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Pump head in pipe with changing diameter/area

Pump head in pipe with changing diameter/area

Pump head in pipe with changing diameter/area

(OP)
I have concentric (annulus) pipes with water flowing through them. The inner pipe is solid, with water obviously flowing between the outside of the innner rod, and the inside diameter of the outer rod.

Currently, there is 10 mm diameter small rod, and a 15 mm inside diameter on the outer rod. The flow rate from our pump is 9 GPM.

We are changing the small rod to 6 mm. How can I solve for the head the pump must have? I haven't worked with head calculations, and I want to make sure the pump will work okay. The rods are completely horizontal (no vertical gradiant).

Possible equations I've dug up:

Bernoulli's Equation:
Head Loss=(Pressure (water fluid)/Spec. Weight) + (Veloc.^2/(2*g)) + Height

Shaft Head=Horsepower/(Specific Weight*Flow Rate)

Though I'm not sure either of these are right! I'm getting a solution for the head in the 600 ft range, and this is entirely too high!

Thanks for any help anyone can offer.

RE: Pump head in pipe with changing diameter/area

(OP)
Actually, my "head loss" using Bernoulli's is 11.xx ft, but my Shaft Head is 661.9 ft, so my "total head" would have to be calculated by subtracting the "head loss" from the shaft head, correct? Still giving me a number of about 650 ft (unrealistic).

RE: Pump head in pipe with changing diameter/area

To calculate the head required from a pump, you use Bernoulli's equation.  Add up the friction losses, velocity head loss, static head (change in height), etc., ahd that's the required head that your pump will need to supply.

Since you are reducing the flow obstruction (the sold inner pipe) you'll have more flow area, and less losses through the pipe overall.  Your flow will therefore increase (usually, but that you can figure out from the pump curve). Are you trying to size a new pump, or just making sure the existing pump has enough head (without running too far off the pipe curve)?   

RE: Pump head in pipe with changing diameter/area

A centrifugal pump delivers head corresponding to the head located at the intersection of the pump's curve and the system curve.


For head loss in your piping system, use a pipe head loss equation, such as Darcy-Wiesbach

with an appropriate friction factor, such as Colebrook-White, Churchill or others.  

Pipe in pipe head losses use the hydraulic radius in lieu of the typical inside diameter.

http://www.lightmypump.com/pump_glossary.htm

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25% to 50% of the total electrical energy usage in certain industrial facilities." - DOE statistic  (Note: Make that 99.99% for pipeline companies) http://virtualpipeline.spaces.live.com/

RE: Pump head in pipe with changing diameter/area

(OP)
Skearse, I'm trying to make sure an existing pump will still work.

The head should be about 150 ft, so the 650 is quite unreasonable.  

RE: Pump head in pipe with changing diameter/area

650 feet of head seems extremely high. However, you have not given enough information to solve the puzzle, such as how long is the tube? and does it discharge to air, reservoir or to another pipe? and what is your suction pressure? and how smooth is the tube? is it steel, glass, wood? and what type of pump is it? centrifugal, axial, screw, positive displacement etc. Also, are you sure you are discharging 9 gpm now? has that been measured and confirmed? and is your current pump sized properly? is it running at max efficiency, or is it running under or over speed?

RE: Pump head in pipe with changing diameter/area

(OP)
cvg, thanks for the reply.

Well the flow tube with the inner rod is only 150 mm. However, there is 5.xx ft of tube the water goes through after the pump, one 45 degree elbow from the pump to that tube, and a 90 degree elbow connecting that tube to the 'cavity' the flow tube/inner rod are in. So, looping that tube back it's about 11 ft of total tube.

The pump is a 1.5 hp rated centrifugal pump. We calculated efficiency, and it's actually being run at about 1.7 hp (so over speed). We are discharging 9 GPM from testing it.

RE: Pump head in pipe with changing diameter/area

(OP)
I feel like I first need to calculate the head loss in the flow tube w/ the inner rod before I analyze the entire circuit, though. Since I am modifying the inside diameter of the flow tube, as well as the inside rod's diameter, I'm trying to see if there is a big change in head loss. If there isn't, then I dont need to look at the whole system because there isn't any justification that I would need a new pump...

RE: Pump head in pipe with changing diameter/area


Have You any actual pressure gauge reading in the piping circuit?
   

At 74th year working on IR-One PhD from UHK  - - -

RE: Pump head in pipe with changing diameter/area

(OP)
No pressure gauge yet, we're working on getting one though.

Would you calculate water pressure by using

P=rho*(V^2)/2

or

P=rho*height*gravity

There is no change in height here, though, so I'm not sure how the second equation would factor in...

RE: Pump head in pipe with changing diameter/area


My old school math used–
   V2 = 2gh
       V= Velocity in ft./sec
       g = 32.174 slugs
       h = ft. H2O
 

At 74th year working on IR-One PhD from UHK  - - -

RE: Pump head in pipe with changing diameter/area

Synchronous electric motor drive - so how are you run over speed - is it VF Drive?   

RE: Pump head in pipe with changing diameter/area

In your early post you mentioned changing the inner rod from 10mm OD to 6 mm OD with the outer pipe having an ID of 15 mm.  Now you mention that the outer pipe is also changing. If you formulate your question so badly, how do you expect any decent replies?

Any fluid mechanics handbook will tell you how to treat an irregular shaped flow path, using the hydraulic radius. As a rough estimate you could use the formula you posted yourself that shows the head loss is proportional to the square of the velocity. Assuming you keep the flow rate constant you can work out the velocity from the change in area, and get the new head as a ratio to the old head.

Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com

RE: Pump head in pipe with changing diameter/area

(OP)
Katmar, my question in this thread is not to solve the problem for me, my question is what formulas would be suggested to solve for an accurate head. Whether I want my inside diameter to be 15 mm, 10 mm, or 1000 mm, I'm looking for the proper equations with variables, not for someone to give me a specific answer. I need to leave that outside ID that was formerly 15 mm a variable so that I can make it a size that will allow for adequate head. So, I didn't change my question at all. I'm not looking for a numerical answer, I'm here to learn how I would go about it getting it accurately.

I did use hydrolic radius in that formula to solve for head, but I don't believe the number I am getting is correct.


Artisi, the number of ~1.7 hp may actually be the 1.5 hp the motor is rated at. The reason for this, is that the formula used to calculate includes a power correction factor'. We do not have a specific value for this, and are estimating it to be 0.8.  

RE: Pump head in pipe with changing diameter/area

Power correction factor or efficiency factor?

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

RE: Pump head in pipe with changing diameter/area

(OP)
I used the Darcy-Weisbach equation:

hl_major=f*L*V^2/(D_h*2*g)

where f is the friction factor, and got a reasonable 250 ft. It's still too high for my pump curve (I expect to be around 175 ft), but is getting there.

In my text books, the formula for turblent flow with an incompressible fluid lists

Head loss= (f*L*V^2)/(D_h*2)

without gravity factored in. Why does one equation account for "g", while the other doesn't?

RE: Pump head in pipe with changing diameter/area

Have a look at the FAQ378-1142: Flow in Annular Space which discusses the determination of the equivalent diameter of the annular space.

The first version you gave of the Darcy-Weisbach formula is correct if you want to calculate the head loss as a height of flowing fluid.  The second version, without the "g" (acceleration of the earth's gravity) should have a density term in the numerator.  This has the effect of giving the head loss in pressure units because for a vertical column of liquid
  pressure = density x gravity x height
If you multiply your first version of D-W (which has units of height) by density x gravity the gravity term cancels and you get
  ΔP = ( ƒM x L x ρ x V2 ) / ( D x 2 )
where
ΔP = pressure drop in Pascal
ƒM = Moody friction factor = about 0.03
L = pipe length = 0.15 metre
ρ = density = 1000 kg/m3
V = velocity = 8.9 m/s for 9 US gpm in a 9 mm pipe
D = pipe equivalent diameter in metre

If we take D to be conservatively 9 mm = 0.009 m then the head loss through the annular pipe is
   ΔP = ( 0.03 x 0.15 x 8.92 x 1000 ) / ( 0.009 x 2 ) = 19800 Pa or 19.8 kPa

This is about 2 metres of head or 6.5 feet

On top of this you would have to make allowance for entrance and exit effects as well as the rest of your pipe circuit.

Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com

RE: Pump head in pipe with changing diameter/area

(OP)
Still not sure why the book had the wrong formula, must have passed through editing. I even scanned it.

Anyway, thanks for the help. What I ended up using was the correct version of the Darcy-Weisbach to calculate major losses:

hl_major=f*L*V^2/(D_h*2*g)

where f (for glass) at a Reynolds number of 191,541 comes out to be ~0.016. Plugging in a D_h of:

D_h = 2*(D_tube/2 - D_rod/2) = 0.005 m (for a 10 mm and 15 mm rod selection)

Velocity is ~ 28.92 m/s.

hl_major came out to be 67.12 ft. Much more reasonable! My colleague said he previously counted minor losses to be 59 ft through the system, making the total head loss 126.1 ft, about where I expected to be. Thanks for all the help guys.

RE: Pump head in pipe with changing diameter/area

In equation 8.30 your author has
  ΔP = h x ρ
instead of the formula in my earlier post.  This is basic physics so I can only guess he has defined some strange units for h.  

Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com

RE: Pump head in pipe with changing diameter/area

I think maybe there is some discrepancy between formulas in U.S. and metric units, especially as I see dimensions given in both feet and meters...

RE: Pump head in pipe with changing diameter/area

Amen to cvg!

Just remembered an old rule of thumb, "For every 1 hp of drive, the equivalent of 1 gpm @ 1500 psi can be produced."

A little numbers playing insinuates that the pressure on the existing 98.17 mm2 is about 250 psi as delivered by a 1.5 hp motor.

Question of KevinH673:
Is the 1.5 hp motor/pump used only for delivering solution through this concentric (annulus) pipe?  
 

At 74th year working on IR-One PhD from UHK  - - -

RE: Pump head in pipe with changing diameter/area

(OP)
Apakrat, no the motor sends fluid through a system that includes some 10' of tube, as well as two or three elbows. The concentric tubes are the ones that are changing in the system though.

RE: Pump head in pipe with changing diameter/area


Evidently, from Your earlier statements above, You need to maintain near the same existing pressure passing thru this concentric tube section.

Without getting into theories, mathematics, etc, nor the worry of calculating possible motor/pump requirement, including cost, would it not be easier, simpler and more economical to just reduce the outer tube sizing and maintain near the same area around the proposed 6 mm inner size, thus maintaining near existing flow and pressure?

Believe You can find stock piping(tubing), fitting near this condition.

 

At 74th year working on IR-One2 PhD from UHK  - - -

RE: Pump head in pipe with changing diameter/area

(OP)
Apakrat, this was precisely my first suggestion to this problem. It makes sense to do, knowing that pressure will be the same, so flow should be okay. However, I first started investigating this further to make sure the heat pulled from the YAG rod would be sufficient. I also don't believe anyone ever optimized the inside diameter of the flow tube, so I set out to do this (and then running into my problem!). Your suggestion does make sense though!

RE: Pump head in pipe with changing diameter/area


Please forgive me, but my interest in this posting is that I am trying to learn also.

Don't actually know why the YAG rod diameter needs to be reduced.  Will guess from prior statements, the present  YAG rod is not performing  a desired heat exchange, using the present flow rate volume.  Exposure to a larger heated medium layer would possibly achieve a desired goal.

Perhaps You might want to look at installing an adjustable Needle Valve in the 15 mm line, down stream of the concentric YAG piping section.
 

At 74th year working on IR-One2 PhD from UHK  - - -

RE: Pump head in pipe with changing diameter/area

(OP)
The YAG rod's diameter is decreasing in order to give a better "mode" burn from the laser (basically, making a more uniform 'circle' from the laser beam for cutting). Doing this will affect the total power, though.

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