On Monday, May 29, 2017 at 7:20:01 PM UTC-4, wrote:


If there is no flow, there will be no pressure differential. Pressure will be the same throughout the volume of liquid from inlet to outlet.

--
Ed Huntress

If you have a column of water, the pressure at the bottom of the column in higher than the pressure at the top. So there is an example of no flow with a pressure differential.

If you take something like say a large nut and tie a string to it and whirl it around your head, then you have a force on the string and a velocity of the nut. But the nut does not go flying off , unless you let go of the string.

In the same way with the outlet blocked off on a centrifugal pump. the water has a velocity , but it just goes around and around. Produces a pressure, but no flow from the pump.

Think of a centrifuge with test tubes in it. The material in the test tubes are subjected to force, but there is no flow.

Dan

On Monday, May 29, 2017 at 7:23:55 PM UTC-4, wrote:
On Monday, May 29, 2017 at 5:37:34 PM UTC-4, wrote:
On Monday, May 29, 2017 at 5:13:41 PM UTC-4, wrote:
On Monday, May 29, 2017 at 3:34:50 PM UTC-4, wrote:
sw

Right. So what is the condition inside of one involute in the second stage? Is it completely full when it's operating?

--
Ed Huntress


It is exactly the same as the first stage.

Dan

And what is that condition? Are the involutes completely filled? And, if so, how is that possible unless the velocity is the same from the input port to the periphery of the wheel?

--
Ed Huntress

First some centrifugal pumps do not have involutes.

Most do, but anything that will accelerate the flow to the periphery will do.


Second are you thinking there could be air in the pump? THa woud be bad as you would have cavitation.

Not necessarily. The air could come from partial filling, which almost certainly is the case at start-up. Once the pump is running, apparently the involute spaces fill.


The pump would be completely filled. How could it not be completely filled?

Easy. Low input pressure; no back pressure; vanes sling the small amount of water to the periphery, with no resistance.

And why would the velocity have to be the same from input port to the periphery?

The velocity doesn't have to be the same. But if it's higher at the periphery, the involute space doesn't fill. If it's higher at the inlet, you have a possible energy-conservation dilemma, unless the volume in each space from center to periphery increases as to the square of distance from the center. I haven't done the math on that but it's easy to tell.

This is the outward radial velocity we're talking about here. The tangential velocity increases by definition.


The flow would be the same, but the passages vary in cross section, so there is no way the velocity could be the same.

Again, it's a matter of which "velocity" you're talking about -- radial or tangential.


Dan

On Monday, May 29, 2017 at 7:39:40 PM UTC-4, wrote:
On Monday, May 29, 2017 at 7:20:01 PM UTC-4, wrote:


If there is no flow, there will be no pressure differential. Pressure will be the same throughout the volume of liquid from inlet to outlet.

--
Ed Huntress

If you have a column of water, the pressure at the bottom of the column in higher than the pressure at the top. So there is an example of no flow with a pressure differential.

Oh, right. I was thinking of externally applied pressure. Gravity or centrifugal force would result in a differential.

Any pressure applied from the inlet, as in the case of a multi-stage centrifugal pump, would be the same throughout the volume within one involute. But the centrifugal force added by the spinning rotor would be greatest at the periphery.


If you take something like say a large nut and tie a string to it and whirl it around your head, then you have a force on the string and a velocity of the nut. But the nut does not go flying off , unless you let go of the string.

In the same way with the outlet blocked off on a centrifugal pump. the water has a velocity , but it just goes around and around. Produces a pressure, but no flow from the pump.

Again, the velocity is the tangential velocity. There is no radial velocity.


Think of a centrifuge with test tubes in it. The material in the test tubes are subjected to force, but there is no flow.

Dan

wrote on 5/29/2017 7:19 PM:
On Monday, May 29, 2017 at 6:21:59 PM UTC-4, Jim Wilkins wrote:
wrote in message
...
On Monday, May 29, 2017 at 5:13:41 PM UTC-4, wrote:
On Monday, May 29, 2017 at 3:34:50 PM UTC-4,
wrote:
sw

Right. So what is the condition inside of one involute in the
second stage? Is it completely full when it's operating?

--
Ed Huntress


It is exactly the same as the first stage.

Dan

And what is that condition? Are the involutes completely filled?
And, if so, how is that possible unless the velocity is the same
from the input port to the periphery of the wheel?

--
Ed Huntress

How did you ever dream up that requirement?

Because if they are not completely filled, there is no physical way to transit any positive pressure at the inlet to the outlet. My further reading suggests they are filled, although some illustrations show them partly filled. Photos taken through transparent windows show them partly filled, but those are illustrations of cavitation. I'm reaching the conclusion that they're completely filled in normal operation.

It isn't a positive
displacement pump. If the outlet valve is closed the input and output
velocities will be zero, yet the periphery is still spinning. The
pressure differential (p) from inlet to outlet will be what the
discharge curve shows for zero flow (q).
-jsw

If there is no flow, there will be no pressure differential. Pressure will be the same throughout the volume of liquid from inlet to outlet.


You need to use your brain to think, Ed.

When the outlet valve is shut off, the impeller is spinning the fluid as
a disk inside the housing.

The impeller is a solid piece (balanced weight) so it does not have the
tendency to fly away from the center when it is spinning at high speed.
The fluid in the disk will try to fly away from the center due to
centrifugal force. The fluid will exert highest pressure on the wall at
the outermost part of the housing while there is no pressure at the
center of the rotating disk.

If the outlet valve is opened at this time, the fluid will be expelled,
and the center of the rotating disk will form a negative pressure to
suck in fluid from the intake valve.

So, if there is no flow, there will definitely be pressure differential
if the impeller is spinning the fluid at high speed.

Another way to prove that there will be pressure differential is by
inspecting the equation for centrifugal force:

https://en.wikipedia.org/wiki/Centrifugal_force#Force

Centrifugal force of a rotating object is proportional to its radius
from the center of rotation. It is obvious that centrifugal force is
zero at the center, and maximum at the farthest point from the center.

centrifugal force = m × ω × ω × r
http://keisan.casio.com/keisan/lib/real/system/2006/1271292951/%B1%F3%BF%B4%CE%CF.gif

You should use angular velocity (ω) instead of linear velocity (v)
because all the fluid molecules in the disk spin at the same angular
velocity regardless of its distance from the center.

wrote on 5/29/2017 7:43 PM:
On Monday, May 29, 2017 at 7:23:55 PM UTC-4, wrote:
On Monday, May 29, 2017 at 5:37:34 PM UTC-4, wrote:
On Monday, May 29, 2017 at 5:13:41 PM UTC-4, wrote:
On Monday, May 29, 2017 at 3:34:50 PM UTC-4, wrote:
sw

Right. So what is the condition inside of one involute in the second stage? Is it completely full when it's operating?

--
Ed Huntress


It is exactly the same as the first stage.

Dan

And what is that condition? Are the involutes completely filled? And, if so, how is that possible unless the velocity is the same from the input port to the periphery of the wheel?

--
Ed Huntress

First some centrifugal pumps do not have involutes.

Most do, but anything that will accelerate the flow to the periphery will do.


Second are you thinking there could be air in the pump? THa woud be bad as you would have cavitation.

Not necessarily. The air could come from partial filling, which almost certainly is the case at start-up. Once the pump is running, apparently the involute spaces fill.


The pump would be completely filled. How could it not be completely filled?

Easy. Low input pressure; no back pressure; vanes sling the small amount of water to the periphery, with no resistance.

And why would the velocity have to be the same from input port to the periphery?

The velocity doesn't have to be the same. But if it's higher at the periphery, the involute space doesn't fill. If it's higher at the inlet, you have a possible energy-conservation dilemma, unless the volume in each space from center to periphery increases as to the square of distance from the center. I haven't done the math on that but it's easy to tell.

This is the outward radial velocity we're talking about here. The tangential velocity increases by definition.


The flow would be the same, but the passages vary in cross section, so there is no way the velocity could be the same.

Again, it's a matter of which "velocity" you're talking about -- radial or tangential.

When you are dealing with centrifugal pump, it is easier to think
"angular velocity" (ω) rather than "linear velocity" (v).

Please let me repeat myself from my other post:

When the outlet valve is shut off, the impeller is spinning the fluid as
a disk inside the housing.

The impeller is a solid piece (balanced weight) so it does not have the
tendency to fly away from the center when it is spinning at high speed.
The fluid in the disk will try to fly away from the center due to
centrifugal force. The fluid will exert highest pressure on the wall at
the outermost part of the housing while there is no pressure at the
center of the rotating disk.

If the outlet valve is opened at this time, the fluid will be expelled,
and the center of the rotating disk will form a negative pressure to
suck in fluid from the intake valve.

So, if there is no flow, there will definitely be pressure differential
if the impeller is spinning the fluid at high speed.

Another way to prove that there will be pressure differential is by
inspecting the equation for centrifugal force:

https://en.wikipedia.org/wiki/Centrifugal_force#Force

Centrifugal force of a rotating object is proportional to its radius
from the center of rotation. It is obvious that centrifugal force is
zero at the center, and maximum at the farthest point from the center.

centrifugal force = m × ω × ω × r
http://keisan.casio.com/keisan/lib/real/system/2006/1271292951/%B1%F3%BF%B4%CE%CF.gif

You should use angular velocity (ω) instead of linear velocity (v)
because all the fluid molecules in the disk spin at the same angular
velocity regardless of its distance from the center.

On 5/29/2017 5:32 PM, wrote:
On Monday, May 29, 2017 at 7:27:02 PM UTC-4, tyre biter wrote:
On 5/29/2017 5:12 PM, wrote:
On Monday, May 29, 2017 at 6:21:04 PM UTC-4, tyre biter wrote:
On 5/29/2017 3:54 PM, wrote:
On Monday, May 29, 2017 at 5:31:58 PM UTC-4, tyre biter wrote:
On 5/29/2017 1:32 PM, wrote:
Your brain must be the size of a pea, Biter.

Your pallid attempts at fighting back are hilarity, crazy eddy.

Ha-ha! What is there to fight?

Then why do you?

If I was fighting, weasel, you'd know it.



You're on your own, weasel. Don't drown in your own drool.


Beg me Crazy Eddy...do it...NOW!

What a trashy, low-class troll you are.

LOLOLOL!!!!

On Monday, May 29, 2017 at 8:51:06 PM UTC-4, tyre biter wrote:
On 5/29/2017 5:32 PM, wrote:
On Monday, May 29, 2017 at 7:27:02 PM UTC-4, tyre biter wrote:
On 5/29/2017 5:12 PM, wrote:
On Monday, May 29, 2017 at 6:21:04 PM UTC-4, tyre biter wrote:
On 5/29/2017 3:54 PM, wrote:
On Monday, May 29, 2017 at 5:31:58 PM UTC-4, tyre biter wrote:
On 5/29/2017 1:32 PM, wrote:
Your brain must be the size of a pea, Biter.

Your pallid attempts at fighting back are hilarity, crazy eddy.

Ha-ha! What is there to fight?

Then why do you?

If I was fighting, weasel, you'd know it.



You're on your own, weasel. Don't drown in your own drool.


Beg me Crazy Eddy...do it...NOW!

What a trashy, low-class troll you are.

LOLOLOL!!!!

....a trashy troll with a Tourette syndrome...sad...

--
Ed Huntress

On Mon, 29 May 2017 14:44:27 -0400, lqDXsb?? ?????? ? ??????? ??ZsAaIo
wrote:

wrote on 5/29/2017 2:03 PM:
On Monday, May 29, 2017 at 1:55:58 PM UTC-4, XCjEwC?? ?????? ? ??????? ??CkYyoU wrote:
wrote on 5/29/2017 1:04 PM:
On Monday, May 29, 2017 at 8:13:36 AM UTC-4, wrote:
On Sunday, May 28, 2017 at 4:04:30 PM UTC-4, wrote:
On Sunday, May 28, 2017 at 11:35:39 AM UTC-4, wrote:



You are over thinking the situation. A multistage pump has a bunch of identical sections all turning at the same speed. Each stage increases the pressure. So you might have a 6 stage pump with each stage increasing the pressure by 10 psi. Which makes for a fairly efficient pump which will supply water at 60 psi. Google it.


From Wik

Multistage centrifugal pumps
Multistage centrifugal pump[5]

A centrifugal pump containing two or more impellers is called a multistage centrifugal pump. The impellers may be mounted on the same shaft or on different shafts. At each stage, the fluid is directed to the center before making its way to the discharge on the outer diameter.

For higher pressures at the outlet, impellers can be connected in series. For higher flow output, impellers can be connected parallel.

A common application of the multistage centrifugal pump is the boiler feedwater pump. For example, a 350 MW unit would require two feedpumps in parallel. Each feedpump is a multistage centrifugal pump producing 150 l/s at 21 MPa.

All energy transferred to the fluid is derived from the mechanical energy driving the impeller. This can be measured at isentropic compression, resulting in a slight temperature increase (in addition to the pressure increase).



Dan

Thanks, Dan. I read that -- and maybe 100 more pages over the past few days. None of them really explain it. To say that the energy is derived from the impeller is axiomatic. It doesn't explain what's going on inside the second stage.

Rather than try to go through it in detail, I'll post something if I find a good explanation.


You don't have enough smarts to process too much information at once,
ED. Please do yourself a favour by looking at this impeller of a simple
centrifugal pump:

https://d2t1xqejof9utc.cloudfront.net/screenshots/pics/bd60670ae9dc5e37fdc08142df0462d2/large.JPG

To give your brain a chance to comprehend, please imagine all the valves
are closed. The fluid is trapped inside the centrifuge, with no way in,
and no way out.

The impeller keeps spinning 'round and 'round. The impeller is trying to
throw the fluid (which is slotted in the empty space in the impeller)
radially outward. Because of this, the fluid is exerting pressure
radially outward against the wall of the casing.

Do you still have difficulty, Ed?

Thanks, Wannabe, but I understand how a centrifugal pump works. What I don't understand is the effect of feeding one with a high pressure head of water.



It is the same. The pump doesn't know (and doesn't care) what's being
fed to it. It will spin and throw the fluid radially outward, resulting
in a higher pressure than it was fed. Imagine the output valve is
closed, in order not to complicate things.


This is true..until the input pressure is higher than the possible
output pressure..and now the pump acts as nothing more than a
restrictor mechanism...and the flow becomes restricted









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On Tuesday, May 30, 2017 at 1:00:01 AM UTC-4, Gunner Asch wrote:



This is true..until the input pressure is higher than the possible
output pressure..and now the pump acts as nothing more than a
restrictor mechanism...and the flow becomes restricted


Sorry, but that never happens. The output pressure will always be higher than the input pressure as long as the pump is running and nothing has failed because of the high pressure.

Dan

On Tuesday, May 30, 2017 at 8:34:06 AM UTC-4, wrote:
On Tuesday, May 30, 2017 at 1:00:01 AM UTC-4, Gunner Asch wrote:



This is true..until the input pressure is higher than the possible
output pressure..and now the pump acts as nothing more than a
restrictor mechanism...and the flow becomes restricted


Sorry, but that never happens. The output pressure will always be higher than the input pressure as long as the pump is running and nothing has failed because of the high pressure.

Dan

And as long as the pump is fully filled. g I finally found the info I was looking for, and it's kind of interesting. Basically, a regular centrifugal pump has to be fully filled to work. Some have priming pumps and other priming gadgets built-in, so they're essentially self-priming, but they only produce more than a small amount of pressure when they're filled. Until they's filled, they're likely to cavitate, so it's only something you can do for seconds at start-up.

From one source: "There are selfpriming centrifugal designs that can lift liquid an average of 15 feet when partially filled (13€� hg vacuum)."

BTW, the internal flow in a centrifugal pump is so complex that they can't model performance with computers. They have to build an actual model and test it.

That satisfies my curiosity. The thing that threw me off is that illustrations of such pumps working often suggest something *less* than complete filling. I suspect that's because it's easier to illustrate the liquid in the pump that way, but it's misleading.

--
Ed Huntress

On Tuesday, May 30, 2017 at 10:27:07 AM UTC-4, wrote:
On Tuesday, May 30, 2017 at 8:34:06 AM UTC-4, wrote:
On Tuesday, May 30, 2017 at 1:00:01 AM UTC-4, Gunner Asch wrote:



This is true..until the input pressure is higher than the possible
output pressure..and now the pump acts as nothing more than a
restrictor mechanism...and the flow becomes restricted


Sorry, but that never happens. The output pressure will always be higher than the input pressure as long as the pump is running and nothing has failed because of the high pressure.

Dan

And as long as the pump is fully filled. g I finally found the info I was looking for, and it's kind of interesting. Basically, a regular centrifugal pump has to be fully filled to work. Some have priming pumps and other priming gadgets built-in, so they're essentially self-priming, but they only produce more than a small amount of pressure when they're filled. Until they's filled, they're likely to cavitate, so it's only something you can do for seconds at start-up.

From one source: "There are selfpriming centrifugal designs that can lift liquid an average of 15 feet when partially filled (13€� hg vacuum).."

BTW, the internal flow in a centrifugal pump is so complex that they can't model performance with computers. They have to build an actual model and test it.

That satisfies my curiosity. The thing that threw me off is that illustrations of such pumps working often suggest something *less* than complete filling. I suspect that's because it's easier to illustrate the liquid in the pump that way, but it's misleading.

--
Ed Huntress

While we're at it, regarding Eric's original question about whether his 10-psi pump with 80 psi input will produce a 90 psi output, the answer is "yes," disregarding friction losses.

--
Ed Huntress

On 5/29/2017 10:39 PM, wrote:
On Monday, May 29, 2017 at 8:51:06 PM UTC-4, tyre biter wrote:
On 5/29/2017 5:32 PM, wrote:
On Monday, May 29, 2017 at 7:27:02 PM UTC-4, tyre biter wrote:
On 5/29/2017 5:12 PM, wrote:
On Monday, May 29, 2017 at 6:21:04 PM UTC-4, tyre biter wrote:
On 5/29/2017 3:54 PM, wrote:
On Monday, May 29, 2017 at 5:31:58 PM UTC-4, tyre biter wrote:
On 5/29/2017 1:32 PM, wrote:
Your brain must be the size of a pea, Biter.

Your pallid attempts at fighting back are hilarity, crazy eddy.

Ha-ha! What is there to fight?

Then why do you?

If I was fighting, weasel, you'd know it.



You're on your own, weasel. Don't drown in your own drool.


Beg me Crazy Eddy...do it...NOW!

What a trashy, low-class troll you are.

LOLOLOL!!!!

...a trashy troll with a Tourette syndrome...sad...

Hey Crazy Eddy...guess what...you can't quit me, brokeback...LOLOL!

On Tuesday, May 30, 2017 at 2:20:43 PM UTC-4, tyre biter wrote:
On 5/29/2017 10:39 PM, wrote:
On Monday, May 29, 2017 at 8:51:06 PM UTC-4, tyre biter wrote:
On 5/29/2017 5:32 PM, wrote:
On Monday, May 29, 2017 at 7:27:02 PM UTC-4, tyre biter wrote:
On 5/29/2017 5:12 PM, wrote:
On Monday, May 29, 2017 at 6:21:04 PM UTC-4, tyre biter wrote:
On 5/29/2017 3:54 PM, wrote:
On Monday, May 29, 2017 at 5:31:58 PM UTC-4, tyre biter wrote:
On 5/29/2017 1:32 PM, wrote:
Your brain must be the size of a pea, Biter.

Your pallid attempts at fighting back are hilarity, crazy eddy.

Ha-ha! What is there to fight?

Then why do you?

If I was fighting, weasel, you'd know it.



You're on your own, weasel. Don't drown in your own drool.


Beg me Crazy Eddy...do it...NOW!

What a trashy, low-class troll you are.

LOLOLOL!!!!

...a trashy troll with a Tourette syndrome...sad...

Hey Crazy Eddy...guess what...you can't quit me, brokeback...LOLOL!

Why would I want to? It's entertaining to see what a cowardly weasel like you can say with a vocabulary of fewer than 1,000 words.

My wife teaches 2nd grade. They have larger vocabularies and they're much more creative than you, but it's encouraging to watch you try.

Tourette can be serious. I hope you're getting treatment for it.

--
Ed Huntress

On 5/30/2017 2:15 PM, wrote:
On Tuesday, May 30, 2017 at 2:20:43 PM UTC-4, tyre biter wrote:
On 5/29/2017 10:39 PM, wrote:
On Monday, May 29, 2017 at 8:51:06 PM UTC-4, tyre biter wrote:
On 5/29/2017 5:32 PM, wrote:
On Monday, May 29, 2017 at 7:27:02 PM UTC-4, tyre biter wrote:
On 5/29/2017 5:12 PM, wrote:
On Monday, May 29, 2017 at 6:21:04 PM UTC-4, tyre biter wrote:
On 5/29/2017 3:54 PM, wrote:
On Monday, May 29, 2017 at 5:31:58 PM UTC-4, tyre biter wrote:
On 5/29/2017 1:32 PM, wrote:
Your brain must be the size of a pea, Biter.

Your pallid attempts at fighting back are hilarity, crazy eddy.

Ha-ha! What is there to fight?

Then why do you?

If I was fighting, weasel, you'd know it.



You're on your own, weasel. Don't drown in your own drool.


Beg me Crazy Eddy...do it...NOW!

What a trashy, low-class troll you are.

LOLOLOL!!!!

...a trashy troll with a Tourette syndrome...sad...

Hey Crazy Eddy...guess what...you can't quit me, brokeback...LOLOL!

Why would I want to?

True, you groove on being humiliated.

It's entertaining to see what a cowardly weasel like you can say with a vocabulary of fewer than 1,000 words.

It's amusing to see least coast lackwit try and keep up.

My wife teaches 2nd grade.

No doubt serving her well in dealing with you...

They have larger vocabularies and they're much more creative than you, but it's encouraging to watch you try.

And the worst thing is you torture a lame analogy like it was the
neighbor's cat.

Tourette can be serious. I hope you're getting treatment for it.


Project much Crazy Eddy?

Laugh, laugh, laugh, laugh!

On Tuesday, May 30, 2017 at 4:49:14 PM UTC-4, tyre biter wrote:
On 5/30/2017 2:15 PM, wrote:


Laugh, laugh, laugh, laugh!

As I said, get treatment for that Tourette syndrome. You could wind up completely debilitated and wind up like tyre biter.

--
Ed Huntress

On 5/30/2017 2:56 PM, wrote:
On Tuesday, May 30, 2017 at 4:49:14 PM UTC-4, tyre biter wrote:
On 5/30/2017 2:15 PM, wrote:


Laugh, laugh, laugh, laugh!

As I said, get treatment for that Tourette syndrome. You could wind up completely debilitated and wind up like tyre biter.


Don't the neighbors have another cat for you to torture?

This tautology was sponsored by Fresh Step Litter - because your
argumentative turds really stink.

LOL!

On Tuesday, May 30, 2017 at 5:11:05 PM UTC-4, tyre biter wrote:
On 5/30/2017 2:56 PM, wrote:
On Tuesday, May 30, 2017 at 4:49:14 PM UTC-4, tyre biter wrote:
On 5/30/2017 2:15 PM, wrote:


Laugh, laugh, laugh, laugh!

As I said, get treatment for that Tourette syndrome. You could wind up completely debilitated and wind up like tyre biter.


Don't the neighbors have another cat for you to torture?

I only bother with weasels, like you. That, and termites.

--
Ed Huntress

On 5/30/2017 3:21 PM, wrote:
On Tuesday, May 30, 2017 at 5:11:05 PM UTC-4, tyre biter wrote:
On 5/30/2017 2:56 PM, wrote:
On Tuesday, May 30, 2017 at 4:49:14 PM UTC-4, tyre biter wrote:
On 5/30/2017 2:15 PM, wrote:


Laugh, laugh, laugh, laugh!

As I said, get treatment for that Tourette syndrome. You could wind up completely debilitated and wind up like tyre biter.


Don't the neighbors have another cat for you to torture?

I only bother with weasels, like you. That, and termites.

--
Ed Huntress


Poor Crazy Eddy, snip and run and all for a lame analogy - again!

You can't quit me, brokeback!

On Tue, 30 May 2017 05:34:01 -0700 (PDT), "
wrote:

On Tuesday, May 30, 2017 at 1:00:01 AM UTC-4, Gunner Asch wrote:



This is true..until the input pressure is higher than the possible
output pressure..and now the pump acts as nothing more than a
restrictor mechanism...and the flow becomes restricted


Sorry, but that never happens. The output pressure will always be higher than the input pressure as long as the pump is running and nothing has failed because of the high pressure.

Dan


Thats hardly true. Stick a fire hose into a 1" pump and let her
rip...what you get out..wont be what you put in. Been there, done
that. Old oil field hand..remember?


---
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On Tue, 30 May 2017 15:46:34 -0700, Gunner Asch
wrote:

On Tue, 30 May 2017 05:34:01 -0700 (PDT), "
wrote:

On Tuesday, May 30, 2017 at 1:00:01 AM UTC-4, Gunner Asch wrote:



This is true..until the input pressure is higher than the possible
output pressure..and now the pump acts as nothing more than a
restrictor mechanism...and the flow becomes restricted


Sorry, but that never happens. The output pressure will always be higher than the input pressure as long as the pump is running and nothing has failed because of the high pressure.

Dan


Thats hardly true. Stick a fire hose into a 1" pump and let her
rip...what you get out..wont be what you put in. Been there, done
that. Old oil field hand..remember?

Actually what we remember is a poor, old, broken down, liar.
--
Cheers,

Schweik

On Wednesday, May 31, 2017 at 11:12:16 AM UTC-4, Neighborhood number 3 wrote:


Thats hardly true. Stick a fire hose into a 1" pump and let her
rip...what you get out..wont be what you put in. Been there, done
that. Old oil field hand..remember?

In 16 years you still haven't figured out how to make water drain into
the ground. Yet now you think there must be someone who'll take advice
from you about fluid dynamics, based on some of your wildly
exaggerated "experience." LOL

Well have to confess I was only thinking of a pump being used to pump.

Incidently you can put water into a centrifugal pump outlet and it will spin the pump.

Dan

On Sunday, May 28, 2017 at 8:35:39 AM UTC-7, wrote:
On Saturday, May 27, 2017 at 6:57:12 PM UTC-4, Ned Simmons wrote:
On Sat, 27 May 2017 12:20:00 -0700 (PDT), wrote:

On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?

That's close. A pump on atmospheric air (1 bar) that makes a 10 bar output,
will be different, because the density of inlet air determines the force
on the motor to spin it up. Feeding in 10bar air, you have higher density,
and the motor torque and work done will rise (because an AC motor operates at
constant speed). Water, being nearly incompressible, doesn't create any
such scaling issues with inlet pressure.

That's exactly what I thought, but Jim's reference to multi-stage pumps threw me. Since water isn't compressible, I don't see how the multi-stage pumps work. For gas, no problem, but I don't get it for liquids.

For liquids, the multistage pumps I've used have pistons, and the low-pressure pistons
stall out when backpressure is high, then the same motor moves smaller high-pressure
pistons to achieve higher fluid pressures with lower volume displacement (but nearly
equal motor power). I suspect there's a differential gear driving all the pistons with
different mechanical advantage, and a reverse-inhibiting clutch on the low pressure pistons.

This is where I have trouble. Assuming these are regular centrifugal turbines, the outlet of the first stage is fed into the axis of the second stage. The pressure from the first-stage outlet is retained at the second-stage inlet, but from there it feeds into the whirling blades of the second stage, the outlet volume of which is LARGER than the inlet volume between any two blades.

Why? Several stages of centrifugal pump in series for a liquid would have same volume flow,
and same work per shaft would deliver the same head, at each stage. So, they'd be the same
size rotors, too. Maybe different shaft seals, though.

On Wednesday, May 31, 2017 at 7:35:13 PM UTC-4, whit3rd wrote:
On Sunday, May 28, 2017 at 8:35:39 AM UTC-7, wrote:
On Saturday, May 27, 2017 at 6:57:12 PM UTC-4, Ned Simmons wrote:
On Sat, 27 May 2017 12:20:00 -0700 (PDT), wrote:

On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?

That's close. A pump on atmospheric air (1 bar) that makes a 10 bar output,
will be different, because the density of inlet air determines the force
on the motor to spin it up. Feeding in 10bar air, you have higher density,
and the motor torque and work done will rise (because an AC motor operates at
constant speed). Water, being nearly incompressible, doesn't create any
such scaling issues with inlet pressure.

That's exactly what I thought, but Jim's reference to multi-stage pumps threw me. Since water isn't compressible, I don't see how the multi-stage pumps work. For gas, no problem, but I don't get it for liquids.

For liquids, the multistage pumps I've used have pistons, and the low-pressure pistons
stall out when backpressure is high, then the same motor moves smaller high-pressure
pistons to achieve higher fluid pressures with lower volume displacement (but nearly
equal motor power). I suspect there's a differential gear driving all the pistons with
different mechanical advantage, and a reverse-inhibiting clutch on the low pressure pistons.

This is where I have trouble. Assuming these are regular centrifugal turbines, the outlet of the first stage is fed into the axis of the second stage. The pressure from the first-stage outlet is retained at the second-stage inlet, but from there it feeds into the whirling blades of the second stage, the outlet volume of which is LARGER than the inlet volume between any two blades.

Why? Several stages of centrifugal pump in series for a liquid would have same volume flow,
and same work per shaft would deliver the same head, at each stage. So, they'd be the same
size rotors, too. Maybe different shaft seals, though.

The issue I was trying to resolve was the question of whether the pump(s) in series were, each, completely filled. Illustrations often show centrifugal pumps as being less than completely filled. But I realized that they have to be completely filled, especially for a multi-stage pump.

If those spaces aren't completely filled, the outlet pressure from one stage, which would be the inlet pressure of the next stage, wouldn't communicate to the output of that following stage. It would be like pushing on the end of a rope.

But I finally found some good explanations.

--
Ed Huntress

wrote on 5/31/2017 8:08 PM:
On Wednesday, May 31, 2017 at 7:35:13 PM UTC-4, whit3rd wrote:
On Sunday, May 28, 2017 at 8:35:39 AM UTC-7, wrote:
On Saturday, May 27, 2017 at 6:57:12 PM UTC-4, Ned Simmons wrote:
On Sat, 27 May 2017 12:20:00 -0700 (PDT), wrote:

On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?

That's close. A pump on atmospheric air (1 bar) that makes a 10 bar output,
will be different, because the density of inlet air determines the force
on the motor to spin it up. Feeding in 10bar air, you have higher density,
and the motor torque and work done will rise (because an AC motor operates at
constant speed). Water, being nearly incompressible, doesn't create any
such scaling issues with inlet pressure.

That's exactly what I thought, but Jim's reference to multi-stage pumps threw me. Since water isn't compressible, I don't see how the multi-stage pumps work. For gas, no problem, but I don't get it for liquids.

For liquids, the multistage pumps I've used have pistons, and the low-pressure pistons
stall out when backpressure is high, then the same motor moves smaller high-pressure
pistons to achieve higher fluid pressures with lower volume displacement (but nearly
equal motor power). I suspect there's a differential gear driving all the pistons with
different mechanical advantage, and a reverse-inhibiting clutch on the low pressure pistons.

This is where I have trouble. Assuming these are regular centrifugal turbines, the outlet of the first stage is fed into the axis of the second stage. The pressure from the first-stage outlet is retained at the second-stage inlet, but from there it feeds into the whirling blades of the second stage, the outlet volume of which is LARGER than the inlet volume between any two blades.

Why? Several stages of centrifugal pump in series for a liquid would have same volume flow,
and same work per shaft would deliver the same head, at each stage. So, they'd be the same
size rotors, too. Maybe different shaft seals, though.

The issue I was trying to resolve was the question of whether the pump(s) in series were, each, completely filled. Illustrations often show centrifugal pumps as being less than completely filled. But I realized that they have to be completely filled, especially for a multi-stage pump.

If those spaces aren't completely filled, the outlet pressure from one stage, which would be the inlet pressure of the next stage, wouldn't communicate to the output of that following stage. It would be like pushing on the end of a rope.

But I finally found some good explanations.


All my posts in this thread should be a good read. The impeller of a
centrifugal pump forms compartments inside the tight fitting casing to
trap incoming fluid so that when the impeller spins, it is forcing the
fluid in all the compartments to go in a circular motion. The
centrifugal force from the rotating fluid generates the pressure
radially outward against the outer wall of the pump.

A very basic impeller:
https://upload.wikimedia.org/wikipedia/commons/thumb/6/6b/Impellerrad.jpg/1280px-Impellerrad.jpg