Pipe Flow Rate 2

[Chairs]

I feel like this should be easy but I'm having a good amount of trouble figuring it out. I'm given the inlet pressure, the length and the inner diameter of the hose and I'm asked to find out the flow rate coming out. I'm also assuming the outlet pressure is atmospheric. Without being given a velocity how would I go about doing this?

P=110 psi
ID=2"
L=48"
Media= 70 deg Water

The hose is straight and rigid so it might as well be a pipe but I'd also be interested to know how it differs when bent.

The point of this is to help outside sales compare the flow rate of smooth bore to convoluted hose.

 

[StoneCold]

Chairs
The answer to your problem is that it is an iterative process. Guess a flow rate, say 400 gallons per min. Crank through the calculations and see if the pressure drop is more or less than 110 psi for a 4ft piece of 2" hose. That will get you to the theoretical answer.

The real answer is "It depends". Imagine if you install a pressure gauge and a second valve upstream of your 4' piece of two inch hose. So we have the configuration, Existing shutoff valve, Gauge on a branch, New Valve. Assume that the valves are 2" full port ball valves so we can ignore the pressure drop there. Open the first valve and the gauge should go to 110psi. When you open the second valve the pressure is going to drop to some steady state flowing value, because the source of the 110 psi water is not infinite. The amount of pressure drop is going to depend on several things, the size of the piping upstream, the size of the pump driving the flow, number of bends in the system, number of valves in the system, etc. So the real world answer of how much flow can you get out of a 2" four foot long hose is dependent on the upstream capacity of the entire system.
If you just want to know can I get more flow out of smooth bore hose than convoluted bore hose? The answer is yes, you should get slightly more flow out of a smooth bore hose than a convoluted bore hose.
To get the actual pressure drops for the convoluted hose you will have to contact the manufacturer. I believe they are all empirical formulas.

Regards
StoneCold

 

[LittleInch]

Are you sure about those numbers? I used a quick app on my phone and got 57m/sec velocity... and 180 m3/hr from a 2 inch pipe.

sometimes you just need to guess and look at the output. many are set up to give you pressure drop so just put what you know in and guess the rest and then adjust to get your differential pressure.

There are any number of websites and apps for fluid flow, but at that sort of velocity and flowrate the accuracy might be abit low...

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[ione]

Littleinch,

I don't mean to be controversial, as this is just a small detail about your post, which I consider extremely wise, but how is it possible to have a velocity of 57 m/s in a 2" pipe with 180 m3/hr flow?


Chairs,

Do not forget to take into account exit losses in your calculations

 

[LittleInch]

Because the OP's pipe is only 48 inches long and has 110 psi across it. this has a flowrate of 50 litres/sec.

As I said - it was a first order number, not including exit losses or any issue with the whole huge velocity issue...

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[ione]

I just wanted to underline that 57 m/s in a 2" pipe do do not correspond to 180 m3/hr....

 

[MikeHalloran]

With a hose that short, the component that's going to dominate is the nipple at either end of the hose, specifically its ID, and that will be affected by the chamfer or radius at each end of each nipple.

I am not aware of a way to terminate a 2" diameter hose without shoving a nipple into its lumen. If there is a way, I'd sure like to hear about it.


Mike Halloran
Pembroke Pines, FL, USA

 

[bimr]

While theoretically a maximum velocity may exist at the speed of sound ie., 1450 m/s, the pressures required would probably exceed the strength of the pipe.

You can use the pipe velocity calculators:

http://irrigation.wsu.edu/Content/Calculators/General/Pipe-Velocity.php

http://www.spiraxsarco.com/resources/calculators/pipes/checking-existing-pipes.asp

However, those pipe flow calculations would be theoretical. In practice, when you have a velocity above 15 ft/sec, the hose material would most likely be eroded away. So the maximum velocity is most likely 15-20 ft/sec irrespective of the pressure.

 

[LittleInch]

Serves me right for not looking at the small print - that was for a DN50 PE pipe which had an Id of only 33mm.

for a PE (smooth bore) pipe with ID of 50mm (DN 75), it is now 145 l/sec, with a velocity of 74m/sec! Flow of 522 m3/hr. There's going to be a host of other exit effects and nozzle impacts which will reduce this and the accuracy of most simple calcs is not good at this sort of extreme flowrate.

How if differs when bent? Not much at that sort of pressure drop and length, but will decrease by maybe 10%. Also you'll have quite a sizeable force on the bend due to the change of direction.

As said above, the convoluted hose is a "special" and needs test data to verify each variation, but you're probably looking at 10-20% more flow, depending on what dimension you are comparing it to.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[iainuts]

Regarding corrugated hose, I posted Hawthorne's paper here:

http://www.eng-tips.com/viewthread.cfm?qid=314904

Latexman also posted another paper regarding flow restriction through corrugaged hose.

In general the flow through corrugated hose is many times more restrictive than pipe. Typically, the restriction through corrugated hose is equal to pipe about 10 times its length.

 

[Chairs]

Ok, thanks for the help but I ran into another problem. I used Darcy-Weisbach's equation, guessed a velocity, solved for Reynold's number, got the friction factor and then solved for the pressure drop. My plan was to do multiple iterations until my deltaP equaled 95.3 psi. (110psi - 14.7psi, My inlet pressure minus atmospheric pressure). The problem is at 4ft OAL I need a flow rate of about 1700 GPM which is ridiculous. I re-did the calulation for 100 ft length and it was much more reasonable at 338 GPM. I think the calculations are correct, the problem is I'm finding the flow rate required to drop the pressure 95 psi instead of what the flow rate will be at that inlet pressure. Is there another way to go about it?
^Smooth bore

For convoluted I'm using the same equation but I'm using a different value for the friction factor.


iainuts I can't open the document on that other thread. Also is there some calculation based on the curvature of the hose/Can you tell me where you got that x10 approximation?

 

[BigInch]

Forget the pipe. That's not a flow friction problem, but a mass and acceleration problem.
110 lbs exerted on the mas of a 48" long column of water will accelerate that column to a certain velocity which will traverse the 48" long pipe in a half a second or so. Therefore the velocity is no more than say 16 fps.

I hate Windowz 8!!!!

 

[77JQX]

These calculations showing high velocities are unrealistic. The exit loss alone - the kinetic energy of the fluid exiting the pipe (calculated as v^2/2g) - limits the maximum velocity of the flow to 38 ft/s (11.5 m/s), assuming 100% of the 95 psi potential energy is converted to kinetic energy. Depending on geometry, the entrance loss could also be significant. I expect that frictional forces along the 4 feet of pipe will be insignificant in comparison.

 

[BigInch]

Now we're getting somewhere.

I hate Windowz 8!!!!

 

[77JQX]

Apparently I can't calculate - the kinetic energy limit is about 120 ft/s.

 

[Chairs]

wouldn't it be 110 times the cross sectional area of the pipe? And it's really that simple?

And 77JQX can you explain that a little more?

 

[BigInch]

yes driving force F = 110 psi x Pipe X-sectional area in in2.

The pipe has to be long enough such that viscous force, shear on the pipe wall x internal pipe circumference x length > = driving force for the frictional flow equations to be valid. Over short lengths the fluid is pretty much just accelerating with a little bit of frictional losses retarding the acceleration. You must consider Newton's theories in addition to Darcy's. Darcy alone won't do.

I hate Windowz 8!!!!

 

[77JQX]

OK - here is my explanation. One of the so-called minor losses included in hydraulic calculations is the exit loss for fluid exiting a pipeline, and represents the kinetic energy of the fluid. It recognises that while firction usually represents the majority of losses in a piping system, a portion of the energy is expended accelerating the fluid up to speed, and isn't recaptured when the fluid exits the system. In calculating minor losses, one common form (where energy is expressed as ft-lb/lb, or more simply feet of head), the minor losses are calculated according to the formula Kv^2/2g, where K is an empirical constant for different fittings, and v is the velocity. For exit losses, K=1. This is exactly analogous to KE=0.5*mv^2. All I did was the hydraulic equivalent of the classical physics problem of solving for velocity by setting mgh=0.5mv^2, but in this case I used: 220 ft= v^2/2g, where 220 feet is equivalent to 95 psi. Which now that I think about it, is probably wrong - presumably his starting P is 110 psig. I think I'll just give up for today.

 

[BigInch]

It's not the numbers that are important in the forum... it's the ideas. This is a liquid bullet in a gun barrel. If the gun barrel was too long... well.... then friction would become much more important.

I hate Windowz 8!!!!

 

[LittleInch]

chairs,

When most people say 110 psi, they mean 110 psig. Atmosphere is 0 psig. Only flow assurance people talk about psia.

your question seems to be changing, but that's ok - just be as specific as you can.

As you can tell above, your short length of tubing (4 feet) with a huge pressure difference is creating something more like a rocket motor / water lance than anything else. I'm not sure that you understand the figures you're using or where they are coming from but rest assured they are much higher than normally seen in fluid flow. if you gave us a bit more of an insight into what you're doing you might find some interesting response.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way