Pump head in pipe with changing diameter/area

[KevinH673]

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.

 

[KevinH673]

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).

 

[skearse]

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)?

 

[BigInch]

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/

 

[KevinH673]

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.

 

[cvg]

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?

 

[KevinH673]

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.

 

[KevinH673]

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...

 

[Apakrat]

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


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

 

[KevinH673]

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...

 

[Apakrat]

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 - - -

 

[Artisi]

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

 

[katmar]

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

 

[KevinH673]

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.

 

[BigInch]

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/

 

[KevinH673]

Power factor. Dealing with an AC motor, using Amp-Volts.

P=E*I*PF

More here...

http://forums.qrz.com/showthread.php?t=189344

 

[KevinH673]

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?

 

[katmar]

Have a look at the faq378-1142 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
[Δ]P = ( [ƒ]_M x L x [ρ] x V² ) / ( D x 2 )
where
[Δ]P = pressure drop in Pascal
[ƒ]_M = 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
[Δ]P = ( 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

 

[KevinH673]

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.

 

[KevinH673]

That .pdf size was rather large, so I took a screen shot of the scanned image, here it is.

http://img88.imageshack.us/img88/4112/wrongequationel1.jpg