressures by 5 if they are all separate from one another contributing to the flow of the larger pipe?

 

[gneill]

Flows add that way. Pressures don't. As a crude analogy, batteries in parallel do not multiply their voltages.

Suppose you had two containers of equal volume. Both contained a gas (or fluid) at the same pressure P. If you opened a valve between the two containers, would you expect the pressure to suddenly double?

 

[Steel Worker]

10-4, I understand

 

[Steel Worker]

Using the equation P2 = PI + 1/2(rho)(V1^2-V2^2)...

If the (5) inlet pipes and the (1) outlet pipe have the same diameter, then P1 = P2?