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.
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
RE: Pump head in pipe with changing diameter/area
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
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
The head should be about 150 ft, so the 650 is quite unreasonable.
RE: Pump head in pipe with changing diameter/area
RE: Pump head in pipe with changing diameter/area
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
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
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
RE: Pump head in pipe with changing diameter/area
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
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
**********************
"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
P=E*I*PF
More here...
http://forums.qrz.com/showthread.php?t=189344
RE: Pump head in pipe with changing diameter/area
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
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
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
http://i
RE: Pump head in pipe with changing diameter/area
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
RE: Pump head in pipe with changing diameter/area
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
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?
At 74th year working on IR-One2 PhD from UHK - - -
RE: Pump head in pipe with changing diameter/area
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