Driving fluid from tank to piping system 1

[Docseven]

I have tried a number of equations and keep getting drastically different results.

With liquid water (say 15C) in a tank connected to a system open to atmosphere with 100 feet of head via an 8 foot section of 2" pipe, if the tank is pressurized with air in order to drive the tank water into the system, such that the air space is regulated to 5 PSIG above the static head pressure at the bottom of the system, what would the flow rate through the 2" pipe be? The tank discharge is level with the bottom of the system and there are 10 feet of liquid above the tank discharge in the tank. Once I figure out a satisfactory solution to this I would like to run it for a 3/4" pipe as well.

Just scratch on the envelope, I tried using Hazen-Williams assuming that the delta P is approx 10 PSIG (10 ft water plus the pressurized surface), a C of 120, and got 371 gpm.

That is,
10 psig / 8 ft = 4.52 * Q^1.85 / (120^1.85 * 2^4.87), solve for Q.

Dropping to a 3/4" pipe resulted in a 28 gpm result.

Is this on the right track?

Photo attached to post.

 

[BronYrAur]

I think you have P2 incorrectly labeled on your drawing. Please see my attachment for clarification. I was very confused about what your are asking.

 

[BronYrAur]

Oops, forgot to attach

 

[Docseven]

The 100 feet static head will not change appreciably during this evolution due to the volumes of the two systems (few thousand vs 100K gallons).

I did have a mislabel on the original drawing but it isn't the way you marked it up. For the sake of this initial go at the problem (calculate time required for the job of pushing a tankerful of new fluid into the existing system), I was told they would keep the air pressure on the tanker at approximately 5 psig over the static head pressure of the system into which they are pumping into which would be the 100 ft of head open to atmospheric pressure. The 10 ft static head in the tank will decrease as the liquid level there is pumped out but I want to look at the initial condition only right now. I marked the drawing correctly now and re-attached.

 

[Latexman]

You have an entrance (K=0.5), 8' of pipe and an exit (K=1).
P1 = 100 ft. of water = 43.3 psig
P2 = P1 + 5 psig ---> P1 = 48.3 psig
P at pipe exit = 48.3 + 10/2.31 = 52.6 psig
dP = 52.6 - 43.3 = 9.3 psi

I got 250 gpm for 2" Sch. 40 and 29 gpm for 3/4" Sch. 40.

Good luck,
Latexman

To a ChE, the glass is always full - 1/2 air and 1/2 water.

 

[Docseven]

I am going to post a third and correct drawing this time! It is an 8 ft length of 2" pipe (and 2nd scenario is going to be 3/4" pipe) and I have finally properly labeled p1 and p2.

 

[LittleInch]

You won't get better than latexmans response.

Just be aware that these sorts of calculations are very sensitive to the ID that you use. So if any of the piping is in fact a hose or other type of 2" pipe then the ID might be quite different. Also whether the bore changes at any valves or connectors. I'm not surprised at the difference in flow between those two different pipes.

Given that the pressure P2 is constant, but the height changes, then as you say the flow will change and reduce as time goes on, so you're in a dynamic situation.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[BronYrAur]

latexman, are your numbers based on Hazen williams? if so, where do your entrance and exit losses come into play?

 

[Docseven]

Thanks very much latexman. I was able to come up with similar results when I switched to Darcy Q=19.64 d^2 sqrt(hl/K) and used approx 20 ft (10 psig+10 ft head in tank initial condition) as hl, along with an entrance loss (k=.5) and straight pipe run (2.8 or .88 for 3/4" and 2" respectively) and an exit (k=1).

 

[Latexman]

BronYrAur,

My handy dandy spreadsheet uses Darcy–Weisbach equation and methodology.

Good luck,
Latexman

To a ChE, the glass is always full - 1/2 air and 1/2 water.

 

[BronYrAur]

Sounds like you guys are on the same page. I don't mean to prolong the thread, but I'm missing something. We don't know GPM or velocity, so I'm trying to juggle the equation. At the initial condition, we have a constant pressure pushing the fluid through an entrance, into the pipe, and then through the exit. So we have a fixed delta-P across the entire "3-stage process" of entrance, pipe, and exit.

Can you lay out the equations for those 3 pieces for me?

 

[ione]

BronYrAur,

It's a trial error process. You adjust the flow rate until it matches the given pressure drop

 

[Latexman]

BronYrAur,

See attached. I characterized the 8' long 2" Sch. 40 pipe with an entrance and exit loss. Then, entered the properties of water. Then I kept entering flow rates until pressure drop was about 9.3 psi. It took about a minute.

Good luck,
Latexman

To a ChE, the glass is always full - 1/2 air and 1/2 water.

 

[bimr]

There are some online calculators that you may use to get within 10% accuracy.

The exit loss from the tank is approximately the same as an orifice loss:

http://www.tlv.com/global/TI/calculator/water-flow-rate-through-orifice.html

Pipeline headloss can also be approximated:

http://irrigation.wsu.edu/Content/Calculators/General/Pipeline-Pressure-Loss.php

 

[georgeverghese]

Latexman,

Based on the last diagram posted, we have

Static head diff = 90ft = 27.3m of water = 38.8psi

Vapor space diff = 0-5 = -5psi

Net differential = 38.8 - 5 = 33.8psi, while you've used 9.3psi ?

 

[LittleInch]

George. The presure isn't 5psi, it's 5psi + P1, so (100/2.31) + 5 = 48.3 pig, not 5.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[georgeverghese]

LI, Okay , I missed out on of the OP's later posts.